IN AN M/M/m PREEMPTIVE-RESUME LCFS QUEUE WITH ...LST of the DF for the ﬁrst passage time from...

International Journal of Pure and Applied Mathematics

Volume 116 No. 2 2017, 501-545ISSN: 1311-8080 (printed version); ISSN: 1314-3395 (on-line version)url: http://www.ijpam.eudoi: 10.12732/ijpam.v116i2.21

PAijpam.eu

DISTRIBUTION OF THE WAITING AND SERVICE TIME

IN AN M/M/m PREEMPTIVE-RESUME LCFS

QUEUE WITH IMPATIENT CUSTOMERS

Hideaki Takagi

University of Tsukuba,Tsukuba Science City, Ibaraki 305-8573, JAPAN

Abstract: We study an M/M/m preemptive-resume last-come, first-served (PR-LCFS)

queue with impatient customers without priority classes. We focus on the time interval from

arrival to either service completion or abandonment, whichever occurs first, of an arbitrary cus-

tomer in the steady state. The problem is formulated as a combination of two one-dimensional

birth-and-death processes, each with two absorbing states. We provide explicit expressions in

terms of Laplace-Stieltjes transform of the distribution function for the first passage time to

service completion and abandonment, which is decomposed into the waiting and service time.

As two special cases, an M/M/m preemptive-loss LCFS system with impatient customers

and an M/M/m preemptive-resume LCFS queue with patient customers only are treated

separately. Some numerical example is presented for computation of theoretical formulas.

AMS Subject Classification: 60K25, 90B22

Key Words: multiserver queue, preemptive-resume, last-come first-served, impatient cus-

tomers, waiting time distribution, birth-and-death process, first passage time, absorbing state

1. Introduction

We consider an M/M/m queueing system with impatient customers withoutexogenous priority classes. Customers arrive according to a Poisson process atrate λ. The service time of each customer is exponentially distributed withmean 1/µ. There are m servers and a waiting room of infinite capacity. We

Received: July 30, 2017

Revised: September 11, 2017

Published: October 7, 2017

c© 2017 Academic Publications, Ltd.

url: www.acadpubl.eu

502 H. Takagi

define a parameter ρ := λ/(mµ), which represents the traffic intensity perserver. At any time, each customer present in the system is either being servedor staying in the waiting room. Each customer in the waiting room leavesthe system (abandons the waiting process) with probability θ∆t within a shorttime interval (t, t + ∆t). That is to say, the patience time for each customeris exponentially distributed with mean 1/θ. We employ another parameterτ := θ/µ, which is the ratio of the mean service time to the mean patiencetime. Customers never leave the system while being served before the serviceis completed.

When a tagged customer arrives to find that not all servers are busy, theservice of this customer is initiated immediately. There are several servicedisciplines with respect to the handling of a tagged customer who arrives whenall servers are busy. Some of them are described as follows:

• First-come, first-served (FCFS): The customer is placed at the tail of thequeue in the waiting room.

• Nonpreemptive last-come, first-served (NP-LCFS): The customer is placedat the head of the queue in the waiting room. This discipline was assumedin the study of priority queues by Durr [1]. In both the FCFS and NP-LCFS disciplines, when one of the servers becomes available, a customerat the head of the queue, if any, is called in for service. In these disciplines,the serving of customers once started is never preempted.

• Preemptive-resume last-come, first-served (PR-LCFS): The arriving cus-tomer preempts the ongoing service of the customer who arrived firstamong those currently being served. The customer whose service is pre-empted is placed at the head of the queue in the waiting room. When oneof the servers becomes available, a customer at the head of the queue, ifany, is called in for his service to be resumed. This discipline is equivalentto the one called “preemptive last-in, first-out” for an M/G/1 queue byWolff [19, p. 456].

• Preemptive-loss last-come, first-served (PL-LCFS): Similar to PR-LCFS,the arriving customer preempts the ongoing service of the customer whoarrived first among those currently being served. However, the customerwhose service is preempted is immediately lost from the system as studiedby Klimov [7, p. 66].

We are concerned with an M/M/m PR-LCFS queue with impatient cus-tomers, and interested in the waiting and service time for an arbitrary customer

DISTRIBUTION OF THE WAITING AND SERVICE TIME... 503

during the interval from his arrival to departure (by either service completion orabandonment of waiting, whichever occurs first) in steady state. An M/M/mPL-LCFS queue can be treated as a special case with θ = ∞ of a PR-LCFSqueue. In our previous work on an M/M/m PR-LCFS queue with impatientcustomers [14, 15], we formulated the problem as a one-dimensional birth-and-death process with two absorbing states and considered the first passage timesin this process. We provided explicit expressions for the probability of servicecompletion and abandonment. Moreover, we obtained sets of computationalformulas for calculating the mean and the second moment of the times untilservice completion and abandonment.

In the present paper, we turn our attention to the distribution of the time toservice completion and abandonment We formulate the problem as a combina-tion of two one-dimensional birth-and-death processes, each with two absorbingstates, and consider the first passage times in the combined process. We pro-vide explicit expressions in terms of Laplace-Stieltjes transform (LST) of thedistribution function (DF) for the waiting time and service time of an arbi-trary customer until departure either by service completion or abandonment ofwaiting. We also consider the initial waiting time until the service begins forthe first time and the subsequent waiting time. Furthermore, the number ofservice preemptions and resumptions that an arbitrary customer experiences inthe system is studied. Two special cases with θ = ∞ (a preemptive-loss sys-tem) and θ = 0 (a system with patient customers only), both with PR-LCFSdiscipline, are analyzed separately. We present some numerical example for themean, second moment, and covariance of conditional waiting and service time.

Queues with impatient (or reneging) customers have been studied exten-sively. Recently, this system has attracted widespread attention as a basicmodel for evaluating the performance of handling inbound calls in telephonecall centers [9, 18] where queues without priority structure are employed. TheM/M/m FCFS nonpreemptive priority queue with impatient customers is stud-ied by Jouini and Roubos [6] and Takagi [12]. The M/M/m LCFS queue isconsidered as well. Riordan [10, p. 112] refers to this model without prioritystructure, providing a difference-differential equation for the density function ofthe waiting time for a customer who arrives to find k other customers alreadywaiting upon arrival. Jagerman [4, p. 226] derives a solution to this equation.Jouini [5] uses the distribution of the busy period duration from Iravani andBalcıoglu [3] to analyze the waiting time. Takagi [13] considers the first pas-sage time until service completion and abandonment, and calculates the meanand second moment of the waiting times in the M/M/m LCFS nonpreemptive(priority and non-priority) queues. The same model is studied by Jouini and

504 H. Takagi

Roubos [6]. To the best of our knowledge, no studies but ours [14, 15, 16]currently exist regarding the PR-LCFS queues in the open literature.

The analytic technique presented in this paper should be interesting in itsown right. However, our framework is common to M/M/m queues with FCFS,NP-LCFS, and PR-LCFS queues. Moreover, it can be applied to the analysis ofM/M/m preemptive-resume priority queues with impatient customers in whichcustomers of the same class are served in either FCFS or LCFS fashion.

2. First Passage Time to Service Preemption and Completion fromState k of Being Served, 0 ≤ k ≤ m − 1

We focus on a tagged customer in state k, signifying that there are k othercustomers who compete with him for service at any time in the steady state,where k = 0, 1, 2, . . .. They are the customers who arrived after the taggedone and have been staying in the system until that time. According to thepreemptive LCFS discipline, an arriving customer always joins the system atstate k = 0 and his service is started immediately.

We first consider a finite-state birth-and-death process of state transitionsfor a tagged customer in state k, 0 ≤ k ≤ m−1, in which he is being served. Theservice to this customer, with probability one, is eventually either preemptedby a customer who arrives after him or completed without preemption.

2.1. Customer Behavior until Service Preemption and Completion

The state transition diagram for the discrete-time one-dimensional birth-and-death process modeling the behavior of a tagged customer in service is shownin Fig. 1. This process has m transient states {0, 1, 2, . . . ,m − 1} and twoabsorbing states denoted by “Pr” (state m) and “Sr”, representing the servicepreemption and completion, respectively. State transitions occur when anothercustomer arrives and the service to one of customers is completed. Therefore,the state transition probabilities and the LST of the DF for the time spent bya tagged customer in state k are given by

αk =kµ

λ+ (k + 1)µ; βk =

µ

λ+ (k + 1)µ,

B∗k(s) =

λ+ (k + 1)µ

s+ λ+ (k + 1)µ.


. . .

. . .✲

✖✕✗✔m−1✛

✲

❄

1−αm−1

−βm−1

✖✕✗✔Pr

✲✛✖✕

✗✔k+1

✲αk+1

✛1−αk−βk

✖✕✗✔

k✲αk

✛1−αk−1

−βk−1

✖✕✗✔k−1

✻ ✻✻

✻ ✻ ✻

✲✛

✲✛ ✖✕

✗✔1

✲α1

✛1−β0

✖✕✗✔0

βm−1 βk+1 βk βk−1 β1 β0

✛✛✲ . . .

. . .

✖✕✗✔Sr

Figure 1: State transition of a tagged customer until service preemptionand completion.

2.2. LST of the DF for the First Passage Time to ServicePreemption and Completion

By H∗k(s,Pr), we denote the joint probability of service preemption and the

LST of the DF for the first passage time from state k to state “Pr” withoutreaching state “Sr”. In addition, we denote by H∗

k(s,Sr) the joint probabilityof service completion and the LST of the DF for the first passage time fromstate k to state “Sr” without reaching state “Pr”. Furthermore, let

H∗k(s) := H∗

k(s,Pr) +H∗k(s,Sr)

be the unconditional LST of the DF for the first passage time from state k tostate either “Pr” or “Sr”, whichever occurs first.

Applying the first step analysis for the discrete-time Markov chain [8,p. 162], [17, p. 116], we have the following finite sets of equations for {H∗

k(s,Pr); 0 ≤k ≤ m− 1}, {H∗

k(s,Sr); 0 ≤ k ≤ m− 1}, and {H∗k(s); 0 ≤ k ≤ m − 1} respec-

tively:

(s+ λ+ µ)H∗0 (s,Pr) = λH∗

1 (s,Pr),

[s+ λ+ (k + 1)µ]H∗k(s,Pr) = kµH∗

k−1(s,Pr) + λH∗k+1(s,Pr)

1 ≤ k ≤ m− 2,

(s + λ+mµ)H∗m−1(s,Pr) = (m− 1)µH∗

m−2(s,Pr) + λ,

(s + λ+ µ)H∗0 (s,Sr) = µ+ λH∗

1 (s,Sr),

[s+ λ+ (k + 1)µ]H∗k(s,Sr) = kµH∗

k−1(s,Sr) + µ+ λH∗k+1(s,Sr)

506 H. Takagi

1 ≤ k ≤ m− 2,

(s+ λ+mµ)H∗m−1(s,Sr) = (m− 1)µH∗

m−2(s,Sr) + µ,

(s+ λ+ µ)H∗0 (s) = µ+ λH∗

1 (s),

[s+ λ+ (k + 1)µ]H∗k (s) = kµH∗

k−1(s) + µ+ λH∗k+1(s)

1 ≤ k ≤ m− 2,

(s+ λ+mµ)H∗m−1(s) = (m− 1)µH∗

m−2(s) + λ+ µ.

In addition, we let H∗m(s,Pr) ≡ 1, H∗

m(s,Sr) ≡ 0, and H∗m(s) ≡ 1. The solution

can be obtained, in terms of functions {h∗k(s); 0 ≤ k ≤ m}, in the form

H∗k(s,Pr) =

h∗k(s)

h∗m(s); H∗

k(s,Sr) =µ

s+ µ

[

1−h∗k(s)

h∗m(s)

]

,

H∗k(s) =

µ

s+ µ+

s

s+ µ·h∗k(s)

h∗m(s)0 ≤ k ≤ m.

2.3. Solution for {h∗

k(s); 0 ≤ k ≤ m}

A finite set of equations for {h∗k(s); 0 ≤ k ≤ m} is given by

h∗0(s) = 1 ; s+ λ+ µ = λh∗1(s),

[s+ λ+ (k + 1)µ]h∗k(s) = kµh∗k−1(s) + λh∗k+1(s)

1 ≤ k ≤ m− 1.

This can be written as the following set of recurrence relations:

h∗k(s) =s+ λ+ kµ

λh∗k−1(s)−

(k − 1)µ

λh∗k−2(s) 2 ≤ k ≤ m.

