Research ArticleEquilibrium and Optimal Strategies in MM1 Queues withWorking Vacations and Vacation Interruptions

Ruiling Tian1 Linmin Hu1 and Xijun Wu2

1College of Sciences Yanshan University Hebei Qinhuangdao 066004 China2Measurement Technology and Instrumentation Key Lab of Hebei Province Yanshan University Hebei Qinhuangdao 066004 China

Correspondence should be addressed to Ruiling Tian tianrlysueducn

Received 6 October 2015 Accepted 5 January 2016

Academic Editor Kishin Sadarangani

Copyright copy 2016 Ruiling Tian et al This is an open access article distributed under the Creative Commons Attribution Licensewhich permits unrestricted use distribution and reproduction in any medium provided the original work is properly cited

We consider the customers equilibrium and socially optimal joining-balking behavior in single-server Markovian queues withmultiple working vacations and vacation interruptions Arriving customers decide whether to join the system or balk based on alinear reward-cost structure that incorporates their desire for service as well as their unwillingness for waiting We consider thatthe system states are observable partially observable and unobservable respectively For these cases we first analyze the stationarybehavior of the system and get the equilibrium strategies of the customers and compare them to socially optimal balking strategiesnumerically

1 Introduction

In the past decades there has been an emerging tendencyin the literature to study queueing systems which are con-cerned with customers decentralized behavior and sociallyoptimal control of arrivals Such an economic analysis ofqueueing systems was pioneered by Naor [1] who studied theobservable MM1 model with a simple linear reward-coststructure It was assumed that an arriving customer observedthe number of customers and then made hisher decisionwhether to join or balkHis studywas complemented byEdel-son and Hildebrand [2] who considered the same queueingsystem as that in [1] but assumed that the customers madetheir decisions without being informed about the state ofthe system Moreover several authors have investigated thesame problem for various queueing systems incorporatingdiverse characteristics such as priorities reneging jockeyingschedules and retrialsThe fundamental results about variousmodels can be found in the comprehensive monograph writ-ten by Hassin and Haviv [3] with extensive bibliographicalreferences

There are some papers in the literature that considered thestrategic behavior of customers in classical vacation queueingmodels Burnetas and Economou [4] studied a Markovian

single-server queueing systemwith setup timesThey derivedequilibrium strategies for the customers under various levelsof information and analyzed the stationary behavior of thesystem under these strategies Sun and Tian [5] considereda queueing model with classical multiple vacations Theyderived the equilibrium and socially optimal joining strate-gies in vacation and busy period of a partially observablequeue respectively The conclusion was summarized thatindividual optimization led to excessive congestion withoutregulation Economou and Kanta [6] considered the Marko-vian single-server queue that alternates between on and offperiods They derived equilibrium threshold strategies infully observable and almost observable queues Guo andHas-sin [7] studied a vacation queue withN-policy and exhaustiveservice They presented the equilibrium and socially optimalstrategies for unobservable and observable queuesThis workwas extended by Guo and Hassin [8] to heterogenous cus-tomers and by Tian et al [9] to MG1 queues Recently Guoand Li [10] studied strategic behavior and social optimizationin partially observable Markovian vacation queues Li et al[11] considered equilibrium strategies in MM1 queues withpartial breakdowns Zhang et al [12] studied a single-serverretrial queue with two types of customers in which the serverwas subject to vacations along with breakdowns and repairs

Hindawi Publishing CorporationMathematical Problems in EngineeringVolume 2016 Article ID 9746962 10 pageshttpdxdoiorg10115520169746962

2 Mathematical Problems in Engineering

They discussed and compared the optimal and equilibriumretrial rates regarding the situations in which the customerswere cooperative or noncooperative respectively

Recently queueing systems with working vacations havebeen studied extensively where the server takes vacationsonce the system becomes empty (ie exhaustive service pol-icy) and can still serve customers at a lower rate than regularone during the vacations Research results of the stationarysystem performance on various working vacation queues canbe consulted in the survey given by Tian et al [13] As forthe work studying customers behavior in queueing systemswith working vacations Zhang et al [14] and Sun and Li[15] obtained equilibrium balking strategies inMM1 queueswith working vacations for four cases with respect to differentlevels of information Sun et al [16] also considered the cus-tomers equilibrium balking behavior in some single-serverMarkovian queues with two-stage working vacations

However we often encounter the situation that the servercan stop the vacation once some indices of the system suchas the number of customers achieve a certain value duringa vacation In many real life congestion situations urgentevents occur during a vacation and the servermust comebackto work rather than continuing to take the residual vacationFor example if the number of customers exceeds the specialvalue during a vacation and the server continues to take thevacation it leads to large cost of waiting customersThereforevacation interruption is more reasonable to the server vaca-tion queues Vacation interruption was introduced by Li andTian [17] and Li et al [18]

In this paper we consider anMM1 queueingmodel withworking vacations and vacation interruptionsWedistinguishthree cases the observable queues the partially observablequeues and the unobservable queues according to the infor-mation levels regarding system states As to the observablecase a customer can observe both the state of the server andthe number of present customers before making decision Inthe unobservable case a customer cannot observe the stateof the system However for the partially observable case acustomer only can observe the state of the server at theirarrival instant and does not observe the number of customerspresentThe customers dilemma is whether to join the systemor balk We study the balking behavior of customers in thesecases and derive Nash equilibrium strategies and sociallyoptimal strategies

This paper is organized as follows Descriptions of themodel are given in Section 2 In Sections 3 4 and 5 wedetermine the equilibrium and socially optimal strategies inobservable queues the partially observable queues and theunobservable queues respectively Conclusions are given inSection 6

2 Model Description

Consider a classical MM1 queue with an arrival rate 120582 anda service rate 120583

1 Upon the completion of service if there is

no customer in the system the server begins a vacation andthe vacation time is assumed to be exponentially distributedwith the parameter 120579 During a vacation period arriving

customers can be served at a mean rate of 1205830 Upon the com-

pletion of a service in the vacation if there are also customersin the queue the server ends the vacation and comes backto the normal working level Otherwise heshe continuesthe vacation until there are customers after a service or avacation ends Meanwhile when a vacation ends if thereare no customers another vacation is taken Otherwise theserver also switches the rate to 120583

1and a regular busy period

starts In this service discipline the server may come backfrom the vacation without completing the vacation And theserver can only go on vacations if there is no customer left inthe system upon the completion of a service Meanwhile thevacation service rate can be only applied to the first customerthat arrived during a vacation period

We assume that the interarrival times the service timesand the working vacation times are mutually independent Inaddition the service discipline is first in first out (FIFO)

Denote by119873(119905) the number of customers in the system attime 119905 Let 119869(119905) = 0 be the state that the server is on workingvacation period at time 119905 and let 119869(119905) = 1 be the state that theserver is busy at time 119905

Arriving customers are assumed to be identical Our int-erest is in the customersrsquo strategic response as they can decidewhether to join or balk upon arrival Assume that a customerrsquosutility consists of a reward for receiving service minus awaiting cost Specifically every customer receives a reward of119877 units for completing service There is a waiting cost of 119862units per time unit that the customer remains in the systemCustomers are risk neutral and maximize their expected netbenefit Finally we assume that there are no retrials of balkingcustomers nor reneging of waiting customers

3 The Observable Queues

We begin with the fully observable case in which the arrivingcustomers know not only the number of present customers119873(119905) but also the state of the server 119869(119905) at arrival time 119905

There exists a balk threshold 119899(119894) such that an arrivingcustomer enters the system at state 119894 (119894 = 0 1) if the numberof the customers present upon arrival does not exceed thespecified threshold So a pure threshold strategy is specifiedby a pair (119899

119890(0) 119899119890(1)) and the balking strategy has the follow-

ing form ldquowhile arriving at time 119905 observe (119873(119905) 119869(119905)) enterif119873(119905) le 119899

119890(119869(119905)) and balk otherwiserdquo

And we have the following result

Theorem 1 In the observable MM1 queue with workingvacations and vacation interruptions there exist thresholds(119899119890(0) 119899119890(1)) which are given by

119899119890(0) = lfloor

1198771205831

119862minus

1205831+ 120579

1205831(1205830+ 120579)

rfloor

119899119890(1) = lfloor

1198771205831

119862rfloor minus 1

(1)

such that a customer who observes the system at state (119873(119905)

119869(119905)) upon his arrival enters if 119873(119905) le 119899119890(119869(119905)) and balks

otherwise

Mathematical Problems in Engineering 3

Proof Based on the reward-cost structure which is imposedon the system we conclude that for an arriving customer thatdecides to enter hisher expected net benefit is

119880 (119899 119894) = 119877 minus 119862119879 (119899 119894) (2)

where 119879(119899 119894) denotes hisher expected mean sojourn timegiven that heshe finds the system at state (119899 119894) upon hisherarrival Then we have the following equations

119879 (0 0) =1

1205830+ 120579

+120579

1205830+ 120579

1

1205831

(3)

119879 (119899 0) =1205830

1205830+ 120579

(1

1205830

+ 119879 (119899 minus 1 1))

+120579

1205830+ 120579

119879 (119899 1)

(4)

119879 (119899 1) =119899 + 1

1205831

(5)

By iterating (4) and taking into account (5) we obtain

119879 (119899 0) =119899

1205831

+1205831+ 120579

1205831(1205830+ 120579)

(6)

We can easily check that 119879(119899 0) is strictly increasing for 119899A customer strictly prefers to enter if the reward for serviceexceeds the expected cost for waiting (ie 119880(119899 119894) gt 0) andis indifferent between entering and balking if the reward isequal to the cost (ie 119880(119899 119894) = 0)

We assume throughout the paper that

119877 gt 119862(1

1205830+ 120579

+120579

1205830+ 120579

1

1205831

) (7)

which ensures that the reward for service exceeds theexpected cost for a customer who finds the system empty Bysolving 119880(119899 119894) ge 0 for 119899 using (5) and (6) we obtain thatthe customer arriving at time 119905 decides to enter if and only if119899 le 119899119890(119869(119905)) where (119899

119890(0) 119899119890(1)) are given by (1)

From Figure 1 we observe that 119899119890(1) is larger than 119899

119890(0)

and 119899119890(0) and 119899

119890(1) are all increasing with 120583

1

For the stationary analysis the system follows a Markovchain and the state space is

Ω119891119900

= 0 0 cup (119899 119894) | 119899 = 1 2 119899 (119894) + 1 119894 = 0 1 (8)

And the transition diagram is illustrated in Figure 2

ne(0)ne(1)

2

4

6

8

10

12

14

n

12 14 16 18 2 22 24 26 28 311205831

Figure 1 Equilibrium threshold strategies for observable case when119877 = 5 119862 = 1 120582 = 1 120583

0= 05 and 120579 = 01

Let 120587119899119894= lim

119905rarrinfin119875119873(119905) = 119899 119869(119905) = 119894 with (119899 119894) isin Ω

119891119900

then 120587119899119894

(119899 119894) isin Ω119891119900 is the stationary distribution of the

process (119873(119905) 119869(119905)) 119905 ge 0In steady state we have the following system equations

12058212058700

= 120583012058710+ 120583112058711 (9)

(120582 + 120579 + 1205830) 1205871198990

= 120582120587119899minus10

119899 = 1 2 119899 (0) (10)

(120579 + 1205830) 120587119899(0)+10

= 120582120587119899(0)0

(11)

(120582 + 1205831) 12058711

= 120583112058721+ 12057912058710+ 120583012058720 (12)

(120582 + 1205831) 1205871198991

= 120582120587119899minus11

+ 1205831120587119899+11

+ 1205791205871198990

+ 1205830120587119899+10

119899 = 2 3 119899 (0)

(13)

(120582 + 1205831) 120587119899(0)+11

= 120582120587119899(0)1

+ 1205831120587119899(0)+2

+ 120579120587119899(0)+10

(14)

(120582 + 1205831) 1205871198991

= 120582120587119899minus11

+ 1205831120587119899+11

119899 = 119899 (0) + 2 119899 (1)

(15)

1205831120587119899(1)+11

= 120582120587119899(1)1

(16)

Define

120588 =120582

1205831

120590 =120582

120582 + 120579 + 1205830

(17)

4 Mathematical Problems in Engineering

1 1 2 1

1 00 0 2 0

120579120579120579120579

1205831 1205831 1205831 1205831 1205831 1205831 1205831

1205830120583012058301205830

1205830

120582120582120582120582 120582

120582120582120582120582120582120582120582middot middot middot

middot middot middot

middot middot middot

1205831

n(0) 1

n(0) 0

n(0) + 1 1

n(0) + 1 0

n(1) 1 n(1) + 1 1

Figure 2 Transition rate diagram for the observable queues

By iterating (10) and (15) taking into account (11) and (16)we obtain

1205871198990

= 12059011989912058700 119899 = 0 1 2 119899 (0)

120587119899(0)+10

=120582120590119899(0)

120579 + 1205830

12058700

1205871198991

= 120588119899minus119899(0)minus1

120587119899(0)+11

119899 = 119899 (0) + 1 119899 (1) + 1

(18)

From (13) it is easy to obtain that 1205871198991

| 1 le 119899 le 119899(0)+1 isa solution of the nonhomogeneous linear difference equation

1205831119909119899+1

minus (120582 + 1205831) 119909119899+ 120582119909119899minus1

= minus1205791205871198990minus 1205830120587119899+10

= minus (120579 + 1205830120590) 12059011989912058700

(19)

So its corresponding characteristic equation is

12058311199092minus (120582 + 120583

1) 119909 + 120582 = 0 (20)

which has two roots at 1 and 120588 Assume that 120588 = 1 thenthe homogeneous solution of (19) is 119909hom

119899= 1198601119899+ 119861120588119899 The

general solution of the nonhomogeneous equation is given as119909gen119899

= 119909hom119899

+119909spec119899

where 119909spec119899

is a specific solution We finda specific solution 119909

spec119899

= 119863120590119899 Substituting it into (19) we

derive

119863 = minus120582 + 120579

120582 + 120579 + 1205830minus 1205831

12058700 (21)

Therefore

119909gen119899

= 119860 + 119861120588119899+ 119863120590119899 119899 = 1 2 119899 (0) + 1 (22)

Substituting (22) into (9) and (12) it follows after some rathertedious algebra that

119860 + 119861120588 = minus120582 (120582 + 120579)

1205831(1205831minus 120582 minus 120579 minus 120583

0)12058700

119860 + 1198611205882= minus

1205822(120582 + 120579)

12058321(1205831minus 120582 minus 120579 minus 120583

0)12058700

(23)

Then we obtain

119860 = 0

119861 = minus120582 + 120579

1205831minus 120582 minus 120579 minus 120583

0

12058700

(24)

Thus from (22) we obtain

1205871198991

= 119861120588119899+ 119863120590119899 119899 = 1 2 119899 (0) + 1 (25)

Thus we have all the stationary probabilities in terms of12058700 The remaining probability 120587

00 can be found from the

normalization condition After some algebraic simplifica-tions we can summarize the results in the following

Theorem 2 In the observable MM1 queues with workingvacations and vacation interruptions the stationary distribu-tion 120587

119899119894| (119899 119894) isin Ω

119891119900 is given as follows

1205871198990

= 12059011989912058700 119899 = 0 1 2 119899 (0)

120587119899(0)+10

=120582120590119899(0)

120579 + 1205830

12058700

1205871198991

= 119861120588119899+ 119863120590119899 119899 = 1 2 119899 (0) + 1

1205871198991

= 120588119899minus119899(0)minus1

(119861120588119899(0)+1

+ 119863120590119899(0)+1

)

119899 = 119899 (0) + 2 119899 (1) + 1

(26)

where 12058700can be solved by the normalization equation

Because the probability of balking is equal to 120587119899(0)+10

+

120587119899(1)+11

the social benefit per time unit when all customers fol-low the threshold policy (119899(0) 119899(1)) equals

119880119904(119899 (0) 119899 (1)) = 120582119877 (1 minus 120587

119899(0)+10minus 120587119899(1)+11

)

minus 119862(

119899(0)+1

sum

119899=0

1198991205871198990+

119899(1)+1

sum

119899=1

1198991205871198991)

(27)

Let (119899lowast(0) 119899lowast(1)) be the socially optimal threshold strategyFrom Figure 3 we obtain (119899

lowast(0) 119899lowast(1)) = (3 4) while the

equilibrium threshold strategy (119899119890(0) 119899119890(1)) = (78 79) We

observe that 119899119890(0) gt 119899

lowast(0) and 119899

119890(1) gt 119899

lowast(1) which indicates

that the individual optimization is inconsistent with the socialoptimization

4 Partially Observable Queues

In this section we turn our attention to the partially observ-able case where arriving customers only can observe the stateof the server at their arrival instant and do not observe thenumber of customers present

A mixed strategy for a customer is specified by a vector(1199020 1199021) where 119902

119894is the probability of joining when the server

is in state 119894 (119894 = 0 1) If all customers follow the same mixed

Mathematical Problems in Engineering 5

2

3

4

5

6

7

8

9

n(1

)

2 3 4 5 6 7 81n(0)

24

68

10

02

46

8

n(1)n(0)

0

10

20

30

40

50

Us(n

(0)n

(1))

68

Figure 3 Social welfare for the observable case when 119877 = 40 119862 = 1 120582 = 15 1205831= 2 120583

0= 05 and 120579 = 05

1 1 2 0 3 1

0 0 1 0 2 0 3 0

n 1

n 0

120579

1205831 1205831

12058301205830

120582q1120582q1

120582q0 120582q0middot middot middot

middot middot middot

middot middot middot

middot middot middot

120579

1205831 1205831

12058301205830

120582q1120582q1

120582q0 120582q0

120579120579

1205831

1205831

1205830

1205830

120582q1

120582q0 120582q0

Figure 4 Transition rate diagram for the partially observable queues

strategy (1199020 1199021) then the system follows a Markov chain

where the arrival rate equals 120582119894= 120582119902119894when the server is in

state 119894 (119894 = 0 1) The state space is Ω119901119900

= 0 0 cup (119899 119894) |

119899 ge 1 119894 = 0 1 and the transition diagram is illustrated inFigure 4

Denote the stationary distribution as120587119899119894= lim119905rarrinfin

119875 119873 (119905) = 119899 119869 (119905) = 119894 (119899 119894) isin Ω119901119900

1205870= 12058700

120587119899= (1205871198990 1205871198991) 119899 ge 1

(28)

Using the lexicographical sequence for the states thetransition rate matrix (generator) Q can be written as thetridiagonal block matrix

Q =((

(

A0C0

B1

A CB A C

B A Cd d d

))

)

(29)

whereA0= minus120582119902

0

B1= (

1205830

1205831

)

C1= (1205821199020 0)

A = (minus (1205821199020+ 120579 + 120583

0) 120579

0 minus (1205821199021+ 1205831))

B = (0 1205830

0 1205831

)

C = (1205821199020

0

0 1205821199021

)

(30)To analyze this QBD process it is necessary to solve for

the minimal nonnegative solution of the matrix quadraticequation

R2B + RA + C = 0 (31)and this solution is called the rate matrix and denoted by R

Lemma 3 If 1205881

= 12058211990211205831

lt 1 the minimal nonnegativesolution of (31) is given by

R = (1205900

1205821199020+ 120579

1205831

1205900

0 1205881

) (32)

where 1205900= 1205821199020(1205821199020+ 120579 + 120583

0)

Proof Because the coefficients of (31) are all upper triangularmatrices we can assume that R has the same structure as

R = (11990311

11990312

0 11990322

) (33)

6 Mathematical Problems in Engineering

Substituting R2 and R into (31) yields the following set ofequations

minus (1205821199020+ 120579 + 120583

0) 11990311+ 1205821199020= 0

12058311199032

22minus (1205821199021+ 1205831) 11990322+ 1205821199021= 0

12058301199032

11+ 120583111990312(11990311+ 11990322) + 120579119903

11minus (1205821199021+ 1205831) 11990312

= 0

(34)

To obtain the minimal nonnegative solution of (31) we take11990322

= 1205881(the other roots are 119903

22= 1) in the second equation

of (34) From the first equation of (34) we obtain 11990311

=

1205821199020(1205821199020+ 120579 + 120583

0) = 1205900 Substituting 120590

0and 1205881into the last

equation of (34) we get 11990312

= ((1205821199020+ 120579)120583

1)1205900 This is the

proof of Lemma 3

Theorem 4 Assuming that 1205881lt 1 the stationary probabilities

120587119899119894

| (119899 119894) isin Ω119901119900 of the partially observable queues are as

follows

1205871198990

= 119870120590119899

0 119899 ge 0

1205871198991

= 1198701205881

1205821199020+ 120579

1205821199020+ 120579 + 120583

0

119899minus1

sum

119895=0

120588119895

1120590119899minus1minus119895

0 119899 ge 1

(35)

where

119870 =120579 + 1205830

1205821199020+ 120579 + 120583

0

(1 +1205821199020+ 120579

1205821199020+ 120579 + 120583

0

1205821199020

1205831minus 1205821199021

)

minus1

(36)

Proof With the matrix-geometric solution method [19] wehave

120587119899= (1205871198990 1205871198991) = (120587

10 12058711)R119899minus1 119899 ge 1 (37)

In addition (12058700 12058710 12058711) satisfies the set of equations

(12058700 12058710 12058711) 119861 [R] = 0 (38)

where

119861 [R] = (A0

C0

B1

A + RB)

= (

minus1205821199020

1205821199020

0

1205830

minus (1205821199020+ 120579 + 120583

0) 1205821199020+ 120579

1205831

0 minus1205831

)

(39)

Then from (38) we derive

minus120582119902012058700+ 120583012058710+ 120583112058711

= 0

120582119902012058700minus (1205821199020+ 120579 + 120583

0) 12058710

(1205821199020+ 120579) 120587

10minus 120583112058711

= 0

(40)

Taking 12058700as a known constant we obtain

12058710

= 120590012058700 (41)

12058711

= 1205881

1205821199020+ 120579

1205821199020+ 120579 + 120583

0

12058700 (42)

Note that

R119899minus1 = (120590119899minus1

0

1205821199020+ 120579

1205831

1205900

119899minus2

sum

119895=0

120590119895

0120588119899minus119895minus2

1

0 120588119899minus1

1

) 119899 ge 2 (43)

By 12058710 12058711 R119899minus1 and (37) we get (35) Finally 120587

00can be

determined by the normalization conditionThis is the proofof Theorem 4

Let 119901119894be the steady-state probability when the server is at

state 119894 (119894 = 0 1) we have

1199010=

infin

sum

119899=0

1205871198990

= 1198701205821199020+ 120579 + 120583

0

120579 + 1205830

(44)

1199011=

infin

sum

119899=1

1205871198991

= 1198701205821199020(1205821199020+ 120579)

(1205831minus 1205821199021) (120579 + 120583

0) (45)

So the effective arrival rate 120582 is given by

120582 = 120582 (11990101199020+ 11990111199021) (46)

We now consider a customer who finds the server at state119894 upon arrival The conditional mean sojourn time of such acustomer that decides to enter given that the others follow thesame mixed strategy (119902

0 1199021) is given by

119882119894(1199020 1199021) =

suminfin

119899=119894119879 (119899 119894) 120587

119899119894

suminfin

119899=119894120587119899119894

(47)

By substituting (5)-(6) and (35) into (44) we obtain

1198820(1199020 1199021) =

1205821199020+ 120579 + 120583

1

1205831(120579 + 120583

0)

1198821(1199020 1199021) =

1

1205831minus 1205821199021

+1205821199020+ 120579 + 120583

0

1205831(1205830+ 120579)

(48)

Based on the reward-cost structure the expected net ben-efit of an arriving customer if heshe finds the server at state119894 and decides to enter is given by

1198800(1199020) = 119877 minus 119862

1205821199020+ 120579 + 120583

1

1205831(120579 + 120583

0)

1198801(1199020 1199021) = 119877 minus 119862(

1

1205831minus 1205821199021

+1205821199020+ 120579 + 120583

0

1205831(1205830+ 120579)

)

(49)

We now can proceed to determine mixed equilibriumstrategies of a customer in the partially observable case andhave the following

Theorem 5 For the partially observable case there exists aunique mixed equilibrium strategy (119902

119890

0 119902119890

1) where the vector

(119902119890

0 119902119890

1) is given as follows

(1) 119862(120579 + 1205831)1205831(120579 + 120583

0) lt 119877 le 119862(120582 + 120579 + 120583

1)1205831(120579 + 120583

0)

(119902119890

0 119902119890

1) =

(1199091 0)

1205831minus 1205830

1205831(120579 + 120583

0)lt

1

120583

(1199091 1199092)

1

120583le

1205831minus 1205830

1205831(120579 + 120583

0)le

1

120583 minus 120582

(1199091 1)

1205831minus 1205830

1205831(120579 + 120583

0)gt

1

120583 minus 120582

(50)

Mathematical Problems in Engineering 7

(2) 119862(120582 + 120579 + 1205831)1205831(120579 + 120583

0) lt 119877

(119902119890

0 119902119890

1) =

(1 0) 119877 lt119862

1205831

+119862 (120582 + 120579 + 120583

0)

1205831(120579 + 120583

0)

(1 1199093)

119862

1205831

+119862 (120582 + 120579 + 120583

0)

1205831(120579 + 120583

0)

le 119877 le119862

1205831minus 120582

+119862 (120582 + 120579 + 120583

0)

1205831(120579 + 120583

0)

(1 1) 119877 gt119862

1205831minus 120582

+119862 (120582 + 120579 + 120583

0)

1205831(120579 + 120583

0)

(51)

where

1199091=

1

120582(1198771205831(120579 + 120583

0)

119862minus 120579 minus 120583

1)

1199092=1205831

120582(1 minus

1205830+ 120579

1205831minus 1205830

)

1199093=

1

120582(1205831minus

1

119877119862 minus (120582 + 120579 + 1205830) 1205831(120579 + 120583

0))

(52)

Proof To prove Theorem 5 we first focus on 1198800(1199020) Con-

dition (1) assures that 1199021198900is positive Therefore we have two

casesCase 1 (119880

0(0) gt 0 and 119880

0(1) le 0) That is 119862(120579 + 120583

1)1205831(120579 +

1205830) lt 119877 le 119862(120582+120579+120583

1)1205831(120579+1205830)) In this case if all customers

who find the system empty enter the system with probability119902119890

0= 1 then the tagged customer suffers a negative expected

benefit if heshe decides to enter Hence 1199021198900= 1 does not lead

to an equilibrium Similarly if all customers use 1199021198900= 0 then

the tagged customer receives a positive benefit from enteringthus 119902119890

0= 0 also cannot be part of an equilibrium mixed

strategy Therefore there exists unique 1199021198900satisfying

119877 minus 119862120582119902119890

0+ 120579 + 120583

1

1205831(120579 + 120583

0)

= 0 (53)

for which customers are indifferent between entering andbalking This is given by

119902119890

0=

1

120582(1198771205831(120579 + 120583

0)

119862minus 120579 minus 120583

1) = 119909

1 (54)

In this situation the expected benefit 1198801is given by

1198801(119902119890

0 1199021) =

119862 (1205831minus 1205830)

1205831(120579 + 120583

0)minus

119862

120583 minus 1205821199021

(55)

Using the standard methodology of equilibrium analysisin unobservable queueing models we derive from (55) thefollowing equilibria

(1) If 1198801(119902119890

0 0) lt 0 then the equilibrium strategy is 119902119890

1=

0(2) If 119880

1(119902119890

0 0) ge 0 and 119880

1(119902119890

0 1) le 0 there exists unique

119902119890

1 satisfying

1198801(119902119890

0 119902119890

1) = 0 (56)

then the equilibrium strategy is 1199021198901= 1199092

(3) If 1198801(119902119890

0 1) gt 0 then the equilibrium strategy is 119902119890

1=

1

Case 2 (1198800(1) ge 0) That is 119862(120582 + 120579 + 120583

1)1205831(120579 + 120583

0) lt 119877) In

this case for every strategy of the other customers the taggedcustomer has a positive expected net benefit if heshe decidesto enter Hence 119902119890

0= 1

In this situation the expected benefit 1198801is given by

1198801(1 1199021) = 119877 minus

119862

120583 minus 1205821199021

minus119862 (120582 + 120579 + 120583

0)

1205831(120579 + 120583

0)

(57)

To find 119902119890

1in equilibrium we also consider the following

subcases

(1) If 1198801(1 0) lt 0 then the equilibrium strategy is 119902119890

1=

0

(2) If 1198801(1 0) ge 0 and 119880

1(1 1) le 0 there exists unique

119902119890

1 satisfying

1198801(1 119902119890

1) = 0 (58)

then the equilibrium strategy is 1199021198901= 1199093

(3) If 1198801(1 1) gt 0 then the equilibrium strategy is 119902119890

1=

1

By rearranging Cases 1 and 2 we can obtain the results ofTheorem 5 This completes the proof