The solution is given by Cramer’s rule as the determinant

h∗k(s) = (−1)k∣

∣

∣H

(k)∣

∣

∣2 ≤ k ≤ m,

where H(k) is a k × k tridiagonal matrix with nonzero elements

H(k)i,i = −

s+ λ+ iµ

λ1 ≤ i ≤ k,

H(k)i,i+1 = 1 1 ≤ i ≤ k − 1,

H(k)i+1,i =

iµ

λ0 ≤ i ≤ k − 1.


Note that h∗k(s) is a kth degree polynomial in s, the coefficient of sk being(1/λ)k, 1 ≤ k ≤ m.

At present, we do not have a simple expression for h∗k(s). However, fors = 0, we get

h∗k(0) =

k∑

j=0

(mρ)j

j!

/

(mρ)k

k!=

1

B(k,mρ)0 ≤ k ≤ m

using Erlang’s B formula:

B(0, a) = 1 ; B(k, a) =ak

k!

/

k∑

j=0

aj

j!k = 1, 2, . . . .

Thus, we obtain the probability of service preemption and completion

pk{Pr} = H∗k(0,Pr) =

h∗k(0)

h∗m(0),

pk{Sr} = H∗k(0,Sr) = 1− pk{Pr} 0 ≤ k ≤ m.

In particular, we have

p0{Pr} = B(m,mρ) ; pm{Pr} = 1,

pm−1{Pr} = ρ[1−B(m,mρ)].

The set of ℓth derivatives {h(ℓ)k (s); ℓ ≤ k ≤ m} satisfies the following set of

recurrence relations:

h(ℓ)0 (s) = h

(ℓ)1 (s) = h

(ℓ)2 (s) = · · · = h

(ℓ)ℓ−1(s) = 0 ; h

(ℓ)ℓ (s) =

ℓ!

λℓ,

h(ℓ)k (s) =

s+ λ+ kµ

λh(ℓ)k−1(s)−

(k − 1)µ

λh(ℓ)k−2(s) +

ℓ

λh(ℓ−1)k−1 (s)

ℓ+ 1 ≤ k ≤ m.

The ℓth derivatives of {h∗k(s); 0 ≤ k ≤ m} with respect to s at s = 0 can becalculated recursively by

h(ℓ)0 (0) = h

(ℓ)1 (0) = · · · = h

(ℓ)ℓ−1(0) = 0,

h(ℓ)k (0) =

ℓ

λ

k∑

j=1

(mρ)j

j!

j−1∑

l=0

h(ℓ−1)l (0)

/

(mρ)k

k!ℓ ≤ k ≤ m.

508 H. Takagi

3. First Passage Time to Service Resumption and Abandonmentfrom State k of Waiting, k ≥ m

We next consider another infinite-state birth-and-death process of state tran-sitions for a tagged customer in state k, k ≥ m, in which he is staying in thewaiting room. With probability one, this customer, eventually, either is calledin to resume his service or abandons waiting.

3.1. Customer Behavior until Service Resumption andAbandonment

The state transition diagram for the discrete-time one-dimensional birth-and-death process modeling the behavior of a tagged customer in the waiting room isshown in Fig. 2. The process has an infinite number of transient states {m,m+1, . . .} and two absorbing states denoted by “Rs” (state m− 1) and “Ab”, rep-resenting the service resumption and abandonment, respectively. State tran-sitions occur when another customer arrives, the service to one of customersis completed, and a waiting customer before the tagged one abandons waiting.Therefore, the state transition probabilities and the LST of the DF for the timespent by a tagged customer in state k are given by

α′k =

mµ+ (k −m)θ

λ+mµ+ (k + 1−m)θ; β′k =

θ

λ+mµ+ (k + 1−m)θ,

B′k(s) =

λ+mµ+ (k + 1−m)θ

s+ λ+mµ+ (k + 1−m)θ.

. . .✲✛✖✕

✗✔k+1

✲α′

k+1

✛1−α′

k

−β′

k

✖✕✗✔

k✲

α′

k

✛1−α′

k−1

−β′

k−1

✖✕✗✔k−1

✲✛ . . .

✲✛✖✕

✗✔m+1

✲α′

m+1

✛1−α′

m

−β′

m

✖✕✗✔m

✻α′

m

✖✕✗✔Rs

❄❄

❄ ❄ ❄

β′

k+1 β′

kβ′

k−1 β′

m+1 β′

m

✖✕✗✔Ab✲ ✛ . . . ✛

Figure 2: State transition of a tagged customer until service resumptionand abandonment.


3.2. LST of the DF for the First Passage Time to ServiceResumption and Abandonment

By W ∗k (s,Rs), we denote the joint probability of service resumption and the

LST of the DF for the first passage time from state k to state “Rs” withoutreaching state “Ab”. In addition, we denote by W ∗

k (s,Ab) the joint probabilityof abandonment and the LST of the DF for the first passage time from state kto state “Ab” without reaching state “Rs”. Furthermore, let

W ∗k (s) :=W ∗

k (s,Rs) +W ∗k (s,Ab)

be the unconditional LST of the DF for the first passage time from state k tostate either “Rs” or “Ab”, whichever occurs first.

Three infinite sets of equations for {W ∗k (s,Rs); k ≥ m}, {W ∗

k (s,Ab); k ≥ m}, and {W ∗

k (s); k ≥ m} are respectively given by

(s+ λ+mµ+ θ)W ∗m(s,Rs) = mµ+ λW ∗

m+1(s,Rs),

[s+ λ+mµ+ (k + 1−m)θ]W ∗k (s,Rs)

= [mµ + (k −m)θ]W ∗k−1(s,Rs) + λW ∗

k+1(s,Rs) k ≥ m+ 1,

(s+ λ+mµ+ θ)W ∗m(s,Ab) = θ + λW ∗

m+1(s,Ab),

[s+ λ+mµ+ (k + 1−m)θ]W ∗k (s,Ab)

= [mµ + (k −m)θ]W ∗k−1(s,Ab) + θ + λW ∗

k+1(s,Ab) k ≥ m+ 1,

(s+ λ+mµ+ θ)W ∗m(s) = mµ+ θ + λW ∗

m+1(s),

[s+ λ+mµ+ (k + 1−m)θ]W ∗k (s)

= [mµ + (k −m)θ]W ∗k−1(s) + θ + λW ∗

k+1(s) k ≥ m+ 1.

The solution is expressed, in terms of the set of functions {G∗k(s);

k ≥ m}, in the form

W ∗k (s,Rs) = G∗

k(s+ θ) ; W ∗k (s,Ab) =

θ

s+ θ[1−G∗

k(s + θ)],

W ∗k (s) =

θ + sG∗k(s+ θ)

s+ θk ≥ m.

Thus we obtain the probability of service resumption and abandonment

pk{Rs} = W ∗k (0,Rs) = G∗

k(θ),

pk{Ab} = W ∗k (0,Ab) = 1−G∗

k(θ) k ≥ m

510 H. Takagi

and the conditional and unconditional mean time to resumption and abandon-ment

E[Wk; Rs] = −G′k(θ) ; E[Wk; Ab] =

1−Gk(θ)

θ+G′

k(θ),

E[Wk] = E[Wk; Rs] + E[Wk; Ab] =1−Gk(θ)

θk ≥ m.

3.3. Busy Period

A busy period started with k (≥ m) customers in an M/M/m queue is the timeinterval, denoted by Gk, from the instant at which there are k customers in thesystem (all servers are busy and k−m customers are waiting) to the first instantat which any one of the servers becomes available. Let us denote by fWk

(t,Rs)and fWk

(t,Ab) the density functions of the time until service resumption andthe time until abandonment, respectively, for a customer in state k, k ≥ m.They are related with the density function fGk

(t) for Gk and the probabilityP{Gk > t} as follows:

fWk(t,Rs) = e−θtfGk

(t) ; fWk(t,Ab) = θe−θtP{Gk > t}.

The function G∗k(s) introduced in Section 3.2 is the LST of the DF for Gk,

k ≥ m. The set of equations for {G∗k(s), k ≥ m} is given by

(s + λ+mµ)G∗m(s) = λG∗

m+1(s) +mµ,

[s+ λ+mµ+ (k −m)θ]G∗k(s)

= [mµ+ (k −m)θ]G∗k−1(s) + λG∗

k+1(s) k ≥ m+ 1.

Subba Rao [11] derives the LST of the DF for the duration of a busy periodin an M/G/1 queue with impatient customers. Iravani and Balcıoglu [3] pro-vides the LST of the DF for the duration of a busy period in an M/M/m queuewith exponentially distributed service times of mean 1/(mµ) by modifying theresult of Subba Rao as follows:

G∗k(s) =

mµ

s+mµ+

∞∑

i=1

(−1)iψi,k−m(λ/θ)

×

i−1∏

j=0

(

1−mµ

s+mµ+ jθ

)

mµ

s+mµ+ iθ

1 +∞∑

i=1

(λ/θ)i

i!

i−1∏

j=0

(

1−mµ

s+mµ+ jθ

)

k ≥ m,


where we define

ψi,k(x) :=i

∑

j=max{0,i−k}

(−x)j

j!

(

k

i− j

)

i ≥ 1, k ≥ 0

with ψ0,k(x) ≡ 1 for k ≥ 0. Thus we get

G∗k(θ) =

∞∑

i=0

i!(−τ/m)iψi,k−m(λ/θ)∏i+1

j=1(1 + jτ/m)

/

∞∑

i=0

ρi∏i

j=0(1 + jτ/m).

In particular, for k = m, since ψi,0(x) = (−x)i/i!, we obtain

G∗m(s) =

mµ

s+mµ+

∞∑

i=1

(λ/θ)i

i!

×

i−1∏

j=0

(

1−mµ

s+mµ+ jθ

)

mµ

s+mµ+ iθ

1 +

∞∑

i=1

(λ/θ)i

i!

i−1∏

j=0

(

1−mµ

s+mµ+ jθ

)

,

1−G∗m(s + θ) =

s+ θ

s+mµ+ θ+

∞∑

i=1

(λ/θ)i

i!

×

i∏

j=1

(

1−mµ

s+mµ+ jθ

)

s+ (i+ 1)θ

s+mµ+ (i+ 1)θ

1 +

∞∑

i=1

(λ/θ)i

i!

i∏

j=1

(

1−mµ

s+mµ+ jθ

)

.

Thus we have

pm{Ab} = 1−G∗m(θ)

=θ

λ

∞∑

i=1

iρi∏i

j=1(1 + jτ/m)

/

∞∑

i=0

ρi∏i

j=0(1 + jτ/m).

The first and second derivatives of G∗k(s) with respect to s at s = θ, k ≥ m, are

given in Appendix.

512 H. Takagi

Alternatively, Jagerman [4, p. 231] shows that

G∗k(s) =

(mµ

θ

)

k−m+1F

(

s

θ, k −m+ 1 +

s+mµ

θ;λ

θ

)

(

s+mµ

θ

)

k−m+1

F

(

s

θ,s+mµ

θ;λ

θ

) k ≥ m

and

G∗m(s)

= mµF

(

s

θ, 1 +

s+mµ

θ;λ

θ

)/[

(s+mµ)F

(

s

θ,s+mµ

θ;λ

θ

)]

,

where the Kummer’s series of the confluent hypergeometric function [2, p. 248]and the ascending factorial are defined by

F (a, b;x) := 1 +∞∑

n=1

(a)n(b)n

xn

n!,

(a)n :=Γ(a+ n)

Γ(a)= a(a+ 1)(a+ 2) · · · (a+ n− 1) n ≥ 1.

We can algebraically confirm that

1 +∞∑

i=1

(λ/θ)i

i!

i−1∏

j=0

(

1−mµ

s+mµ+ jθ

)

= F

(

s

θ,s+mµ

θ;λ

θ

)

,

mµ

s+mµ+

∞∑

i=1

(λ/θ)i

i!

i−1∏

j=0

(

1−mµ

s+mµ+ jθ

)

mµ

s+mµ+ iθ

=mµ

s+mµF

(

s

θ, 1 +

s+mµ

θ;λ

θ

)

,

and numerically confirm thatmµ

s+mµ

+

∞∑

i=1

(−1)iψi,k−m(λ/θ)

i−1∏

j=0

(

1−mµ

s+mµ+ jθ

)

mµ

s+mµ+ iθ

=

[

(mµ

θ

)

k−m+1

/(

s+mµ

θ

)

k−m+1

]

× F

(

s

θ, k −m+ 1 +

s+mµ

θ;λ

θ

)

k ≥ m.


4. Joint Distribution of the Waiting and Service Time untilDeparture

We are now in a position to consider the distribution of the time until departure(either abandonment or service completion) for the tagged customer in a com-bination of two birth-and-death processes whose state transitions are depictedin Figs. 1 and 2. We note that state “Pr” in Fig. 1 and state m in Fig. 2 areactually the common single state, so are state “Rs” in Fig. 2 and state m− 1in Fig. 1.

The time until departure consists of the waiting time (the time that thecustomer spends staying in the waiting room) and the service time (the timeduring which the customer is being served). The waiting time and the servicetime are not independent of each other. Therefore, we will derive the joint LSTof the DF for the waiting and service time for a customer who abandons waiting,denoted by T ∗

k (s, s′,Ab), and for a customer who gets served until completion,

denoted by T ∗k (s, s

′,Sr). Then, we obtain the probability of abandonment andservice completion, the marginal LST of the DF for the waiting time, the servicetime, and the total time spent in the system as follows:

Pk{Ab} := T ∗k (0, 0,Ab) ; Pk{Sr} := T ∗

k (0, 0,Sr),

W∗k(s,Ab) := T ∗

k (s, 0,Ab) ; H∗k(s,Ab) := T ∗

k (0, s,Ab),

W∗k(s,Sr) := T ∗

k (s, 0,Sr) ; H∗k(s,Sr) := T ∗

k (0, s,Sr),

T ∗k (s,Ab) := T ∗

k (s, s,Ab) ; T ∗k (s,Sr) := T ∗

k (s, s,Sr).

4.1. Waiting and Service Time until Abandonment

We first consider the waiting and service time until abandonment for an arbi-trary customer who abandons waiting.

(1) For a customer being served in state k, 0 ≤ k ≤ m− 1, the first passageto abandonment (“Ab”) during his waiting time consists of the followingpassages:

(i) the initial passage from state k to state “Pr” in Fig. 1 (which is statem in Fig. 2),

(ii) no or several cycles of transitions from statem to state “Rs” in Fig. 2and from state m− 1 back to state “Pr” in Fig. 1, followed by

(iii) the final passage from state m to state “Ab” in Fig. 2.

514 H. Takagi

Owing to the Markovian property of state transitions, the times to takethese passages in succession are independent of each other. Therefore, weget

T ∗k (s, s

′,Ab) = H∗k(s

′,Pr)W ∗m(s,Ab)

+H∗k(s

′,Pr)[W ∗m(s,Rs)H∗

m−1(s′,Pr)]W ∗

m(s,Ab)

+H∗k(s

′,Pr)[W ∗m(s,Rs)H∗

m−1(s′,Pr)]2W ∗

m(s,Ab) + · · ·

= H∗k(s

′,Pr)W ∗m(s,Ab)

∞∑

n=0

[W ∗m(s,Rs)H∗

m−1(s′,Pr)]n

=H∗

k(s′,Pr)W ∗

m(s,Ab)

1−W ∗m(s,Rs)H∗

m−1(s′,Pr)

=θ

s+ θ·

h∗k(s′)[1−G∗

m(s+ θ)]

h∗m(s′)− h∗m−1(s′)G∗

m(s + θ).

This joint distribution leads to the marginal distribution

W∗k(s,Ab) =

pk{Pr}W∗m(s,Ab)

1− pm−1{Pr}W ∗m(s,Rs)

=θ

s+ θ·

h∗k(0)[1 −G∗m(s+ θ)]

h∗m(0)− h∗m−1(0)G∗m(s+ θ)

,

H∗k(s,Ab) =

pm{Ab}H∗k(s,Pr)

1− pm{Rs}H∗m−1(s,Pr)

=h∗k(s)[1−G∗

m(θ)]

h∗m(s)− h∗m−1(s)G∗m(θ)

,

T ∗k (s,Ab) =

H∗k(s,Pr)W

∗m(s,Ab)


m−1(s,Pr)

=θ

s+ θ·

h∗k(s)[1 −G∗m(s+ θ)]

h∗m(s)− h∗m−1(s)G∗m(s+ θ)

.

Then we obtain the probability of abandonment

Pk{Ab} =pk{Pr}pm{Ab}

1− pm{Rs}pm−1{Pr}=

h∗k(0)[1 −G∗m(θ)]

h∗m(0)− h∗m−1(0)G∗m(θ)

,

the mean waiting and service time

E[Wk,Ab] =1

θPk{Ab}+

h∗k(0)[h∗m(0)− h∗m−1(0)]G

′m(θ)

[h∗m(0)− h∗m−1(0)G∗m(θ)]2

,


E[Hk,Ab] = −[1−G∗m(θ)]

{

h′k(0)

h∗m(0)− h∗m−1(0)G∗m(θ)

−h∗k(0)[h

′m(0)− h′m−1(0)G

∗m(θ)]

[h∗m(0) − h∗m−1(0)G∗m(θ)]2

}

,

E[Tk,Ab] =1

θPk{Ab} −

h′k(0)[1 −G∗m(θ)]

h∗m(0) − h∗m−1(0)G∗m(θ)

+

h∗k(0)

{

[h∗m(0) − h∗m−1(0)]G′m(θ)

+ [h′m(0)− h′m−1(0)G∗m(θ)][1−G∗

m(θ)]

}

[h∗m(0)− h∗m−1(0)G∗m(θ)]2

,

and the second moment of the waiting and service time

E[W2k ,Ab] =

2

θE[Wk,Ab]− h∗k(0)[h

∗m(0) − h∗m−1(0)]

×

{

G′′m(θ)

[h∗m(0) − h∗m−1(0)G∗m(θ)]2

+2h∗m−1(0)[G

′m(θ)]2

[h∗m(0)− h∗m−1(0)G∗m(θ)]3

}

,

E[H2k,Ab] = [1−G∗

m(θ)]

{

h′′k(0)

h∗m(0)− h∗m−1(0)G∗m(θ)

−2h′k(0)[h

′m(0) − h′m−1(0)G

∗m(θ)]

[h∗m(0) − h∗m−1(0)G∗m(θ)]2

−h∗k(0)[h

′′m(0)− h′′m−1(0)G

∗m(θ)]

[h∗m(0) − h∗m−1(0)G∗m(θ)]2

+2h∗k(0)[h

′m(0) − h′m−1(0)G

∗m(θ)]2

[h∗m(0)− h∗m−1(0)G∗m(θ)]3

}

,

E[WkHk,Ab] =1

θE[Hk,Ab]

+ G′m(θ)

{

2h∗k(0)[h∗m(0) − h∗m−1(0)][h

′m(0) − h′m−1(0)G

∗m(θ)]

[h∗m(0) − h∗m−1(0)G∗m(θ)]3

−h∗k(0)[h

′m(0) − h′m−1(0)] + h′k(0)[h

∗m(0)− h∗m−1(0)]

[h∗m(0) − h∗m−1(0)G∗m(θ)]2

}

.

For the total time spent in the system, we can write

T ∗k (s,Ab) =

θ

s+ θ· h∗k(s)U

∗(s),

where U∗(s) :=1−G∗

m(s+ θ)


.

516 H. Takagi

Then, we have

Pk{Ab} = h∗k(0)U∗(0),

E[T ℓk ,Ab] =

ℓ!

θℓ

ℓ∑

l=0

(−θ)l

l!

l∑

n=0

(

l

n

)

h(n)k (0)U (l−n)(0)

ℓ = 1, 2, . . . ,

where the derivatives of U∗(s) with respect to s at s = 0 are given by

U∗(0) =1−G∗

m(θ)

h∗m(0) − h∗m−1(0)G∗m(θ)

,

U ′(0) = −

{

[h∗m(0)− h∗m−1(0)]G′m(θ)

+ [h′m(0)− h′m−1(0)G∗m(θ)][1−G∗

m(θ)]

}

[h∗m(0)− h∗m−1(0)G∗m(θ)]2

,

U ′′(0) = −

[h∗m(0)− h∗m−1(0)]G′′m(θ)

+[h′′m(0)−h′′m−1(0)G∗m(θ)−2h′m−1(0)G

′m(θ)]

× [1−G∗m(θ)]

[h∗m(0)− h∗m−1(0)G∗m(θ)]2

+

2

[h′m(0)− h′m−1(0)G∗m(θ)− h∗m−1(0)G

′m(θ)]

×{[h∗m(0)− h∗m−1(0)]G′m(θ)

+ [h′m(0)− h′m−1(0)G∗m(θ)][1−G∗

m(θ)]}

[h∗m(0)− h∗m−1(0)G∗m(θ)]3

.

(2) For a customer waiting in state k, k ≥ m, the first passage to abandon-ment (“Ab”) is either

(a) a direct passage from state k to state “Ab” without reaching state”Rs” in Fig. 2, or

(b) a sequence of the following passages:

(i) the initial passage from state k to state “Rs” in Fig. 2 (which isstate m− 1 in Fig. 1),

(ii) no or several cycles of transitions from state m−1 to state “Pr”in Fig. 1 and from state m back to state “Rs” in Fig. 2,

(iii) the passage from state m−1 to state “Pr” in Fig. 1, followed by


(iv) the final passage from state m to state “Ab” in Fig. 2.

Therefore, from the Markovian property of state transitions, we get

T ∗k (s, s

′,Ab) =W ∗k (s,Ab)

+W ∗k (s,Rs)H

∗m−1(s

′,Pr)W ∗m(s,Ab)

+W ∗k (s,Rs)[H

∗m−1(s

′,Pr)W ∗m(s,Rs)]H∗

m−1(s′,Pr)W ∗

m(s,Ab)

+W ∗k (s,Rs)[H

∗m−1(s

′,Pr)W ∗m(s,Rs)]2H∗

m−1(s′,Pr)W ∗

m(s,Ab)

+ · · ·

=W ∗k (s,Ab) +W ∗

k (s,Rs)H∗m−1(s

′,Pr)W ∗m(s,Ab)

×∞∑

n=0

[H∗m−1(s

′,Pr)W ∗m(s,Rs)]n

=W ∗k (s,Ab) +

W ∗k (s,Rs)H

∗m−1(s

′,Pr)W ∗m(s,Ab)


m−1(s′,Pr)

=θ

s+ θ

{

1−[h∗m(s′)− h∗m−1(s

′)]G∗k(s+ θ)

h∗m(s′)− h∗m−1(s′)G∗

m(s + θ)

}

.


W∗k(s,Ab) = W ∗

k (s,Ab) +pm−1{Pr}W

∗k (s,Rs)W

∗m(s,Ab)


=θ

s+ θ

{

1−[h∗m(0) − h∗m−1(0)]G

∗k(s+ θ)

h∗m(0) − h∗m−1(0)G∗m(s+ θ)

}

,

H∗k(s,Ab) = pk{Ab}+

pk{Rs}pm{Ab}H∗m−1(s,Pr)


= 1−[h∗m(s)− h∗m−1(s)]G

∗k(θ)


,

T ∗k (s,Ab) = W ∗

k (s,Ab) +W ∗

k (s,Rs)H∗m−1(s,Pr)W

∗m(s,Ab)


m−1(s,Pr)

=θ

s+ θ

{

1−[h∗m(s)− h∗m−1(s)]G

∗k(s + θ)


}

.