From Theorem 5 we can compare 119902119890

0with 119902

119890

1 Figure 5

shows that 1199021198900is not always less than 119902

119890

1 Namely sometimes

the information that the server takes a working vacation doesnot make the customers less willing to enter the systembecause in the vacation the server provides a lower serviceto customers But when the vacation time and waiting timebecome longer customers do not want to enter the system invacation

Then fromTheorem 4 themean queue length is given by

119864 [119871] =

infin

sum

119899=0

1198991205871198990+

infin

sum

119899=0

1198991205871198991

= 1198701205900

(1 minus 1205900)2(1 +

1205821199020+ 120579

1205831

1 minus 12058811205900

(1 minus 1205881)2)

(59)

8 Mathematical Problems in Engineering

03

04

05

06

07

08

09

1

Equi

libriu

m jo

inin

g pr

obab

ility

qe0qe1 (120579 = 01 C = 2)

qe0qe1 (120579 = 05 C = 3)

0 04 06 08 1 12 14 16 1802Arrival rate 120582

Figure 5 Equilibrium mixed strategies for the partially observablecase when 119877 = 5 120583

1= 2 and 120583

0= 05

0 04 06 08 1 12 14 16 1802Arrival rate 120582

0

01

02

03

04

05

06

07

08

09

1

Join

ing

prob

abili

ty

qe0qe1

qlowast0qlowast1

Figure 6 Equilibrium and socially optimal mixed strategies for thepartially observable case when 119877 = 5 119862 = 2 120583

1= 2 120583

0= 05 and

120579 = 01

And the social benefit when all customers follow a mixedpolicy (119902

0 1199021) can now be easily computed as follows

119880119904(1199020 1199021) = 120582119877 minus 119862119864 [119871]

= 120582 (11990101199020+ 11990111199021) 119877

minus1198621198701205900

(1 minus 1205900)2(1 +

1205821199020+ 120579

1205831

1 minus 12058811205900

(1 minus 1205881)2)

(60)

The goal of a social planner is to maximize overall socialwelfare Let (119902lowast

0 119902lowast

1) be the socially optimal mixed strategy

Figure 6 compares customers equilibrium mixed strategy(119902119890

0 119902119890

1) and their socially optimal mixed strategy (119902lowast

0 119902lowast

1) We

also observe that 1199021198900gt 119902lowast

0and 1199021198901gt 119902lowast

1This ordering is typical

when customers make individual decisions maximizing theirown profitThen they ignore those negative externalities thatthey impose on later arrivals and they tend to overuse thesystem It is clear that these externalities should be taken intoaccount when we aim to maximize the total revenue

5 The Unobservable Queues

In this section we consider the fully unobservable case wherearriving customers cannot observe the system state

There are two pure strategies available for a customer thatis to join the queue or not to join the queue A pure or mixedstrategy can be described by a fraction 119902 (0 le 119902 le 1) whichis the probability of joining and the effective arrival rate orjoining rate is 120582119902 The transition diagram is illustrated inFigure 7

To identify the equilibrium strategies of the customerswe should first investigate the stationary distribution of thesystem when all customers follow a given strategy 119902 whichcan be obtained by Theorem 4 by taking 119902

0= 1199021= 119902 So we

have

1205871198990

= 1198701120590119899

119902 119899 ge 0

1205871198991

= 1198701120588119902

120582119902 + 120579

120582119902 + 120579 + 1205830

119899minus1

sum

119895=0

120588119895

119902120590119899minus1minus119895

119902 119899 ge 1

(61)

where 120588119902= 120582119902120583

1 120590119902= 120582119902(120582119902 + 120579 + 120583

0) and

1198701=

120579 + 1205830

120582119902 + 120579 + 1205830

(1 +120582119902 + 120579

120582119902 + 120579 + 1205830

120582119902

1205831minus 120582119902

)

minus1

(62)

Then the mean number of the customers in the system is

119864 [119871] =

infin

sum

119899=0

1198991205871198990+

infin

sum

119899=0

1198991205871198991

=1198701120590119902

(1 minus 120590119902)2(1 +

120582119902 + 120579

1205831

1 minus 120588119902120590119902

(1 minus 120588119902)2)

(63)

Hence the mean sojourn time of a customer who decides toenter upon hisher arrival can be obtained by using Littlersquoslaw

119864 [119882] =1198701120590119902

120582119902 (1 minus 120590119902)2(1 +

120582119902 + 120579

1205831

1 minus 120588119902120590119902

(1 minus 120588119902)2)

=1

120579 + 1205830

(1

+1205831(120579 + 120583

0)

1205831minus 120582119902

1

1205830((1205831minus 120582119902) (120582119902 + 120579)) + 120583

1

)

(64)

And if 120582 lt 1205831 119864[119882] is strictly increasing for 119902 isin [0 1]

Mathematical Problems in Engineering 9

1 1 2 0 3 1

0 0 1 0 2 0 3 0

n 1

n 0

120579

1205831 1205831

12058301205830

120582q120582q

120582q 120582qmiddot middot middot

middot middot middot

middot middot middot

middot middot middot

120579

1205831 1205831

12058301205830

120582q120582q

120582q 120582q

120579120579

1205831

1205831

1205830

1205830

120582q

120582q 120582q

Figure 7 Transition rate diagram for the unobservable queues

We consider a tagged customer If heshe decides to enterthe system hisher expected net benefit is

119880 (119902) = 119877 minus 119862119864 [119882] = 119877 minus119862

120579 + 1205830

(1

+1205831(120579 + 120583

0)

1205831minus 120582119902

1

1205830((1205831minus 120582119902) (120582119902 + 120579)) + 120583

1

)

(65)

We note that 119880(119902) is strictly decreasing in 119902 and it has aunique root 119902lowast

119890 So there exists a unique mixed equilibrium

strategy 119902119890and 119902119890= min119902lowast

119890 1

We now turn our attention to social optimization Thesocial benefit per time unit can now be easily computed as

119880119904(119902) = 120582119902 (119877 minus 119862119864 [119882]) = 120582119902(119877 minus

119862

120579 + 1205830

(1

+1205831(120579 + 120583

0)

1205831minus 120582119902

1

1205830((1205831minus 120582119902) (120582119902 + 120579)) + 120583

1

))

(66)

As for the equilibrium social welfare per time unit119880119904(119902119890)

Figure 8 shows that 119880119904(119902119890) first increases and then decreases

with 120582 and the social benefit achieves a maximum forintermediate values of this parameter The reason for thisbehavior is that when the arrival rate is small the system israrely crowded therefore as more customers arrive they areserved and the social benefit improves However with anysmall increase of the arrival rate 120582 the expected waiting timeincreases which has a detrimental effect on the social benefit

Let 119909lowast be the root of the equation 1198781015840(119902) = 0 and let 119902lowast be

the optimal joining probability Then we have the followingconclusions if 0 lt 119909

lowastlt 1 and 119878

10158401015840(119902) le 0 then 119902

lowast= 119909lowast if

0 lt 119909lowastlt 1 and 119878

10158401015840(119902) gt 0 or 119909lowast ge 1 then 119902

lowast= 1

From Figure 9 we observe that 119902119890

gt 119902lowast As a result

individual optimization leads to queues that are longer thansocially desired Therefore it is clear that the social plannerwants a toll to discourage arrivals

6 Conclusion

In this paper we analyzed the customer strategic behaviorin the MM1 queueing system with working vacations andvacation interruptions where arriving customers have optionto decide whether to join the system or balk Three differentcases with respect to the levels of information provided toarriving customers have been investigated extensively and theequilibrium strategies for each case were derived And wefound that the equilibrium joining probability of the busy

02 04 06 08 1 12 14 16 18 20Arrival rate 120582

Equi

libriu

m so

cial

wel

fare

Us(qe)

0

5

10

15

Figure 8 Equilibrium social welfare for the unobservable casewhen119877 = 10 119862 = 1 120583

1= 2 120583

0= 1 and 120579 = 1

03

04

05

06

07

08

09

1

Join

ing

prob

abili

ty

0 1 15 205Arrival rate 120582

qeqlowast

Figure 9 Equilibrium and socially optimal strategies for the unob-servable case when 119877 = 3 119862 = 2 120583

1= 2 120583

0= 05 and 120579 = 001

period may be smaller than that in vacation time We alsocompared the equilibrium strategy with the socially optimalstrategy numerically and we observed that customers makeindividual decisions maximizing their own profit and tendto overuse the system For the social planner a toll will beadopted to discourage customers from joining

10 Mathematical Problems in Engineering

Furthermore an interesting extension would be to incor-porate the present model in a profit maximizing frameworkwhere the owner or manager of the system imposes anentrance feeThe problem is to find the optimal fee that max-imizes the owners total profit subject to the constraint thatcustomers independently decide whether to enter or not andtake into account the fee in addition to the service reward andwaiting cost Another direction for future work is to concernthe non-Markovian queues with various vacation policies

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Acknowledgments

This work is partially supported by the Natural Foundation ofHebei Province (nos A2014203096 andC2014203212) Chinaand the Young Faculty Autonomous Research Project of Yan-shan University (nos 13LGB032 13LGA017 and 13LGB013)China

References

[1] P Naor ldquoThe regulation of queue size by levying tollsrdquo Econo-metrica vol 37 no 1 pp 15ndash24 1969

[2] N M Edelson and D K Hildebrand ldquoCongestion tolls forPoisson queuing processesrdquo Econometrica vol 43 no 1 pp 81ndash92 1975

[3] R Hassin and M Haviv Equilibrium Behavior in Queueing Sys-tems To Queue or Not to Queue Kluwer Academic PublishersDordrecht The Netherlands 2003

[4] A Burnetas and A Economou ldquoEquilibrium customer strate-gies in a single server Markovian queue with setup timesrdquoQueueing Systems vol 56 no 3-4 pp 213ndash228 2007

[5] W Sun and N Tian ldquoContrast of the equilibrium and sociallyoptimal strategies in a queue with vacationsrdquo Journal of Compu-tational Information Systems vol 4 no 5 pp 2167ndash2172 2008

[6] A Economou and S Kanta ldquoEquilibrium balking strategiesin the observable single-server queue with breakdowns andrepairsrdquoOperations Research Letters vol 36 no 6 pp 696ndash6992008

[7] P Guo and R Hassin ldquoStrategic behavior and social optimiza-tion in Markovian vacation queuesrdquo Operations Research vol59 no 4 pp 986ndash997 2011

[8] P Guo and R Hassin ldquoStrategic behavior and social optimiza-tion in Markovian vacation queues the case of heterogeneouscustomersrdquo European Journal of Operational Research vol 222no 2 pp 278ndash286 2012

[9] R Tian D Yue and W Yue ldquoOptimal balking strategies inan MG1 queueing system with a removable server under N-policyrdquo Journal of Industrial andManagementOptimization vol11 no 3 pp 715ndash731 2015

[10] P Guo and Q Li ldquoStrategic behavior and social optimizationin partially-observableMarkovian vacation queuesrdquoOperationsResearch Letters vol 41 no 3 pp 277ndash284 2013

[11] L Li J Wang and F Zhang ldquoEquilibrium customer strategiesin Markovian queues with partial breakdownsrdquo Computers ampIndustrial Engineering vol 66 no 4 pp 751ndash757 2013

[12] F Zhang J Wang and B Liu ldquoOn the optimal and equilibriumretrial rates in an unreliable retrial queue with vacationsrdquo Jour-nal of Industrial andManagement Optimization vol 8 no 4 pp861ndash875 2012

[13] N Tian J Li and Z G Zhang ldquoMatrix analysis methodand working vacation queuesa surveyrdquo International Journal ofInformation and Management Sciences vol 20 pp 603ndash6332009

[14] F Zhang J Wang and B Liu ldquoEquilibrium balking strategiesin Markovian queues with working vacationsrdquo Applied Mathe-matical Modelling vol 37 no 16-17 pp 8264ndash8282 2013

[15] W Sun and S Li ldquoEquilibrium and optimal behavior of cus-tomers in Markovian queues with multiple working vacationsrdquoTOP vol 22 no 2 pp 694ndash715 2014

[16] W Sun S Li and Q-L Li ldquoEquilibrium balking strategiesof customers in Markovian queues with two-stage workingvacationsrdquo Applied Mathematics and Computation vol 248 pp195ndash214 2014

[17] J Li andNTian ldquoTheMM1 queuewithworking vacations andvacation interruptionsrdquo Journal of Systems Science and SystemsEngineering vol 16 no 1 pp 121ndash127 2007

[18] J-H Li N-S Tian and Z-Y Ma ldquoPerformance analysis ofGIM1 queue with working vacations and vacation interrup-tionrdquo Applied Mathematical Modelling vol 32 no 12 pp 2715ndash2730 2008

[19] M Neuts Matrix-Geometric Solution in Stochastic ModelsJohns Hopkins University Press Baltimore Md USA 1981

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Mathematical Problems in Engineering

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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

2 Mathematical Problems in Engineering

They discussed and compared the optimal and equilibriumretrial rates regarding the situations in which the customerswere cooperative or noncooperative respectively

Recently queueing systems with working vacations havebeen studied extensively where the server takes vacationsonce the system becomes empty (ie exhaustive service pol-icy) and can still serve customers at a lower rate than regularone during the vacations Research results of the stationarysystem performance on various working vacation queues canbe consulted in the survey given by Tian et al [13] As forthe work studying customers behavior in queueing systemswith working vacations Zhang et al [14] and Sun and Li[15] obtained equilibrium balking strategies inMM1 queueswith working vacations for four cases with respect to differentlevels of information Sun et al [16] also considered the cus-tomers equilibrium balking behavior in some single-serverMarkovian queues with two-stage working vacations

However we often encounter the situation that the servercan stop the vacation once some indices of the system suchas the number of customers achieve a certain value duringa vacation In many real life congestion situations urgentevents occur during a vacation and the servermust comebackto work rather than continuing to take the residual vacationFor example if the number of customers exceeds the specialvalue during a vacation and the server continues to take thevacation it leads to large cost of waiting customersThereforevacation interruption is more reasonable to the server vaca-tion queues Vacation interruption was introduced by Li andTian [17] and Li et al [18]

In this paper we consider anMM1 queueingmodel withworking vacations and vacation interruptionsWedistinguishthree cases the observable queues the partially observablequeues and the unobservable queues according to the infor-mation levels regarding system states As to the observablecase a customer can observe both the state of the server andthe number of present customers before making decision Inthe unobservable case a customer cannot observe the stateof the system However for the partially observable case acustomer only can observe the state of the server at theirarrival instant and does not observe the number of customerspresentThe customers dilemma is whether to join the systemor balk We study the balking behavior of customers in thesecases and derive Nash equilibrium strategies and sociallyoptimal strategies

This paper is organized as follows Descriptions of themodel are given in Section 2 In Sections 3 4 and 5 wedetermine the equilibrium and socially optimal strategies inobservable queues the partially observable queues and theunobservable queues respectively Conclusions are given inSection 6

2 Model Description

Consider a classical MM1 queue with an arrival rate 120582 anda service rate 120583

1 Upon the completion of service if there is

no customer in the system the server begins a vacation andthe vacation time is assumed to be exponentially distributedwith the parameter 120579 During a vacation period arriving

customers can be served at a mean rate of 1205830 Upon the com-

pletion of a service in the vacation if there are also customersin the queue the server ends the vacation and comes backto the normal working level Otherwise heshe continuesthe vacation until there are customers after a service or avacation ends Meanwhile when a vacation ends if thereare no customers another vacation is taken Otherwise theserver also switches the rate to 120583

1and a regular busy period

starts In this service discipline the server may come backfrom the vacation without completing the vacation And theserver can only go on vacations if there is no customer left inthe system upon the completion of a service Meanwhile thevacation service rate can be only applied to the first customerthat arrived during a vacation period

We assume that the interarrival times the service timesand the working vacation times are mutually independent Inaddition the service discipline is first in first out (FIFO)

Denote by119873(119905) the number of customers in the system attime 119905 Let 119869(119905) = 0 be the state that the server is on workingvacation period at time 119905 and let 119869(119905) = 1 be the state that theserver is busy at time 119905

Arriving customers are assumed to be identical Our int-erest is in the customersrsquo strategic response as they can decidewhether to join or balk upon arrival Assume that a customerrsquosutility consists of a reward for receiving service minus awaiting cost Specifically every customer receives a reward of119877 units for completing service There is a waiting cost of 119862units per time unit that the customer remains in the systemCustomers are risk neutral and maximize their expected netbenefit Finally we assume that there are no retrials of balkingcustomers nor reneging of waiting customers

3 The Observable Queues

We begin with the fully observable case in which the arrivingcustomers know not only the number of present customers119873(119905) but also the state of the server 119869(119905) at arrival time 119905

There exists a balk threshold 119899(119894) such that an arrivingcustomer enters the system at state 119894 (119894 = 0 1) if the numberof the customers present upon arrival does not exceed thespecified threshold So a pure threshold strategy is specifiedby a pair (119899

119890(0) 119899119890(1)) and the balking strategy has the follow-

ing form ldquowhile arriving at time 119905 observe (119873(119905) 119869(119905)) enterif119873(119905) le 119899

119890(119869(119905)) and balk otherwiserdquo

And we have the following result

Theorem 1 In the observable MM1 queue with workingvacations and vacation interruptions there exist thresholds(119899119890(0) 119899119890(1)) which are given by

119899119890(0) = lfloor

1198771205831

119862minus

1205831+ 120579

1205831(1205830+ 120579)

rfloor

119899119890(1) = lfloor

1198771205831

119862rfloor minus 1

(1)

such that a customer who observes the system at state (119873(119905)

119869(119905)) upon his arrival enters if 119873(119905) le 119899119890(119869(119905)) and balks

otherwise

Mathematical Problems in Engineering 3

Proof Based on the reward-cost structure which is imposedon the system we conclude that for an arriving customer thatdecides to enter hisher expected net benefit is

119880 (119899 119894) = 119877 minus 119862119879 (119899 119894) (2)

where 119879(119899 119894) denotes hisher expected mean sojourn timegiven that heshe finds the system at state (119899 119894) upon hisherarrival Then we have the following equations

119879 (0 0) =1

1205830+ 120579

+120579

1205830+ 120579

1

1205831

(3)

119879 (119899 0) =1205830

1205830+ 120579

(1

1205830

+ 119879 (119899 minus 1 1))

+120579

1205830+ 120579

119879 (119899 1)

(4)

119879 (119899 1) =119899 + 1

1205831

(5)

By iterating (4) and taking into account (5) we obtain

119879 (119899 0) =119899

1205831

+1205831+ 120579

1205831(1205830+ 120579)

(6)

We can easily check that 119879(119899 0) is strictly increasing for 119899A customer strictly prefers to enter if the reward for serviceexceeds the expected cost for waiting (ie 119880(119899 119894) gt 0) andis indifferent between entering and balking if the reward isequal to the cost (ie 119880(119899 119894) = 0)

We assume throughout the paper that

119877 gt 119862(1

1205830+ 120579

+120579

1205830+ 120579

1

1205831

) (7)

which ensures that the reward for service exceeds theexpected cost for a customer who finds the system empty Bysolving 119880(119899 119894) ge 0 for 119899 using (5) and (6) we obtain thatthe customer arriving at time 119905 decides to enter if and only if119899 le 119899119890(119869(119905)) where (119899

119890(0) 119899119890(1)) are given by (1)

From Figure 1 we observe that 119899119890(1) is larger than 119899

119890(0)

and 119899119890(0) and 119899

119890(1) are all increasing with 120583

1

For the stationary analysis the system follows a Markovchain and the state space is

Ω119891119900

= 0 0 cup (119899 119894) | 119899 = 1 2 119899 (119894) + 1 119894 = 0 1 (8)

And the transition diagram is illustrated in Figure 2

ne(0)ne(1)

2

4

6

8

10

12

14

n

12 14 16 18 2 22 24 26 28 311205831

Figure 1 Equilibrium threshold strategies for observable case when119877 = 5 119862 = 1 120582 = 1 120583

0= 05 and 120579 = 01

Let 120587119899119894= lim

119905rarrinfin119875119873(119905) = 119899 119869(119905) = 119894 with (119899 119894) isin Ω

119891119900

then 120587119899119894

(119899 119894) isin Ω119891119900 is the stationary distribution of the

process (119873(119905) 119869(119905)) 119905 ge 0In steady state we have the following system equations

12058212058700

= 120583012058710+ 120583112058711 (9)

(120582 + 120579 + 1205830) 1205871198990

= 120582120587119899minus10

119899 = 1 2 119899 (0) (10)

(120579 + 1205830) 120587119899(0)+10

= 120582120587119899(0)0

(11)

(120582 + 1205831) 12058711

= 120583112058721+ 12057912058710+ 120583012058720 (12)

(120582 + 1205831) 1205871198991

= 120582120587119899minus11

+ 1205831120587119899+11

+ 1205791205871198990

+ 1205830120587119899+10

119899 = 2 3 119899 (0)

(13)

(120582 + 1205831) 120587119899(0)+11

= 120582120587119899(0)1

+ 1205831120587119899(0)+2

+ 120579120587119899(0)+10

(14)

(120582 + 1205831) 1205871198991

= 120582120587119899minus11

+ 1205831120587119899+11

119899 = 119899 (0) + 2 119899 (1)

(15)

1205831120587119899(1)+11

= 120582120587119899(1)1

(16)

Define

120588 =120582

1205831

120590 =120582

120582 + 120579 + 1205830

(17)

4 Mathematical Problems in Engineering

1 1 2 1

1 00 0 2 0

120579120579120579120579

1205831 1205831 1205831 1205831 1205831 1205831 1205831

1205830120583012058301205830

1205830

120582120582120582120582 120582

120582120582120582120582120582120582120582middot middot middot

middot middot middot

middot middot middot

1205831

n(0) 1

n(0) 0

n(0) + 1 1

n(0) + 1 0

n(1) 1 n(1) + 1 1

Figure 2 Transition rate diagram for the observable queues

By iterating (10) and (15) taking into account (11) and (16)we obtain

1205871198990

= 12059011989912058700 119899 = 0 1 2 119899 (0)

120587119899(0)+10

=120582120590119899(0)

120579 + 1205830

12058700

1205871198991

= 120588119899minus119899(0)minus1

120587119899(0)+11

119899 = 119899 (0) + 1 119899 (1) + 1

(18)

From (13) it is easy to obtain that 1205871198991

| 1 le 119899 le 119899(0)+1 isa solution of the nonhomogeneous linear difference equation

1205831119909119899+1

minus (120582 + 1205831) 119909119899+ 120582119909119899minus1

= minus1205791205871198990minus 1205830120587119899+10

= minus (120579 + 1205830120590) 12059011989912058700

(19)

So its corresponding characteristic equation is

12058311199092minus (120582 + 120583

1) 119909 + 120582 = 0 (20)

which has two roots at 1 and 120588 Assume that 120588 = 1 thenthe homogeneous solution of (19) is 119909hom

119899= 1198601119899+ 119861120588119899 The

general solution of the nonhomogeneous equation is given as119909gen119899

= 119909hom119899

+119909spec119899

where 119909spec119899

is a specific solution We finda specific solution 119909

spec119899

= 119863120590119899 Substituting it into (19) we

derive

119863 = minus120582 + 120579

120582 + 120579 + 1205830minus 1205831

12058700 (21)

Therefore

119909gen119899

= 119860 + 119861120588119899+ 119863120590119899 119899 = 1 2 119899 (0) + 1 (22)

Substituting (22) into (9) and (12) it follows after some rathertedious algebra that

119860 + 119861120588 = minus120582 (120582 + 120579)

1205831(1205831minus 120582 minus 120579 minus 120583

0)12058700

119860 + 1198611205882= minus

1205822(120582 + 120579)

12058321(1205831minus 120582 minus 120579 minus 120583

0)12058700

(23)

Then we obtain

119860 = 0

119861 = minus120582 + 120579

1205831minus 120582 minus 120579 minus 120583

0

12058700

(24)

Thus from (22) we obtain

1205871198991

= 119861120588119899+ 119863120590119899 119899 = 1 2 119899 (0) + 1 (25)

Thus we have all the stationary probabilities in terms of12058700 The remaining probability 120587

00 can be found from the

normalization condition After some algebraic simplifica-tions we can summarize the results in the following

Theorem 2 In the observable MM1 queues with workingvacations and vacation interruptions the stationary distribu-tion 120587

119899119894| (119899 119894) isin Ω

119891119900 is given as follows

1205871198990

= 12059011989912058700 119899 = 0 1 2 119899 (0)

120587119899(0)+10

=120582120590119899(0)

120579 + 1205830

12058700

1205871198991

= 119861120588119899+ 119863120590119899 119899 = 1 2 119899 (0) + 1

1205871198991

= 120588119899minus119899(0)minus1

(119861120588119899(0)+1

+ 119863120590119899(0)+1

)

119899 = 119899 (0) + 2 119899 (1) + 1

(26)

where 12058700can be solved by the normalization equation

Because the probability of balking is equal to 120587119899(0)+10

+

120587119899(1)+11

the social benefit per time unit when all customers fol-low the threshold policy (119899(0) 119899(1)) equals

119880119904(119899 (0) 119899 (1)) = 120582119877 (1 minus 120587

119899(0)+10minus 120587119899(1)+11

)

minus 119862(

119899(0)+1

sum

119899=0

1198991205871198990+

119899(1)+1

sum

119899=1

1198991205871198991)

(27)

Let (119899lowast(0) 119899lowast(1)) be the socially optimal threshold strategyFrom Figure 3 we obtain (119899

lowast(0) 119899lowast(1)) = (3 4) while the

equilibrium threshold strategy (119899119890(0) 119899119890(1)) = (78 79) We

observe that 119899119890(0) gt 119899

lowast(0) and 119899

119890(1) gt 119899

lowast(1) which indicates

that the individual optimization is inconsistent with the socialoptimization

4 Partially Observable Queues

In this section we turn our attention to the partially observ-able case where arriving customers only can observe the stateof the server at their arrival instant and do not observe thenumber of customers present

A mixed strategy for a customer is specified by a vector(1199020 1199021) where 119902

119894is the probability of joining when the server

is in state 119894 (119894 = 0 1) If all customers follow the same mixed

Mathematical Problems in Engineering 5

2

3

4

5

6

7

8

9

n(1

)

2 3 4 5 6 7 81n(0)

24

68

10

02

46

8

n(1)n(0)

0

10

20

30

40

50

Us(n

(0)n

(1))

68

Figure 3 Social welfare for the observable case when 119877 = 40 119862 = 1 120582 = 15 1205831= 2 120583

0= 05 and 120579 = 05

1 1 2 0 3 1

0 0 1 0 2 0 3 0

n 1

n 0

120579

1205831 1205831

12058301205830

120582q1120582q1

120582q0 120582q0middot middot middot

middot middot middot

middot middot middot

middot middot middot

120579

1205831 1205831

12058301205830

120582q1120582q1

120582q0 120582q0

120579120579

1205831

1205831

1205830

1205830

120582q1

120582q0 120582q0

Figure 4 Transition rate diagram for the partially observable queues

strategy (1199020 1199021) then the system follows a Markov chain

where the arrival rate equals 120582119894= 120582119902119894when the server is in

state 119894 (119894 = 0 1) The state space is Ω119901119900

= 0 0 cup (119899 119894) |

119899 ge 1 119894 = 0 1 and the transition diagram is illustrated inFigure 4

Denote the stationary distribution as120587119899119894= lim119905rarrinfin

119875 119873 (119905) = 119899 119869 (119905) = 119894 (119899 119894) isin Ω119901119900

1205870= 12058700

120587119899= (1205871198990 1205871198991) 119899 ge 1

(28)

Using the lexicographical sequence for the states thetransition rate matrix (generator) Q can be written as thetridiagonal block matrix

Q =((

(

A0C0

B1

A CB A C

B A Cd d d

))

)

(29)

whereA0= minus120582119902

0

B1= (

1205830

1205831

)

C1= (1205821199020 0)

A = (minus (1205821199020+ 120579 + 120583

0) 120579

0 minus (1205821199021+ 1205831))

B = (0 1205830

0 1205831

)

C = (1205821199020

0

0 1205821199021

)

(30)To analyze this QBD process it is necessary to solve for

the minimal nonnegative solution of the matrix quadraticequation

R2B + RA + C = 0 (31)and this solution is called the rate matrix and denoted by R

Lemma 3 If 1205881

= 12058211990211205831

lt 1 the minimal nonnegativesolution of (31) is given by

R = (1205900

1205821199020+ 120579

1205831

1205900

0 1205881

) (32)

where 1205900= 1205821199020(1205821199020+ 120579 + 120583

0)

Proof Because the coefficients of (31) are all upper triangularmatrices we can assume that R has the same structure as

R = (11990311

11990312

0 11990322

) (33)