Therefore, we obtain the probability of abandonment

Pk{Ab} = pk{Ab}+pk{Rs}pm−1{Pr}pm{Ab}

1− pm−1{Pr}pm{Rs}

= 1−[h∗m(0)− h∗m−1(0)]G

∗k(θ)

h∗m(0)− h∗m−1(0)G∗m(θ)

,

518 H. Takagi


E[Wk,Ab] =1

θPk{Ab}+ [h∗m(0) − h∗m−1(0)]

×

{

G′k(θ)

h∗m(0) − h∗m−1(0)G∗m(θ)

+G∗

k(θ)h∗m−1(0)G

′m(θ)

[h∗m(0) − h∗m−1(0)G∗m(θ)]2

}

,

E[Hk,Ab]=G∗

k(θ)[h′m(0)h∗m−1(0)−h

∗m(0)h′m−1(0)][1−G

∗m(θ)]

[h∗m(0)− h∗m−1(0)G∗m(θ)]2

,

E[Tk,Ab] =1

θPk{Ab}+

G′k(θ)[h

∗m(0)− h∗m−1(0)]

h∗m(0)− h∗m−1(0)G∗m(θ)

+

G∗k(θ)

{

[h∗m(0) − h∗m−1(0)]h∗m−1(0)G

′m(θ)

+ [h′m(0)h∗m−1(0)− h∗m(0)h′m−1(0)][1 −G∗m(θ)]

}

[h∗m(0)− h∗m−1(0)G∗m(θ)]2

,


E[W2k ,Ab] =

2

θE[Wk,Ab]− [h∗m(0)− h∗m−1(0)]

×

{

G′′k(θ)

h∗m(0)− h∗m−1(0)G∗m(θ)

+2G∗

k(θ)[h∗m−1(0)G

′m(θ)]2

[h∗m(0)− h∗m−1(0)G∗m(θ)]3

+h∗m−1(0)[G

∗k(θ)G

′′m(θ) + 2G′

k(θ)G′m(θ)]

[h∗m(0)− h∗m−1(0)G∗m(θ)]2

}

,

E[H2k,Ab] = G∗

k(θ)[1−G∗m(θ)]

×

{

h∗m(0)h′′m−1(0)− h′′m(0)h∗m−1(0)

[h∗m(0) − h∗m−1(0)G∗m(θ)]2

−2[h∗m(0)h′m−1(0)−h

′m(0)h∗m−1(0)][h

′m(0)−h′m−1(0)G

∗m(θ)]

[h∗m(0)− h∗m−1(0)G∗m(θ)]3

}

,

E[WkHk,Ab] =1

θE[Hk,Ab]

+ [h′m(0)h∗m−1(0) − h′m−1(0)h∗m(0)]

×

{

G∗k(θ)G

′m(θ)[h∗m(0)− 2h∗m−1(0) + h∗m−1(0)G

∗m(θ)]

[h∗m(0) − h∗m−1(0)G∗m(θ)]3

−G′

k(θ)[1−G∗m(θ)]

[h∗m(0) − h∗m−1(0)G∗m(θ)]2

}

.


T ∗k (s,Ab) =

θ

s+ θ[1−G∗

k(s+ θ)V ∗(s)],


where V ∗(s) :=h∗m(s)− h∗m−1(s)


.

Then, we have

Pk{Ab} = 1−G∗k(θ)V

∗(0),

E[T ℓk ,Ab] =

ℓ!

θℓ

[

1−ℓ

∑

l=0

(−θ)l

l!

l∑

n=0

(

l

n

)

G(n)k (θ)V (l−n)(0)

]

ℓ = 1, 2, . . . ,

where the derivatives of V ∗(s) with respect to s at s = 0 are given by

V ∗(0) =h∗m(0) − h∗m−1(0)

h∗m(0) − h∗m−1(0)G∗m(θ)

,

V ′(0) =

{

[h∗m(0) − h∗m−1(0)]h∗m−1(0)G

′m(θ)

+ [h′m(0)h∗m−1(0)− h∗m(0)h′m−1(0)][1 −G∗m(θ)]

}

[h∗m(0) − h∗m−1(0)G∗m(θ)]2

,

V ′′(0) =

[h∗m(0)−h∗m−1(0)]

×[h∗m−1(0)G′′m(θ)+2h′m−1(0)G

′m(θ)]

+ [h′′m(0)h∗m−1(0)− h∗m(0)h′′m−1(0)][1 −G∗m(θ)]

[h∗m(0)− h∗m−1(0)G∗m(θ)]2

−

2

[h′m(0)− h′m−1(0)G∗m(θ)− h∗m−1(0)G

′m(θ)]

× {[h∗m(0) − h∗m−1(0)]h∗m−1(0)G

′m(θ)

+ [h′m(0)h∗m−1(0) − h∗m(0)h′m−1(0)][1 −G∗m(θ)]}

[h∗m(0) − h∗m−1(0)G∗m(θ)]3

.

From the relation h∗m(s)U∗(s) = 1−G∗m(s+ θ)V ∗(s), we have two equiv-

alent expressions

T ∗m(s,Ab) =

θ

s+ θ· h∗m(s)U∗(s)

=θ

s+ θ[1−G∗

m(s+ θ)V ∗(s)]

520 H. Takagi

=θ

s+ θ·

h∗m(s)[1−G∗m(s+ θ)]


with

l∑

n=0

(

l

n

)

h(n)m (0)U (l−n)(0) +

l∑

n=0

(

l

n

)

G(n)m (θ)V (l−n)(0) = 0

l = 1, 2, . . . .

4.2. Waiting and Service Time until Service Completion

We next consider the waiting and service time until service completion for anarbitrary customer who gets served.

(1) For a customer being served in state k, 0 ≤ k ≤ m − 1, by an argumentsimilar to the one in Section 4.1 (2), we get

T ∗k (s, s

′,Sr) = H∗k(s

′,Sr)

+H∗k(s

′,Pr)W ∗m(s,Rs)H∗

m−1(s′,Sr)

+H∗k(s

′,Pr)W ∗m(s,Rs)[H∗

m−1(s′,Pr)W ∗

m(s,Rs)]H∗m−1(s

′,Sr)

+H∗k(s

′,Pr)W ∗m(s,Rs)[H∗

m−1(s′,Pr)W ∗

m(s,Rs)]2H∗m−1(s

′,Sr)

+ · · ·

= H∗k(s

′,Sr) +H∗k(s

′,Pr)W ∗m(s,Rs)H∗

m−1(s′,Sr)

×∞∑

n=0

[H∗m−1(s


= H∗k(s

′,Sr) +H∗

k(s′,Pr)W ∗

m(s,Rs)H∗m−1(s

′,Sr)


m−1(s′,Pr)

=µ

s′ + µ

{

1−h∗k(s

′)[1−G∗m(s+ θ)]

h∗m(s′)− h∗m−1(s′)G∗

m(s+ θ)

}

.


W∗k(s,Sr) = pk{Sr}+

pk{Pr}pm−1{Sr}W∗m(s,Rs)


= 1−h∗k(0)[1 −G∗

m(s+ θ)]

h∗m(0) − h∗m−1(0)G∗m(s+ θ)

,

H∗k(s,Sr) = H∗

k(s,Sr) +pm{Rs}H∗

k(s,Pr)H∗m−1(s,Sr)



=µ

s+ µ

{

1−h∗k(s)[1 −G∗

m(θ)]


}

,

T ∗k (s,Sr) = H∗

k(s,Sr) +H∗

k(s,Pr)W∗m(s,Rs)H∗

m−1(s,Sr)


m−1(s,Pr)

=µ

s+ µ

{

1−h∗k(s)[1−G∗

m(s+ θ)]


}

.

Then we obtain the probability of service completion

Pk{Sr} = pk{Sr}+pk{Pr}pm−1{Sr}pm{Rs}


= 1−h∗k(0)[1 −G∗

m(θ)]

h∗m(0)− h∗m−1(0)G∗m(θ)

= 1− Pk{Ab},


E[Wk,Sr] = −h∗k(0)[h

∗m(0)− h∗m−1(0)]G

′m(θ)

[h∗m(0) − h∗m−1(0)G∗m(θ)]2

,

E[Hk,Sr] =1

µPk{Sr}+ [1−G∗

m(θ)]

×

{

h′k(0)

h∗m(0) − h∗m−1(0)G∗m(θ)

−h∗k(0)[h

′m(0) − h′m−1(0)G

∗m(θ)]

[h∗m(0)− h∗m−1(0)G∗m(θ)]2

}

,

E[Tk,Sr] =1

µPk{Sr}+

h′k(0)[1 −G∗m(θ)]

h∗m(0)− h∗m−1(0)G∗m(θ)

−

h∗k(0)

{

[h∗m(0)− h∗m−1(0)]G′m(θ)

+ [h′m(0) − h′m−1(0)G∗m(θ)][1−G∗

m(θ)]

}

[h∗m(0)− h∗m−1(0)G∗m(θ)]2

,


E[W2k ,Sr] = h∗k(0)[h

∗m(0)− h∗m−1(0)]

×

{

G′′m(θ)

[h∗m(0) − h∗m−1(0)G∗m(θ)]2

+2h∗m−1(0)[G

′m(θ)]2

[h∗m(0)− h∗m−1(0)G∗m(θ)]3

}

,

E[H2k,Sr] =

2

µE[Hk,Sr]− [1−G∗

m(θ)]

×

{

h′′k(0)

h∗m(0) − h∗m−1(0)G∗m(θ)

−h∗k(0)[h

′′m(0) − h′′m−1(0)G

∗m(θ)]

[h∗m(0)− h∗m−1(0)G∗m(θ)]2

522 H. Takagi

−2h′k(0)[h

′m(0) − h′′m−1(0)G

∗m(θ)]

[h∗m(0)− h∗m−1(0)G∗m(θ)]2

+2h∗k(0)[h

′m(0)− h′m−1(0)G

∗m(θ)]2

[h∗m(0)− h∗m−1(0)G∗m(θ)]3

}

,

E[WkHk,Sr] =1

µE[Wk,Sr] +G′

m(θ)

×

{

h∗k(0)[h′m(0)− h′m−1(0)] + h′k(0)[h

∗m(0) − h∗m−1(0)]

[h∗m(0)− h∗m−1(0)G∗m(θ)]2

−2h∗k(0)[h

∗m(0) − h∗m−1(0)][h

′m(0) − h′m−1(0)G

∗m(θ)]

[h∗m(0) − h∗m−1(0)G∗m(θ)]3

}

.


T ∗k (s,Sr) =

µ

s+ µ[1− h∗k(s)U

∗(s)],

where U∗(s) is given in Section 4.1(1). Then, we get

Pk{Sr} = 1− h∗k(0)U∗(0),

E[T ℓk ,Sr] =

ℓ!

µℓ

[

1−ℓ

∑

l=0

(−µ)l

l!

l∑

n=0

(

l

n

)

h(n)k (0)U (l−n)(0)

]

ℓ = 1, 2, . . . .

(2) For a customer waiting in state k, k ≥ m, by an argument similar to theone in Section 4.1(1), we get

T ∗k (s, s

′,Sr) =W ∗k (s,Rs)H

∗m−1(s

′,Sr)

+W ∗k (s,Rs)[H

∗m−1(s

′,Pr)W ∗m(s,Rs)]H∗

m−1(s,Sr)

+W ∗k (s,Rs)[H

∗m−1(s

′,Pr)W ∗m(s,Rs)]2H∗

m−1(s′,Sr) + · · ·

=W ∗k (s,Rs)H

∗m−1(s

′,Sr)

∞∑

n=0

[H∗m−1(s


=W ∗

k (s,Rs)H∗m−1(s

′,Sr)


m−1(s′,Pr)

=µ

s′ + µ·[h∗m(s′)− h∗m−1(s

′)]G∗k(s+ θ)

h∗m(s′)− h∗m−1(s′)G∗

m(s + θ).



W∗k(s,Sr) =

pm−1{Sr}W∗k (s,Rs)


=[h∗m(0)− h∗m−1(0)]G

∗k(s+ θ)

h∗m(0)− h∗m−1(0)G∗m(s+ θ)

,

H∗k(s,Sr) =

pk{Rs}H∗m−1(s,Sr)


=µ

s+ µ·[h∗m(s)− h∗m−1(s)]G

∗k(θ)


,

T ∗k (s,Sr) =

W ∗k (s,Rs)H

∗m−1(s,Sr)


m−1(s,Pr),

=µ

s+ µ·[h∗m(s)− h∗m−1(s)]G

∗k(s+ θ)


.