6 Mathematical Problems in Engineering

Substituting R2 and R into (31) yields the following set ofequations

minus (1205821199020+ 120579 + 120583

0) 11990311+ 1205821199020= 0

12058311199032

22minus (1205821199021+ 1205831) 11990322+ 1205821199021= 0

12058301199032

11+ 120583111990312(11990311+ 11990322) + 120579119903

11minus (1205821199021+ 1205831) 11990312

= 0

(34)

To obtain the minimal nonnegative solution of (31) we take11990322

= 1205881(the other roots are 119903

22= 1) in the second equation

of (34) From the first equation of (34) we obtain 11990311

=

1205821199020(1205821199020+ 120579 + 120583

0) = 1205900 Substituting 120590

0and 1205881into the last

equation of (34) we get 11990312

= ((1205821199020+ 120579)120583

1)1205900 This is the

proof of Lemma 3

Theorem 4 Assuming that 1205881lt 1 the stationary probabilities

120587119899119894

| (119899 119894) isin Ω119901119900 of the partially observable queues are as

follows

1205871198990

= 119870120590119899

0 119899 ge 0

1205871198991

= 1198701205881

1205821199020+ 120579

1205821199020+ 120579 + 120583

0

119899minus1

sum

119895=0

120588119895

1120590119899minus1minus119895

0 119899 ge 1

(35)

where

119870 =120579 + 1205830

1205821199020+ 120579 + 120583

0

(1 +1205821199020+ 120579

1205821199020+ 120579 + 120583

0

1205821199020

1205831minus 1205821199021

)

minus1

(36)

Proof With the matrix-geometric solution method [19] wehave

120587119899= (1205871198990 1205871198991) = (120587

10 12058711)R119899minus1 119899 ge 1 (37)

In addition (12058700 12058710 12058711) satisfies the set of equations

(12058700 12058710 12058711) 119861 [R] = 0 (38)

where

119861 [R] = (A0

C0

B1

A + RB)

= (

minus1205821199020

1205821199020

0

1205830

minus (1205821199020+ 120579 + 120583

0) 1205821199020+ 120579

1205831

0 minus1205831

)

(39)

Then from (38) we derive

minus120582119902012058700+ 120583012058710+ 120583112058711

= 0

120582119902012058700minus (1205821199020+ 120579 + 120583

0) 12058710

(1205821199020+ 120579) 120587

10minus 120583112058711

= 0

(40)

Taking 12058700as a known constant we obtain

12058710

= 120590012058700 (41)

12058711

= 1205881

1205821199020+ 120579

1205821199020+ 120579 + 120583

0

12058700 (42)

Note that

R119899minus1 = (120590119899minus1

0

1205821199020+ 120579

1205831

1205900

119899minus2

sum

119895=0

120590119895

0120588119899minus119895minus2

1

0 120588119899minus1

1

) 119899 ge 2 (43)

By 12058710 12058711 R119899minus1 and (37) we get (35) Finally 120587

00can be

determined by the normalization conditionThis is the proofof Theorem 4

Let 119901119894be the steady-state probability when the server is at

state 119894 (119894 = 0 1) we have

1199010=

infin

sum

119899=0

1205871198990

= 1198701205821199020+ 120579 + 120583

0

120579 + 1205830

(44)

1199011=

infin

sum

119899=1

1205871198991

= 1198701205821199020(1205821199020+ 120579)

(1205831minus 1205821199021) (120579 + 120583

0) (45)

So the effective arrival rate 120582 is given by

120582 = 120582 (11990101199020+ 11990111199021) (46)

We now consider a customer who finds the server at state119894 upon arrival The conditional mean sojourn time of such acustomer that decides to enter given that the others follow thesame mixed strategy (119902

0 1199021) is given by

119882119894(1199020 1199021) =

suminfin

119899=119894119879 (119899 119894) 120587

119899119894

suminfin

119899=119894120587119899119894

(47)

By substituting (5)-(6) and (35) into (44) we obtain

1198820(1199020 1199021) =

1205821199020+ 120579 + 120583

1

1205831(120579 + 120583

0)

1198821(1199020 1199021) =

1

1205831minus 1205821199021

+1205821199020+ 120579 + 120583

0

1205831(1205830+ 120579)

(48)

Based on the reward-cost structure the expected net ben-efit of an arriving customer if heshe finds the server at state119894 and decides to enter is given by

1198800(1199020) = 119877 minus 119862

1205821199020+ 120579 + 120583

1

1205831(120579 + 120583

0)

1198801(1199020 1199021) = 119877 minus 119862(

1

1205831minus 1205821199021

+1205821199020+ 120579 + 120583

0

1205831(1205830+ 120579)

)

(49)

We now can proceed to determine mixed equilibriumstrategies of a customer in the partially observable case andhave the following

Theorem 5 For the partially observable case there exists aunique mixed equilibrium strategy (119902

119890

0 119902119890

1) where the vector

(119902119890

0 119902119890

1) is given as follows

(1) 119862(120579 + 1205831)1205831(120579 + 120583

0) lt 119877 le 119862(120582 + 120579 + 120583

1)1205831(120579 + 120583

0)

(119902119890

0 119902119890

1) =

(1199091 0)

1205831minus 1205830

1205831(120579 + 120583

0)lt

1

120583

(1199091 1199092)

1

120583le

1205831minus 1205830

1205831(120579 + 120583

0)le

1

120583 minus 120582

(1199091 1)

1205831minus 1205830

1205831(120579 + 120583

0)gt

1

120583 minus 120582

(50)

Mathematical Problems in Engineering 7

(2) 119862(120582 + 120579 + 1205831)1205831(120579 + 120583

0) lt 119877

(119902119890

0 119902119890

1) =

(1 0) 119877 lt119862

1205831

+119862 (120582 + 120579 + 120583

0)

1205831(120579 + 120583

0)

(1 1199093)

119862

1205831

+119862 (120582 + 120579 + 120583

0)

1205831(120579 + 120583

0)

le 119877 le119862

1205831minus 120582

+119862 (120582 + 120579 + 120583

0)

1205831(120579 + 120583

0)

(1 1) 119877 gt119862

1205831minus 120582

+119862 (120582 + 120579 + 120583

0)

1205831(120579 + 120583

0)

(51)

where

1199091=

1

120582(1198771205831(120579 + 120583

0)

119862minus 120579 minus 120583

1)

1199092=1205831

120582(1 minus

1205830+ 120579

1205831minus 1205830

)

1199093=

1

120582(1205831minus

1

119877119862 minus (120582 + 120579 + 1205830) 1205831(120579 + 120583

0))

(52)

Proof To prove Theorem 5 we first focus on 1198800(1199020) Con-

dition (1) assures that 1199021198900is positive Therefore we have two

casesCase 1 (119880

0(0) gt 0 and 119880

0(1) le 0) That is 119862(120579 + 120583

1)1205831(120579 +

1205830) lt 119877 le 119862(120582+120579+120583

1)1205831(120579+1205830)) In this case if all customers

who find the system empty enter the system with probability119902119890

0= 1 then the tagged customer suffers a negative expected

benefit if heshe decides to enter Hence 1199021198900= 1 does not lead

to an equilibrium Similarly if all customers use 1199021198900= 0 then

the tagged customer receives a positive benefit from enteringthus 119902119890

0= 0 also cannot be part of an equilibrium mixed

strategy Therefore there exists unique 1199021198900satisfying

119877 minus 119862120582119902119890

0+ 120579 + 120583

1

1205831(120579 + 120583

0)

= 0 (53)

for which customers are indifferent between entering andbalking This is given by

119902119890

0=

1

120582(1198771205831(120579 + 120583

0)

119862minus 120579 minus 120583

1) = 119909

1 (54)

In this situation the expected benefit 1198801is given by

1198801(119902119890

0 1199021) =

119862 (1205831minus 1205830)

1205831(120579 + 120583

0)minus

119862

120583 minus 1205821199021

(55)

Using the standard methodology of equilibrium analysisin unobservable queueing models we derive from (55) thefollowing equilibria

(1) If 1198801(119902119890

0 0) lt 0 then the equilibrium strategy is 119902119890

1=

0(2) If 119880

1(119902119890

0 0) ge 0 and 119880

1(119902119890

0 1) le 0 there exists unique

119902119890

1 satisfying

1198801(119902119890

0 119902119890

1) = 0 (56)

then the equilibrium strategy is 1199021198901= 1199092

(3) If 1198801(119902119890

0 1) gt 0 then the equilibrium strategy is 119902119890

1=

1

Case 2 (1198800(1) ge 0) That is 119862(120582 + 120579 + 120583

1)1205831(120579 + 120583

0) lt 119877) In

this case for every strategy of the other customers the taggedcustomer has a positive expected net benefit if heshe decidesto enter Hence 119902119890

0= 1

In this situation the expected benefit 1198801is given by

1198801(1 1199021) = 119877 minus

119862

120583 minus 1205821199021

minus119862 (120582 + 120579 + 120583

0)

1205831(120579 + 120583

0)

(57)

To find 119902119890

1in equilibrium we also consider the following

subcases

(1) If 1198801(1 0) lt 0 then the equilibrium strategy is 119902119890

1=

0

(2) If 1198801(1 0) ge 0 and 119880

1(1 1) le 0 there exists unique

119902119890

1 satisfying

1198801(1 119902119890

1) = 0 (58)

then the equilibrium strategy is 1199021198901= 1199093

(3) If 1198801(1 1) gt 0 then the equilibrium strategy is 119902119890

1=

1

By rearranging Cases 1 and 2 we can obtain the results ofTheorem 5 This completes the proof

From Theorem 5 we can compare 119902119890

0with 119902

119890

1 Figure 5

shows that 1199021198900is not always less than 119902

119890

1 Namely sometimes

the information that the server takes a working vacation doesnot make the customers less willing to enter the systembecause in the vacation the server provides a lower serviceto customers But when the vacation time and waiting timebecome longer customers do not want to enter the system invacation

Then fromTheorem 4 themean queue length is given by

119864 [119871] =

infin

sum

119899=0

1198991205871198990+

infin

sum

119899=0

1198991205871198991

= 1198701205900

(1 minus 1205900)2(1 +

1205821199020+ 120579

1205831

1 minus 12058811205900

(1 minus 1205881)2)

(59)

8 Mathematical Problems in Engineering

03

04

05

06

07

08

09

1

Equi

libriu

m jo

inin

g pr

obab

ility

qe0qe1 (120579 = 01 C = 2)

qe0qe1 (120579 = 05 C = 3)

0 04 06 08 1 12 14 16 1802Arrival rate 120582

Figure 5 Equilibrium mixed strategies for the partially observablecase when 119877 = 5 120583

1= 2 and 120583

0= 05

0 04 06 08 1 12 14 16 1802Arrival rate 120582

0

01

02

03

04

05

06

07

08

09

1

Join

ing

prob

abili

ty

qe0qe1

qlowast0qlowast1

Figure 6 Equilibrium and socially optimal mixed strategies for thepartially observable case when 119877 = 5 119862 = 2 120583

1= 2 120583

0= 05 and

120579 = 01

And the social benefit when all customers follow a mixedpolicy (119902

0 1199021) can now be easily computed as follows

119880119904(1199020 1199021) = 120582119877 minus 119862119864 [119871]

= 120582 (11990101199020+ 11990111199021) 119877

minus1198621198701205900

(1 minus 1205900)2(1 +

1205821199020+ 120579

1205831

1 minus 12058811205900

(1 minus 1205881)2)

(60)

The goal of a social planner is to maximize overall socialwelfare Let (119902lowast

0 119902lowast

1) be the socially optimal mixed strategy

Figure 6 compares customers equilibrium mixed strategy(119902119890

0 119902119890

1) and their socially optimal mixed strategy (119902lowast

0 119902lowast

1) We

also observe that 1199021198900gt 119902lowast

0and 1199021198901gt 119902lowast

1This ordering is typical

when customers make individual decisions maximizing theirown profitThen they ignore those negative externalities thatthey impose on later arrivals and they tend to overuse thesystem It is clear that these externalities should be taken intoaccount when we aim to maximize the total revenue

5 The Unobservable Queues

In this section we consider the fully unobservable case wherearriving customers cannot observe the system state

There are two pure strategies available for a customer thatis to join the queue or not to join the queue A pure or mixedstrategy can be described by a fraction 119902 (0 le 119902 le 1) whichis the probability of joining and the effective arrival rate orjoining rate is 120582119902 The transition diagram is illustrated inFigure 7

To identify the equilibrium strategies of the customerswe should first investigate the stationary distribution of thesystem when all customers follow a given strategy 119902 whichcan be obtained by Theorem 4 by taking 119902

0= 1199021= 119902 So we

have

1205871198990

= 1198701120590119899

119902 119899 ge 0

1205871198991

= 1198701120588119902

120582119902 + 120579

120582119902 + 120579 + 1205830

119899minus1

sum

119895=0

120588119895

119902120590119899minus1minus119895

119902 119899 ge 1

(61)

where 120588119902= 120582119902120583

1 120590119902= 120582119902(120582119902 + 120579 + 120583

0) and

1198701=

120579 + 1205830

120582119902 + 120579 + 1205830

(1 +120582119902 + 120579

120582119902 + 120579 + 1205830

120582119902

1205831minus 120582119902

)

minus1

(62)

Then the mean number of the customers in the system is

119864 [119871] =

infin

sum

119899=0

1198991205871198990+

infin

sum

119899=0

1198991205871198991

=1198701120590119902

(1 minus 120590119902)2(1 +

120582119902 + 120579

1205831

1 minus 120588119902120590119902

(1 minus 120588119902)2)

(63)

Hence the mean sojourn time of a customer who decides toenter upon hisher arrival can be obtained by using Littlersquoslaw

119864 [119882] =1198701120590119902

120582119902 (1 minus 120590119902)2(1 +

120582119902 + 120579

1205831

1 minus 120588119902120590119902

(1 minus 120588119902)2)

=1

120579 + 1205830

(1

+1205831(120579 + 120583

0)

1205831minus 120582119902

1

1205830((1205831minus 120582119902) (120582119902 + 120579)) + 120583

1

)

(64)

And if 120582 lt 1205831 119864[119882] is strictly increasing for 119902 isin [0 1]

Mathematical Problems in Engineering 9

1 1 2 0 3 1

0 0 1 0 2 0 3 0

n 1

n 0

120579

1205831 1205831

12058301205830

120582q120582q

120582q 120582qmiddot middot middot

middot middot middot

middot middot middot

middot middot middot

120579

1205831 1205831

12058301205830

120582q120582q

120582q 120582q

120579120579

1205831

1205831

1205830

1205830

120582q

120582q 120582q

Figure 7 Transition rate diagram for the unobservable queues

We consider a tagged customer If heshe decides to enterthe system hisher expected net benefit is

119880 (119902) = 119877 minus 119862119864 [119882] = 119877 minus119862

120579 + 1205830

(1

+1205831(120579 + 120583

0)

1205831minus 120582119902

1

1205830((1205831minus 120582119902) (120582119902 + 120579)) + 120583

1

)

(65)

We note that 119880(119902) is strictly decreasing in 119902 and it has aunique root 119902lowast

119890 So there exists a unique mixed equilibrium

strategy 119902119890and 119902119890= min119902lowast

119890 1

We now turn our attention to social optimization Thesocial benefit per time unit can now be easily computed as

119880119904(119902) = 120582119902 (119877 minus 119862119864 [119882]) = 120582119902(119877 minus

119862

120579 + 1205830

(1

+1205831(120579 + 120583

0)

1205831minus 120582119902

1

1205830((1205831minus 120582119902) (120582119902 + 120579)) + 120583

1

))

(66)

As for the equilibrium social welfare per time unit119880119904(119902119890)

Figure 8 shows that 119880119904(119902119890) first increases and then decreases

with 120582 and the social benefit achieves a maximum forintermediate values of this parameter The reason for thisbehavior is that when the arrival rate is small the system israrely crowded therefore as more customers arrive they areserved and the social benefit improves However with anysmall increase of the arrival rate 120582 the expected waiting timeincreases which has a detrimental effect on the social benefit

Let 119909lowast be the root of the equation 1198781015840(119902) = 0 and let 119902lowast be

the optimal joining probability Then we have the followingconclusions if 0 lt 119909

lowastlt 1 and 119878

10158401015840(119902) le 0 then 119902

lowast= 119909lowast if

0 lt 119909lowastlt 1 and 119878

10158401015840(119902) gt 0 or 119909lowast ge 1 then 119902

lowast= 1

From Figure 9 we observe that 119902119890

gt 119902lowast As a result

individual optimization leads to queues that are longer thansocially desired Therefore it is clear that the social plannerwants a toll to discourage arrivals

6 Conclusion

In this paper we analyzed the customer strategic behaviorin the MM1 queueing system with working vacations andvacation interruptions where arriving customers have optionto decide whether to join the system or balk Three differentcases with respect to the levels of information provided toarriving customers have been investigated extensively and theequilibrium strategies for each case were derived And wefound that the equilibrium joining probability of the busy

02 04 06 08 1 12 14 16 18 20Arrival rate 120582

Equi

libriu

m so

cial

wel

fare

Us(qe)

0

5

10

15

Figure 8 Equilibrium social welfare for the unobservable casewhen119877 = 10 119862 = 1 120583

1= 2 120583

0= 1 and 120579 = 1

03

04

05

06

07

08

09

1

Join

ing

prob

abili

ty

0 1 15 205Arrival rate 120582

qeqlowast

Figure 9 Equilibrium and socially optimal strategies for the unob-servable case when 119877 = 3 119862 = 2 120583

1= 2 120583

0= 05 and 120579 = 001

period may be smaller than that in vacation time We alsocompared the equilibrium strategy with the socially optimalstrategy numerically and we observed that customers makeindividual decisions maximizing their own profit and tendto overuse the system For the social planner a toll will beadopted to discourage customers from joining

10 Mathematical Problems in Engineering

Furthermore an interesting extension would be to incor-porate the present model in a profit maximizing frameworkwhere the owner or manager of the system imposes anentrance feeThe problem is to find the optimal fee that max-imizes the owners total profit subject to the constraint thatcustomers independently decide whether to enter or not andtake into account the fee in addition to the service reward andwaiting cost Another direction for future work is to concernthe non-Markovian queues with various vacation policies

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Acknowledgments

This work is partially supported by the Natural Foundation ofHebei Province (nos A2014203096 andC2014203212) Chinaand the Young Faculty Autonomous Research Project of Yan-shan University (nos 13LGB032 13LGA017 and 13LGB013)China

References

[1] P Naor ldquoThe regulation of queue size by levying tollsrdquo Econo-metrica vol 37 no 1 pp 15ndash24 1969

[2] N M Edelson and D K Hildebrand ldquoCongestion tolls forPoisson queuing processesrdquo Econometrica vol 43 no 1 pp 81ndash92 1975

[3] R Hassin and M Haviv Equilibrium Behavior in Queueing Sys-tems To Queue or Not to Queue Kluwer Academic PublishersDordrecht The Netherlands 2003

[4] A Burnetas and A Economou ldquoEquilibrium customer strate-gies in a single server Markovian queue with setup timesrdquoQueueing Systems vol 56 no 3-4 pp 213ndash228 2007

[5] W Sun and N Tian ldquoContrast of the equilibrium and sociallyoptimal strategies in a queue with vacationsrdquo Journal of Compu-tational Information Systems vol 4 no 5 pp 2167ndash2172 2008

[6] A Economou and S Kanta ldquoEquilibrium balking strategiesin the observable single-server queue with breakdowns andrepairsrdquoOperations Research Letters vol 36 no 6 pp 696ndash6992008

[7] P Guo and R Hassin ldquoStrategic behavior and social optimiza-tion in Markovian vacation queuesrdquo Operations Research vol59 no 4 pp 986ndash997 2011

[8] P Guo and R Hassin ldquoStrategic behavior and social optimiza-tion in Markovian vacation queues the case of heterogeneouscustomersrdquo European Journal of Operational Research vol 222no 2 pp 278ndash286 2012

[9] R Tian D Yue and W Yue ldquoOptimal balking strategies inan MG1 queueing system with a removable server under N-policyrdquo Journal of Industrial andManagementOptimization vol11 no 3 pp 715ndash731 2015

[10] P Guo and Q Li ldquoStrategic behavior and social optimizationin partially-observableMarkovian vacation queuesrdquoOperationsResearch Letters vol 41 no 3 pp 277ndash284 2013

[11] L Li J Wang and F Zhang ldquoEquilibrium customer strategiesin Markovian queues with partial breakdownsrdquo Computers ampIndustrial Engineering vol 66 no 4 pp 751ndash757 2013

[12] F Zhang J Wang and B Liu ldquoOn the optimal and equilibriumretrial rates in an unreliable retrial queue with vacationsrdquo Jour-nal of Industrial andManagement Optimization vol 8 no 4 pp861ndash875 2012

[13] N Tian J Li and Z G Zhang ldquoMatrix analysis methodand working vacation queuesa surveyrdquo International Journal ofInformation and Management Sciences vol 20 pp 603ndash6332009

[14] F Zhang J Wang and B Liu ldquoEquilibrium balking strategiesin Markovian queues with working vacationsrdquo Applied Mathe-matical Modelling vol 37 no 16-17 pp 8264ndash8282 2013

[15] W Sun and S Li ldquoEquilibrium and optimal behavior of cus-tomers in Markovian queues with multiple working vacationsrdquoTOP vol 22 no 2 pp 694ndash715 2014

[16] W Sun S Li and Q-L Li ldquoEquilibrium balking strategiesof customers in Markovian queues with two-stage workingvacationsrdquo Applied Mathematics and Computation vol 248 pp195ndash214 2014

[17] J Li andNTian ldquoTheMM1 queuewithworking vacations andvacation interruptionsrdquo Journal of Systems Science and SystemsEngineering vol 16 no 1 pp 121ndash127 2007

[18] J-H Li N-S Tian and Z-Y Ma ldquoPerformance analysis ofGIM1 queue with working vacations and vacation interrup-tionrdquo Applied Mathematical Modelling vol 32 no 12 pp 2715ndash2730 2008

[19] M Neuts Matrix-Geometric Solution in Stochastic ModelsJohns Hopkins University Press Baltimore Md USA 1981

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

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CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

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The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Mathematical Problems in Engineering 3

Proof Based on the reward-cost structure which is imposedon the system we conclude that for an arriving customer thatdecides to enter hisher expected net benefit is

119880 (119899 119894) = 119877 minus 119862119879 (119899 119894) (2)

where 119879(119899 119894) denotes hisher expected mean sojourn timegiven that heshe finds the system at state (119899 119894) upon hisherarrival Then we have the following equations

119879 (0 0) =1

1205830+ 120579

+120579

1205830+ 120579

1

1205831

(3)

119879 (119899 0) =1205830

1205830+ 120579

(1

1205830

+ 119879 (119899 minus 1 1))

+120579

1205830+ 120579

119879 (119899 1)

(4)

119879 (119899 1) =119899 + 1

1205831

(5)

By iterating (4) and taking into account (5) we obtain

119879 (119899 0) =119899

1205831

+1205831+ 120579

1205831(1205830+ 120579)

(6)

We can easily check that 119879(119899 0) is strictly increasing for 119899A customer strictly prefers to enter if the reward for serviceexceeds the expected cost for waiting (ie 119880(119899 119894) gt 0) andis indifferent between entering and balking if the reward isequal to the cost (ie 119880(119899 119894) = 0)

We assume throughout the paper that

119877 gt 119862(1

1205830+ 120579

+120579

1205830+ 120579

1

1205831

) (7)

which ensures that the reward for service exceeds theexpected cost for a customer who finds the system empty Bysolving 119880(119899 119894) ge 0 for 119899 using (5) and (6) we obtain thatthe customer arriving at time 119905 decides to enter if and only if119899 le 119899119890(119869(119905)) where (119899

119890(0) 119899119890(1)) are given by (1)

From Figure 1 we observe that 119899119890(1) is larger than 119899

119890(0)

and 119899119890(0) and 119899

119890(1) are all increasing with 120583

1

For the stationary analysis the system follows a Markovchain and the state space is

Ω119891119900

= 0 0 cup (119899 119894) | 119899 = 1 2 119899 (119894) + 1 119894 = 0 1 (8)

And the transition diagram is illustrated in Figure 2

ne(0)ne(1)

2

4

6

8

10

12

14

n

12 14 16 18 2 22 24 26 28 311205831

Figure 1 Equilibrium threshold strategies for observable case when119877 = 5 119862 = 1 120582 = 1 120583

0= 05 and 120579 = 01

Let 120587119899119894= lim

119905rarrinfin119875119873(119905) = 119899 119869(119905) = 119894 with (119899 119894) isin Ω

119891119900

then 120587119899119894

(119899 119894) isin Ω119891119900 is the stationary distribution of the

process (119873(119905) 119869(119905)) 119905 ge 0In steady state we have the following system equations

12058212058700

= 120583012058710+ 120583112058711 (9)

(120582 + 120579 + 1205830) 1205871198990

= 120582120587119899minus10

119899 = 1 2 119899 (0) (10)

(120579 + 1205830) 120587119899(0)+10

= 120582120587119899(0)0

(11)

(120582 + 1205831) 12058711

= 120583112058721+ 12057912058710+ 120583012058720 (12)

(120582 + 1205831) 1205871198991

= 120582120587119899minus11

+ 1205831120587119899+11

+ 1205791205871198990

+ 1205830120587119899+10

119899 = 2 3 119899 (0)

(13)

(120582 + 1205831) 120587119899(0)+11

= 120582120587119899(0)1

+ 1205831120587119899(0)+2

+ 120579120587119899(0)+10

(14)

(120582 + 1205831) 1205871198991

= 120582120587119899minus11

+ 1205831120587119899+11

119899 = 119899 (0) + 2 119899 (1)

(15)

1205831120587119899(1)+11

= 120582120587119899(1)1

(16)

Define

120588 =120582

1205831

120590 =120582

120582 + 120579 + 1205830

(17)

4 Mathematical Problems in Engineering

1 1 2 1

1 00 0 2 0

120579120579120579120579

1205831 1205831 1205831 1205831 1205831 1205831 1205831

1205830120583012058301205830

1205830

120582120582120582120582 120582

120582120582120582120582120582120582120582middot middot middot

middot middot middot

middot middot middot

1205831

n(0) 1

n(0) 0

n(0) + 1 1

n(0) + 1 0

n(1) 1 n(1) + 1 1

Figure 2 Transition rate diagram for the observable queues

By iterating (10) and (15) taking into account (11) and (16)we obtain

1205871198990

= 12059011989912058700 119899 = 0 1 2 119899 (0)

120587119899(0)+10

=120582120590119899(0)

120579 + 1205830

12058700

1205871198991

= 120588119899minus119899(0)minus1

120587119899(0)+11

119899 = 119899 (0) + 1 119899 (1) + 1

(18)

From (13) it is easy to obtain that 1205871198991

| 1 le 119899 le 119899(0)+1 isa solution of the nonhomogeneous linear difference equation

1205831119909119899+1

minus (120582 + 1205831) 119909119899+ 120582119909119899minus1

= minus1205791205871198990minus 1205830120587119899+10

= minus (120579 + 1205830120590) 12059011989912058700

(19)

So its corresponding characteristic equation is

12058311199092minus (120582 + 120583

1) 119909 + 120582 = 0 (20)

which has two roots at 1 and 120588 Assume that 120588 = 1 thenthe homogeneous solution of (19) is 119909hom

119899= 1198601119899+ 119861120588119899 The

general solution of the nonhomogeneous equation is given as119909gen119899

= 119909hom119899

+119909spec119899

where 119909spec119899

is a specific solution We finda specific solution 119909

spec119899

= 119863120590119899 Substituting it into (19) we

derive

119863 = minus120582 + 120579

120582 + 120579 + 1205830minus 1205831

12058700 (21)

Therefore

119909gen119899

= 119860 + 119861120588119899+ 119863120590119899 119899 = 1 2 119899 (0) + 1 (22)

Substituting (22) into (9) and (12) it follows after some rathertedious algebra that

119860 + 119861120588 = minus120582 (120582 + 120579)

1205831(1205831minus 120582 minus 120579 minus 120583

0)12058700

119860 + 1198611205882= minus

1205822(120582 + 120579)

12058321(1205831minus 120582 minus 120579 minus 120583

0)12058700

(23)

Then we obtain

119860 = 0

119861 = minus120582 + 120579

1205831minus 120582 minus 120579 minus 120583

0

12058700

(24)

Thus from (22) we obtain

1205871198991

= 119861120588119899+ 119863120590119899 119899 = 1 2 119899 (0) + 1 (25)

Thus we have all the stationary probabilities in terms of12058700 The remaining probability 120587

00 can be found from the

normalization condition After some algebraic simplifica-tions we can summarize the results in the following

Theorem 2 In the observable MM1 queues with workingvacations and vacation interruptions the stationary distribu-tion 120587

119899119894| (119899 119894) isin Ω

119891119900 is given as follows

1205871198990

= 12059011989912058700 119899 = 0 1 2 119899 (0)

120587119899(0)+10

=120582120590119899(0)

120579 + 1205830

12058700

1205871198991

= 119861120588119899+ 119863120590119899 119899 = 1 2 119899 (0) + 1

1205871198991

= 120588119899minus119899(0)minus1

(119861120588119899(0)+1

+ 119863120590119899(0)+1

)

119899 = 119899 (0) + 2 119899 (1) + 1

(26)

where 12058700can be solved by the normalization equation

Because the probability of balking is equal to 120587119899(0)+10

+

120587119899(1)+11

the social benefit per time unit when all customers fol-low the threshold policy (119899(0) 119899(1)) equals

119880119904(119899 (0) 119899 (1)) = 120582119877 (1 minus 120587

119899(0)+10minus 120587119899(1)+11

)

minus 119862(

119899(0)+1

sum

119899=0

1198991205871198990+

119899(1)+1

sum

119899=1

1198991205871198991)

(27)

Let (119899lowast(0) 119899lowast(1)) be the socially optimal threshold strategyFrom Figure 3 we obtain (119899

lowast(0) 119899lowast(1)) = (3 4) while the

equilibrium threshold strategy (119899119890(0) 119899119890(1)) = (78 79) We

observe that 119899119890(0) gt 119899

lowast(0) and 119899

119890(1) gt 119899

lowast(1) which indicates

that the individual optimization is inconsistent with the socialoptimization

4 Partially Observable Queues

In this section we turn our attention to the partially observ-able case where arriving customers only can observe the stateof the server at their arrival instant and do not observe thenumber of customers present

A mixed strategy for a customer is specified by a vector(1199020 1199021) where 119902

119894is the probability of joining when the server

is in state 119894 (119894 = 0 1) If all customers follow the same mixed

Mathematical Problems in Engineering 5

2

3

4

5

6

7

8

9

n(1

)

2 3 4 5 6 7 81n(0)

24

68

10

02

46

8

n(1)n(0)

0

10

20

30

40

50

Us(n

(0)n

(1))

68

Figure 3 Social welfare for the observable case when 119877 = 40 119862 = 1 120582 = 15 1205831= 2 120583

0= 05 and 120579 = 05

1 1 2 0 3 1

0 0 1 0 2 0 3 0

n 1

n 0

120579

1205831 1205831

12058301205830

120582q1120582q1

120582q0 120582q0middot middot middot

middot middot middot

middot middot middot

middot middot middot

120579

1205831 1205831

12058301205830

120582q1120582q1

120582q0 120582q0

120579120579

1205831

1205831

1205830

1205830

120582q1

120582q0 120582q0

Figure 4 Transition rate diagram for the partially observable queues

strategy (1199020 1199021) then the system follows a Markov chain

where the arrival rate equals 120582119894= 120582119902119894when the server is in

state 119894 (119894 = 0 1) The state space is Ω119901119900

= 0 0 cup (119899 119894) |

119899 ge 1 119894 = 0 1 and the transition diagram is illustrated inFigure 4

Denote the stationary distribution as120587119899119894= lim119905rarrinfin

119875 119873 (119905) = 119899 119869 (119905) = 119894 (119899 119894) isin Ω119901119900

1205870= 12058700

120587119899= (1205871198990 1205871198991) 119899 ge 1

(28)

Using the lexicographical sequence for the states thetransition rate matrix (generator) Q can be written as thetridiagonal block matrix

Q =((

(

A0C0

B1

A CB A C

B A Cd d d

))

)

(29)

whereA0= minus120582119902

0

B1= (

1205830

1205831

)

C1= (1205821199020 0)

A = (minus (1205821199020+ 120579 + 120583

0) 120579

0 minus (1205821199021+ 1205831))

B = (0 1205830

0 1205831

)

C = (1205821199020

0

0 1205821199021

)

(30)To analyze this QBD process it is necessary to solve for

the minimal nonnegative solution of the matrix quadraticequation

R2B + RA + C = 0 (31)and this solution is called the rate matrix and denoted by R

Lemma 3 If 1205881

= 12058211990211205831

lt 1 the minimal nonnegativesolution of (31) is given by

R = (1205900

1205821199020+ 120579

1205831

1205900

0 1205881

) (32)

where 1205900= 1205821199020(1205821199020+ 120579 + 120583

0)

Proof Because the coefficients of (31) are all upper triangularmatrices we can assume that R has the same structure as

R = (11990311

11990312

0 11990322

) (33)

6 Mathematical Problems in Engineering

Substituting R2 and R into (31) yields the following set ofequations

minus (1205821199020+ 120579 + 120583

0) 11990311+ 1205821199020= 0

12058311199032

22minus (1205821199021+ 1205831) 11990322+ 1205821199021= 0

12058301199032

11+ 120583111990312(11990311+ 11990322) + 120579119903

11minus (1205821199021+ 1205831) 11990312

= 0

(34)

To obtain the minimal nonnegative solution of (31) we take11990322

= 1205881(the other roots are 119903

22= 1) in the second equation

of (34) From the first equation of (34) we obtain 11990311

=

1205821199020(1205821199020+ 120579 + 120583

0) = 1205900 Substituting 120590

0and 1205881into the last

equation of (34) we get 11990312

= ((1205821199020+ 120579)120583

1)1205900 This is the

proof of Lemma 3

Theorem 4 Assuming that 1205881lt 1 the stationary probabilities

120587119899119894

| (119899 119894) isin Ω119901119900 of the partially observable queues are as

follows

1205871198990

= 119870120590119899

0 119899 ge 0

1205871198991

= 1198701205881

1205821199020+ 120579

1205821199020+ 120579 + 120583

0

119899minus1

sum

119895=0

120588119895

1120590119899minus1minus119895

0 119899 ge 1

(35)

where

119870 =120579 + 1205830

1205821199020+ 120579 + 120583

0

(1 +1205821199020+ 120579

1205821199020+ 120579 + 120583

0

1205821199020

1205831minus 1205821199021

)

minus1

(36)

Proof With the matrix-geometric solution method [19] wehave

120587119899= (1205871198990 1205871198991) = (120587

10 12058711)R119899minus1 119899 ge 1 (37)

In addition (12058700 12058710 12058711) satisfies the set of equations

(12058700 12058710 12058711) 119861 [R] = 0 (38)

where

119861 [R] = (A0

C0

B1

A + RB)

= (

minus1205821199020

1205821199020

0

1205830

minus (1205821199020+ 120579 + 120583

0) 1205821199020+ 120579

1205831

0 minus1205831

)

(39)

Then from (38) we derive

minus120582119902012058700+ 120583012058710+ 120583112058711

= 0

120582119902012058700minus (1205821199020+ 120579 + 120583

0) 12058710

(1205821199020+ 120579) 120587

10minus 120583112058711

= 0

(40)

Taking 12058700as a known constant we obtain

12058710

= 120590012058700 (41)

12058711

= 1205881

1205821199020+ 120579

1205821199020+ 120579 + 120583

0

12058700 (42)

Note that

R119899minus1 = (120590119899minus1

0

1205821199020+ 120579

1205831

1205900

119899minus2

sum

119895=0

120590119895

0120588119899minus119895minus2

1

0 120588119899minus1

1

) 119899 ge 2 (43)

By 12058710 12058711 R119899minus1 and (37) we get (35) Finally 120587

00can be

determined by the normalization conditionThis is the proofof Theorem 4

Let 119901119894be the steady-state probability when the server is at

state 119894 (119894 = 0 1) we have

1199010=

infin

sum

119899=0

1205871198990

= 1198701205821199020+ 120579 + 120583

0

120579 + 1205830

(44)

1199011=

infin

sum

119899=1

1205871198991

= 1198701205821199020(1205821199020+ 120579)

(1205831minus 1205821199021) (120579 + 120583

0) (45)

So the effective arrival rate 120582 is given by

120582 = 120582 (11990101199020+ 11990111199021) (46)

We now consider a customer who finds the server at state119894 upon arrival The conditional mean sojourn time of such acustomer that decides to enter given that the others follow thesame mixed strategy (119902

0 1199021) is given by

119882119894(1199020 1199021) =

suminfin

119899=119894119879 (119899 119894) 120587

119899119894

suminfin

119899=119894120587119899119894

(47)

By substituting (5)-(6) and (35) into (44) we obtain

1198820(1199020 1199021) =

1205821199020+ 120579 + 120583

1

1205831(120579 + 120583

0)

1198821(1199020 1199021) =

1

1205831minus 1205821199021

+1205821199020+ 120579 + 120583

0

1205831(1205830+ 120579)

(48)

Based on the reward-cost structure the expected net ben-efit of an arriving customer if heshe finds the server at state119894 and decides to enter is given by

1198800(1199020) = 119877 minus 119862

1205821199020+ 120579 + 120583

1

1205831(120579 + 120583

0)

1198801(1199020 1199021) = 119877 minus 119862(

1

1205831minus 1205821199021

+1205821199020+ 120579 + 120583

0

1205831(1205830+ 120579)

)

(49)

We now can proceed to determine mixed equilibriumstrategies of a customer in the partially observable case andhave the following

Theorem 5 For the partially observable case there exists aunique mixed equilibrium strategy (119902

119890

0 119902119890

1) where the vector

(119902119890

0 119902119890

1) is given as follows

(1) 119862(120579 + 1205831)1205831(120579 + 120583

0) lt 119877 le 119862(120582 + 120579 + 120583

1)1205831(120579 + 120583

0)

(119902119890

0 119902119890

1) =

(1199091 0)

1205831minus 1205830

1205831(120579 + 120583

0)lt

1

120583

(1199091 1199092)

1

120583le

1205831minus 1205830

1205831(120579 + 120583

0)le

1

120583 minus 120582

(1199091 1)

1205831minus 1205830

1205831(120579 + 120583

0)gt

1

120583 minus 120582

(50)

Mathematical Problems in Engineering 7

(2) 119862(120582 + 120579 + 1205831)1205831(120579 + 120583

0) lt 119877

(119902119890

0 119902119890

1) =

(1 0) 119877 lt119862

1205831

+119862 (120582 + 120579 + 120583

0)

1205831(120579 + 120583

0)

(1 1199093)

119862

1205831

+119862 (120582 + 120579 + 120583

0)

1205831(120579 + 120583

0)

le 119877 le119862

1205831minus 120582

+119862 (120582 + 120579 + 120583

0)

1205831(120579 + 120583

0)

(1 1) 119877 gt119862

1205831minus 120582

+119862 (120582 + 120579 + 120583

0)

1205831(120579 + 120583

0)

(51)

where

1199091=

1

120582(1198771205831(120579 + 120583

0)

119862minus 120579 minus 120583

1)

1199092=1205831

120582(1 minus

1205830+ 120579

1205831minus 1205830

)

1199093=

1

120582(1205831minus

1

119877119862 minus (120582 + 120579 + 1205830) 1205831(120579 + 120583

0))

(52)

Proof To prove Theorem 5 we first focus on 1198800(1199020) Con-

dition (1) assures that 1199021198900is positive Therefore we have two

casesCase 1 (119880

0(0) gt 0 and 119880

0(1) le 0) That is 119862(120579 + 120583

1)1205831(120579 +

1205830) lt 119877 le 119862(120582+120579+120583

1)1205831(120579+1205830)) In this case if all customers

who find the system empty enter the system with probability119902119890

0= 1 then the tagged customer suffers a negative expected

benefit if heshe decides to enter Hence 1199021198900= 1 does not lead

to an equilibrium Similarly if all customers use 1199021198900= 0 then

the tagged customer receives a positive benefit from enteringthus 119902119890

0= 0 also cannot be part of an equilibrium mixed

strategy Therefore there exists unique 1199021198900satisfying

119877 minus 119862120582119902119890

0+ 120579 + 120583

1

1205831(120579 + 120583

0)

= 0 (53)

for which customers are indifferent between entering andbalking This is given by

119902119890

0=

1

120582(1198771205831(120579 + 120583

0)

119862minus 120579 minus 120583

1) = 119909

1 (54)

In this situation the expected benefit 1198801is given by

1198801(119902119890

0 1199021) =

119862 (1205831minus 1205830)

1205831(120579 + 120583

0)minus

119862

120583 minus 1205821199021

(55)

Using the standard methodology of equilibrium analysisin unobservable queueing models we derive from (55) thefollowing equilibria

(1) If 1198801(119902119890

0 0) lt 0 then the equilibrium strategy is 119902119890

1=

0(2) If 119880

1(119902119890

0 0) ge 0 and 119880

1(119902119890

0 1) le 0 there exists unique

119902119890

1 satisfying

1198801(119902119890

0 119902119890

1) = 0 (56)

then the equilibrium strategy is 1199021198901= 1199092

(3) If 1198801(119902119890

0 1) gt 0 then the equilibrium strategy is 119902119890

1=

1

Case 2 (1198800(1) ge 0) That is 119862(120582 + 120579 + 120583

1)1205831(120579 + 120583

0) lt 119877) In

this case for every strategy of the other customers the taggedcustomer has a positive expected net benefit if heshe decidesto enter Hence 119902119890

0= 1

In this situation the expected benefit 1198801is given by

1198801(1 1199021) = 119877 minus

119862

120583 minus 1205821199021

minus119862 (120582 + 120579 + 120583

0)

1205831(120579 + 120583

0)

(57)

To find 119902119890

1in equilibrium we also consider the following

subcases

(1) If 1198801(1 0) lt 0 then the equilibrium strategy is 119902119890

1=

0

(2) If 1198801(1 0) ge 0 and 119880

1(1 1) le 0 there exists unique

119902119890

1 satisfying

1198801(1 119902119890

1) = 0 (58)

then the equilibrium strategy is 1199021198901= 1199093

(3) If 1198801(1 1) gt 0 then the equilibrium strategy is 119902119890

1=

1

By rearranging Cases 1 and 2 we can obtain the results ofTheorem 5 This completes the proof

From Theorem 5 we can compare 119902119890

0with 119902

119890

1 Figure 5

shows that 1199021198900is not always less than 119902

119890

1 Namely sometimes

the information that the server takes a working vacation doesnot make the customers less willing to enter the systembecause in the vacation the server provides a lower serviceto customers But when the vacation time and waiting timebecome longer customers do not want to enter the system invacation

Then fromTheorem 4 themean queue length is given by

119864 [119871] =

infin

sum

119899=0

1198991205871198990+

infin

sum

119899=0

1198991205871198991

= 1198701205900

(1 minus 1205900)2(1 +

1205821199020+ 120579

1205831

1 minus 12058811205900

(1 minus 1205881)2)

(59)

8 Mathematical Problems in Engineering

03

04

05

06

07

08

09

1

Equi

libriu

m jo

inin

g pr

obab

ility

qe0qe1 (120579 = 01 C = 2)

qe0qe1 (120579 = 05 C = 3)

0 04 06 08 1 12 14 16 1802Arrival rate 120582

Figure 5 Equilibrium mixed strategies for the partially observablecase when 119877 = 5 120583

1= 2 and 120583

0= 05

0 04 06 08 1 12 14 16 1802Arrival rate 120582

0

01

02

03

04

05

06

07

08

09

1

Join

ing

prob

abili

ty

qe0qe1

qlowast0qlowast1

Figure 6 Equilibrium and socially optimal mixed strategies for thepartially observable case when 119877 = 5 119862 = 2 120583

1= 2 120583

0= 05 and

120579 = 01

And the social benefit when all customers follow a mixedpolicy (119902

0 1199021) can now be easily computed as follows

119880119904(1199020 1199021) = 120582119877 minus 119862119864 [119871]

= 120582 (11990101199020+ 11990111199021) 119877

minus1198621198701205900

(1 minus 1205900)2(1 +

1205821199020+ 120579

1205831

1 minus 12058811205900

(1 minus 1205881)2)

(60)

The goal of a social planner is to maximize overall socialwelfare Let (119902lowast

0 119902lowast

1) be the socially optimal mixed strategy

Figure 6 compares customers equilibrium mixed strategy(119902119890

0 119902119890

1) and their socially optimal mixed strategy (119902lowast

0 119902lowast

1) We

also observe that 1199021198900gt 119902lowast

0and 1199021198901gt 119902lowast

1This ordering is typical

when customers make individual decisions maximizing theirown profitThen they ignore those negative externalities thatthey impose on later arrivals and they tend to overuse thesystem It is clear that these externalities should be taken intoaccount when we aim to maximize the total revenue

5 The Unobservable Queues

In this section we consider the fully unobservable case wherearriving customers cannot observe the system state

There are two pure strategies available for a customer thatis to join the queue or not to join the queue A pure or mixedstrategy can be described by a fraction 119902 (0 le 119902 le 1) whichis the probability of joining and the effective arrival rate orjoining rate is 120582119902 The transition diagram is illustrated inFigure 7

To identify the equilibrium strategies of the customerswe should first investigate the stationary distribution of thesystem when all customers follow a given strategy 119902 whichcan be obtained by Theorem 4 by taking 119902

0= 1199021= 119902 So we

have

1205871198990

= 1198701120590119899

119902 119899 ge 0

1205871198991

= 1198701120588119902

120582119902 + 120579

120582119902 + 120579 + 1205830

119899minus1

sum

119895=0

120588119895

119902120590119899minus1minus119895

119902 119899 ge 1

(61)

where 120588119902= 120582119902120583

1 120590119902= 120582119902(120582119902 + 120579 + 120583

0) and

1198701=

120579 + 1205830

120582119902 + 120579 + 1205830

(1 +120582119902 + 120579

120582119902 + 120579 + 1205830

120582119902

1205831minus 120582119902

)

minus1

(62)

Then the mean number of the customers in the system is

119864 [119871] =

infin

sum

119899=0

1198991205871198990+

infin

sum

119899=0

1198991205871198991

=1198701120590119902

(1 minus 120590119902)2(1 +

120582119902 + 120579

1205831

1 minus 120588119902120590119902

(1 minus 120588119902)2)

(63)

Hence the mean sojourn time of a customer who decides toenter upon hisher arrival can be obtained by using Littlersquoslaw

119864 [119882] =1198701120590119902

120582119902 (1 minus 120590119902)2(1 +

120582119902 + 120579

1205831

1 minus 120588119902120590119902

(1 minus 120588119902)2)

=1

120579 + 1205830

(1

+1205831(120579 + 120583

0)

1205831minus 120582119902

1

1205830((1205831minus 120582119902) (120582119902 + 120579)) + 120583

1

)

(64)

And if 120582 lt 1205831 119864[119882] is strictly increasing for 119902 isin [0 1]

Mathematical Problems in Engineering 9

1 1 2 0 3 1

0 0 1 0 2 0 3 0

n 1

n 0

120579

1205831 1205831

12058301205830

120582q120582q

120582q 120582qmiddot middot middot

middot middot middot

middot middot middot

middot middot middot

120579

1205831 1205831

12058301205830

120582q120582q

120582q 120582q

120579120579

1205831

1205831

1205830

1205830

120582q

120582q 120582q

Figure 7 Transition rate diagram for the unobservable queues

We consider a tagged customer If heshe decides to enterthe system hisher expected net benefit is

119880 (119902) = 119877 minus 119862119864 [119882] = 119877 minus119862

120579 + 1205830

(1

+1205831(120579 + 120583

0)

1205831minus 120582119902

1

1205830((1205831minus 120582119902) (120582119902 + 120579)) + 120583

1

)

(65)

We note that 119880(119902) is strictly decreasing in 119902 and it has aunique root 119902lowast

119890 So there exists a unique mixed equilibrium

strategy 119902119890and 119902119890= min119902lowast

119890 1

We now turn our attention to social optimization Thesocial benefit per time unit can now be easily computed as

119880119904(119902) = 120582119902 (119877 minus 119862119864 [119882]) = 120582119902(119877 minus

119862

120579 + 1205830

(1

+1205831(120579 + 120583

0)

1205831minus 120582119902

1

1205830((1205831minus 120582119902) (120582119902 + 120579)) + 120583

1

))

(66)

As for the equilibrium social welfare per time unit119880119904(119902119890)

Figure 8 shows that 119880119904(119902119890) first increases and then decreases

with 120582 and the social benefit achieves a maximum forintermediate values of this parameter The reason for thisbehavior is that when the arrival rate is small the system israrely crowded therefore as more customers arrive they areserved and the social benefit improves However with anysmall increase of the arrival rate 120582 the expected waiting timeincreases which has a detrimental effect on the social benefit

Let 119909lowast be the root of the equation 1198781015840(119902) = 0 and let 119902lowast be

the optimal joining probability Then we have the followingconclusions if 0 lt 119909

lowastlt 1 and 119878

10158401015840(119902) le 0 then 119902

lowast= 119909lowast if

0 lt 119909lowastlt 1 and 119878

10158401015840(119902) gt 0 or 119909lowast ge 1 then 119902

lowast= 1

From Figure 9 we observe that 119902119890

gt 119902lowast As a result

individual optimization leads to queues that are longer thansocially desired Therefore it is clear that the social plannerwants a toll to discourage arrivals

6 Conclusion

In this paper we analyzed the customer strategic behaviorin the MM1 queueing system with working vacations andvacation interruptions where arriving customers have optionto decide whether to join the system or balk Three differentcases with respect to the levels of information provided toarriving customers have been investigated extensively and theequilibrium strategies for each case were derived And wefound that the equilibrium joining probability of the busy

02 04 06 08 1 12 14 16 18 20Arrival rate 120582

Equi

libriu

m so

cial

wel

fare

Us(qe)

0

5

10

15

Figure 8 Equilibrium social welfare for the unobservable casewhen119877 = 10 119862 = 1 120583

1= 2 120583

0= 1 and 120579 = 1

03

04

05

06

07

08

09

1

Join

ing

prob

abili

ty

0 1 15 205Arrival rate 120582

qeqlowast

Figure 9 Equilibrium and socially optimal strategies for the unob-servable case when 119877 = 3 119862 = 2 120583

1= 2 120583

0= 05 and 120579 = 001

period may be smaller than that in vacation time We alsocompared the equilibrium strategy with the socially optimalstrategy numerically and we observed that customers makeindividual decisions maximizing their own profit and tendto overuse the system For the social planner a toll will beadopted to discourage customers from joining

10 Mathematical Problems in Engineering

Furthermore an interesting extension would be to incor-porate the present model in a profit maximizing frameworkwhere the owner or manager of the system imposes anentrance feeThe problem is to find the optimal fee that max-imizes the owners total profit subject to the constraint thatcustomers independently decide whether to enter or not andtake into account the fee in addition to the service reward andwaiting cost Another direction for future work is to concernthe non-Markovian queues with various vacation policies

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Acknowledgments

This work is partially supported by the Natural Foundation ofHebei Province (nos A2014203096 andC2014203212) Chinaand the Young Faculty Autonomous Research Project of Yan-shan University (nos 13LGB032 13LGA017 and 13LGB013)China

References

[1] P Naor ldquoThe regulation of queue size by levying tollsrdquo Econo-metrica vol 37 no 1 pp 15ndash24 1969

[2] N M Edelson and D K Hildebrand ldquoCongestion tolls forPoisson queuing processesrdquo Econometrica vol 43 no 1 pp 81ndash92 1975

[3] R Hassin and M Haviv Equilibrium Behavior in Queueing Sys-tems To Queue or Not to Queue Kluwer Academic PublishersDordrecht The Netherlands 2003

[4] A Burnetas and A Economou ldquoEquilibrium customer strate-gies in a single server Markovian queue with setup timesrdquoQueueing Systems vol 56 no 3-4 pp 213ndash228 2007

[5] W Sun and N Tian ldquoContrast of the equilibrium and sociallyoptimal strategies in a queue with vacationsrdquo Journal of Compu-tational Information Systems vol 4 no 5 pp 2167ndash2172 2008

[6] A Economou and S Kanta ldquoEquilibrium balking strategiesin the observable single-server queue with breakdowns andrepairsrdquoOperations Research Letters vol 36 no 6 pp 696ndash6992008

[7] P Guo and R Hassin ldquoStrategic behavior and social optimiza-tion in Markovian vacation queuesrdquo Operations Research vol59 no 4 pp 986ndash997 2011

[8] P Guo and R Hassin ldquoStrategic behavior and social optimiza-tion in Markovian vacation queues the case of heterogeneouscustomersrdquo European Journal of Operational Research vol 222no 2 pp 278ndash286 2012

[9] R Tian D Yue and W Yue ldquoOptimal balking strategies inan MG1 queueing system with a removable server under N-policyrdquo Journal of Industrial andManagementOptimization vol11 no 3 pp 715ndash731 2015

[10] P Guo and Q Li ldquoStrategic behavior and social optimizationin partially-observableMarkovian vacation queuesrdquoOperationsResearch Letters vol 41 no 3 pp 277ndash284 2013

[11] L Li J Wang and F Zhang ldquoEquilibrium customer strategiesin Markovian queues with partial breakdownsrdquo Computers ampIndustrial Engineering vol 66 no 4 pp 751ndash757 2013

[12] F Zhang J Wang and B Liu ldquoOn the optimal and equilibriumretrial rates in an unreliable retrial queue with vacationsrdquo Jour-nal of Industrial andManagement Optimization vol 8 no 4 pp861ndash875 2012

[13] N Tian J Li and Z G Zhang ldquoMatrix analysis methodand working vacation queuesa surveyrdquo International Journal ofInformation and Management Sciences vol 20 pp 603ndash6332009

[14] F Zhang J Wang and B Liu ldquoEquilibrium balking strategiesin Markovian queues with working vacationsrdquo Applied Mathe-matical Modelling vol 37 no 16-17 pp 8264ndash8282 2013

[15] W Sun and S Li ldquoEquilibrium and optimal behavior of cus-tomers in Markovian queues with multiple working vacationsrdquoTOP vol 22 no 2 pp 694ndash715 2014

[16] W Sun S Li and Q-L Li ldquoEquilibrium balking strategiesof customers in Markovian queues with two-stage workingvacationsrdquo Applied Mathematics and Computation vol 248 pp195ndash214 2014

[17] J Li andNTian ldquoTheMM1 queuewithworking vacations andvacation interruptionsrdquo Journal of Systems Science and SystemsEngineering vol 16 no 1 pp 121ndash127 2007

[18] J-H Li N-S Tian and Z-Y Ma ldquoPerformance analysis ofGIM1 queue with working vacations and vacation interrup-tionrdquo Applied Mathematical Modelling vol 32 no 12 pp 2715ndash2730 2008

[19] M Neuts Matrix-Geometric Solution in Stochastic ModelsJohns Hopkins University Press Baltimore Md USA 1981

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Mathematical Problems in Engineering

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Stochastic AnalysisInternational Journal of

4 Mathematical Problems in Engineering

1 1 2 1

1 00 0 2 0

120579120579120579120579

1205831 1205831 1205831 1205831 1205831 1205831 1205831

1205830120583012058301205830

1205830

120582120582120582120582 120582

120582120582120582120582120582120582120582middot middot middot

middot middot middot

middot middot middot

1205831

n(0) 1

n(0) 0

n(0) + 1 1

n(0) + 1 0

n(1) 1 n(1) + 1 1

Figure 2 Transition rate diagram for the observable queues

By iterating (10) and (15) taking into account (11) and (16)we obtain

1205871198990

= 12059011989912058700 119899 = 0 1 2 119899 (0)

120587119899(0)+10

=120582120590119899(0)

120579 + 1205830

12058700

1205871198991

= 120588119899minus119899(0)minus1

120587119899(0)+11

119899 = 119899 (0) + 1 119899 (1) + 1

(18)

From (13) it is easy to obtain that 1205871198991

| 1 le 119899 le 119899(0)+1 isa solution of the nonhomogeneous linear difference equation

1205831119909119899+1

minus (120582 + 1205831) 119909119899+ 120582119909119899minus1

= minus1205791205871198990minus 1205830120587119899+10

= minus (120579 + 1205830120590) 12059011989912058700

(19)

So its corresponding characteristic equation is

12058311199092minus (120582 + 120583

1) 119909 + 120582 = 0 (20)

which has two roots at 1 and 120588 Assume that 120588 = 1 thenthe homogeneous solution of (19) is 119909hom

119899= 1198601119899+ 119861120588119899 The

general solution of the nonhomogeneous equation is given as119909gen119899

= 119909hom119899

+119909spec119899

where 119909spec119899

is a specific solution We finda specific solution 119909

spec119899

= 119863120590119899 Substituting it into (19) we

derive

119863 = minus120582 + 120579

120582 + 120579 + 1205830minus 1205831

12058700 (21)

Therefore

119909gen119899

= 119860 + 119861120588119899+ 119863120590119899 119899 = 1 2 119899 (0) + 1 (22)

Substituting (22) into (9) and (12) it follows after some rathertedious algebra that

119860 + 119861120588 = minus120582 (120582 + 120579)

1205831(1205831minus 120582 minus 120579 minus 120583

0)12058700

119860 + 1198611205882= minus

1205822(120582 + 120579)

12058321(1205831minus 120582 minus 120579 minus 120583

0)12058700

(23)