Then we get the probability of service completion

Pk{Sr} =pk{Rs}pm−1{Sr}

1− pm{Rs}pm−1{Pr}

=[h∗m(0) − h∗m−1(0)]G

∗k(θ)

h∗m(0) − h∗m−1(0)G∗m(θ)

= 1− Pk{Ab},


E[Wk,Sr] = −[h∗m(0) − h∗m−1(0)]

×

{

G′k(θ)

h∗m(0) − h∗m−1(0)G∗m(θ)

+G∗

k(θ)h∗m−1(0)G

′m(θ)

[h∗m(0)− h∗m−1(0)G∗m(θ)]2

}

,

E[Hk,Sr] =1

µPk{Sr}

−G∗

k(θ)[h′m(0)h∗m−1(0)− h∗m(0)h′m−1(0)][1 −G∗

m(θ)]

[h∗m(0) − h∗m−1(0)G∗m(θ)]2

,

E[Tk,Sr] =1

µPk{Sr} −

G′k(θ)[h

∗m(0)− h∗m−1(0)]

h∗m(0)− h∗m−1(0)G∗m(θ)

−

G∗k(θ)

{

[h∗m(0) − h∗m−1(0)]h∗m−1(0)G

′m(θ)

+[h′m(0)h∗m−1(0) − h∗m(0)h′m−1(0)][1 −G∗m(θ)]

}

[h∗m(0)− h∗m−1(0)G∗m(θ)]2

,

524 H. Takagi


E[W2k ,Sr] = [h∗m(0)− h∗m−1(0)]

{

G′′k(θ)

h∗m(0)− h∗m−1(0)G∗m(θ)

+h∗m−1(0)[G

∗k(θ)G

′′m(θ) + 2G′

k(θ)G′m(θ)]

[h∗m(0) − h∗m−1(0)G∗m(θ)]2

+2G∗

k(θ)[h∗m−1(0)G

′m(θ)]2

[h∗m(0)− h∗m−1(0)G∗m(θ)]3

}

,

E[H2k,Sr] =

2

µE[Hk,Sr]−G∗

k(θ)[1−G∗m(θ)]

×

{

h∗m(0)h′′m−1(0) − h′′m(0)h∗m−1(0)

[h∗m(0)− h∗m−1(0)G∗m(θ)]2

−2[h∗m(0)h′m−1(0)−h

′m(0)h∗m−1(0)][h

′m(0)−h′m−1(0)G

∗m(θ)]

[h∗m(0) − h∗m−1(0)G∗m(θ)]3

}

,

E[WkHk,Sr] =1

µE[Wk,Sr]−[h′m(0)h∗m−1(0)−h

′m−1(0)h

∗m(0)]

×

{

G∗k(θ)G

′m(θ)[h∗m(0) − 2h∗m−1(0) + h∗m−1(0)G

∗m(θ)]

[h∗m(0)− h∗m−1(0)G∗m(θ)]3

−G′

k(θ)[1−G∗m(θ)]

[h∗m(0)− h∗m−1(0)G∗m(θ)]2

}

.


T ∗k (s,Sr) =

µ

s+ µG∗

k(s + θ)V ∗(s),

where V ∗(s) is given in Section 4.1(2). Then, we have

Pk{Sr} = G∗k(θ)V

∗(0),

E[T ℓk ,Sr] =

ℓ!

µℓ

ℓ∑

l=0

(−µ)l

l!

l∑

n=0

(

l

n

)

G(n)k (θ)V (l−n)(0)

ℓ = 1, 2, . . . .

We note that

T ∗m(s,Sr) =

µ

s+ µ[1− h∗m(s)U∗(s)] =

µ

s+ µG∗

m(s+ θ)V ∗(s)

=µ

s+ µ·[h∗m(s)− h∗m−1(s)]G

∗m(s+ θ)]


.


4.3. Waiting and Service Time until Departure

We finally consider the waiting and service time until departure (either aban-donment or service completion) for an arbitrary customer in state k (k ≥ 0).Let

T ∗k (s, s

′) := T ∗k (s, s

′,Ab) + T ∗k (s, s

′,Sr) k ≥ 0

be the unconditional joint LST of the DF for the waiting and service time fora customer in state k. Then, we obtain the marginal LST of the DF for thewaiting time, the service time, and the total time spent in the system as followsfor k ≥ 0:

W∗k(s) := T ∗

k (s, 0) ; H∗k(s) := T ∗

k (0, s) ; T ∗k (s) := T ∗

k (s, s).

(1) For a customer being served in state k, 0 ≤ k ≤ m− 1, we have

T ∗k (s, s

′) =µ

s′ + µ+

(

θ

s+ θ−

µ

s′ + µ

)

×h∗k(s

′)[1−G∗m(s+ θ)]

h∗m(s′)− h∗m−1(s′)G∗

m(s+ θ).


W∗k(s) = W∗

k(s,Ab) +W∗k(s,Sr)

= 1−s

s+ θ·

h∗k(0)[1 −G∗m(s+ θ)]

h∗m(0) − h∗m−1(0)G∗m(s+ θ)

= 1−s

θW∗

k(s,Ab) =θ

s+ θ+

s

s+ θW∗

k(s,Sr),

H∗k(s) = H∗

k(s,Ab) +H∗k(s,Sr)

=µ

s+ µ+

s

s+ µ·

h∗k(s)[1−G∗m(θ)]


=µ

s+ µ+

s

s+ µH∗

k(s,Ab) = 1−s

µH∗

k(s,Sr),

T ∗k (s) = T ∗

k (s,Ab) + T ∗k (s,Sr)

= 1−s

θT ∗k (s,Ab)−

s

µT ∗k (s,Sr)

=µ

s+ µ+

(

θ

s+ θ−

µ

s+ µ

)

×h∗k(s)[1 −G∗

m(s+ θ)]


526 H. Takagi

=µ

s+ µ+

(

θ

s+ θ−

µ

s+ µ

)

h∗k(s)U∗(s),

where U∗(s) is given in Section 4.1(1). Then, we obtain the mean waitingand service time

E[Wk] =h∗k(0)[1 −G∗

m(θ)]

θ[h∗m(0)− h∗m−1(0)G∗m(θ)]

,

E[Hk] =1

µ

{

1−h∗k(0)[1 −G∗

m(θ)]

h∗m(0)− h∗m−1(0)G∗m(θ)

}

,

E[Tk] = E[Wk] + E[Hk]

=1

µ+

(

1

θ−

1

µ

)

h∗k(0)[1 −G∗m(θ)]

h∗m(0) − h∗m−1(0)G∗m(θ)

,


E[W2k ] =

2h∗k(0)[1 −G∗m(θ)]

θ2[h∗m(0)− h∗m−1(s)G∗m(θ)]

+2h∗k(0)[h

∗m(0)− h∗m−1(0)]G

′m(θ)

θ[h∗m(0)− h∗m−1(s)G∗m(θ)]2

,

E[H2k] =

2

µ2−

2[1−G∗m(θ)]

µ2

{

h∗k(0) − µh′k(0)

h∗m(0)− h∗m−1(0)G∗m(θ)

+h∗k(0)[h

′m(0) − h′m−1(0)G

∗m(θ)]

[h∗m(0) − h∗m−1(0)G∗m(θ)]2

}

,

E[T 2k ] =

2

µ2+ 2

(

1

θ2−

1

µ2

)

h∗k(0)[1 −G∗m(θ)]

h∗m(0)− h∗m−1(s)G∗m(θ)

+ 2

(

1

θ−

1

µ

)

h∗k(0)[h∗m(0) − h∗m−1(0)]G

′m(θ)

[h∗m(0) − h∗m−1(s)G∗m(θ)]2

− 2

(

1

θ−

1

µ

)

[1−G∗m(θ)]

{

h′k(0)

h∗m(0) − h∗m−1(s)G∗m(θ)

−h∗k(0)[h

′m(0) − h′m−1(0)G

∗m(θ)]

[h∗m(0) − h∗m−1(s)G∗m(θ)]2

}

,

E[WkHk] = −h∗k(0)[h

∗m(0) − h∗m−1(0)]G

′m(θ)

µ[h∗m(0)− h∗m−1(s)G∗m(θ)]2

−1−G∗

m(θ)

θ

{

h′k(0)

h∗m(0)− h∗m−1(s)G∗m(θ)

−h∗k(0)[h

′m(0) − h′m−1(0)G

∗m(θ)]

[h∗m(0)− h∗m−1(s)G∗m(θ)]2

}

.


We generally have

E[T ℓk ] =

ℓ!

µℓ+ (−1)ℓ

ℓ∑

l=1

(−1)l(

ℓ

l

)

l!

(

1

θl−

1

µl

)

ℓ−l∑

n=0

(

ℓ− l

n

)

h(n)k (0)U (ℓ−l−n)(0) ℓ = 1, 2, . . . .

(2) For a customer waiting in state k, k ≥ m, we have

T ∗k (s, s

′) =θ

s+ θ+

(

µ

s′ + µ−

θ

s+ θ

)

×[h∗m(s′)− h∗m−1(s

′)]G∗k(s+ θ)

h∗m(s′)− h∗m−1(s′)G∗

m(s+ θ).


W∗k(s) =

θ

s+ θ+

s

s+ θ·[h∗m(0)− h∗m−1(0)]G

∗k(s+ θ)

h∗m(0)− h∗m−1(0)G∗m(s+ θ)

= 1−s

θW∗

k(s,Ab) =θ

s+ θ+

s

s+ θW∗

k(s,Sr),

H∗k(s) = 1−

s

s+ µ·[h∗m(s)− h∗m−1(s)]G

∗k(θ)


=µ

s+ µ+

s

s+ µH∗

k(s,Ab) = 1−s

µH∗

k(s,Sr),

T ∗k (s) = T ∗

k (s,Ab) + T ∗k (s,Sr)

= 1−s

θT ∗k (s,Ab)−

s

µT ∗k (s,Sr)

=θ

s+ θ+

(

µ

s+ µ−

θ

s+ θ

)

×[h∗m(s)− h∗m−1(s)]G

∗k(s + θ)


=θ

s+ θ+

(

µ

s+ µ−

θ

s+ θ

)

G∗k(s + θ)V ∗(s),

where V ∗(s) is given in Section 4.1(2). Then, we obtain the mean waitingand service time

E[Wk] =1

θ

{

1−h∗m(0)− h∗m−1(0)]G

∗k(θ)

h∗m(0) − h∗m−1(0)G∗m(θ)

}

,

528 H. Takagi

E[Hk] =[h∗m(0)− h∗m−1(0)]G

∗k(θ)

µ[h∗m(0) − h∗m−1(0)G∗m(θ)]

,

E[Tk] = E[Wk] + E[Hk]

=1

θ+

(

1

µ−

1

θ

)

h∗m(0)− h∗m−1(0)]G∗k(θ)

h∗m(0)− h∗m−1(0)G∗m(θ)

,


E[W2k ] =

2

θ2

{

1−[h∗m(0) − h∗m−1(0)][G

∗k(θ)− θG′

k(θ)]

h∗m(0)− h∗m−1(0)G∗m(θ)

}

+2G∗

k(θ)[h∗m(0)− h∗m−1(0)]h

∗m−1(0)G

′m(θ)

θ[h∗m(0)− h∗m−1(s)G∗m(θ)]2

,

E[H2k] = 2G∗

k(θ)

{

h∗m(0) − h∗m−1(0)

µ2[h∗m(0) − h∗m−1(0)G∗m(θ)]

−[h′m(0)h∗m−1(0)− h∗m(0)h′m−1(0)][1 −G∗

m(θ)]

µ[h∗m(0)− h∗m−1(0)G∗m(θ)]2

}

,

E[T 2k ] =

2

θ2+ 2

(

1

µ2−

1

θ2

)

[h∗m(0)− h∗m−1(0)]G∗k(θ)

h∗m(0) − h∗m−1(0)G∗m(θ)

+ 2

(

1

θ−

1

µ

)

[h∗m(0)− h∗m−1(0)]G′k(θ)

h∗m(0) − h∗m−1(0)G∗m(θ)

+ 2

(

1

θ−1

µ

)

{

G∗k(θ)[h

∗m(0)−h∗m−1(0)]h

∗m−1(0)G

′m(θ)

+[h′m(0)h∗m−1(0)−h∗m(0)h′m−1(0)][1−G

∗m(θ)]

}

[h∗m(0) − h∗m−1(0)G∗m(θ)]2

,

E[WkHk] = −[h∗m(0)− h∗m−1(0)]G

′k(θ)

µ[h∗m(0) − h∗m−1(0)G∗m(θ)]

−G∗

k(θ)[h∗m(0) − h∗m−1(0)]h

∗m−1(0)G

′m(θ)

µ[h∗m(0)− h∗m−1(0)G∗m(θ)]2

+G∗

k(θ)[h′m(0)h∗m−1(0)− h∗m(0)h′m−1(0)][1 −G∗

m(θ)]

θ[h∗m(0)− h∗m−1(0)G∗m(θ)]2

.

We generally have

E[T ℓk ] =

ℓ!

θℓ+ (−1)ℓ

ℓ∑

l=1

(−1)l(

ℓ

l

)

l!

(

1

µl−

1

θl

)

×ℓ−l∑

n=0

(

ℓ− l

n

)

G(n)k (θ)V (ℓ−l−n)(0) ℓ = 1, 2, . . . .