Then we obtain

119860 = 0

119861 = minus120582 + 120579

1205831minus 120582 minus 120579 minus 120583

0

12058700

(24)

Thus from (22) we obtain

1205871198991

= 119861120588119899+ 119863120590119899 119899 = 1 2 119899 (0) + 1 (25)

Thus we have all the stationary probabilities in terms of12058700 The remaining probability 120587

00 can be found from the

normalization condition After some algebraic simplifica-tions we can summarize the results in the following

Theorem 2 In the observable MM1 queues with workingvacations and vacation interruptions the stationary distribu-tion 120587

119899119894| (119899 119894) isin Ω

119891119900 is given as follows

1205871198990

= 12059011989912058700 119899 = 0 1 2 119899 (0)

120587119899(0)+10

=120582120590119899(0)

120579 + 1205830

12058700

1205871198991

= 119861120588119899+ 119863120590119899 119899 = 1 2 119899 (0) + 1

1205871198991

= 120588119899minus119899(0)minus1

(119861120588119899(0)+1

+ 119863120590119899(0)+1

)

119899 = 119899 (0) + 2 119899 (1) + 1

(26)

where 12058700can be solved by the normalization equation

Because the probability of balking is equal to 120587119899(0)+10

+

120587119899(1)+11

the social benefit per time unit when all customers fol-low the threshold policy (119899(0) 119899(1)) equals

119880119904(119899 (0) 119899 (1)) = 120582119877 (1 minus 120587

119899(0)+10minus 120587119899(1)+11

)

minus 119862(

119899(0)+1

sum

119899=0

1198991205871198990+

119899(1)+1

sum

119899=1

1198991205871198991)

(27)

Let (119899lowast(0) 119899lowast(1)) be the socially optimal threshold strategyFrom Figure 3 we obtain (119899

lowast(0) 119899lowast(1)) = (3 4) while the

equilibrium threshold strategy (119899119890(0) 119899119890(1)) = (78 79) We

observe that 119899119890(0) gt 119899

lowast(0) and 119899

119890(1) gt 119899

lowast(1) which indicates

that the individual optimization is inconsistent with the socialoptimization

4 Partially Observable Queues

In this section we turn our attention to the partially observ-able case where arriving customers only can observe the stateof the server at their arrival instant and do not observe thenumber of customers present

A mixed strategy for a customer is specified by a vector(1199020 1199021) where 119902

119894is the probability of joining when the server

is in state 119894 (119894 = 0 1) If all customers follow the same mixed

Mathematical Problems in Engineering 5

2

3

4

5

6

7

8

9

n(1

)

2 3 4 5 6 7 81n(0)

24

68

10

02

46

8

n(1)n(0)

0

10

20

30

40

50

Us(n

(0)n

(1))

68

Figure 3 Social welfare for the observable case when 119877 = 40 119862 = 1 120582 = 15 1205831= 2 120583

0= 05 and 120579 = 05

1 1 2 0 3 1

0 0 1 0 2 0 3 0

n 1

n 0

120579

1205831 1205831

12058301205830

120582q1120582q1

120582q0 120582q0middot middot middot

middot middot middot

middot middot middot

middot middot middot

120579

1205831 1205831

12058301205830

120582q1120582q1

120582q0 120582q0

120579120579

1205831

1205831

1205830

1205830

120582q1

120582q0 120582q0

Figure 4 Transition rate diagram for the partially observable queues

strategy (1199020 1199021) then the system follows a Markov chain

where the arrival rate equals 120582119894= 120582119902119894when the server is in

state 119894 (119894 = 0 1) The state space is Ω119901119900

= 0 0 cup (119899 119894) |

119899 ge 1 119894 = 0 1 and the transition diagram is illustrated inFigure 4

Denote the stationary distribution as120587119899119894= lim119905rarrinfin

119875 119873 (119905) = 119899 119869 (119905) = 119894 (119899 119894) isin Ω119901119900

1205870= 12058700

120587119899= (1205871198990 1205871198991) 119899 ge 1

(28)

Using the lexicographical sequence for the states thetransition rate matrix (generator) Q can be written as thetridiagonal block matrix

Q =((

(

A0C0

B1

A CB A C

B A Cd d d

))

)

(29)

whereA0= minus120582119902

0

B1= (

1205830

1205831

)

C1= (1205821199020 0)

A = (minus (1205821199020+ 120579 + 120583

0) 120579

0 minus (1205821199021+ 1205831))

B = (0 1205830

0 1205831

)

C = (1205821199020

0

0 1205821199021

)

(30)To analyze this QBD process it is necessary to solve for

the minimal nonnegative solution of the matrix quadraticequation

R2B + RA + C = 0 (31)and this solution is called the rate matrix and denoted by R

Lemma 3 If 1205881

= 12058211990211205831

lt 1 the minimal nonnegativesolution of (31) is given by

R = (1205900

1205821199020+ 120579

1205831

1205900

0 1205881

) (32)

where 1205900= 1205821199020(1205821199020+ 120579 + 120583

0)

Proof Because the coefficients of (31) are all upper triangularmatrices we can assume that R has the same structure as

R = (11990311

11990312

0 11990322

) (33)

6 Mathematical Problems in Engineering

Substituting R2 and R into (31) yields the following set ofequations

minus (1205821199020+ 120579 + 120583

0) 11990311+ 1205821199020= 0

12058311199032

22minus (1205821199021+ 1205831) 11990322+ 1205821199021= 0

12058301199032

11+ 120583111990312(11990311+ 11990322) + 120579119903

11minus (1205821199021+ 1205831) 11990312

= 0

(34)

To obtain the minimal nonnegative solution of (31) we take11990322

= 1205881(the other roots are 119903

22= 1) in the second equation

of (34) From the first equation of (34) we obtain 11990311

=

1205821199020(1205821199020+ 120579 + 120583

0) = 1205900 Substituting 120590

0and 1205881into the last

equation of (34) we get 11990312

= ((1205821199020+ 120579)120583

1)1205900 This is the

proof of Lemma 3

Theorem 4 Assuming that 1205881lt 1 the stationary probabilities

120587119899119894

| (119899 119894) isin Ω119901119900 of the partially observable queues are as

follows

1205871198990

= 119870120590119899

0 119899 ge 0

1205871198991

= 1198701205881

1205821199020+ 120579

1205821199020+ 120579 + 120583

0

119899minus1

sum

119895=0

120588119895

1120590119899minus1minus119895

0 119899 ge 1

(35)

where

119870 =120579 + 1205830

1205821199020+ 120579 + 120583

0

(1 +1205821199020+ 120579

1205821199020+ 120579 + 120583

0

1205821199020

1205831minus 1205821199021

)

minus1

(36)

Proof With the matrix-geometric solution method [19] wehave

120587119899= (1205871198990 1205871198991) = (120587

10 12058711)R119899minus1 119899 ge 1 (37)

In addition (12058700 12058710 12058711) satisfies the set of equations

(12058700 12058710 12058711) 119861 [R] = 0 (38)

where

119861 [R] = (A0

C0

B1

A + RB)

= (

minus1205821199020

1205821199020

0

1205830

minus (1205821199020+ 120579 + 120583

0) 1205821199020+ 120579

1205831

0 minus1205831

)

(39)

Then from (38) we derive

minus120582119902012058700+ 120583012058710+ 120583112058711

= 0

120582119902012058700minus (1205821199020+ 120579 + 120583

0) 12058710

(1205821199020+ 120579) 120587

10minus 120583112058711

= 0

(40)

Taking 12058700as a known constant we obtain

12058710

= 120590012058700 (41)

12058711

= 1205881

1205821199020+ 120579

1205821199020+ 120579 + 120583

0

12058700 (42)

Note that

R119899minus1 = (120590119899minus1

0

1205821199020+ 120579

1205831

1205900

119899minus2

sum

119895=0

120590119895

0120588119899minus119895minus2

1

0 120588119899minus1

1

) 119899 ge 2 (43)

By 12058710 12058711 R119899minus1 and (37) we get (35) Finally 120587

00can be

determined by the normalization conditionThis is the proofof Theorem 4

Let 119901119894be the steady-state probability when the server is at

state 119894 (119894 = 0 1) we have

1199010=

infin

sum

119899=0

1205871198990

= 1198701205821199020+ 120579 + 120583

0

120579 + 1205830

(44)

1199011=

infin

sum

119899=1

1205871198991

= 1198701205821199020(1205821199020+ 120579)

(1205831minus 1205821199021) (120579 + 120583

0) (45)

So the effective arrival rate 120582 is given by

120582 = 120582 (11990101199020+ 11990111199021) (46)

We now consider a customer who finds the server at state119894 upon arrival The conditional mean sojourn time of such acustomer that decides to enter given that the others follow thesame mixed strategy (119902

0 1199021) is given by

119882119894(1199020 1199021) =

suminfin

119899=119894119879 (119899 119894) 120587

119899119894

suminfin

119899=119894120587119899119894

(47)

By substituting (5)-(6) and (35) into (44) we obtain

1198820(1199020 1199021) =

1205821199020+ 120579 + 120583

1

1205831(120579 + 120583

0)

1198821(1199020 1199021) =

1

1205831minus 1205821199021

+1205821199020+ 120579 + 120583

0

1205831(1205830+ 120579)

(48)

Based on the reward-cost structure the expected net ben-efit of an arriving customer if heshe finds the server at state119894 and decides to enter is given by

1198800(1199020) = 119877 minus 119862

1205821199020+ 120579 + 120583

1

1205831(120579 + 120583

0)

1198801(1199020 1199021) = 119877 minus 119862(

1

1205831minus 1205821199021

+1205821199020+ 120579 + 120583

0

1205831(1205830+ 120579)

)

(49)

We now can proceed to determine mixed equilibriumstrategies of a customer in the partially observable case andhave the following

Theorem 5 For the partially observable case there exists aunique mixed equilibrium strategy (119902

119890

0 119902119890

1) where the vector

(119902119890

0 119902119890

1) is given as follows

(1) 119862(120579 + 1205831)1205831(120579 + 120583

0) lt 119877 le 119862(120582 + 120579 + 120583

1)1205831(120579 + 120583

0)

(119902119890

0 119902119890

1) =

(1199091 0)

1205831minus 1205830

1205831(120579 + 120583

0)lt

1

120583

(1199091 1199092)

1

120583le

1205831minus 1205830

1205831(120579 + 120583

0)le

1

120583 minus 120582

(1199091 1)

1205831minus 1205830

1205831(120579 + 120583

0)gt

1

120583 minus 120582

(50)

Mathematical Problems in Engineering 7

(2) 119862(120582 + 120579 + 1205831)1205831(120579 + 120583

0) lt 119877

(119902119890

0 119902119890

1) =

(1 0) 119877 lt119862

1205831

+119862 (120582 + 120579 + 120583

0)

1205831(120579 + 120583

0)

(1 1199093)

119862

1205831

+119862 (120582 + 120579 + 120583

0)

1205831(120579 + 120583

0)

le 119877 le119862

1205831minus 120582

+119862 (120582 + 120579 + 120583

0)

1205831(120579 + 120583

0)

(1 1) 119877 gt119862

1205831minus 120582

+119862 (120582 + 120579 + 120583

0)

1205831(120579 + 120583

0)

(51)

where

1199091=

1

120582(1198771205831(120579 + 120583

0)

119862minus 120579 minus 120583

1)

1199092=1205831

120582(1 minus

1205830+ 120579

1205831minus 1205830

)

1199093=

1

120582(1205831minus

1

119877119862 minus (120582 + 120579 + 1205830) 1205831(120579 + 120583

0))

(52)

Proof To prove Theorem 5 we first focus on 1198800(1199020) Con-

dition (1) assures that 1199021198900is positive Therefore we have two

casesCase 1 (119880

0(0) gt 0 and 119880

0(1) le 0) That is 119862(120579 + 120583

1)1205831(120579 +

1205830) lt 119877 le 119862(120582+120579+120583

1)1205831(120579+1205830)) In this case if all customers

who find the system empty enter the system with probability119902119890

0= 1 then the tagged customer suffers a negative expected

benefit if heshe decides to enter Hence 1199021198900= 1 does not lead

to an equilibrium Similarly if all customers use 1199021198900= 0 then

the tagged customer receives a positive benefit from enteringthus 119902119890

0= 0 also cannot be part of an equilibrium mixed

strategy Therefore there exists unique 1199021198900satisfying

119877 minus 119862120582119902119890

0+ 120579 + 120583

1

1205831(120579 + 120583

0)

= 0 (53)

for which customers are indifferent between entering andbalking This is given by

119902119890

0=

1

120582(1198771205831(120579 + 120583

0)

119862minus 120579 minus 120583

1) = 119909

1 (54)

In this situation the expected benefit 1198801is given by

1198801(119902119890

0 1199021) =

119862 (1205831minus 1205830)

1205831(120579 + 120583

0)minus

119862

120583 minus 1205821199021

(55)

Using the standard methodology of equilibrium analysisin unobservable queueing models we derive from (55) thefollowing equilibria

(1) If 1198801(119902119890

0 0) lt 0 then the equilibrium strategy is 119902119890

1=

0(2) If 119880

1(119902119890

0 0) ge 0 and 119880

1(119902119890

0 1) le 0 there exists unique

119902119890

1 satisfying

1198801(119902119890

0 119902119890

1) = 0 (56)

then the equilibrium strategy is 1199021198901= 1199092

(3) If 1198801(119902119890

0 1) gt 0 then the equilibrium strategy is 119902119890

1=

1

Case 2 (1198800(1) ge 0) That is 119862(120582 + 120579 + 120583

1)1205831(120579 + 120583

0) lt 119877) In

this case for every strategy of the other customers the taggedcustomer has a positive expected net benefit if heshe decidesto enter Hence 119902119890

0= 1

In this situation the expected benefit 1198801is given by

1198801(1 1199021) = 119877 minus

119862

120583 minus 1205821199021

minus119862 (120582 + 120579 + 120583

0)

1205831(120579 + 120583

0)

(57)

To find 119902119890

1in equilibrium we also consider the following

subcases

(1) If 1198801(1 0) lt 0 then the equilibrium strategy is 119902119890

1=

0

(2) If 1198801(1 0) ge 0 and 119880

1(1 1) le 0 there exists unique

119902119890

1 satisfying

1198801(1 119902119890

1) = 0 (58)

then the equilibrium strategy is 1199021198901= 1199093

(3) If 1198801(1 1) gt 0 then the equilibrium strategy is 119902119890

1=

1

By rearranging Cases 1 and 2 we can obtain the results ofTheorem 5 This completes the proof

From Theorem 5 we can compare 119902119890

0with 119902

119890

1 Figure 5

shows that 1199021198900is not always less than 119902

119890

1 Namely sometimes

the information that the server takes a working vacation doesnot make the customers less willing to enter the systembecause in the vacation the server provides a lower serviceto customers But when the vacation time and waiting timebecome longer customers do not want to enter the system invacation

Then fromTheorem 4 themean queue length is given by

119864 [119871] =

infin

sum

119899=0

1198991205871198990+

infin

sum

119899=0

1198991205871198991

= 1198701205900

(1 minus 1205900)2(1 +

1205821199020+ 120579

1205831

1 minus 12058811205900

(1 minus 1205881)2)

(59)

8 Mathematical Problems in Engineering

03

04

05

06

07

08

09

1

Equi

libriu

m jo

inin

g pr

obab

ility

qe0qe1 (120579 = 01 C = 2)

qe0qe1 (120579 = 05 C = 3)

0 04 06 08 1 12 14 16 1802Arrival rate 120582

Figure 5 Equilibrium mixed strategies for the partially observablecase when 119877 = 5 120583

1= 2 and 120583

0= 05

0 04 06 08 1 12 14 16 1802Arrival rate 120582

0

01

02

03

04

05

06

07

08

09

1

Join

ing

prob

abili

ty

qe0qe1

qlowast0qlowast1

Figure 6 Equilibrium and socially optimal mixed strategies for thepartially observable case when 119877 = 5 119862 = 2 120583

1= 2 120583

0= 05 and

120579 = 01

And the social benefit when all customers follow a mixedpolicy (119902

0 1199021) can now be easily computed as follows

119880119904(1199020 1199021) = 120582119877 minus 119862119864 [119871]

= 120582 (11990101199020+ 11990111199021) 119877

minus1198621198701205900

(1 minus 1205900)2(1 +

1205821199020+ 120579

1205831

1 minus 12058811205900

(1 minus 1205881)2)

(60)

The goal of a social planner is to maximize overall socialwelfare Let (119902lowast

0 119902lowast

1) be the socially optimal mixed strategy

Figure 6 compares customers equilibrium mixed strategy(119902119890

0 119902119890

1) and their socially optimal mixed strategy (119902lowast

0 119902lowast

1) We

also observe that 1199021198900gt 119902lowast

0and 1199021198901gt 119902lowast

1This ordering is typical

when customers make individual decisions maximizing theirown profitThen they ignore those negative externalities thatthey impose on later arrivals and they tend to overuse thesystem It is clear that these externalities should be taken intoaccount when we aim to maximize the total revenue

5 The Unobservable Queues

In this section we consider the fully unobservable case wherearriving customers cannot observe the system state

There are two pure strategies available for a customer thatis to join the queue or not to join the queue A pure or mixedstrategy can be described by a fraction 119902 (0 le 119902 le 1) whichis the probability of joining and the effective arrival rate orjoining rate is 120582119902 The transition diagram is illustrated inFigure 7

To identify the equilibrium strategies of the customerswe should first investigate the stationary distribution of thesystem when all customers follow a given strategy 119902 whichcan be obtained by Theorem 4 by taking 119902

0= 1199021= 119902 So we

have

1205871198990

= 1198701120590119899

119902 119899 ge 0

1205871198991

= 1198701120588119902

120582119902 + 120579

120582119902 + 120579 + 1205830

119899minus1

sum

119895=0

120588119895

119902120590119899minus1minus119895

119902 119899 ge 1

(61)

where 120588119902= 120582119902120583

1 120590119902= 120582119902(120582119902 + 120579 + 120583

0) and

1198701=

120579 + 1205830

120582119902 + 120579 + 1205830

(1 +120582119902 + 120579

120582119902 + 120579 + 1205830

120582119902

1205831minus 120582119902

)

minus1

(62)

Then the mean number of the customers in the system is

119864 [119871] =

infin

sum

119899=0

1198991205871198990+

infin

sum

119899=0

1198991205871198991

=1198701120590119902

(1 minus 120590119902)2(1 +

120582119902 + 120579

1205831

1 minus 120588119902120590119902

(1 minus 120588119902)2)

(63)

Hence the mean sojourn time of a customer who decides toenter upon hisher arrival can be obtained by using Littlersquoslaw

119864 [119882] =1198701120590119902

120582119902 (1 minus 120590119902)2(1 +

120582119902 + 120579

1205831

1 minus 120588119902120590119902

(1 minus 120588119902)2)

=1

120579 + 1205830

(1

+1205831(120579 + 120583

0)

1205831minus 120582119902

1

1205830((1205831minus 120582119902) (120582119902 + 120579)) + 120583

1

)

(64)

And if 120582 lt 1205831 119864[119882] is strictly increasing for 119902 isin [0 1]

Mathematical Problems in Engineering 9

1 1 2 0 3 1

0 0 1 0 2 0 3 0

n 1

n 0

120579

1205831 1205831

12058301205830

120582q120582q

120582q 120582qmiddot middot middot

middot middot middot

middot middot middot

middot middot middot

120579

1205831 1205831

12058301205830

120582q120582q

120582q 120582q

120579120579

1205831

1205831

1205830

1205830

120582q

120582q 120582q

Figure 7 Transition rate diagram for the unobservable queues

We consider a tagged customer If heshe decides to enterthe system hisher expected net benefit is

119880 (119902) = 119877 minus 119862119864 [119882] = 119877 minus119862

120579 + 1205830

(1

+1205831(120579 + 120583

0)

1205831minus 120582119902

1

1205830((1205831minus 120582119902) (120582119902 + 120579)) + 120583

1

)

(65)

We note that 119880(119902) is strictly decreasing in 119902 and it has aunique root 119902lowast

119890 So there exists a unique mixed equilibrium

strategy 119902119890and 119902119890= min119902lowast

119890 1

We now turn our attention to social optimization Thesocial benefit per time unit can now be easily computed as

119880119904(119902) = 120582119902 (119877 minus 119862119864 [119882]) = 120582119902(119877 minus

119862

120579 + 1205830

(1

+1205831(120579 + 120583

0)

1205831minus 120582119902

1

1205830((1205831minus 120582119902) (120582119902 + 120579)) + 120583

1

))

(66)

As for the equilibrium social welfare per time unit119880119904(119902119890)

Figure 8 shows that 119880119904(119902119890) first increases and then decreases

with 120582 and the social benefit achieves a maximum forintermediate values of this parameter The reason for thisbehavior is that when the arrival rate is small the system israrely crowded therefore as more customers arrive they areserved and the social benefit improves However with anysmall increase of the arrival rate 120582 the expected waiting timeincreases which has a detrimental effect on the social benefit

Let 119909lowast be the root of the equation 1198781015840(119902) = 0 and let 119902lowast be

the optimal joining probability Then we have the followingconclusions if 0 lt 119909

lowastlt 1 and 119878

10158401015840(119902) le 0 then 119902

lowast= 119909lowast if

0 lt 119909lowastlt 1 and 119878

10158401015840(119902) gt 0 or 119909lowast ge 1 then 119902

lowast= 1

From Figure 9 we observe that 119902119890

gt 119902lowast As a result

individual optimization leads to queues that are longer thansocially desired Therefore it is clear that the social plannerwants a toll to discourage arrivals

6 Conclusion

In this paper we analyzed the customer strategic behaviorin the MM1 queueing system with working vacations andvacation interruptions where arriving customers have optionto decide whether to join the system or balk Three differentcases with respect to the levels of information provided toarriving customers have been investigated extensively and theequilibrium strategies for each case were derived And wefound that the equilibrium joining probability of the busy

02 04 06 08 1 12 14 16 18 20Arrival rate 120582

Equi

libriu

m so

cial

wel

fare

Us(qe)

0

5

10

15

Figure 8 Equilibrium social welfare for the unobservable casewhen119877 = 10 119862 = 1 120583

1= 2 120583

0= 1 and 120579 = 1

03

04

05

06

07

08

09

1

Join

ing

prob

abili

ty

0 1 15 205Arrival rate 120582

qeqlowast

Figure 9 Equilibrium and socially optimal strategies for the unob-servable case when 119877 = 3 119862 = 2 120583

1= 2 120583

0= 05 and 120579 = 001

period may be smaller than that in vacation time We alsocompared the equilibrium strategy with the socially optimalstrategy numerically and we observed that customers makeindividual decisions maximizing their own profit and tendto overuse the system For the social planner a toll will beadopted to discourage customers from joining

10 Mathematical Problems in Engineering

Furthermore an interesting extension would be to incor-porate the present model in a profit maximizing frameworkwhere the owner or manager of the system imposes anentrance feeThe problem is to find the optimal fee that max-imizes the owners total profit subject to the constraint thatcustomers independently decide whether to enter or not andtake into account the fee in addition to the service reward andwaiting cost Another direction for future work is to concernthe non-Markovian queues with various vacation policies

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Acknowledgments

This work is partially supported by the Natural Foundation ofHebei Province (nos A2014203096 andC2014203212) Chinaand the Young Faculty Autonomous Research Project of Yan-shan University (nos 13LGB032 13LGA017 and 13LGB013)China

References

[1] P Naor ldquoThe regulation of queue size by levying tollsrdquo Econo-metrica vol 37 no 1 pp 15ndash24 1969

[2] N M Edelson and D K Hildebrand ldquoCongestion tolls forPoisson queuing processesrdquo Econometrica vol 43 no 1 pp 81ndash92 1975

[3] R Hassin and M Haviv Equilibrium Behavior in Queueing Sys-tems To Queue or Not to Queue Kluwer Academic PublishersDordrecht The Netherlands 2003

[4] A Burnetas and A Economou ldquoEquilibrium customer strate-gies in a single server Markovian queue with setup timesrdquoQueueing Systems vol 56 no 3-4 pp 213ndash228 2007

[5] W Sun and N Tian ldquoContrast of the equilibrium and sociallyoptimal strategies in a queue with vacationsrdquo Journal of Compu-tational Information Systems vol 4 no 5 pp 2167ndash2172 2008

[6] A Economou and S Kanta ldquoEquilibrium balking strategiesin the observable single-server queue with breakdowns andrepairsrdquoOperations Research Letters vol 36 no 6 pp 696ndash6992008

[7] P Guo and R Hassin ldquoStrategic behavior and social optimiza-tion in Markovian vacation queuesrdquo Operations Research vol59 no 4 pp 986ndash997 2011

[8] P Guo and R Hassin ldquoStrategic behavior and social optimiza-tion in Markovian vacation queues the case of heterogeneouscustomersrdquo European Journal of Operational Research vol 222no 2 pp 278ndash286 2012

[9] R Tian D Yue and W Yue ldquoOptimal balking strategies inan MG1 queueing system with a removable server under N-policyrdquo Journal of Industrial andManagementOptimization vol11 no 3 pp 715ndash731 2015

[10] P Guo and Q Li ldquoStrategic behavior and social optimizationin partially-observableMarkovian vacation queuesrdquoOperationsResearch Letters vol 41 no 3 pp 277ndash284 2013

[11] L Li J Wang and F Zhang ldquoEquilibrium customer strategiesin Markovian queues with partial breakdownsrdquo Computers ampIndustrial Engineering vol 66 no 4 pp 751ndash757 2013

[12] F Zhang J Wang and B Liu ldquoOn the optimal and equilibriumretrial rates in an unreliable retrial queue with vacationsrdquo Jour-nal of Industrial andManagement Optimization vol 8 no 4 pp861ndash875 2012

[13] N Tian J Li and Z G Zhang ldquoMatrix analysis methodand working vacation queuesa surveyrdquo International Journal ofInformation and Management Sciences vol 20 pp 603ndash6332009

[14] F Zhang J Wang and B Liu ldquoEquilibrium balking strategiesin Markovian queues with working vacationsrdquo Applied Mathe-matical Modelling vol 37 no 16-17 pp 8264ndash8282 2013

[15] W Sun and S Li ldquoEquilibrium and optimal behavior of cus-tomers in Markovian queues with multiple working vacationsrdquoTOP vol 22 no 2 pp 694ndash715 2014

[16] W Sun S Li and Q-L Li ldquoEquilibrium balking strategiesof customers in Markovian queues with two-stage workingvacationsrdquo Applied Mathematics and Computation vol 248 pp195ndash214 2014

[17] J Li andNTian ldquoTheMM1 queuewithworking vacations andvacation interruptionsrdquo Journal of Systems Science and SystemsEngineering vol 16 no 1 pp 121ndash127 2007

[18] J-H Li N-S Tian and Z-Y Ma ldquoPerformance analysis ofGIM1 queue with working vacations and vacation interrup-tionrdquo Applied Mathematical Modelling vol 32 no 12 pp 2715ndash2730 2008

[19] M Neuts Matrix-Geometric Solution in Stochastic ModelsJohns Hopkins University Press Baltimore Md USA 1981

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

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Mathematical PhysicsAdvances in

Complex AnalysisJournal of

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OptimizationJournal of

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CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

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Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Mathematical Problems in Engineering 5

2

3

4

5

6

7

8

9

n(1

)

2 3 4 5 6 7 81n(0)

24

68

10

02

46

8

n(1)n(0)

0

10

20

30

40

50

Us(n

(0)n

(1))

68

Figure 3 Social welfare for the observable case when 119877 = 40 119862 = 1 120582 = 15 1205831= 2 120583

0= 05 and 120579 = 05

1 1 2 0 3 1

0 0 1 0 2 0 3 0

n 1

n 0

120579

1205831 1205831

12058301205830

120582q1120582q1

120582q0 120582q0middot middot middot

middot middot middot

middot middot middot

middot middot middot

120579

1205831 1205831

12058301205830

120582q1120582q1

120582q0 120582q0

120579120579

1205831

1205831

1205830

1205830

120582q1

120582q0 120582q0

Figure 4 Transition rate diagram for the partially observable queues

strategy (1199020 1199021) then the system follows a Markov chain

where the arrival rate equals 120582119894= 120582119902119894when the server is in

state 119894 (119894 = 0 1) The state space is Ω119901119900

= 0 0 cup (119899 119894) |

119899 ge 1 119894 = 0 1 and the transition diagram is illustrated inFigure 4

Denote the stationary distribution as120587119899119894= lim119905rarrinfin

119875 119873 (119905) = 119899 119869 (119905) = 119894 (119899 119894) isin Ω119901119900

1205870= 12058700

120587119899= (1205871198990 1205871198991) 119899 ge 1

(28)

Using the lexicographical sequence for the states thetransition rate matrix (generator) Q can be written as thetridiagonal block matrix

Q =((

(

A0C0

B1

A CB A C

B A Cd d d

))

)

(29)

whereA0= minus120582119902

0

B1= (

1205830

1205831

)

C1= (1205821199020 0)

A = (minus (1205821199020+ 120579 + 120583

0) 120579

0 minus (1205821199021+ 1205831))

B = (0 1205830

0 1205831

)

C = (1205821199020

0

0 1205821199021

)

(30)To analyze this QBD process it is necessary to solve for

the minimal nonnegative solution of the matrix quadraticequation

R2B + RA + C = 0 (31)and this solution is called the rate matrix and denoted by R

Lemma 3 If 1205881

= 12058211990211205831

lt 1 the minimal nonnegativesolution of (31) is given by

R = (1205900

1205821199020+ 120579

1205831

1205900

0 1205881

) (32)

where 1205900= 1205821199020(1205821199020+ 120579 + 120583

0)

Proof Because the coefficients of (31) are all upper triangularmatrices we can assume that R has the same structure as

R = (11990311

11990312

0 11990322

) (33)

6 Mathematical Problems in Engineering

Substituting R2 and R into (31) yields the following set ofequations

minus (1205821199020+ 120579 + 120583

0) 11990311+ 1205821199020= 0

12058311199032

22minus (1205821199021+ 1205831) 11990322+ 1205821199021= 0

12058301199032

11+ 120583111990312(11990311+ 11990322) + 120579119903

11minus (1205821199021+ 1205831) 11990312

= 0

(34)

To obtain the minimal nonnegative solution of (31) we take11990322

= 1205881(the other roots are 119903

22= 1) in the second equation

of (34) From the first equation of (34) we obtain 11990311

=

1205821199020(1205821199020+ 120579 + 120583

0) = 1205900 Substituting 120590

0and 1205881into the last

equation of (34) we get 11990312

= ((1205821199020+ 120579)120583

1)1205900 This is the

proof of Lemma 3

Theorem 4 Assuming that 1205881lt 1 the stationary probabilities

120587119899119894

| (119899 119894) isin Ω119901119900 of the partially observable queues are as

follows

1205871198990

= 119870120590119899

0 119899 ge 0

1205871198991

= 1198701205881

1205821199020+ 120579

1205821199020+ 120579 + 120583

0

119899minus1

sum

119895=0

120588119895

1120590119899minus1minus119895

0 119899 ge 1

(35)

where

119870 =120579 + 1205830

1205821199020+ 120579 + 120583

0

(1 +1205821199020+ 120579

1205821199020+ 120579 + 120583

0

1205821199020

1205831minus 1205821199021

)

minus1

(36)

Proof With the matrix-geometric solution method [19] wehave

120587119899= (1205871198990 1205871198991) = (120587

10 12058711)R119899minus1 119899 ge 1 (37)

In addition (12058700 12058710 12058711) satisfies the set of equations

(12058700 12058710 12058711) 119861 [R] = 0 (38)

where

119861 [R] = (A0

C0

B1

A + RB)

= (

minus1205821199020

1205821199020

0

1205830

minus (1205821199020+ 120579 + 120583

0) 1205821199020+ 120579

1205831

0 minus1205831

)

(39)

Then from (38) we derive

minus120582119902012058700+ 120583012058710+ 120583112058711

= 0

120582119902012058700minus (1205821199020+ 120579 + 120583

0) 12058710

(1205821199020+ 120579) 120587

10minus 120583112058711

= 0

(40)

Taking 12058700as a known constant we obtain

12058710

= 120590012058700 (41)

12058711

= 1205881

1205821199020+ 120579

1205821199020+ 120579 + 120583

0

12058700 (42)

Note that

R119899minus1 = (120590119899minus1

0

1205821199020+ 120579

1205831

1205900

119899minus2

sum

119895=0

120590119895

0120588119899minus119895minus2

1

0 120588119899minus1

1

) 119899 ge 2 (43)

By 12058710 12058711 R119899minus1 and (37) we get (35) Finally 120587

00can be

determined by the normalization conditionThis is the proofof Theorem 4

Let 119901119894be the steady-state probability when the server is at

state 119894 (119894 = 0 1) we have

1199010=

infin

sum

119899=0

1205871198990

= 1198701205821199020+ 120579 + 120583

0

120579 + 1205830

(44)

1199011=

infin

sum

119899=1

1205871198991

= 1198701205821199020(1205821199020+ 120579)

(1205831minus 1205821199021) (120579 + 120583

0) (45)

So the effective arrival rate 120582 is given by

120582 = 120582 (11990101199020+ 11990111199021) (46)

We now consider a customer who finds the server at state119894 upon arrival The conditional mean sojourn time of such acustomer that decides to enter given that the others follow thesame mixed strategy (119902

0 1199021) is given by

119882119894(1199020 1199021) =

suminfin

119899=119894119879 (119899 119894) 120587

119899119894

suminfin

119899=119894120587119899119894

(47)

By substituting (5)-(6) and (35) into (44) we obtain

1198820(1199020 1199021) =

1205821199020+ 120579 + 120583

1

1205831(120579 + 120583

0)

1198821(1199020 1199021) =

1

1205831minus 1205821199021

+1205821199020+ 120579 + 120583

0

1205831(1205830+ 120579)

(48)

Based on the reward-cost structure the expected net ben-efit of an arriving customer if heshe finds the server at state119894 and decides to enter is given by

1198800(1199020) = 119877 minus 119862

1205821199020+ 120579 + 120583

1

1205831(120579 + 120583

0)

1198801(1199020 1199021) = 119877 minus 119862(

1

1205831minus 1205821199021

+1205821199020+ 120579 + 120583

0

1205831(1205830+ 120579)

)

(49)

We now can proceed to determine mixed equilibriumstrategies of a customer in the partially observable case andhave the following

Theorem 5 For the partially observable case there exists aunique mixed equilibrium strategy (119902

119890

0 119902119890

1) where the vector

(119902119890

0 119902119890

1) is given as follows

(1) 119862(120579 + 1205831)1205831(120579 + 120583

0) lt 119877 le 119862(120582 + 120579 + 120583

1)1205831(120579 + 120583

0)

(119902119890

0 119902119890

1) =

(1199091 0)

1205831minus 1205830

1205831(120579 + 120583

0)lt

1

120583

(1199091 1199092)

1

120583le

1205831minus 1205830

1205831(120579 + 120583

0)le

1

120583 minus 120582

(1199091 1)

1205831minus 1205830

1205831(120579 + 120583

0)gt

1

120583 minus 120582

(50)

Mathematical Problems in Engineering 7

(2) 119862(120582 + 120579 + 1205831)1205831(120579 + 120583

0) lt 119877

(119902119890

0 119902119890

1) =

(1 0) 119877 lt119862

1205831

+119862 (120582 + 120579 + 120583

0)

1205831(120579 + 120583

0)

(1 1199093)

119862

1205831

+119862 (120582 + 120579 + 120583

0)

1205831(120579 + 120583

0)

le 119877 le119862

1205831minus 120582

+119862 (120582 + 120579 + 120583

0)

1205831(120579 + 120583

0)

(1 1) 119877 gt119862

1205831minus 120582

+119862 (120582 + 120579 + 120583

0)

1205831(120579 + 120583

0)

(51)

where

1199091=

1

120582(1198771205831(120579 + 120583

0)

119862minus 120579 minus 120583

1)

1199092=1205831

120582(1 minus

1205830+ 120579

1205831minus 1205830

)

1199093=

1

120582(1205831minus

1

119877119862 minus (120582 + 120579 + 1205830) 1205831(120579 + 120583

0))

(52)

Proof To prove Theorem 5 we first focus on 1198800(1199020) Con-

dition (1) assures that 1199021198900is positive Therefore we have two

casesCase 1 (119880

0(0) gt 0 and 119880

0(1) le 0) That is 119862(120579 + 120583

1)1205831(120579 +

1205830) lt 119877 le 119862(120582+120579+120583

1)1205831(120579+1205830)) In this case if all customers

who find the system empty enter the system with probability119902119890

0= 1 then the tagged customer suffers a negative expected

benefit if heshe decides to enter Hence 1199021198900= 1 does not lead

to an equilibrium Similarly if all customers use 1199021198900= 0 then

the tagged customer receives a positive benefit from enteringthus 119902119890

0= 0 also cannot be part of an equilibrium mixed

strategy Therefore there exists unique 1199021198900satisfying

119877 minus 119862120582119902119890

0+ 120579 + 120583

1

1205831(120579 + 120583

0)

= 0 (53)

for which customers are indifferent between entering andbalking This is given by

119902119890

0=

1

120582(1198771205831(120579 + 120583

0)

119862minus 120579 minus 120583

1) = 119909

1 (54)

In this situation the expected benefit 1198801is given by

1198801(119902119890

0 1199021) =

119862 (1205831minus 1205830)

1205831(120579 + 120583

0)minus

119862

120583 minus 1205821199021

(55)

Using the standard methodology of equilibrium analysisin unobservable queueing models we derive from (55) thefollowing equilibria

(1) If 1198801(119902119890

0 0) lt 0 then the equilibrium strategy is 119902119890

1=

0(2) If 119880

1(119902119890

0 0) ge 0 and 119880

1(119902119890

0 1) le 0 there exists unique

119902119890

1 satisfying

1198801(119902119890

0 119902119890

1) = 0 (56)

then the equilibrium strategy is 1199021198901= 1199092

(3) If 1198801(119902119890

0 1) gt 0 then the equilibrium strategy is 119902119890

1=

1

Case 2 (1198800(1) ge 0) That is 119862(120582 + 120579 + 120583

1)1205831(120579 + 120583

0) lt 119877) In

this case for every strategy of the other customers the taggedcustomer has a positive expected net benefit if heshe decidesto enter Hence 119902119890

0= 1

In this situation the expected benefit 1198801is given by

1198801(1 1199021) = 119877 minus

119862

120583 minus 1205821199021

minus119862 (120582 + 120579 + 120583

0)

1205831(120579 + 120583

0)

(57)

To find 119902119890

1in equilibrium we also consider the following

subcases

(1) If 1198801(1 0) lt 0 then the equilibrium strategy is 119902119890

1=

0

(2) If 1198801(1 0) ge 0 and 119880

1(1 1) le 0 there exists unique

119902119890

1 satisfying

1198801(1 119902119890

1) = 0 (58)

then the equilibrium strategy is 1199021198901= 1199093

(3) If 1198801(1 1) gt 0 then the equilibrium strategy is 119902119890

1=

1

By rearranging Cases 1 and 2 we can obtain the results ofTheorem 5 This completes the proof

From Theorem 5 we can compare 119902119890

0with 119902

119890

1 Figure 5

shows that 1199021198900is not always less than 119902

119890

1 Namely sometimes

the information that the server takes a working vacation doesnot make the customers less willing to enter the systembecause in the vacation the server provides a lower serviceto customers But when the vacation time and waiting timebecome longer customers do not want to enter the system invacation

Then fromTheorem 4 themean queue length is given by

119864 [119871] =

infin

sum

119899=0

1198991205871198990+

infin

sum

119899=0

1198991205871198991

= 1198701205900

(1 minus 1205900)2(1 +

1205821199020+ 120579

1205831

1 minus 12058811205900

(1 minus 1205881)2)

(59)

8 Mathematical Problems in Engineering

03

04

05

06

07

08

09

1

Equi

libriu

m jo

inin

g pr

obab

ility

qe0qe1 (120579 = 01 C = 2)

qe0qe1 (120579 = 05 C = 3)

0 04 06 08 1 12 14 16 1802Arrival rate 120582

Figure 5 Equilibrium mixed strategies for the partially observablecase when 119877 = 5 120583

1= 2 and 120583

0= 05

0 04 06 08 1 12 14 16 1802Arrival rate 120582

0

01

02

03

04

05

06

07

08

09

1

Join

ing

prob

abili

ty

qe0qe1

qlowast0qlowast1

Figure 6 Equilibrium and socially optimal mixed strategies for thepartially observable case when 119877 = 5 119862 = 2 120583

1= 2 120583

0= 05 and

120579 = 01

And the social benefit when all customers follow a mixedpolicy (119902

0 1199021) can now be easily computed as follows

119880119904(1199020 1199021) = 120582119877 minus 119862119864 [119871]

= 120582 (11990101199020+ 11990111199021) 119877

minus1198621198701205900

(1 minus 1205900)2(1 +

1205821199020+ 120579

1205831

1 minus 12058811205900

(1 minus 1205881)2)

(60)

The goal of a social planner is to maximize overall socialwelfare Let (119902lowast

0 119902lowast

1) be the socially optimal mixed strategy

Figure 6 compares customers equilibrium mixed strategy(119902119890

0 119902119890

1) and their socially optimal mixed strategy (119902lowast

0 119902lowast

1) We

also observe that 1199021198900gt 119902lowast

0and 1199021198901gt 119902lowast

1This ordering is typical

when customers make individual decisions maximizing theirown profitThen they ignore those negative externalities thatthey impose on later arrivals and they tend to overuse thesystem It is clear that these externalities should be taken intoaccount when we aim to maximize the total revenue

5 The Unobservable Queues

In this section we consider the fully unobservable case wherearriving customers cannot observe the system state

There are two pure strategies available for a customer thatis to join the queue or not to join the queue A pure or mixedstrategy can be described by a fraction 119902 (0 le 119902 le 1) whichis the probability of joining and the effective arrival rate orjoining rate is 120582119902 The transition diagram is illustrated inFigure 7

To identify the equilibrium strategies of the customerswe should first investigate the stationary distribution of thesystem when all customers follow a given strategy 119902 whichcan be obtained by Theorem 4 by taking 119902

0= 1199021= 119902 So we

have

1205871198990

= 1198701120590119899

119902 119899 ge 0

1205871198991

= 1198701120588119902

120582119902 + 120579

120582119902 + 120579 + 1205830

119899minus1

sum

119895=0

120588119895

119902120590119899minus1minus119895

119902 119899 ge 1

(61)

where 120588119902= 120582119902120583

1 120590119902= 120582119902(120582119902 + 120579 + 120583

0) and

1198701=

120579 + 1205830

120582119902 + 120579 + 1205830

(1 +120582119902 + 120579

120582119902 + 120579 + 1205830

120582119902

1205831minus 120582119902

)

minus1

(62)

Then the mean number of the customers in the system is

119864 [119871] =

infin

sum

119899=0

1198991205871198990+

infin

sum

119899=0

1198991205871198991

=1198701120590119902

(1 minus 120590119902)2(1 +

120582119902 + 120579

1205831

1 minus 120588119902120590119902

(1 minus 120588119902)2)

(63)

Hence the mean sojourn time of a customer who decides toenter upon hisher arrival can be obtained by using Littlersquoslaw

119864 [119882] =1198701120590119902

120582119902 (1 minus 120590119902)2(1 +

120582119902 + 120579

1205831

1 minus 120588119902120590119902

(1 minus 120588119902)2)

=1

120579 + 1205830

(1

+1205831(120579 + 120583

0)

1205831minus 120582119902

1

1205830((1205831minus 120582119902) (120582119902 + 120579)) + 120583

1

)

(64)

And if 120582 lt 1205831 119864[119882] is strictly increasing for 119902 isin [0 1]

Mathematical Problems in Engineering 9

1 1 2 0 3 1

0 0 1 0 2 0 3 0

n 1

n 0

120579

1205831 1205831

12058301205830

120582q120582q

120582q 120582qmiddot middot middot

middot middot middot

middot middot middot

middot middot middot

120579

1205831 1205831

12058301205830

120582q120582q

120582q 120582q

120579120579

1205831

1205831

1205830

1205830

120582q

120582q 120582q

Figure 7 Transition rate diagram for the unobservable queues

We consider a tagged customer If heshe decides to enterthe system hisher expected net benefit is

119880 (119902) = 119877 minus 119862119864 [119882] = 119877 minus119862

120579 + 1205830

(1

+1205831(120579 + 120583

0)

1205831minus 120582119902

1

1205830((1205831minus 120582119902) (120582119902 + 120579)) + 120583

1

)

(65)

We note that 119880(119902) is strictly decreasing in 119902 and it has aunique root 119902lowast

119890 So there exists a unique mixed equilibrium

strategy 119902119890and 119902119890= min119902lowast

119890 1

We now turn our attention to social optimization Thesocial benefit per time unit can now be easily computed as

119880119904(119902) = 120582119902 (119877 minus 119862119864 [119882]) = 120582119902(119877 minus

119862

120579 + 1205830

(1

+1205831(120579 + 120583

0)

1205831minus 120582119902

1

1205830((1205831minus 120582119902) (120582119902 + 120579)) + 120583

1

))

(66)

As for the equilibrium social welfare per time unit119880119904(119902119890)

Figure 8 shows that 119880119904(119902119890) first increases and then decreases

with 120582 and the social benefit achieves a maximum forintermediate values of this parameter The reason for thisbehavior is that when the arrival rate is small the system israrely crowded therefore as more customers arrive they areserved and the social benefit improves However with anysmall increase of the arrival rate 120582 the expected waiting timeincreases which has a detrimental effect on the social benefit

Let 119909lowast be the root of the equation 1198781015840(119902) = 0 and let 119902lowast be

the optimal joining probability Then we have the followingconclusions if 0 lt 119909

lowastlt 1 and 119878

10158401015840(119902) le 0 then 119902

lowast= 119909lowast if

0 lt 119909lowastlt 1 and 119878

10158401015840(119902) gt 0 or 119909lowast ge 1 then 119902

lowast= 1

From Figure 9 we observe that 119902119890

gt 119902lowast As a result

individual optimization leads to queues that are longer thansocially desired Therefore it is clear that the social plannerwants a toll to discourage arrivals

6 Conclusion

In this paper we analyzed the customer strategic behaviorin the MM1 queueing system with working vacations andvacation interruptions where arriving customers have optionto decide whether to join the system or balk Three differentcases with respect to the levels of information provided toarriving customers have been investigated extensively and theequilibrium strategies for each case were derived And wefound that the equilibrium joining probability of the busy

02 04 06 08 1 12 14 16 18 20Arrival rate 120582

Equi

libriu

m so

cial

wel

fare

Us(qe)

0

5

10

15

Figure 8 Equilibrium social welfare for the unobservable casewhen119877 = 10 119862 = 1 120583

1= 2 120583

0= 1 and 120579 = 1

03

04

05

06

07

08

09

1

Join

ing

prob

abili

ty

0 1 15 205Arrival rate 120582

qeqlowast

Figure 9 Equilibrium and socially optimal strategies for the unob-servable case when 119877 = 3 119862 = 2 120583

1= 2 120583

0= 05 and 120579 = 001

period may be smaller than that in vacation time We alsocompared the equilibrium strategy with the socially optimalstrategy numerically and we observed that customers makeindividual decisions maximizing their own profit and tendto overuse the system For the social planner a toll will beadopted to discourage customers from joining

10 Mathematical Problems in Engineering

Furthermore an interesting extension would be to incor-porate the present model in a profit maximizing frameworkwhere the owner or manager of the system imposes anentrance feeThe problem is to find the optimal fee that max-imizes the owners total profit subject to the constraint thatcustomers independently decide whether to enter or not andtake into account the fee in addition to the service reward andwaiting cost Another direction for future work is to concernthe non-Markovian queues with various vacation policies

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Acknowledgments

This work is partially supported by the Natural Foundation ofHebei Province (nos A2014203096 andC2014203212) Chinaand the Young Faculty Autonomous Research Project of Yan-shan University (nos 13LGB032 13LGA017 and 13LGB013)China

References

[1] P Naor ldquoThe regulation of queue size by levying tollsrdquo Econo-metrica vol 37 no 1 pp 15ndash24 1969

[2] N M Edelson and D K Hildebrand ldquoCongestion tolls forPoisson queuing processesrdquo Econometrica vol 43 no 1 pp 81ndash92 1975

[3] R Hassin and M Haviv Equilibrium Behavior in Queueing Sys-tems To Queue or Not to Queue Kluwer Academic PublishersDordrecht The Netherlands 2003

[4] A Burnetas and A Economou ldquoEquilibrium customer strate-gies in a single server Markovian queue with setup timesrdquoQueueing Systems vol 56 no 3-4 pp 213ndash228 2007

[5] W Sun and N Tian ldquoContrast of the equilibrium and sociallyoptimal strategies in a queue with vacationsrdquo Journal of Compu-tational Information Systems vol 4 no 5 pp 2167ndash2172 2008

[6] A Economou and S Kanta ldquoEquilibrium balking strategiesin the observable single-server queue with breakdowns andrepairsrdquoOperations Research Letters vol 36 no 6 pp 696ndash6992008

[7] P Guo and R Hassin ldquoStrategic behavior and social optimiza-tion in Markovian vacation queuesrdquo Operations Research vol59 no 4 pp 986ndash997 2011

[8] P Guo and R Hassin ldquoStrategic behavior and social optimiza-tion in Markovian vacation queues the case of heterogeneouscustomersrdquo European Journal of Operational Research vol 222no 2 pp 278ndash286 2012

[9] R Tian D Yue and W Yue ldquoOptimal balking strategies inan MG1 queueing system with a removable server under N-policyrdquo Journal of Industrial andManagementOptimization vol11 no 3 pp 715ndash731 2015

[10] P Guo and Q Li ldquoStrategic behavior and social optimizationin partially-observableMarkovian vacation queuesrdquoOperationsResearch Letters vol 41 no 3 pp 277ndash284 2013

[11] L Li J Wang and F Zhang ldquoEquilibrium customer strategiesin Markovian queues with partial breakdownsrdquo Computers ampIndustrial Engineering vol 66 no 4 pp 751ndash757 2013

[12] F Zhang J Wang and B Liu ldquoOn the optimal and equilibriumretrial rates in an unreliable retrial queue with vacationsrdquo Jour-nal of Industrial andManagement Optimization vol 8 no 4 pp861ndash875 2012

[13] N Tian J Li and Z G Zhang ldquoMatrix analysis methodand working vacation queuesa surveyrdquo International Journal ofInformation and Management Sciences vol 20 pp 603ndash6332009

[14] F Zhang J Wang and B Liu ldquoEquilibrium balking strategiesin Markovian queues with working vacationsrdquo Applied Mathe-matical Modelling vol 37 no 16-17 pp 8264ndash8282 2013

[15] W Sun and S Li ldquoEquilibrium and optimal behavior of cus-tomers in Markovian queues with multiple working vacationsrdquoTOP vol 22 no 2 pp 694ndash715 2014

[16] W Sun S Li and Q-L Li ldquoEquilibrium balking strategiesof customers in Markovian queues with two-stage workingvacationsrdquo Applied Mathematics and Computation vol 248 pp195ndash214 2014

[17] J Li andNTian ldquoTheMM1 queuewithworking vacations andvacation interruptionsrdquo Journal of Systems Science and SystemsEngineering vol 16 no 1 pp 121ndash127 2007

[18] J-H Li N-S Tian and Z-Y Ma ldquoPerformance analysis ofGIM1 queue with working vacations and vacation interrup-tionrdquo Applied Mathematical Modelling vol 32 no 12 pp 2715ndash2730 2008

[19] M Neuts Matrix-Geometric Solution in Stochastic ModelsJohns Hopkins University Press Baltimore Md USA 1981

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

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Mathematical PhysicsAdvances in

Complex AnalysisJournal of

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OptimizationJournal of

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CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

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Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

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The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

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Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

6 Mathematical Problems in Engineering

Substituting R2 and R into (31) yields the following set ofequations

minus (1205821199020+ 120579 + 120583

0) 11990311+ 1205821199020= 0

12058311199032

22minus (1205821199021+ 1205831) 11990322+ 1205821199021= 0

12058301199032

11+ 120583111990312(11990311+ 11990322) + 120579119903

11minus (1205821199021+ 1205831) 11990312

= 0

(34)

To obtain the minimal nonnegative solution of (31) we take11990322

= 1205881(the other roots are 119903

22= 1) in the second equation

of (34) From the first equation of (34) we obtain 11990311

=

1205821199020(1205821199020+ 120579 + 120583

0) = 1205900 Substituting 120590

0and 1205881into the last

equation of (34) we get 11990312

= ((1205821199020+ 120579)120583

1)1205900 This is the

proof of Lemma 3

Theorem 4 Assuming that 1205881lt 1 the stationary probabilities

120587119899119894

| (119899 119894) isin Ω119901119900 of the partially observable queues are as

follows

1205871198990

= 119870120590119899

0 119899 ge 0

1205871198991

= 1198701205881

1205821199020+ 120579

1205821199020+ 120579 + 120583

0

119899minus1

sum

119895=0

120588119895

1120590119899minus1minus119895

0 119899 ge 1

(35)

where

119870 =120579 + 1205830

1205821199020+ 120579 + 120583

0

(1 +1205821199020+ 120579

1205821199020+ 120579 + 120583

0

1205821199020

1205831minus 1205821199021

)

minus1

(36)

Proof With the matrix-geometric solution method [19] wehave

120587119899= (1205871198990 1205871198991) = (120587

10 12058711)R119899minus1 119899 ge 1 (37)

In addition (12058700 12058710 12058711) satisfies the set of equations

(12058700 12058710 12058711) 119861 [R] = 0 (38)

where

119861 [R] = (A0

C0

B1

A + RB)

= (

minus1205821199020

1205821199020

0

1205830

minus (1205821199020+ 120579 + 120583

0) 1205821199020+ 120579

1205831

0 minus1205831

)

(39)

Then from (38) we derive

minus120582119902012058700+ 120583012058710+ 120583112058711

= 0

120582119902012058700minus (1205821199020+ 120579 + 120583

0) 12058710

(1205821199020+ 120579) 120587

10minus 120583112058711

= 0

(40)

Taking 12058700as a known constant we obtain

12058710

= 120590012058700 (41)

12058711

= 1205881

1205821199020+ 120579

1205821199020+ 120579 + 120583

0

12058700 (42)

Note that

R119899minus1 = (120590119899minus1

0

1205821199020+ 120579

1205831

1205900

119899minus2

sum

119895=0

120590119895

0120588119899minus119895minus2

1

0 120588119899minus1

1

) 119899 ge 2 (43)

By 12058710 12058711 R119899minus1 and (37) we get (35) Finally 120587

00can be

determined by the normalization conditionThis is the proofof Theorem 4

Let 119901119894be the steady-state probability when the server is at

state 119894 (119894 = 0 1) we have

1199010=

infin

sum

119899=0

1205871198990

= 1198701205821199020+ 120579 + 120583

0

120579 + 1205830

(44)

1199011=

infin

sum

119899=1

1205871198991

= 1198701205821199020(1205821199020+ 120579)

(1205831minus 1205821199021) (120579 + 120583

0) (45)

So the effective arrival rate 120582 is given by

120582 = 120582 (11990101199020+ 11990111199021) (46)

We now consider a customer who finds the server at state119894 upon arrival The conditional mean sojourn time of such acustomer that decides to enter given that the others follow thesame mixed strategy (119902

0 1199021) is given by

119882119894(1199020 1199021) =

suminfin

119899=119894119879 (119899 119894) 120587

119899119894

suminfin

119899=119894120587119899119894

(47)

By substituting (5)-(6) and (35) into (44) we obtain

1198820(1199020 1199021) =

1205821199020+ 120579 + 120583

1

1205831(120579 + 120583

0)

1198821(1199020 1199021) =

1

1205831minus 1205821199021

+1205821199020+ 120579 + 120583

0

1205831(1205830+ 120579)

(48)

Based on the reward-cost structure the expected net ben-efit of an arriving customer if heshe finds the server at state119894 and decides to enter is given by

1198800(1199020) = 119877 minus 119862

1205821199020+ 120579 + 120583

1

1205831(120579 + 120583

0)

1198801(1199020 1199021) = 119877 minus 119862(

1

1205831minus 1205821199021

+1205821199020+ 120579 + 120583

0

1205831(1205830+ 120579)

)

(49)

We now can proceed to determine mixed equilibriumstrategies of a customer in the partially observable case andhave the following

Theorem 5 For the partially observable case there exists aunique mixed equilibrium strategy (119902

119890

0 119902119890

1) where the vector

(119902119890

0 119902119890

1) is given as follows

(1) 119862(120579 + 1205831)1205831(120579 + 120583

0) lt 119877 le 119862(120582 + 120579 + 120583

1)1205831(120579 + 120583

0)

(119902119890

0 119902119890

1) =

(1199091 0)

1205831minus 1205830

1205831(120579 + 120583

0)lt

1

120583

(1199091 1199092)

1

120583le

1205831minus 1205830

1205831(120579 + 120583

0)le

1

120583 minus 120582

(1199091 1)

1205831minus 1205830

1205831(120579 + 120583

0)gt

1

120583 minus 120582

(50)

Mathematical Problems in Engineering 7

(2) 119862(120582 + 120579 + 1205831)1205831(120579 + 120583

0) lt 119877

(119902119890

0 119902119890

1) =

(1 0) 119877 lt119862

1205831

+119862 (120582 + 120579 + 120583

0)

1205831(120579 + 120583

0)