(3) Recursive relation among moments of distribution for the waiting andservice time

From the explicit expressions for T ∗k (s, s

′,Ab), T ∗k (s, s

′,Sr), and T ∗k (s, s

′)given above, it can be shown that the unconditional and conditional jointLST of the DF for the waiting and service time until departure for acustomer in state k satisfy the following same relation in both cases 0 ≤k ≤ m− 1 and k ≥ m:

T ∗k (s, s

′) = 1−s

θT ∗k (s, s

′,Ab)−s′

µT ∗k (s, s

′,Sr) k ≥ 0.

This yields the recursive relation among unconditional and conditionalmoments as follows:

E[WℓkH

ℓ′

k ] =ℓ

θE[Wℓ−1

k Hℓ′

k ,Ab] +ℓ′

µE[Wℓ

kHℓ′−1k ,Sr]

ℓ, ℓ′ = 0, 1, 2, . . . .

In particular, we get

E[Wk] =1

θPk{Ab} ; E[Hk] =

1

µPk{Sr},

E[WkHk] =1

θE[Hk,Ab] +

1

µE[Wk,Sr],

E[Wℓk] =

ℓ

θE[Wℓ−1

k ,Ab] ; E[Hℓk] =

ℓ

µE[Hℓ−1

k ,Sr]

ℓ = 2, 3, . . . .

Furthermore, it follows from the relation

T ∗k (s) = 1−

s

θT ∗k (s,Ab)−

s

µT ∗k (s,Sr) k ≥ 0

(or from Tk = Wk +Hk) that

E[Tk] =1

θPk{Ab}+

1

µPk{Sr},

E[T ℓk ] =

ℓ

θE[T ℓ−1

k ,Ab] +ℓ

µE[T ℓ−1

k ,Sr] ℓ = 2, 3, . . . .

530 H. Takagi

4.4. Waiting and Service Time of an Arriving Customer

According to the preemptive LCFS discipline, an arriving customer always joinsthe system at state k = 0. Therefore, the mean of his waiting and service timeis given by

E[W0] =1

θP0{Ab} =

E[L]

λ; E[H0] =

1

µP0{Sr} =

E[S]

λ

with an instance of Little’s theorem [19, p. 235]

E[T0] =1

θP0{Ab}+

1

µP0{Sr} =

E[N ]

λ,

where E[L], E[S], and E[N ] are the mean number of customers present in thewaiting room, the service facility, and the entire system, respectively, in thesteady state [15].

An arbitrary tagged customer is affected, regardless of whether he is beingserved (0 ≤ k ≤ m − 1) or waiting (k ≥ m), by only those customers whoarrive after him. None of the customers present in the system at his arrivaltime competes for service with him at any time. Therefore, for an arbitraryarriving customer, we have

T ∗(s, s′,Sr) = T ∗0 (s, s

′,Sr)

=µ

s′ + µ

[

1−1−G∗

m(s+ θ)

h∗m(s′)− h∗m−1(s′)G∗

m(s+ θ)

]

,

T ∗(s, s′,Ab) = T ∗0 (s, s

′,Ab)

=θ

s+ θ·

1−G∗m(s + θ)

h∗m(s′)− h∗m−1(s′)G∗

m(s+ θ),

which lead to

P{Sr} = 1−1−G∗

m(θ)

h∗m(0)− h∗m−1(0)G∗m(θ)

,

P{Ab} =1−G∗

m(θ)

h∗m(0) − h∗m−1(0)G∗m(θ)

,

W∗(s,Sr) = 1−1−G∗

m(s+ θ)

h∗m(0)− h∗m−1(0)G∗m(s+ θ)

,

H∗(s,Sr) =µ

s+ µ

[

1−1−G∗

m(θ)


]

,


T ∗(s,Sr) =µ

s+ µ

[

1−1−G∗

m(s+ θ)

h∗m(s)− h∗m−1(s)G∗m(s + θ)

]

,

W∗(s,Ab) =θ

s+ θ·

1−G∗m(s+ θ)

h∗m(0)− h∗m−1(0)G∗m(s + θ)

,

H∗(s,Ab) =1−G∗

m(θ)


,

T ∗(s,Ab) =θ

s+ θ·

1−G∗m(s+ θ)


.

4.5. Initial and Subsequent Waiting Time

In addition, for a customer waiting in state k, k ≥ m, let us call a portion ofthe waiting time until the service begins for the first time the initial waiting

time, denoted W◦k , and call the rest of the waiting time the subsequent waiting

time, denoted by W•k . We can find the joint distribution of W◦

k and W•k , and

observe that W◦k and W•

k are not independent.

Let W∗k(s, s

′,Ab) and W∗k(s, s

′,Sr) be the joint LSTs of the DF for W◦k and

W•k for a tagged customer in state k, k ≥ m, who abandons waiting and who

gets served, respectively.

(1) Waiting time until abandonment

By an analysis similar to the one in Section 4.1(2), we obtain

W∗k(s, s

′,Ab) = W ∗k (s,Ab) +

W ∗k (s,Rs)pm−1{Pr}W

∗m(s′,Ab)

1− pm−1{Pr}W ∗m(s′,Rs)

=θ

s+ θ[1−G∗

k(s+ θ)]

+θ

s′ + θ·h∗m−1(0)G

∗k(s+ θ)[1−G∗

m(s′ + θ)]

h∗m(0)− h∗m−1(0)G∗m(s′ + θ)

.

Then, the LST of the DF for the total waiting time of a customer whoabandons waiting is given by W∗

k(s,Ab) = W∗k(s, s,Ab), which agrees

with the result in Section 4.1(2). The marginal distribution, mean, andsecond moment of W◦

k and W•k are given as follows:

W◦k(s,Ab) = W∗

k(s, 0,Ab) =θ

s+ θ[1−G∗

k(s+ θ)]

+h∗m−1(0)G

∗k(s+ θ)[1−G∗

m(θ)]

h∗m(0)− h∗m−1(0)G∗m(θ)

,

532 H. Takagi

E[W◦k ,Ab] =

1−G∗k(θ)

θ+

[h∗m(0)− h∗m−1(0)]G′k(θ)

h∗m(0)− h∗m−1(0)G∗m(θ)

,

E[(W◦k )

2,Ab] =2[1−G∗

k(θ)]

θ2+

2G′k(θ)

θ

−[h∗m(0)− h∗m−1(0)]G

′′k(θ)

h∗m(0)− h∗m−1(0)G∗m(θ)

,

W•k(s,Ab) = W∗

k(0, s,Ab) = 1−G∗k(θ)

+θ

s+ θ·h∗m−1(0)G

∗k(θ)[1−G∗

m(s + θ)]

h∗m(0)− h∗m−1(0)G∗m(s+ θ)

,

E[W•k ,Ab] = h∗m−1(0)G

∗k(θ)

{

1−G∗m(θ)

θ[h∗m(0)− h∗m−1(0)G∗m(θ)]

+[h∗m(0)− h∗m−1(0)]G

′m(θ)

[h∗m(0)− h∗m−1(0)G∗m(θ)]2

}

,

E[(W•k )

2,Ab] = h∗m−1(0)G∗k(θ)

{

2[1−G∗m(θ)]

θ2[h∗m(0) − h∗m−1(0)G∗m(θ)]

+[h∗m(0)− h∗m−1(0)][2G

′m(θ)− θG′′

m(θ)]

θ[h∗m(0)− h∗m−1(0)G∗m(θ)]2

−2[h∗m(0) − h∗m−1(0)]hm−1(0)[G

′m(θ)]2

[h∗m(0) − h∗m−1(0)G∗m(θ)]3

}

,

E[W◦kW

•k ,Ab] = −h∗m−1(0)G

′k(θ)

{

1−G∗m(θ)

θ[h∗m(0) − h∗m−1(0)G∗m(θ)]

+[h∗m(0)− h∗m−1(0)]G

′m(θ)

[h∗m(0)− h∗m−1(0)G∗m(θ)]2

}

.

Since W∗k(s, s

′ |Ab) 6= W◦k(s |Ab)W

•k(s

′ |Ab), W◦k and W•

k are not inde-pendent for a customer who abandons waiting.

(2) Waiting time until service completion

By an analysis similar to the one in Section 4.2(2), we obtain

W∗k(s, s

′,Sr) =W ∗

k (s,Rs)pm−1{Sr}

1− pm−1{Pr}W ∗m(s′,Rs)

=[h∗m(0) − h∗m−1(0)]G

∗k(s+ θ)

h∗m(0)− h∗m−1(0)G∗m(s′ + θ)

.

Then, the LST of the DF for the total waiting time of a customer whogets served is given by W∗

k(s,Sr) = W∗k(s, s,Sr), which agrees with the


result in Section 4.2(2). The marginal distribution, mean, and secondmoment of W◦

k and W•k are given as follows:

W◦k(s,Sr) = W∗

k(s, 0,Sr) =[h∗m(0)− h∗m−1(0)]G

∗k(s+ θ)

h∗m(0)− h∗m−1(0)G∗m(θ)

,

E[W◦k ,Sr] = −

[h∗m(0) − h∗m−1(0)]G′k(θ)

h∗m(0) − h∗m−1(0)G∗m(θ)

,

E[(W◦k )

2,Sr] =[h∗m(0)− h∗m−1(0)]G

′′k(θ)

h∗m(0)− h∗m−1(0)G∗m(θ)

,

W•k(s,Sr) = W∗

k(0, s,Sr) =[h∗m(0) − h∗m−1(0)]G

∗k(θ)

h∗m(0)− h∗m−1(0)G∗m(s+ θ)

,

E[W•k ,Sr] = −

[h∗m(0) − h∗m−1(0)]h∗m−1(0)G

∗k(θ)G

′m(θ)

[h∗m(0) − h∗m−1(0)G∗m(θ)]2

,

E[(W•k )

2,Sr] = [h∗m(0)− h∗m−1(0)]G∗k(θ)

×

{

h∗m−1(0)G′′m(θ)

[h∗m(0) − h∗m−1(0)G∗m(θ)]2

+2[h∗m−1(0)G

′m(θ)]2

[h∗m(0)− h∗m−1(0)G∗m(θ)]3

}

,

E[W◦kW

•k ,Sr] =

[h∗m(0) − h∗m−1(0)]h∗m−1(0)G

′k(θ)G

′m(θ)

[h∗m(0) − h∗m−1(0)G∗m(θ)]2

.

Since W∗k(s, s

′ |Sr) 6= W◦k(s |Sr)W

•k(s

′ |Sr), W◦k and W•

k are not indepen-dent for a customer who gets served.

4.6. Number of Service Preemptions and Resumptions

We are interested in the number of service preemptions and resumptions thatan arbitrary customer experiences before he departs from the system. LetJk and Kk be the number of preemptions and resumptions, respectively, untildeparture (either by service completion or abandonment) for a customer in statek (≥ 0). According to the Markovian property of transitions, these numbersare geometrically distributed.

(1) Preemptions and resumptions until abandonment

(i) For a customer in state k, 0 ≤ k ≤ m− 1, we have

P{Jk = n,Ab} = pk{Pr}pm{Ab}[pm{Rs}pm−1{Pr}]n−1

n = 1, 2, . . . .

534 H. Takagi

The probability distribution and the mean of Jk for a customer whoabandons are given by

P{Jk = n | Ab} =P{Jk = n,Ab}

Pk{Ab}

= (1− pm{Rs}pm−1{Pr})[pm{Rs}pm−1{Pr}]n−1,

E[Jk | Ab] =∞∑

n=1

nP{Jk = n | Ab}

=1

1− pm{Rs}pm−1{Pr}=

h∗m(0)

h∗m(0)− h∗m−1(0)G∗m(θ)

.

Since Jk = Kk + 1, we have

P{Kk = n,Ab} = P{Jk = n+ 1,Ab}

= pk{Pr}pm{Ab}[pm{Rs}pm−1{Pr}]n

n = 0, 1, 2, . . . ,

E[Kk | Ab] =pm{Rs}pm−1{Pr}

[1− pm{Rs}pm−1{Pr}]2

=h∗m−1(0)G

∗m(θ)

h∗m(0)− h∗m−1(0)G∗m(θ)

= E[Jk | Ab]− 1.

The distribution of Jk and Kk for a customer who is being served andeventually abandons does not depend on k, since it is only related tothe number of preemptions and resumptions before abandonment.