(1 1199093)

119862

1205831

+119862 (120582 + 120579 + 120583

0)

1205831(120579 + 120583

0)

le 119877 le119862

1205831minus 120582

+119862 (120582 + 120579 + 120583

0)

1205831(120579 + 120583

0)

(1 1) 119877 gt119862

1205831minus 120582

+119862 (120582 + 120579 + 120583

0)

1205831(120579 + 120583

0)

(51)

where

1199091=

1

120582(1198771205831(120579 + 120583

0)

119862minus 120579 minus 120583

1)

1199092=1205831

120582(1 minus

1205830+ 120579

1205831minus 1205830

)

1199093=

1

120582(1205831minus

1

119877119862 minus (120582 + 120579 + 1205830) 1205831(120579 + 120583

0))

(52)

Proof To prove Theorem 5 we first focus on 1198800(1199020) Con-

dition (1) assures that 1199021198900is positive Therefore we have two

casesCase 1 (119880

0(0) gt 0 and 119880

0(1) le 0) That is 119862(120579 + 120583

1)1205831(120579 +

1205830) lt 119877 le 119862(120582+120579+120583

1)1205831(120579+1205830)) In this case if all customers

who find the system empty enter the system with probability119902119890

0= 1 then the tagged customer suffers a negative expected

benefit if heshe decides to enter Hence 1199021198900= 1 does not lead

to an equilibrium Similarly if all customers use 1199021198900= 0 then

the tagged customer receives a positive benefit from enteringthus 119902119890

0= 0 also cannot be part of an equilibrium mixed

strategy Therefore there exists unique 1199021198900satisfying

119877 minus 119862120582119902119890

0+ 120579 + 120583

1

1205831(120579 + 120583

0)

= 0 (53)

for which customers are indifferent between entering andbalking This is given by

119902119890

0=

1

120582(1198771205831(120579 + 120583

0)

119862minus 120579 minus 120583

1) = 119909

1 (54)

In this situation the expected benefit 1198801is given by

1198801(119902119890

0 1199021) =

119862 (1205831minus 1205830)

1205831(120579 + 120583

0)minus

119862

120583 minus 1205821199021

(55)

Using the standard methodology of equilibrium analysisin unobservable queueing models we derive from (55) thefollowing equilibria

(1) If 1198801(119902119890

0 0) lt 0 then the equilibrium strategy is 119902119890

1=

0(2) If 119880

1(119902119890

0 0) ge 0 and 119880

1(119902119890

0 1) le 0 there exists unique

119902119890

1 satisfying

1198801(119902119890

0 119902119890

1) = 0 (56)

then the equilibrium strategy is 1199021198901= 1199092

(3) If 1198801(119902119890

0 1) gt 0 then the equilibrium strategy is 119902119890

1=

1

Case 2 (1198800(1) ge 0) That is 119862(120582 + 120579 + 120583

1)1205831(120579 + 120583

0) lt 119877) In

this case for every strategy of the other customers the taggedcustomer has a positive expected net benefit if heshe decidesto enter Hence 119902119890

0= 1

In this situation the expected benefit 1198801is given by

1198801(1 1199021) = 119877 minus

119862

120583 minus 1205821199021

minus119862 (120582 + 120579 + 120583

0)

1205831(120579 + 120583

0)

(57)

To find 119902119890

1in equilibrium we also consider the following

subcases

(1) If 1198801(1 0) lt 0 then the equilibrium strategy is 119902119890

1=

0

(2) If 1198801(1 0) ge 0 and 119880

1(1 1) le 0 there exists unique

119902119890

1 satisfying

1198801(1 119902119890

1) = 0 (58)

then the equilibrium strategy is 1199021198901= 1199093

(3) If 1198801(1 1) gt 0 then the equilibrium strategy is 119902119890

1=

1

By rearranging Cases 1 and 2 we can obtain the results ofTheorem 5 This completes the proof

From Theorem 5 we can compare 119902119890

0with 119902

119890

1 Figure 5

shows that 1199021198900is not always less than 119902

119890

1 Namely sometimes

the information that the server takes a working vacation doesnot make the customers less willing to enter the systembecause in the vacation the server provides a lower serviceto customers But when the vacation time and waiting timebecome longer customers do not want to enter the system invacation

Then fromTheorem 4 themean queue length is given by

119864 [119871] =

infin

sum

119899=0

1198991205871198990+

infin

sum

119899=0

1198991205871198991

= 1198701205900

(1 minus 1205900)2(1 +

1205821199020+ 120579

1205831

1 minus 12058811205900

(1 minus 1205881)2)

(59)

8 Mathematical Problems in Engineering

03

04

05

06

07

08

09

1

Equi

libriu

m jo

inin

g pr

obab

ility

qe0qe1 (120579 = 01 C = 2)

qe0qe1 (120579 = 05 C = 3)

0 04 06 08 1 12 14 16 1802Arrival rate 120582

Figure 5 Equilibrium mixed strategies for the partially observablecase when 119877 = 5 120583

1= 2 and 120583

0= 05

0 04 06 08 1 12 14 16 1802Arrival rate 120582

0

01

02

03

04

05

06

07

08

09

1

Join

ing

prob

abili

ty

qe0qe1

qlowast0qlowast1

Figure 6 Equilibrium and socially optimal mixed strategies for thepartially observable case when 119877 = 5 119862 = 2 120583

1= 2 120583

0= 05 and

120579 = 01

And the social benefit when all customers follow a mixedpolicy (119902

0 1199021) can now be easily computed as follows

119880119904(1199020 1199021) = 120582119877 minus 119862119864 [119871]

= 120582 (11990101199020+ 11990111199021) 119877

minus1198621198701205900

(1 minus 1205900)2(1 +

1205821199020+ 120579

1205831

1 minus 12058811205900

(1 minus 1205881)2)

(60)

The goal of a social planner is to maximize overall socialwelfare Let (119902lowast

0 119902lowast

1) be the socially optimal mixed strategy

Figure 6 compares customers equilibrium mixed strategy(119902119890

0 119902119890

1) and their socially optimal mixed strategy (119902lowast

0 119902lowast

1) We

also observe that 1199021198900gt 119902lowast

0and 1199021198901gt 119902lowast

1This ordering is typical

when customers make individual decisions maximizing theirown profitThen they ignore those negative externalities thatthey impose on later arrivals and they tend to overuse thesystem It is clear that these externalities should be taken intoaccount when we aim to maximize the total revenue

5 The Unobservable Queues

In this section we consider the fully unobservable case wherearriving customers cannot observe the system state

There are two pure strategies available for a customer thatis to join the queue or not to join the queue A pure or mixedstrategy can be described by a fraction 119902 (0 le 119902 le 1) whichis the probability of joining and the effective arrival rate orjoining rate is 120582119902 The transition diagram is illustrated inFigure 7

To identify the equilibrium strategies of the customerswe should first investigate the stationary distribution of thesystem when all customers follow a given strategy 119902 whichcan be obtained by Theorem 4 by taking 119902

0= 1199021= 119902 So we

have

1205871198990

= 1198701120590119899

119902 119899 ge 0

1205871198991

= 1198701120588119902

120582119902 + 120579

120582119902 + 120579 + 1205830

119899minus1

sum

119895=0

120588119895

119902120590119899minus1minus119895

119902 119899 ge 1

(61)

where 120588119902= 120582119902120583

1 120590119902= 120582119902(120582119902 + 120579 + 120583

0) and

1198701=

120579 + 1205830

120582119902 + 120579 + 1205830

(1 +120582119902 + 120579

120582119902 + 120579 + 1205830

120582119902

1205831minus 120582119902

)

minus1

(62)

Then the mean number of the customers in the system is

119864 [119871] =

infin

sum

119899=0

1198991205871198990+

infin

sum

119899=0

1198991205871198991

=1198701120590119902

(1 minus 120590119902)2(1 +

120582119902 + 120579

1205831

1 minus 120588119902120590119902

(1 minus 120588119902)2)

(63)

Hence the mean sojourn time of a customer who decides toenter upon hisher arrival can be obtained by using Littlersquoslaw

119864 [119882] =1198701120590119902

120582119902 (1 minus 120590119902)2(1 +

120582119902 + 120579

1205831

1 minus 120588119902120590119902

(1 minus 120588119902)2)

=1

120579 + 1205830

(1

+1205831(120579 + 120583

0)

1205831minus 120582119902

1

1205830((1205831minus 120582119902) (120582119902 + 120579)) + 120583

1

)

(64)

And if 120582 lt 1205831 119864[119882] is strictly increasing for 119902 isin [0 1]

Mathematical Problems in Engineering 9

1 1 2 0 3 1

0 0 1 0 2 0 3 0

n 1

n 0

120579

1205831 1205831

12058301205830

120582q120582q

120582q 120582qmiddot middot middot

middot middot middot

middot middot middot

middot middot middot

120579

1205831 1205831

12058301205830

120582q120582q

120582q 120582q

120579120579

1205831

1205831

1205830

1205830

120582q

120582q 120582q

Figure 7 Transition rate diagram for the unobservable queues

We consider a tagged customer If heshe decides to enterthe system hisher expected net benefit is

119880 (119902) = 119877 minus 119862119864 [119882] = 119877 minus119862

120579 + 1205830

(1

+1205831(120579 + 120583

0)

1205831minus 120582119902

1

1205830((1205831minus 120582119902) (120582119902 + 120579)) + 120583

1

)

(65)

We note that 119880(119902) is strictly decreasing in 119902 and it has aunique root 119902lowast

119890 So there exists a unique mixed equilibrium

strategy 119902119890and 119902119890= min119902lowast

119890 1

We now turn our attention to social optimization Thesocial benefit per time unit can now be easily computed as

119880119904(119902) = 120582119902 (119877 minus 119862119864 [119882]) = 120582119902(119877 minus

119862

120579 + 1205830

(1

+1205831(120579 + 120583

0)

1205831minus 120582119902

1

1205830((1205831minus 120582119902) (120582119902 + 120579)) + 120583

1

))

(66)

As for the equilibrium social welfare per time unit119880119904(119902119890)

Figure 8 shows that 119880119904(119902119890) first increases and then decreases

with 120582 and the social benefit achieves a maximum forintermediate values of this parameter The reason for thisbehavior is that when the arrival rate is small the system israrely crowded therefore as more customers arrive they areserved and the social benefit improves However with anysmall increase of the arrival rate 120582 the expected waiting timeincreases which has a detrimental effect on the social benefit

Let 119909lowast be the root of the equation 1198781015840(119902) = 0 and let 119902lowast be

the optimal joining probability Then we have the followingconclusions if 0 lt 119909

lowastlt 1 and 119878

10158401015840(119902) le 0 then 119902

lowast= 119909lowast if

0 lt 119909lowastlt 1 and 119878

10158401015840(119902) gt 0 or 119909lowast ge 1 then 119902

lowast= 1

From Figure 9 we observe that 119902119890

gt 119902lowast As a result

individual optimization leads to queues that are longer thansocially desired Therefore it is clear that the social plannerwants a toll to discourage arrivals

6 Conclusion

In this paper we analyzed the customer strategic behaviorin the MM1 queueing system with working vacations andvacation interruptions where arriving customers have optionto decide whether to join the system or balk Three differentcases with respect to the levels of information provided toarriving customers have been investigated extensively and theequilibrium strategies for each case were derived And wefound that the equilibrium joining probability of the busy

02 04 06 08 1 12 14 16 18 20Arrival rate 120582

Equi

libriu

m so

cial

wel

fare

Us(qe)

0

5

10

15

Figure 8 Equilibrium social welfare for the unobservable casewhen119877 = 10 119862 = 1 120583

1= 2 120583

0= 1 and 120579 = 1

03

04

05

06

07

08

09

1

Join

ing

prob

abili

ty

0 1 15 205Arrival rate 120582

qeqlowast

Figure 9 Equilibrium and socially optimal strategies for the unob-servable case when 119877 = 3 119862 = 2 120583

1= 2 120583

0= 05 and 120579 = 001

period may be smaller than that in vacation time We alsocompared the equilibrium strategy with the socially optimalstrategy numerically and we observed that customers makeindividual decisions maximizing their own profit and tendto overuse the system For the social planner a toll will beadopted to discourage customers from joining

10 Mathematical Problems in Engineering

Furthermore an interesting extension would be to incor-porate the present model in a profit maximizing frameworkwhere the owner or manager of the system imposes anentrance feeThe problem is to find the optimal fee that max-imizes the owners total profit subject to the constraint thatcustomers independently decide whether to enter or not andtake into account the fee in addition to the service reward andwaiting cost Another direction for future work is to concernthe non-Markovian queues with various vacation policies

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Acknowledgments

This work is partially supported by the Natural Foundation ofHebei Province (nos A2014203096 andC2014203212) Chinaand the Young Faculty Autonomous Research Project of Yan-shan University (nos 13LGB032 13LGA017 and 13LGB013)China

References

[1] P Naor ldquoThe regulation of queue size by levying tollsrdquo Econo-metrica vol 37 no 1 pp 15ndash24 1969

[2] N M Edelson and D K Hildebrand ldquoCongestion tolls forPoisson queuing processesrdquo Econometrica vol 43 no 1 pp 81ndash92 1975

[3] R Hassin and M Haviv Equilibrium Behavior in Queueing Sys-tems To Queue or Not to Queue Kluwer Academic PublishersDordrecht The Netherlands 2003

[4] A Burnetas and A Economou ldquoEquilibrium customer strate-gies in a single server Markovian queue with setup timesrdquoQueueing Systems vol 56 no 3-4 pp 213ndash228 2007

[5] W Sun and N Tian ldquoContrast of the equilibrium and sociallyoptimal strategies in a queue with vacationsrdquo Journal of Compu-tational Information Systems vol 4 no 5 pp 2167ndash2172 2008

[6] A Economou and S Kanta ldquoEquilibrium balking strategiesin the observable single-server queue with breakdowns andrepairsrdquoOperations Research Letters vol 36 no 6 pp 696ndash6992008

[7] P Guo and R Hassin ldquoStrategic behavior and social optimiza-tion in Markovian vacation queuesrdquo Operations Research vol59 no 4 pp 986ndash997 2011

[8] P Guo and R Hassin ldquoStrategic behavior and social optimiza-tion in Markovian vacation queues the case of heterogeneouscustomersrdquo European Journal of Operational Research vol 222no 2 pp 278ndash286 2012

[9] R Tian D Yue and W Yue ldquoOptimal balking strategies inan MG1 queueing system with a removable server under N-policyrdquo Journal of Industrial andManagementOptimization vol11 no 3 pp 715ndash731 2015

[10] P Guo and Q Li ldquoStrategic behavior and social optimizationin partially-observableMarkovian vacation queuesrdquoOperationsResearch Letters vol 41 no 3 pp 277ndash284 2013

[11] L Li J Wang and F Zhang ldquoEquilibrium customer strategiesin Markovian queues with partial breakdownsrdquo Computers ampIndustrial Engineering vol 66 no 4 pp 751ndash757 2013

[12] F Zhang J Wang and B Liu ldquoOn the optimal and equilibriumretrial rates in an unreliable retrial queue with vacationsrdquo Jour-nal of Industrial andManagement Optimization vol 8 no 4 pp861ndash875 2012

[13] N Tian J Li and Z G Zhang ldquoMatrix analysis methodand working vacation queuesa surveyrdquo International Journal ofInformation and Management Sciences vol 20 pp 603ndash6332009

[14] F Zhang J Wang and B Liu ldquoEquilibrium balking strategiesin Markovian queues with working vacationsrdquo Applied Mathe-matical Modelling vol 37 no 16-17 pp 8264ndash8282 2013

[15] W Sun and S Li ldquoEquilibrium and optimal behavior of cus-tomers in Markovian queues with multiple working vacationsrdquoTOP vol 22 no 2 pp 694ndash715 2014

[16] W Sun S Li and Q-L Li ldquoEquilibrium balking strategiesof customers in Markovian queues with two-stage workingvacationsrdquo Applied Mathematics and Computation vol 248 pp195ndash214 2014

[17] J Li andNTian ldquoTheMM1 queuewithworking vacations andvacation interruptionsrdquo Journal of Systems Science and SystemsEngineering vol 16 no 1 pp 121ndash127 2007

[18] J-H Li N-S Tian and Z-Y Ma ldquoPerformance analysis ofGIM1 queue with working vacations and vacation interrup-tionrdquo Applied Mathematical Modelling vol 32 no 12 pp 2715ndash2730 2008

[19] M Neuts Matrix-Geometric Solution in Stochastic ModelsJohns Hopkins University Press Baltimore Md USA 1981

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

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CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

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Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Mathematical Problems in Engineering 7

(2) 119862(120582 + 120579 + 1205831)1205831(120579 + 120583

0) lt 119877

(119902119890

0 119902119890

1) =

(1 0) 119877 lt119862

1205831

+119862 (120582 + 120579 + 120583

0)

1205831(120579 + 120583

0)

(1 1199093)

119862

1205831

+119862 (120582 + 120579 + 120583

0)

1205831(120579 + 120583

0)

le 119877 le119862

1205831minus 120582

+119862 (120582 + 120579 + 120583

0)

1205831(120579 + 120583

0)

(1 1) 119877 gt119862

1205831minus 120582

+119862 (120582 + 120579 + 120583

0)

1205831(120579 + 120583

0)

(51)

where

1199091=

1

120582(1198771205831(120579 + 120583

0)

119862minus 120579 minus 120583

1)

1199092=1205831

120582(1 minus

1205830+ 120579

1205831minus 1205830

)

1199093=

1

120582(1205831minus

1

119877119862 minus (120582 + 120579 + 1205830) 1205831(120579 + 120583

0))

(52)

Proof To prove Theorem 5 we first focus on 1198800(1199020) Con-

dition (1) assures that 1199021198900is positive Therefore we have two

casesCase 1 (119880

0(0) gt 0 and 119880

0(1) le 0) That is 119862(120579 + 120583

1)1205831(120579 +

1205830) lt 119877 le 119862(120582+120579+120583

1)1205831(120579+1205830)) In this case if all customers

who find the system empty enter the system with probability119902119890

0= 1 then the tagged customer suffers a negative expected

benefit if heshe decides to enter Hence 1199021198900= 1 does not lead

to an equilibrium Similarly if all customers use 1199021198900= 0 then

the tagged customer receives a positive benefit from enteringthus 119902119890

0= 0 also cannot be part of an equilibrium mixed

strategy Therefore there exists unique 1199021198900satisfying

119877 minus 119862120582119902119890

0+ 120579 + 120583

1

1205831(120579 + 120583

0)

= 0 (53)

for which customers are indifferent between entering andbalking This is given by

119902119890

0=

1

120582(1198771205831(120579 + 120583

0)

119862minus 120579 minus 120583

1) = 119909

1 (54)

In this situation the expected benefit 1198801is given by

1198801(119902119890

0 1199021) =

119862 (1205831minus 1205830)

1205831(120579 + 120583

0)minus

119862

120583 minus 1205821199021

(55)

Using the standard methodology of equilibrium analysisin unobservable queueing models we derive from (55) thefollowing equilibria

(1) If 1198801(119902119890

0 0) lt 0 then the equilibrium strategy is 119902119890

1=

0(2) If 119880

1(119902119890

0 0) ge 0 and 119880

1(119902119890

0 1) le 0 there exists unique

119902119890

1 satisfying

1198801(119902119890

0 119902119890

1) = 0 (56)

then the equilibrium strategy is 1199021198901= 1199092

(3) If 1198801(119902119890

0 1) gt 0 then the equilibrium strategy is 119902119890

1=

1

Case 2 (1198800(1) ge 0) That is 119862(120582 + 120579 + 120583

1)1205831(120579 + 120583

0) lt 119877) In

this case for every strategy of the other customers the taggedcustomer has a positive expected net benefit if heshe decidesto enter Hence 119902119890

0= 1

In this situation the expected benefit 1198801is given by

1198801(1 1199021) = 119877 minus

119862

120583 minus 1205821199021

minus119862 (120582 + 120579 + 120583

0)

1205831(120579 + 120583

0)

(57)

To find 119902119890

1in equilibrium we also consider the following

subcases

(1) If 1198801(1 0) lt 0 then the equilibrium strategy is 119902119890

1=

0

(2) If 1198801(1 0) ge 0 and 119880

1(1 1) le 0 there exists unique

119902119890

1 satisfying

1198801(1 119902119890

1) = 0 (58)

then the equilibrium strategy is 1199021198901= 1199093

(3) If 1198801(1 1) gt 0 then the equilibrium strategy is 119902119890

1=

1

By rearranging Cases 1 and 2 we can obtain the results ofTheorem 5 This completes the proof

From Theorem 5 we can compare 119902119890

0with 119902

119890

1 Figure 5

shows that 1199021198900is not always less than 119902

119890

1 Namely sometimes

the information that the server takes a working vacation doesnot make the customers less willing to enter the systembecause in the vacation the server provides a lower serviceto customers But when the vacation time and waiting timebecome longer customers do not want to enter the system invacation

Then fromTheorem 4 themean queue length is given by

119864 [119871] =

infin

sum

119899=0

1198991205871198990+

infin

sum

119899=0

1198991205871198991

= 1198701205900

(1 minus 1205900)2(1 +

1205821199020+ 120579

1205831

1 minus 12058811205900

(1 minus 1205881)2)

(59)

8 Mathematical Problems in Engineering

03

04

05

06

07

08

09

1

Equi

libriu

m jo

inin

g pr

obab

ility

qe0qe1 (120579 = 01 C = 2)

qe0qe1 (120579 = 05 C = 3)

0 04 06 08 1 12 14 16 1802Arrival rate 120582

Figure 5 Equilibrium mixed strategies for the partially observablecase when 119877 = 5 120583

1= 2 and 120583

0= 05

0 04 06 08 1 12 14 16 1802Arrival rate 120582

0

01

02

03

04

05

06

07

08

09

1

Join

ing

prob

abili

ty

qe0qe1

qlowast0qlowast1

Figure 6 Equilibrium and socially optimal mixed strategies for thepartially observable case when 119877 = 5 119862 = 2 120583

1= 2 120583

0= 05 and

120579 = 01

And the social benefit when all customers follow a mixedpolicy (119902

0 1199021) can now be easily computed as follows

119880119904(1199020 1199021) = 120582119877 minus 119862119864 [119871]

= 120582 (11990101199020+ 11990111199021) 119877

minus1198621198701205900

(1 minus 1205900)2(1 +

1205821199020+ 120579

1205831

1 minus 12058811205900

(1 minus 1205881)2)

(60)

The goal of a social planner is to maximize overall socialwelfare Let (119902lowast

0 119902lowast

1) be the socially optimal mixed strategy

Figure 6 compares customers equilibrium mixed strategy(119902119890

0 119902119890

1) and their socially optimal mixed strategy (119902lowast

0 119902lowast

1) We

also observe that 1199021198900gt 119902lowast

0and 1199021198901gt 119902lowast

1This ordering is typical

when customers make individual decisions maximizing theirown profitThen they ignore those negative externalities thatthey impose on later arrivals and they tend to overuse thesystem It is clear that these externalities should be taken intoaccount when we aim to maximize the total revenue

5 The Unobservable Queues

In this section we consider the fully unobservable case wherearriving customers cannot observe the system state

There are two pure strategies available for a customer thatis to join the queue or not to join the queue A pure or mixedstrategy can be described by a fraction 119902 (0 le 119902 le 1) whichis the probability of joining and the effective arrival rate orjoining rate is 120582119902 The transition diagram is illustrated inFigure 7

To identify the equilibrium strategies of the customerswe should first investigate the stationary distribution of thesystem when all customers follow a given strategy 119902 whichcan be obtained by Theorem 4 by taking 119902

0= 1199021= 119902 So we

have

1205871198990

= 1198701120590119899

119902 119899 ge 0

1205871198991

= 1198701120588119902

120582119902 + 120579

120582119902 + 120579 + 1205830

119899minus1

sum

119895=0

120588119895

119902120590119899minus1minus119895

119902 119899 ge 1

(61)

where 120588119902= 120582119902120583

1 120590119902= 120582119902(120582119902 + 120579 + 120583

0) and

1198701=

120579 + 1205830

120582119902 + 120579 + 1205830

(1 +120582119902 + 120579

120582119902 + 120579 + 1205830

120582119902

1205831minus 120582119902

)

minus1

(62)

Then the mean number of the customers in the system is

119864 [119871] =

infin

sum

119899=0

1198991205871198990+

infin

sum

119899=0

1198991205871198991

=1198701120590119902

(1 minus 120590119902)2(1 +

120582119902 + 120579

1205831

1 minus 120588119902120590119902

(1 minus 120588119902)2)

(63)

Hence the mean sojourn time of a customer who decides toenter upon hisher arrival can be obtained by using Littlersquoslaw

119864 [119882] =1198701120590119902

120582119902 (1 minus 120590119902)2(1 +

120582119902 + 120579

1205831

1 minus 120588119902120590119902

(1 minus 120588119902)2)

=1

120579 + 1205830

(1

+1205831(120579 + 120583

0)

1205831minus 120582119902

1

1205830((1205831minus 120582119902) (120582119902 + 120579)) + 120583

1

)

(64)

And if 120582 lt 1205831 119864[119882] is strictly increasing for 119902 isin [0 1]

Mathematical Problems in Engineering 9

1 1 2 0 3 1

0 0 1 0 2 0 3 0

n 1

n 0

120579

1205831 1205831

12058301205830

120582q120582q

120582q 120582qmiddot middot middot

middot middot middot

middot middot middot

middot middot middot

120579

1205831 1205831

12058301205830

120582q120582q

120582q 120582q

120579120579

1205831

1205831

1205830

1205830

120582q

120582q 120582q

Figure 7 Transition rate diagram for the unobservable queues

We consider a tagged customer If heshe decides to enterthe system hisher expected net benefit is

119880 (119902) = 119877 minus 119862119864 [119882] = 119877 minus119862

120579 + 1205830

(1

+1205831(120579 + 120583

0)

1205831minus 120582119902

1

1205830((1205831minus 120582119902) (120582119902 + 120579)) + 120583

1

)

(65)

We note that 119880(119902) is strictly decreasing in 119902 and it has aunique root 119902lowast

119890 So there exists a unique mixed equilibrium

strategy 119902119890and 119902119890= min119902lowast

119890 1

We now turn our attention to social optimization Thesocial benefit per time unit can now be easily computed as

119880119904(119902) = 120582119902 (119877 minus 119862119864 [119882]) = 120582119902(119877 minus

119862

120579 + 1205830

(1

+1205831(120579 + 120583

0)

1205831minus 120582119902

1

1205830((1205831minus 120582119902) (120582119902 + 120579)) + 120583

1

))

(66)

As for the equilibrium social welfare per time unit119880119904(119902119890)

Figure 8 shows that 119880119904(119902119890) first increases and then decreases

with 120582 and the social benefit achieves a maximum forintermediate values of this parameter The reason for thisbehavior is that when the arrival rate is small the system israrely crowded therefore as more customers arrive they areserved and the social benefit improves However with anysmall increase of the arrival rate 120582 the expected waiting timeincreases which has a detrimental effect on the social benefit

Let 119909lowast be the root of the equation 1198781015840(119902) = 0 and let 119902lowast be

the optimal joining probability Then we have the followingconclusions if 0 lt 119909

lowastlt 1 and 119878

10158401015840(119902) le 0 then 119902

lowast= 119909lowast if

0 lt 119909lowastlt 1 and 119878

10158401015840(119902) gt 0 or 119909lowast ge 1 then 119902

lowast= 1

From Figure 9 we observe that 119902119890

gt 119902lowast As a result

individual optimization leads to queues that are longer thansocially desired Therefore it is clear that the social plannerwants a toll to discourage arrivals

6 Conclusion

In this paper we analyzed the customer strategic behaviorin the MM1 queueing system with working vacations andvacation interruptions where arriving customers have optionto decide whether to join the system or balk Three differentcases with respect to the levels of information provided toarriving customers have been investigated extensively and theequilibrium strategies for each case were derived And wefound that the equilibrium joining probability of the busy

02 04 06 08 1 12 14 16 18 20Arrival rate 120582

Equi

libriu

m so

cial

wel

fare

Us(qe)

0

5

10

15

Figure 8 Equilibrium social welfare for the unobservable casewhen119877 = 10 119862 = 1 120583

1= 2 120583

0= 1 and 120579 = 1

03

04

05

06

07

08

09

1

Join

ing

prob

abili

ty

0 1 15 205Arrival rate 120582

qeqlowast

Figure 9 Equilibrium and socially optimal strategies for the unob-servable case when 119877 = 3 119862 = 2 120583

1= 2 120583

0= 05 and 120579 = 001

period may be smaller than that in vacation time We alsocompared the equilibrium strategy with the socially optimalstrategy numerically and we observed that customers makeindividual decisions maximizing their own profit and tendto overuse the system For the social planner a toll will beadopted to discourage customers from joining