(ii) For a customer in state k, k ≥ m, we have Jk = Kk and

P{Jk = n,Ab} = P{Kk = n,Ab}

=

pk{Ab} n = 0

pk{Rs}pm−1{Pr}pm{Ab}[pm−1{Pr}pm{Rs}]n−1

n = 1, 2, . . . ,

E[Jk,Ab] = E[Kk,Ab] =pk{Rs}pm−1{Pr}pm{Ab}

[1− pm−1{Pr}pm{Rs}]2

=h∗m−1(0)h

∗m(0)G∗

k(θ)[1−G∗m(θ)]

[h∗m(0)− h∗m−1(0)G∗m(θ)]2

.

In this case, the distribution of Jk and Kk for a customrer who iswaiting and abandons, P{Jk = n | Ab} = P{Kk = n | Ab} =P{Jk = n,Ab}/Pk{Ab}, does depend on k, since the possibilty ofreaching “Ab” before reaching “Rs” depends on k in Fig. 2.


(2) Preemptions and resumptions until service completion

(i) For a customer in state k, 0 ≤ k ≤ m− 1, we have Jk = Kk and

P{Jk = n,Sr} = P{Kk = n,Sr}

=

pk{Sr} n = 0

pk{Pr}pm{Rs}pm−1{Sr}[pm−1{Pr}pm{Rs}]n−1

n = 1, 2, . . . ,

E[Jk,Sr] = E[Kk,Sr] =pk{Pr}pm{Rs}pm−1{Sr}

[1 − pm{Rs}pm−1{Pr}]2

=h∗k(0)[h

∗m(0)− h∗m−1(0)]G

∗m(θ)

[h∗m(0) − h∗m−1(0)G∗m(θ)]2

.

(ii) For a customer in state k, k ≥ m, we have

P{Kk = n,Sr} = pk{Rs}pm−1{Sr}[pm−1{Pr}pm{Rs}]n−1

n = 1, 2, . . .

The distribution and the mean of Kk for a customer who gets servedare given by

P{Kk = n | Sr} =P{Kk = n,Sr}

Pk{Sr}

= (1− pm−1{Pr}pm{Rs})[pm−1{Pr}pm{Rs}]n−1

E[Kk | Sr] =∞∑

n=1

nP{Kk = n | Sr}

=1

1− pm−1{Pr}pm{Rs}=

h∗m(0)

h∗m(0)− h∗m−1(0)G∗m(θ)

.

Since Kk = Jk + 1, we have

P{Jk = n,Sr} = P{Kk = n+ 1,Sr}

= pk{Rs}pm−1{Sr}[pm−1{Pr}pm{Rs}]n

n = 0, 1, 2, . . . ,

E[Jk | Sr] =pm−1{Pr}pm{Rs}


=h∗m−1(0)G

∗m(θ)

h∗m(0)− h∗m−1(0)G∗m(θ)

= E[Kk | Sr]− 1.

536 H. Takagi

The distribution of Jk and Kk for a customer who is waiting andeventually gets served does not depend on k for the reason similarto the one in (1)(i).

(3) Preemptions and resumptions until departure

The unconditional mean number of preemptions and resumptions untildeparture is given as follows.

(i) For a customer in state k, 0 ≤ k ≤ m− 1, we have

E[Jk] =h∗k(0)

h∗m(0) − h∗m−1(0)G∗m(θ)

,

E[Kk] = E[Jk]−Pk{Ab} =h∗k(0)G

∗m(θ)

h∗m(0)− h∗m−1(0)G∗m(θ)

.

(ii) For a customer in state k, k ≥ m, we have

E[Jk] =h∗m−1(0)G

∗k(θ)

h∗m(0)− h∗m−1(0)G∗m(θ)

,

E[Kk] = E[Jk] + Pk{Sr} =h∗m(0)G∗

k(θ)

h∗m(0)− h∗m−1(0)G∗m(θ)

.

5. Numerical Example

In Tables 1–5, we present some numerical example of our theoretical formulas.We assume thatm = 5, µ = 1, θ = 2, and λ = 10 (ρ = 2, τ = 2). Our experiencereveals that the numerical computation using the distribution formulas in thispaper is much faster and more stable than that using recursive relations ofmoments given in our previous work [14, 15].

6. Special Cases

We consider two special cases of our model with respect to θ. Recursive compu-tational formulas for calculating the moments of the time until service comple-tion and abandonment in these special cases, as well as numerical results, areprovided in [15]. Direct numerical calculation, using the distribution formulasbelow results in the same values. Therefore, in this section, we only show theexplicit LST of the DF for the time until service completion and abandonment.


Table 1: Numerical example.

k Pk{Ab} Pk{Sr} E[Wk,Ab] E[Hk,Ab] E[Tk,Ab] E[Wk,Sr]

0 0.51270 0.48730 0.24299 0.30992 0.55291 0.013361 0.56396 0.43604 0.26729 0.28965 0.55693 0.014702 0.62549 0.37451 0.29645 0.26019 0.55663 0.016303 0.70034 0.29966 0.33192 0.21777 0.54969 0.018254 0.79283 0.20717 0.37576 0.15678 0.53254 0.020665 0.90911 0.09089 0.43087 0.06878 0.49965 0.023696 0.94907 0.05093 0.45368 0.03854 0.49222 0.020857 0.96686 0.03314 0.46548 0.02508 0.49056 0.017958 0.97624 0.02376 0.47252 0.01798 0.49050 0.015609 0.98181 0.01819 0.47713 0.01377 0.49090 0.01378

10 0.98541 0.01459 0.48038 0.01104 0.49142 0.0123315 0.99293 0.00707 0.48827 0.00535 0.49362 0.0081920 0.99541 0.00459 0.49146 0.00347 0.49493 0.0062430 0.99732 0.00268 0.49432 0.00203 0.49635 0.00434

k E[Hk, Sr] E[Tk, Sr] E[Wk] E[Hk] E[Tk] E[WkHk; Ab]

0 0.17738 0.19074 0.25635 0.48730 0.74365 0.157001 0.14639 0.16108 0.28198 0.43604 0.71802 0.148402 0.11432 0.13062 0.31274 0.37451 0.68726 0.135653 0.08189 0.10014 0.35017 0.29966 0.64983 0.117024 0.05039 0.07105 0.39642 0.20717 0.60358 0.089945 0.02211 0.04579 0.45456 0.09089 0.54544 0.050536 0.01239 0.03324 0.47454 0.05093 0.52546 0.034057 0.00806 0.02601 0.48343 0.03314 0.51567 0.025478 0.00578 0.02138 0.48812 0.02376 0.51188 0.020339 0.00442 0.01820 0.49090 0.01819 0.50910 0.01695

10 0.00355 0.01588 0.49270 0.01459 0.50730 0.0145615 0.00172 0.00991 0.49647 0.00707 0.50353 0.0087420 0.00112 0.00736 0.49771 0.00459 0.50229 0.0063730 0.00065 0.00499 0.49866 0.00268 0.50134 0.00424

538 H. Takagi

Table 2: Numerical example – continued.

k E[WkHk; Sr] E[WkHk] E[W2k ,Ab] E[H2

k,Ab] E[T 2k ,Ab]

0 0.01132 0.16832 0.23301 0.24230 0.789301 0.01112 0.15952 0.25631 0.20454 0.757652 0.01074 0.14639 0.28427 0.16329 0.718863 0.01011 0.12713 0.31829 0.11933 0.671664 0.00911 0.09905 0.36033 0.07452 0.614735 0.00755 0.05808 0.41317 0.03269 0.546926 0.00608 0.04012 0.43606 0.01832 0.522477 0.00502 0.03049 0.44855 0.01192 0.511418 0.00426 0.02459 0.45641 0.00855 0.505629 0.00371 0.02066 0.46184 0.00654 0.50227

10 0.00329 0.01785 0.46583 0.00524 0.5002015 0.00213 0.01087 0.47650 0.00254 0.4865220 0.00161 0.00798 0.48413 0.00165 0.4958330 0.00111 0.00535 0.48638 0.00096 0.49583

k E[W2k ,Sr] E[H2

k,Sr] E[T 2k ,Sr] E[T 2

k ] E[W2k ] E[H2

k]

0 0.00998 0.11247 0.14509 0.93439 0.22499 0.354761 0.01098 0.08824 0.12145 0.87910 0.26729 0.292782 0.01217 0.06536 0.09902 0.81788 0.29645 0.228653 0.01363 0.04445 0.07831 0.74997 0.33192 0.163784 0.01543 0.02625 0.05990 0.67463 0.37576 0.100775 0.01769 0.01152 0.04432 0.59124 0.43087 0.044216 0.01762 0.00645 0.03623 0.55870 0.45368 0.024777 0.01693 0.00420 0.03117 0.54257 0.46548 0.016128 0.01611 0.00301 0.02764 0.53326 0.47252 0.011569 0.01530 0.00230 0.02502 0.52729 0.47713 0.00885

10 0.01455 0.00185 0.02297 0.52317 0.48037 0.0071015 0.01177 0.00090 0.01693 0.51345 0.48827 0.0034420 0.01003 0.00058 0.01383 0.50965 0.49146 0.0022330 0.00794 0.00034 0.01050 0.50633 0.49432 0.00130


Table 3: Numerical example – continued.

k E[WkHk] E[W2kHk] E[WkH

2k] E[W3

k ] E[H3k] E[T 3

k ]

0 0.16832 0.16697 0.14830 0.34952 0.33740 1.619231 0.15952 0.15937 0.12451 0.38447 0.26471 1.500832 0.14639 0.14782 0.10313 0.42641 0.19608 1.375353 0.12713 0.13065 0.07989 0.47744 0.13336 1.242424 0.09905 0.10537 0.05548 0.54059 0.07875 1.101795 0.05808 0.06822 0.03145 0.61976 0.03545 0.953336 0.04012 0.05167 0.02131 0.65409 0.01936 0.892407 0.03049 0.04240 0.01600 0.67283 0.01260 0.860618 0.02459 0.03644 0.01280 0.68462 0.00903 0.841369 0.02066 0.03224 0.01069 0.69275 0.00691 0.82847

10 0.01785 0.02911 0.00920 0.69874 0.00554 0.8192115 0.01087 0.02050 0.00554 0.71476 0.00269 0.7955620 0.00798 0.01640 0.00404 0.72215 0.00174 0.7852230 0.00535 0.01218 0.00270 0.72957 0.00102 0.77524

6.1. Special Case θ = ∞, a Preemptive-Loss System

We first consider an extreme case θ = ∞, in which customers who are pushedout of service leave the system immediately. This service discipline can bereferred to as preemptive-loss [7, p. 66]. We still assume the LCFS discipline.

In this special case, there are no customers in the waiting room. Therefore,the time Tk spent by an arbitrary customer in state k, 0 ≤ k ≤ m − 1, isequivalent to the first passage time from state k until service preemption andcompletion, which is considered in Section 2. From the results in Section 2.2,we simply have

T ∗k (s,Ab) =

h∗k(s)

h∗m(s); T ∗

k (s,Sr) =µ

s+ µ

[

1−h∗k(s)

h∗m(s)

]

,

T ∗k (s) = T ∗

k (s,Ab) + T ∗k (s,Sr) =

µ

s+ µ+

s

s+ µ·h∗k(s)

h∗m(s)

0 ≤ k ≤ m,

where the set of functions {h∗k(s); 0 ≤ k ≤ m} is given in Section 2.3. Theprobability that a customer in state k is lost is given by

Pk{Ab} = T ∗k (0,Ab) =

h∗k(0)

h∗m(0)=B(m,mρ)

B(k,mρ).

540 H. Takagi

Table 4: Numerical example for the initial and subsequent waiting time.

k E[W◦

k ,Ab] E[W•

k ,Ab] E[(W◦

k )2,Ab] E[(W•

k )2,Ab] E[W◦

kW•

k ,Ab]

5 0.26601 0.16485 0.20204 0.15808 0.026536 0.36131 0.09237 0.29026 0.08858 0.028617 0.40538 0.06011 0.33780 0.05764 0.026568 0.42943 0.04309 0.36708 0.04232 0.024009 0.44414 0.03299 0.38681 0.03164 0.0216910 0.45391 0.02646 0.40100 0.02538 0.0197215 0.47546 0.01282 0.43704 0.01229 0.0135820 0.48314 0.00832 0.45246 0.00798 0.0104930 0.48974 0.00486 0.46696 0.00466 0.00738

k E[W◦

k ,Sr] E[W•

k ,Sr] E[(W◦

k )2,Sr] E[(W•

k )2,Sr] E[W◦

kW•

k ,Sr]

5 0.01463 0.00906 0.00801 0.00677 0.001466 0.01577 0.00508 0.01069 0.00379 0.001577 0.01464 0.00330 0.01154 0.00247 0.001468 0.01323 0.00237 0.01170 0.00177 0.001319 0.01196 0.00181 0.01156 0.00135 0.00119

10 0.01087 0.00145 0.01129 0.00109 0.0010815 0.00749 0.00070 0.00975 0.00053 0.0007520 0.00579 0.00046 0.00853 0.00034 0.0005830 0.00407 0.00027 0.00693 0.00020 0.00041

6.2. Special Case θ = 0, in which All Customers are Patient

We next consider another extreme case θ = 0, in which all customers are verypatient in an M/M/m PR-LCFS queue. No customers abandon waiting. Thatis, all customers are served until completion. Therefore, Pk{Sr} = 1 for k ≥ 0.The same queue with FCFS discipline is an ordinary M/M/m queue.

The LST G∗k(s), k ≥ m, of the DF for the busy period Gk, considered in

Section 3.3, now satisfies the following set of recursive equations:

λG∗k+1(s)− (s + λ+ µ)G∗

k(s) +mµG∗k−1(s) = 0 k ≥ m.

We note that Gk is equivalent to the busy period started with k customers inan M/M/1 queue with mean service time 1/(mµ). The solution is given in theform

G∗k(s) = [G∗(s)]k−m+1 k ≥ m− 1,


Table 5: Numerical example for the number of service preemptions andresumptions.

k E[Jk | Ab] E[Kk | Ab] E[Jk | Sr] E[Kk | Sr] E[Jk] E[Kk]

0 1.61971 0.61971 0.17037 0.17037 0.91344 0.400741 1.61971 0.61971 0.20944 0.20944 1.00478 0.440822 1.61971 0.61971 0.27045 0.27045 1.11440 0.488913 1.61971 0.61971 0.37846 0.37846 1.24776 0.547424 1.61971 0.61971 0.61971 0.61971 1.41254 0.619715 0.61971 0.61971 0.61971 1.61971 1.41254 0.619716 0.33262 0.33262 0.61971 1.61971 0.61971 0.710607 0.21246 0.21246 0.61971 1.61971 0.22595 0.259098 0.15085 0.15085 0.61971 1.61971 0.16199 0.185759 0.11484 0.11484 0.61971 1.61971 0.12403 0.14222

10 0.09178 0.09178 0.61971 1.61971 0.09948 0.1140715 0.04411 0.04411 0.61971 1.61971 0.04818 0.0552520 0.02858 0.02858 0.61971 1.61971 0.03129 0.0358830 0.01665 0.01665 0.61971 1.61971 0.01826 0.02094

where

G∗(s) :=s+ λ+mµ−

√

(s+ λ+mµ)2 − 4mλµ

2λ

with G∗(0) = 1 and

G∗(0) = 1, G′(0) = −1

mµ(1− ρ), G′′(0) =

2

(mµ)2(1− ρ)3,

G′′′(0) = −6(1 + ρ)

(mµ)3(1− ρ)5.

For a customer being served in state k, 0 ≤ k ≤ m− 1, from the result inSection 4.2(1), we have

T ∗k (s, s

′) =µ

s′ + µ

{

1−h∗k(s

′)[1 −G∗(s)]

h∗m(s′)− h∗m−1(s′)G∗(s)

}

,

W∗k(s) = 1−

h∗k(0)[1 −G∗(s)]

h∗m(0)− h∗m−1(0)G∗(s)

; H∗k(s) =

µ

s+ µ,

T ∗k (s) =

µ

s+ µ

{

1−h∗k(s)[1−G∗(s)]

h∗m(s)− h∗m−1(s)G∗(s)

}

,

where the set of functions {h∗k(s); 0 ≤ k ≤ m} is given in Section 2.3.

542 H. Takagi

For a customer waiting in state k, k ≥ m, from the result in Section 4.2(2),we have

T ∗k (s, s

′) =µ

s′ + µ·[h∗m(s′)− h∗m−1(s

′)][G∗(s)]k−m+1

h∗m(s′)− h∗m−1(s′)G∗(s)

,

W∗k(s) =

[h∗m(0)− h∗m−1(0)][G∗(s)]k−m+1

h∗m(0) − h∗m−1(0)G∗(s)

; H∗k(s) =

µ

s+ µ,

T ∗k (s) =

µ

s+ µ·[h∗m(s)− h∗m−1(s)][G

∗(s)]k−m+1

h∗m(s)− h∗m−1(s)G∗(s)

.

We note that the time interval in which a customer is being served has the sameexponential distribution as the original service time requirement. However it isnot independent of the waiting time, because T ∗

k (s, s′) 6= W∗

k(s)H∗k(s

′) for k ≥ 0.

Appendix: Derivatives of function G∗

k(s) at s = θ

The function G∗k(s) is given in Section 3.3 for k ≥ m. We show the first and

second derivatives of G∗k(s) with respect to s at s = θ for k ≥ m and k = m

particularly. For the brevity of notation, let us use the convention

A := A(m,µ, θ, λ) =∞∑

i=0

ρi∏i

j=0(1 + jτ/m)ρ :=

λ

mµ,

I(i) := I(i,m, µ, θ) =i

∑

j=1

1

(1 + jτ/m)jτ/mτ :=

θ

µ, i ≥ 0.

The function ψi,k(x), i ≥ 0, k ≥ 0, is defined in Section 3.3.

(1) G′

k(θ) := [dG∗

k(s)/ds]s=θ, k ≥ m

mµG′

k(θ) =1

A

[

∞∑

i=1

i!(−τ/m)iψi,k−m(λ/θ)I(i)∏i+1

j=1(1 + jτ/m)

−∞∑

i=0

i!(−τ/m)iψi,k−m(λ/θ)

[1 + (i + 1)τ/m]∏i+1

j=1(1 + jτ/m)

]

−1

A2

[

∞∑

i=0


j=1(1 + jτ/m)

]

∞∑

i=1

ρiI(i)∏i

j=1(1 + jτ/m)

.


(2) G′

m(θ) := [dG∗

m(s)/ds]s=θ

mµG′

m(θ)

=1

A

[

∞∑

i=1

ρiI(i)∏i+1

j=1(1 + jτ/m)

−∞∑

i=0

ρi

[1+(i+1)τ/m]∏i+1

j=1(1+jτ/m)

]

−1

A2

[

∞∑

i=0

ρi∏i+1

j=1(1 + jτ/m)

]

∞∑

i=1

ρiI(i)∏i

j=1(1 + jτ/m)

.

(3) G′′

k(θ) := [d2G∗

k(s)/ds2]s=θ, k ≥ m

(mµ)2G′′

k(θ) =1

A

∞∑

i=0


j=1(1 + jτ/m)

×

[I(i)]2 − 2

i∑

j=1

1

(1 + jτ/m)2jτ/m−

i∑

j=1

1

[(1 + jτ/m)jτ/m]2

−2I(i)

1 + (i + 1)τ/m+

2

[1 + (i+ 1)τ/m]2

}

−1

A2

⟨

2

{

∞∑

i=0


j=1(1 + jτ/m)

[

I(i)−1

1 + (i+ 1)τ/m

]

}

×

[

∞∑

i=1

ρiI(i)]∏i

j=1(1 + jτ/m)

]

+

[

∞∑

i=0


j=1(1 + jτ/m)

]

×∞∑

i=1

ρi∏i

j=1(1 + jτ/m)

[I(i)]2 −i

∑

j=1

1 + 2jτ/m

[(1 + jτ/m)jτ/m]2

⟩

+2

A3

[

∞∑

i=0


j=1(1 + jτ/m)

][

∞∑

i=1

ρiI(i)∏i

j=1(1 + jτ/m)

]2

.

(4) G′′

m(θ) := [d2G∗

m(s)/ds2]s=θ

(mµ)2G′′

m(θ) =1

A

∞∑

i=0

ρi∏i+1

j=1(1 + jτ/m)

×

[I(i)]2 − 2i

∑

j=1

1

(1 + jτ/m)2jτ/m−

i∑

j=1

1

[(1 + jτ/m)jτ/m]2

−2I(i)

1 + (i+ 1)τ/m+

2

[1 + (i + 1)τ/m]2

}

−1

A2

⟨

2

{

∞∑

i=0

ρi∏i+1

j=1(1 + jτ/m)

[

I(i)−1

1 + (i + 1)τ/m

]

}

544 H. Takagi

×

[

∞∑

i=1

ρiI(i)∏i

j=1(1 + jτ/m)

]

+

[

∞∑

i=0

ρi∏i+1

j=1(1 + jτ/m)

]

×∞∑

i=1

ρi∏i

j=1(1 + jτ/m)

[I(i)]2 −i

∑

j=1

1 + 2jτ/m

[(1 + jτ/m)jτ/m]2

⟩

+2

A3

[

∞∑

i=0

ρi∏i+1

j=1(1 + jτ/m)

][

∞∑

i=1

ρiI(i)∏i

j=1(1 + jτ/m)

]2

.

Acknowledgments

The author is supported by the Grant-in-Aid for Scientific Research (C) No.26330354 from the Japan Society for the Promotion of Science (JSPS) in theacademic year 2016.

References

[1] L. Durr, A single-server priority queuing system with general holding times, Poisson in-put, and reverse-order-of-arrival queuing discipline. Operations Research, 17, No.2 (1969),351-358, doi: 10.1287/opre.17.2.351.

[2] A. Erdelyi, W. Magnus, F. Oberhettinger, and F. G. Tricomi (editors), Higher Transcen-

dental Functions, Volume I, McGrawHill, New York (1953).

[3] F. Iravani and B. Balcıoglu, On priority queues with impatient customers, Queueing

Systems, 58, No.4 (2008), 239-260, doi: 10.1007/s11134-008-9069-6.

[4] D. L. Jagerman, Differential Equations with Applications to Queues, Marcel Dekker, NewYork (2000).

[5] O. Jouini, Analysis of a last come first served queueing system with customer aban-donment, Computers & Operations Research, 39, No.12 (2012), 3040-3045, doi:

10.1016/j.cor.2012.03.009.

[6] O. Jouini and A. Roubos, On multiple priority multi-server queues with impa-tience, Journal of the Operational Research Society, 65, No.5 (2014), 616-632, doi:

10.1057/jors.2012.153.

[7] G. P. Klimow, Bedienungsprozesse, Birkhauser, Basel, Switzerland (1979).

[8] V. G. Kulkarni, Modeling and Analysis of Stochastic Systems, Chapman & Hall, BocaRaton, Florida (1995).

[9] A. Mandelbaum and S. Zeltyn, Service engineering in action: The Palm/Erlang-A queue,with applications to call centers. In: Advances in Services Innovations, Springer, Berlin(2007), 17-45.


[10] J. Riordan, Stochastic Service Systems, John Wiley & Sons, New York (1962).

[11] S. Subba Rao, Queuing with balking and reneging in M/G/1 systems, Metrika, 12, No.1(1967), 173-188, doi: 10.1007/BF02613493.

[12] H. Takagi, Waiting time in the M/M/m FCFS nonpreemptive priority queue with im-patient customers, International Journal of Pure and Applied Mathematics, 97, No.3(2014), 311-344, doi: 10.12732/ijpam.v97i3.5.

[13] H. Takagi, Waiting time in the M/M/m LCFS nonpreemptive priority queue with im-patient customers, Annals of Operations Research, 247, No.1 (2016), 257-289, doi:

10.1007/s10479-015-1876-7.

[14] H. Takagi, Times to service completion and abandonment in the M/M/m preemptiveLCFS queue with impatient customers, QTNA’ 2016, December 13–15, 2016, Wellington,New Zealand, doi: 10.1145/3016032.3016036.

[15] H. Takagi, Times until service completion and abandonment in an M/M/m preemptive-resume LCFS queue with impatient customers, to appear in Journal of Industrial and

Management Optimization.

[16] H. Takagi, Waiting and service time of a unique customer in an M/M/m preemptiveLCFS queue with impatient customers, presented at QTNA’ 2017, August 21–23, 2017,Qinhuangdao, China.

[17] H. M. Taylor and S. Karlin, An Introduction to Stochastic Modeling, Third edition, Aca-demic Press, San Diego, California (1998).

[18] W. Whitt, Engineering solution of a basic call-center model, Management Science, 51,No.2 (2005), 221-235, doi: 10.1287/mnsc.1040.0302.

[19] R. W. Wolff, Stochastic Modeling and the Theory of Queues, Prentice Hall, EnglewoodCliffs, New Jersey (1989).

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