10 Mathematical Problems in Engineering

Furthermore an interesting extension would be to incor-porate the present model in a profit maximizing frameworkwhere the owner or manager of the system imposes anentrance feeThe problem is to find the optimal fee that max-imizes the owners total profit subject to the constraint thatcustomers independently decide whether to enter or not andtake into account the fee in addition to the service reward andwaiting cost Another direction for future work is to concernthe non-Markovian queues with various vacation policies

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Acknowledgments

This work is partially supported by the Natural Foundation ofHebei Province (nos A2014203096 andC2014203212) Chinaand the Young Faculty Autonomous Research Project of Yan-shan University (nos 13LGB032 13LGA017 and 13LGB013)China

References

[1] P Naor ldquoThe regulation of queue size by levying tollsrdquo Econo-metrica vol 37 no 1 pp 15ndash24 1969

[2] N M Edelson and D K Hildebrand ldquoCongestion tolls forPoisson queuing processesrdquo Econometrica vol 43 no 1 pp 81ndash92 1975

[3] R Hassin and M Haviv Equilibrium Behavior in Queueing Sys-tems To Queue or Not to Queue Kluwer Academic PublishersDordrecht The Netherlands 2003

[4] A Burnetas and A Economou ldquoEquilibrium customer strate-gies in a single server Markovian queue with setup timesrdquoQueueing Systems vol 56 no 3-4 pp 213ndash228 2007

[5] W Sun and N Tian ldquoContrast of the equilibrium and sociallyoptimal strategies in a queue with vacationsrdquo Journal of Compu-tational Information Systems vol 4 no 5 pp 2167ndash2172 2008

[6] A Economou and S Kanta ldquoEquilibrium balking strategiesin the observable single-server queue with breakdowns andrepairsrdquoOperations Research Letters vol 36 no 6 pp 696ndash6992008

[7] P Guo and R Hassin ldquoStrategic behavior and social optimiza-tion in Markovian vacation queuesrdquo Operations Research vol59 no 4 pp 986ndash997 2011

[8] P Guo and R Hassin ldquoStrategic behavior and social optimiza-tion in Markovian vacation queues the case of heterogeneouscustomersrdquo European Journal of Operational Research vol 222no 2 pp 278ndash286 2012

[9] R Tian D Yue and W Yue ldquoOptimal balking strategies inan MG1 queueing system with a removable server under N-policyrdquo Journal of Industrial andManagementOptimization vol11 no 3 pp 715ndash731 2015

[10] P Guo and Q Li ldquoStrategic behavior and social optimizationin partially-observableMarkovian vacation queuesrdquoOperationsResearch Letters vol 41 no 3 pp 277ndash284 2013

[11] L Li J Wang and F Zhang ldquoEquilibrium customer strategiesin Markovian queues with partial breakdownsrdquo Computers ampIndustrial Engineering vol 66 no 4 pp 751ndash757 2013

[12] F Zhang J Wang and B Liu ldquoOn the optimal and equilibriumretrial rates in an unreliable retrial queue with vacationsrdquo Jour-nal of Industrial andManagement Optimization vol 8 no 4 pp861ndash875 2012

[13] N Tian J Li and Z G Zhang ldquoMatrix analysis methodand working vacation queuesa surveyrdquo International Journal ofInformation and Management Sciences vol 20 pp 603ndash6332009

[14] F Zhang J Wang and B Liu ldquoEquilibrium balking strategiesin Markovian queues with working vacationsrdquo Applied Mathe-matical Modelling vol 37 no 16-17 pp 8264ndash8282 2013

[15] W Sun and S Li ldquoEquilibrium and optimal behavior of cus-tomers in Markovian queues with multiple working vacationsrdquoTOP vol 22 no 2 pp 694ndash715 2014

[16] W Sun S Li and Q-L Li ldquoEquilibrium balking strategiesof customers in Markovian queues with two-stage workingvacationsrdquo Applied Mathematics and Computation vol 248 pp195ndash214 2014

[17] J Li andNTian ldquoTheMM1 queuewithworking vacations andvacation interruptionsrdquo Journal of Systems Science and SystemsEngineering vol 16 no 1 pp 121ndash127 2007

[18] J-H Li N-S Tian and Z-Y Ma ldquoPerformance analysis ofGIM1 queue with working vacations and vacation interrup-tionrdquo Applied Mathematical Modelling vol 32 no 12 pp 2715ndash2730 2008

[19] M Neuts Matrix-Geometric Solution in Stochastic ModelsJohns Hopkins University Press Baltimore Md USA 1981

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

8 Mathematical Problems in Engineering

03

04

05

06

07

08

09

1

Equi

libriu

m jo

inin

g pr

obab

ility

qe0qe1 (120579 = 01 C = 2)

qe0qe1 (120579 = 05 C = 3)

0 04 06 08 1 12 14 16 1802Arrival rate 120582

Figure 5 Equilibrium mixed strategies for the partially observablecase when 119877 = 5 120583

1= 2 and 120583

0= 05

0 04 06 08 1 12 14 16 1802Arrival rate 120582

0

01

02

03

04

05

06

07

08

09

1

Join

ing

prob

abili

ty

qe0qe1

qlowast0qlowast1

Figure 6 Equilibrium and socially optimal mixed strategies for thepartially observable case when 119877 = 5 119862 = 2 120583

1= 2 120583

0= 05 and

120579 = 01

And the social benefit when all customers follow a mixedpolicy (119902

0 1199021) can now be easily computed as follows

119880119904(1199020 1199021) = 120582119877 minus 119862119864 [119871]

= 120582 (11990101199020+ 11990111199021) 119877

minus1198621198701205900

(1 minus 1205900)2(1 +

1205821199020+ 120579

1205831

1 minus 12058811205900

(1 minus 1205881)2)

(60)

The goal of a social planner is to maximize overall socialwelfare Let (119902lowast

0 119902lowast

1) be the socially optimal mixed strategy

Figure 6 compares customers equilibrium mixed strategy(119902119890

0 119902119890

1) and their socially optimal mixed strategy (119902lowast

0 119902lowast

1) We

also observe that 1199021198900gt 119902lowast

0and 1199021198901gt 119902lowast

1This ordering is typical

when customers make individual decisions maximizing theirown profitThen they ignore those negative externalities thatthey impose on later arrivals and they tend to overuse thesystem It is clear that these externalities should be taken intoaccount when we aim to maximize the total revenue

5 The Unobservable Queues

In this section we consider the fully unobservable case wherearriving customers cannot observe the system state

There are two pure strategies available for a customer thatis to join the queue or not to join the queue A pure or mixedstrategy can be described by a fraction 119902 (0 le 119902 le 1) whichis the probability of joining and the effective arrival rate orjoining rate is 120582119902 The transition diagram is illustrated inFigure 7

To identify the equilibrium strategies of the customerswe should first investigate the stationary distribution of thesystem when all customers follow a given strategy 119902 whichcan be obtained by Theorem 4 by taking 119902

0= 1199021= 119902 So we

have

1205871198990

= 1198701120590119899

119902 119899 ge 0

1205871198991

= 1198701120588119902

120582119902 + 120579

120582119902 + 120579 + 1205830

119899minus1

sum

119895=0

120588119895

119902120590119899minus1minus119895

119902 119899 ge 1

(61)

where 120588119902= 120582119902120583

1 120590119902= 120582119902(120582119902 + 120579 + 120583

0) and

1198701=

120579 + 1205830

120582119902 + 120579 + 1205830

(1 +120582119902 + 120579

120582119902 + 120579 + 1205830

120582119902

1205831minus 120582119902

)

minus1

(62)

Then the mean number of the customers in the system is

119864 [119871] =

infin

sum

119899=0

1198991205871198990+

infin

sum

119899=0

1198991205871198991

=1198701120590119902

(1 minus 120590119902)2(1 +

120582119902 + 120579

1205831

1 minus 120588119902120590119902

(1 minus 120588119902)2)

(63)

Hence the mean sojourn time of a customer who decides toenter upon hisher arrival can be obtained by using Littlersquoslaw

119864 [119882] =1198701120590119902

120582119902 (1 minus 120590119902)2(1 +

120582119902 + 120579

1205831

1 minus 120588119902120590119902

(1 minus 120588119902)2)

=1

120579 + 1205830

(1

+1205831(120579 + 120583

0)

1205831minus 120582119902

1

1205830((1205831minus 120582119902) (120582119902 + 120579)) + 120583

1

)

(64)

And if 120582 lt 1205831 119864[119882] is strictly increasing for 119902 isin [0 1]

Mathematical Problems in Engineering 9

1 1 2 0 3 1

0 0 1 0 2 0 3 0

n 1

n 0

120579

1205831 1205831

12058301205830

120582q120582q

120582q 120582qmiddot middot middot

middot middot middot

middot middot middot

middot middot middot

120579

1205831 1205831

12058301205830

120582q120582q

120582q 120582q

120579120579

1205831

1205831

1205830

1205830

120582q

120582q 120582q

Figure 7 Transition rate diagram for the unobservable queues

We consider a tagged customer If heshe decides to enterthe system hisher expected net benefit is

119880 (119902) = 119877 minus 119862119864 [119882] = 119877 minus119862

120579 + 1205830

(1

+1205831(120579 + 120583

0)

1205831minus 120582119902

1

1205830((1205831minus 120582119902) (120582119902 + 120579)) + 120583

1

)

(65)

We note that 119880(119902) is strictly decreasing in 119902 and it has aunique root 119902lowast

119890 So there exists a unique mixed equilibrium

strategy 119902119890and 119902119890= min119902lowast

119890 1

We now turn our attention to social optimization Thesocial benefit per time unit can now be easily computed as

119880119904(119902) = 120582119902 (119877 minus 119862119864 [119882]) = 120582119902(119877 minus

119862

120579 + 1205830

(1

+1205831(120579 + 120583

0)

1205831minus 120582119902

1

1205830((1205831minus 120582119902) (120582119902 + 120579)) + 120583

1

))

(66)

As for the equilibrium social welfare per time unit119880119904(119902119890)

Figure 8 shows that 119880119904(119902119890) first increases and then decreases

with 120582 and the social benefit achieves a maximum forintermediate values of this parameter The reason for thisbehavior is that when the arrival rate is small the system israrely crowded therefore as more customers arrive they areserved and the social benefit improves However with anysmall increase of the arrival rate 120582 the expected waiting timeincreases which has a detrimental effect on the social benefit

Let 119909lowast be the root of the equation 1198781015840(119902) = 0 and let 119902lowast be

the optimal joining probability Then we have the followingconclusions if 0 lt 119909

lowastlt 1 and 119878

10158401015840(119902) le 0 then 119902

lowast= 119909lowast if

0 lt 119909lowastlt 1 and 119878

10158401015840(119902) gt 0 or 119909lowast ge 1 then 119902

lowast= 1

From Figure 9 we observe that 119902119890

gt 119902lowast As a result

individual optimization leads to queues that are longer thansocially desired Therefore it is clear that the social plannerwants a toll to discourage arrivals

6 Conclusion

In this paper we analyzed the customer strategic behaviorin the MM1 queueing system with working vacations andvacation interruptions where arriving customers have optionto decide whether to join the system or balk Three differentcases with respect to the levels of information provided toarriving customers have been investigated extensively and theequilibrium strategies for each case were derived And wefound that the equilibrium joining probability of the busy

02 04 06 08 1 12 14 16 18 20Arrival rate 120582

Equi

libriu

m so

cial

wel

fare

Us(qe)

0

5

10

15

Figure 8 Equilibrium social welfare for the unobservable casewhen119877 = 10 119862 = 1 120583

1= 2 120583

0= 1 and 120579 = 1

03

04

05

06

07

08

09

1

Join

ing

prob

abili

ty

0 1 15 205Arrival rate 120582

qeqlowast

Figure 9 Equilibrium and socially optimal strategies for the unob-servable case when 119877 = 3 119862 = 2 120583

1= 2 120583

0= 05 and 120579 = 001

period may be smaller than that in vacation time We alsocompared the equilibrium strategy with the socially optimalstrategy numerically and we observed that customers makeindividual decisions maximizing their own profit and tendto overuse the system For the social planner a toll will beadopted to discourage customers from joining

10 Mathematical Problems in Engineering

Furthermore an interesting extension would be to incor-porate the present model in a profit maximizing frameworkwhere the owner or manager of the system imposes anentrance feeThe problem is to find the optimal fee that max-imizes the owners total profit subject to the constraint thatcustomers independently decide whether to enter or not andtake into account the fee in addition to the service reward andwaiting cost Another direction for future work is to concernthe non-Markovian queues with various vacation policies

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Acknowledgments

This work is partially supported by the Natural Foundation ofHebei Province (nos A2014203096 andC2014203212) Chinaand the Young Faculty Autonomous Research Project of Yan-shan University (nos 13LGB032 13LGA017 and 13LGB013)China

References

[1] P Naor ldquoThe regulation of queue size by levying tollsrdquo Econo-metrica vol 37 no 1 pp 15ndash24 1969

[2] N M Edelson and D K Hildebrand ldquoCongestion tolls forPoisson queuing processesrdquo Econometrica vol 43 no 1 pp 81ndash92 1975

[3] R Hassin and M Haviv Equilibrium Behavior in Queueing Sys-tems To Queue or Not to Queue Kluwer Academic PublishersDordrecht The Netherlands 2003

[4] A Burnetas and A Economou ldquoEquilibrium customer strate-gies in a single server Markovian queue with setup timesrdquoQueueing Systems vol 56 no 3-4 pp 213ndash228 2007

[5] W Sun and N Tian ldquoContrast of the equilibrium and sociallyoptimal strategies in a queue with vacationsrdquo Journal of Compu-tational Information Systems vol 4 no 5 pp 2167ndash2172 2008

[6] A Economou and S Kanta ldquoEquilibrium balking strategiesin the observable single-server queue with breakdowns andrepairsrdquoOperations Research Letters vol 36 no 6 pp 696ndash6992008

[7] P Guo and R Hassin ldquoStrategic behavior and social optimiza-tion in Markovian vacation queuesrdquo Operations Research vol59 no 4 pp 986ndash997 2011

[8] P Guo and R Hassin ldquoStrategic behavior and social optimiza-tion in Markovian vacation queues the case of heterogeneouscustomersrdquo European Journal of Operational Research vol 222no 2 pp 278ndash286 2012

[9] R Tian D Yue and W Yue ldquoOptimal balking strategies inan MG1 queueing system with a removable server under N-policyrdquo Journal of Industrial andManagementOptimization vol11 no 3 pp 715ndash731 2015

[10] P Guo and Q Li ldquoStrategic behavior and social optimizationin partially-observableMarkovian vacation queuesrdquoOperationsResearch Letters vol 41 no 3 pp 277ndash284 2013

[11] L Li J Wang and F Zhang ldquoEquilibrium customer strategiesin Markovian queues with partial breakdownsrdquo Computers ampIndustrial Engineering vol 66 no 4 pp 751ndash757 2013

[12] F Zhang J Wang and B Liu ldquoOn the optimal and equilibriumretrial rates in an unreliable retrial queue with vacationsrdquo Jour-nal of Industrial andManagement Optimization vol 8 no 4 pp861ndash875 2012

[13] N Tian J Li and Z G Zhang ldquoMatrix analysis methodand working vacation queuesa surveyrdquo International Journal ofInformation and Management Sciences vol 20 pp 603ndash6332009

[14] F Zhang J Wang and B Liu ldquoEquilibrium balking strategiesin Markovian queues with working vacationsrdquo Applied Mathe-matical Modelling vol 37 no 16-17 pp 8264ndash8282 2013

[15] W Sun and S Li ldquoEquilibrium and optimal behavior of cus-tomers in Markovian queues with multiple working vacationsrdquoTOP vol 22 no 2 pp 694ndash715 2014

[16] W Sun S Li and Q-L Li ldquoEquilibrium balking strategiesof customers in Markovian queues with two-stage workingvacationsrdquo Applied Mathematics and Computation vol 248 pp195ndash214 2014

[17] J Li andNTian ldquoTheMM1 queuewithworking vacations andvacation interruptionsrdquo Journal of Systems Science and SystemsEngineering vol 16 no 1 pp 121ndash127 2007

[18] J-H Li N-S Tian and Z-Y Ma ldquoPerformance analysis ofGIM1 queue with working vacations and vacation interrup-tionrdquo Applied Mathematical Modelling vol 32 no 12 pp 2715ndash2730 2008

[19] M Neuts Matrix-Geometric Solution in Stochastic ModelsJohns Hopkins University Press Baltimore Md USA 1981

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Mathematical Problems in Engineering 9

1 1 2 0 3 1

0 0 1 0 2 0 3 0

n 1

n 0

120579

1205831 1205831

12058301205830

120582q120582q

120582q 120582qmiddot middot middot

middot middot middot

middot middot middot

middot middot middot

120579

1205831 1205831

12058301205830

120582q120582q

120582q 120582q

120579120579

1205831

1205831

1205830

1205830

120582q

120582q 120582q

Figure 7 Transition rate diagram for the unobservable queues

We consider a tagged customer If heshe decides to enterthe system hisher expected net benefit is

119880 (119902) = 119877 minus 119862119864 [119882] = 119877 minus119862

120579 + 1205830

(1

+1205831(120579 + 120583

0)

1205831minus 120582119902

1

1205830((1205831minus 120582119902) (120582119902 + 120579)) + 120583

1

)

(65)

We note that 119880(119902) is strictly decreasing in 119902 and it has aunique root 119902lowast

119890 So there exists a unique mixed equilibrium

strategy 119902119890and 119902119890= min119902lowast

119890 1

We now turn our attention to social optimization Thesocial benefit per time unit can now be easily computed as

119880119904(119902) = 120582119902 (119877 minus 119862119864 [119882]) = 120582119902(119877 minus

119862

120579 + 1205830

(1

+1205831(120579 + 120583

0)

1205831minus 120582119902

1

1205830((1205831minus 120582119902) (120582119902 + 120579)) + 120583

1

))

(66)

As for the equilibrium social welfare per time unit119880119904(119902119890)

Figure 8 shows that 119880119904(119902119890) first increases and then decreases

with 120582 and the social benefit achieves a maximum forintermediate values of this parameter The reason for thisbehavior is that when the arrival rate is small the system israrely crowded therefore as more customers arrive they areserved and the social benefit improves However with anysmall increase of the arrival rate 120582 the expected waiting timeincreases which has a detrimental effect on the social benefit

Let 119909lowast be the root of the equation 1198781015840(119902) = 0 and let 119902lowast be

the optimal joining probability Then we have the followingconclusions if 0 lt 119909

lowastlt 1 and 119878

10158401015840(119902) le 0 then 119902

lowast= 119909lowast if

0 lt 119909lowastlt 1 and 119878

10158401015840(119902) gt 0 or 119909lowast ge 1 then 119902

lowast= 1

From Figure 9 we observe that 119902119890

gt 119902lowast As a result

individual optimization leads to queues that are longer thansocially desired Therefore it is clear that the social plannerwants a toll to discourage arrivals

6 Conclusion

In this paper we analyzed the customer strategic behaviorin the MM1 queueing system with working vacations andvacation interruptions where arriving customers have optionto decide whether to join the system or balk Three differentcases with respect to the levels of information provided toarriving customers have been investigated extensively and theequilibrium strategies for each case were derived And wefound that the equilibrium joining probability of the busy

02 04 06 08 1 12 14 16 18 20Arrival rate 120582

Equi

libriu

m so

cial

wel

fare

Us(qe)

0

5

10

15

Figure 8 Equilibrium social welfare for the unobservable casewhen119877 = 10 119862 = 1 120583

1= 2 120583

0= 1 and 120579 = 1

03

04

05

06

07

08

09

1

Join

ing

prob

abili

ty

0 1 15 205Arrival rate 120582

qeqlowast

Figure 9 Equilibrium and socially optimal strategies for the unob-servable case when 119877 = 3 119862 = 2 120583

1= 2 120583

0= 05 and 120579 = 001

period may be smaller than that in vacation time We alsocompared the equilibrium strategy with the socially optimalstrategy numerically and we observed that customers makeindividual decisions maximizing their own profit and tendto overuse the system For the social planner a toll will beadopted to discourage customers from joining

10 Mathematical Problems in Engineering

Furthermore an interesting extension would be to incor-porate the present model in a profit maximizing frameworkwhere the owner or manager of the system imposes anentrance feeThe problem is to find the optimal fee that max-imizes the owners total profit subject to the constraint thatcustomers independently decide whether to enter or not andtake into account the fee in addition to the service reward andwaiting cost Another direction for future work is to concernthe non-Markovian queues with various vacation policies

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Acknowledgments

This work is partially supported by the Natural Foundation ofHebei Province (nos A2014203096 andC2014203212) Chinaand the Young Faculty Autonomous Research Project of Yan-shan University (nos 13LGB032 13LGA017 and 13LGB013)China

References

[1] P Naor ldquoThe regulation of queue size by levying tollsrdquo Econo-metrica vol 37 no 1 pp 15ndash24 1969

[2] N M Edelson and D K Hildebrand ldquoCongestion tolls forPoisson queuing processesrdquo Econometrica vol 43 no 1 pp 81ndash92 1975

[3] R Hassin and M Haviv Equilibrium Behavior in Queueing Sys-tems To Queue or Not to Queue Kluwer Academic PublishersDordrecht The Netherlands 2003

[4] A Burnetas and A Economou ldquoEquilibrium customer strate-gies in a single server Markovian queue with setup timesrdquoQueueing Systems vol 56 no 3-4 pp 213ndash228 2007

[5] W Sun and N Tian ldquoContrast of the equilibrium and sociallyoptimal strategies in a queue with vacationsrdquo Journal of Compu-tational Information Systems vol 4 no 5 pp 2167ndash2172 2008

[6] A Economou and S Kanta ldquoEquilibrium balking strategiesin the observable single-server queue with breakdowns andrepairsrdquoOperations Research Letters vol 36 no 6 pp 696ndash6992008

[7] P Guo and R Hassin ldquoStrategic behavior and social optimiza-tion in Markovian vacation queuesrdquo Operations Research vol59 no 4 pp 986ndash997 2011

[8] P Guo and R Hassin ldquoStrategic behavior and social optimiza-tion in Markovian vacation queues the case of heterogeneouscustomersrdquo European Journal of Operational Research vol 222no 2 pp 278ndash286 2012

[9] R Tian D Yue and W Yue ldquoOptimal balking strategies inan MG1 queueing system with a removable server under N-policyrdquo Journal of Industrial andManagementOptimization vol11 no 3 pp 715ndash731 2015

[10] P Guo and Q Li ldquoStrategic behavior and social optimizationin partially-observableMarkovian vacation queuesrdquoOperationsResearch Letters vol 41 no 3 pp 277ndash284 2013

[11] L Li J Wang and F Zhang ldquoEquilibrium customer strategiesin Markovian queues with partial breakdownsrdquo Computers ampIndustrial Engineering vol 66 no 4 pp 751ndash757 2013

[12] F Zhang J Wang and B Liu ldquoOn the optimal and equilibriumretrial rates in an unreliable retrial queue with vacationsrdquo Jour-nal of Industrial andManagement Optimization vol 8 no 4 pp861ndash875 2012

[13] N Tian J Li and Z G Zhang ldquoMatrix analysis methodand working vacation queuesa surveyrdquo International Journal ofInformation and Management Sciences vol 20 pp 603ndash6332009

[14] F Zhang J Wang and B Liu ldquoEquilibrium balking strategiesin Markovian queues with working vacationsrdquo Applied Mathe-matical Modelling vol 37 no 16-17 pp 8264ndash8282 2013

[15] W Sun and S Li ldquoEquilibrium and optimal behavior of cus-tomers in Markovian queues with multiple working vacationsrdquoTOP vol 22 no 2 pp 694ndash715 2014

[16] W Sun S Li and Q-L Li ldquoEquilibrium balking strategiesof customers in Markovian queues with two-stage workingvacationsrdquo Applied Mathematics and Computation vol 248 pp195ndash214 2014

[17] J Li andNTian ldquoTheMM1 queuewithworking vacations andvacation interruptionsrdquo Journal of Systems Science and SystemsEngineering vol 16 no 1 pp 121ndash127 2007

[18] J-H Li N-S Tian and Z-Y Ma ldquoPerformance analysis ofGIM1 queue with working vacations and vacation interrup-tionrdquo Applied Mathematical Modelling vol 32 no 12 pp 2715ndash2730 2008

[19] M Neuts Matrix-Geometric Solution in Stochastic ModelsJohns Hopkins University Press Baltimore Md USA 1981

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

10 Mathematical Problems in Engineering

Furthermore an interesting extension would be to incor-porate the present model in a profit maximizing frameworkwhere the owner or manager of the system imposes anentrance feeThe problem is to find the optimal fee that max-imizes the owners total profit subject to the constraint thatcustomers independently decide whether to enter or not andtake into account the fee in addition to the service reward andwaiting cost Another direction for future work is to concernthe non-Markovian queues with various vacation policies

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Acknowledgments

This work is partially supported by the Natural Foundation ofHebei Province (nos A2014203096 andC2014203212) Chinaand the Young Faculty Autonomous Research Project of Yan-shan University (nos 13LGB032 13LGA017 and 13LGB013)China

References

[1] P Naor ldquoThe regulation of queue size by levying tollsrdquo Econo-metrica vol 37 no 1 pp 15ndash24 1969

[2] N M Edelson and D K Hildebrand ldquoCongestion tolls forPoisson queuing processesrdquo Econometrica vol 43 no 1 pp 81ndash92 1975

[3] R Hassin and M Haviv Equilibrium Behavior in Queueing Sys-tems To Queue or Not to Queue Kluwer Academic PublishersDordrecht The Netherlands 2003

[4] A Burnetas and A Economou ldquoEquilibrium customer strate-gies in a single server Markovian queue with setup timesrdquoQueueing Systems vol 56 no 3-4 pp 213ndash228 2007

[5] W Sun and N Tian ldquoContrast of the equilibrium and sociallyoptimal strategies in a queue with vacationsrdquo Journal of Compu-tational Information Systems vol 4 no 5 pp 2167ndash2172 2008

[6] A Economou and S Kanta ldquoEquilibrium balking strategiesin the observable single-server queue with breakdowns andrepairsrdquoOperations Research Letters vol 36 no 6 pp 696ndash6992008

[7] P Guo and R Hassin ldquoStrategic behavior and social optimiza-tion in Markovian vacation queuesrdquo Operations Research vol59 no 4 pp 986ndash997 2011

[8] P Guo and R Hassin ldquoStrategic behavior and social optimiza-tion in Markovian vacation queues the case of heterogeneouscustomersrdquo European Journal of Operational Research vol 222no 2 pp 278ndash286 2012

[9] R Tian D Yue and W Yue ldquoOptimal balking strategies inan MG1 queueing system with a removable server under N-policyrdquo Journal of Industrial andManagementOptimization vol11 no 3 pp 715ndash731 2015

[10] P Guo and Q Li ldquoStrategic behavior and social optimizationin partially-observableMarkovian vacation queuesrdquoOperationsResearch Letters vol 41 no 3 pp 277ndash284 2013

[11] L Li J Wang and F Zhang ldquoEquilibrium customer strategiesin Markovian queues with partial breakdownsrdquo Computers ampIndustrial Engineering vol 66 no 4 pp 751ndash757 2013

[12] F Zhang J Wang and B Liu ldquoOn the optimal and equilibriumretrial rates in an unreliable retrial queue with vacationsrdquo Jour-nal of Industrial andManagement Optimization vol 8 no 4 pp861ndash875 2012

[13] N Tian J Li and Z G Zhang ldquoMatrix analysis methodand working vacation queuesa surveyrdquo International Journal ofInformation and Management Sciences vol 20 pp 603ndash6332009

[14] F Zhang J Wang and B Liu ldquoEquilibrium balking strategiesin Markovian queues with working vacationsrdquo Applied Mathe-matical Modelling vol 37 no 16-17 pp 8264ndash8282 2013

[15] W Sun and S Li ldquoEquilibrium and optimal behavior of cus-tomers in Markovian queues with multiple working vacationsrdquoTOP vol 22 no 2 pp 694ndash715 2014

[16] W Sun S Li and Q-L Li ldquoEquilibrium balking strategiesof customers in Markovian queues with two-stage workingvacationsrdquo Applied Mathematics and Computation vol 248 pp195ndash214 2014

[17] J Li andNTian ldquoTheMM1 queuewithworking vacations andvacation interruptionsrdquo Journal of Systems Science and SystemsEngineering vol 16 no 1 pp 121ndash127 2007

[18] J-H Li N-S Tian and Z-Y Ma ldquoPerformance analysis ofGIM1 queue with working vacations and vacation interrup-tionrdquo Applied Mathematical Modelling vol 32 no 12 pp 2715ndash2730 2008

[19] M Neuts Matrix-Geometric Solution in Stochastic ModelsJohns Hopkins University Press Baltimore Md USA 1981

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of