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Research Papers

Bhavin Patel, Pravin Bhathawala / International Journal of Engineering Research and

Applications (IJERA) ISSN: 2248-9622 www.ijera.com

Vol. 2, Issue 4, July-August 2012, pp.776-778

776 | P a g e

M/M/c queueing model for bed-occupancy management

Bhavin Patel1 and Pravin Bhathawala

2

1Assistant Professor, Humanities Department, Sankalchand Patel College of Engineering, Visnagar, Gujarat,

India; 2 Professor & Head, Department of Mathematics, BIT, Baroda, Gujarat, India.

Abstract:

The present paper describes the

movement of patients in a hospital by using

queuing model with exponential arrival and

service time distributions. A queuing model is

used to determine the optimal number of beds in

order to improve hospital care. Also, we describe

a way of optimising the average cost per day by

balancing costs of empty beds against costs of

delayed patients.

Keywords: bed occupancy; Poisson distribution;

queuing theory

The theoretical model We consider a M/M/c queue in which the

number of beds is fixed. We assume that, patient

arrivals follow a Poisson process with arrival rate

and the service time follows Poisson distribution

with mean . Here, is the average length of stay

per patient.

The average number 𝑎 of arrivals during an

average length of stay is

a , known as the

offered load. From queuing theory, we see that the

probability that, j beds are occupied is given by

𝑃 𝑗 𝑜𝑐𝑐𝑢𝑝𝑖𝑒𝑑 𝑏𝑒𝑑𝑠 = 𝑃𝑗 =

𝑎𝑗

𝑗 !

𝑎𝑘

𝑘 !𝑐𝑘=0

,

𝑗 = 0,1,2, … . . , 𝑐. (1)

and 𝑃𝑗 = 0 , for 𝑗 > 𝑐.

The probability that, all 𝑐 beds are

occupied is given by

𝑃𝑐 =

𝑎𝑐

𝑐!

𝑎𝑘

𝑘!𝑐𝑘=0

From the above formula, we deduce that

the probability that, all 𝑐 beds are occupied or the

fraction of arrivals that is lost is given by

𝐵 𝑐, 𝑎 =𝑎𝑐

𝑐!

𝑎𝑘

𝑘 !𝑐𝑘=0

, (2)

called the Erlang’s loss formula.

The mean number of occupied beds is

given by

𝑎′ = 𝑎[1 − 𝐵 𝑐, 𝑎 ] (3)

which is also known as the carried load. The carried

load 𝑎′ equals that portion of the offered load 𝑎 that

is not lost from the system. The offered load 𝑎 is the

carried load 𝑎′, if the number of beds are infinite i.e.

when 𝑐 = ∞, 𝑎 = 𝑎′. The lost load (𝐿𝐿), that is, the

offered load 𝑎 that is lost from the system and is

given by

𝐿𝐿 = 𝑎𝐵(𝑐, 𝑎). (4)

We see that, the proportion of arrivals that is lost, is

the ratio of the lost load to the offered load, that is

𝐵 𝑐, 𝑎 =𝑎𝐵 (𝑐 ,𝑎)

𝑎.

We define the bed occupancy as

𝜌 =𝑎 ′

𝑐. (5)

Assuming that, the system is in steady state, we

must have 𝜌 ≤ 1.

Application of the model The loss model (optimising the number of

beds)

We consider the model where the arrival

rate is 25 patients per day and the mean

length of stay 10 days. In table-1, we present a

table of various system characteristics as a function

of the number of beds 𝑐 for a hospital. When we

have 100 beds, the probability that, the fraction of

arrivals that is lost i.e. lost demand probability, is

near to zero and the bed occupancy is %5.2 i.e. on

average there are 5.2 patients in a hospital. We can

see from this table that, the bed occupancy is

decreased as the number of beds increase. So, we

can conclude that, the more number of patients can

be admitted to the hospital with maximum of 100

beds.




777 | P a g e

Table-1 The main service features ( 25

patients per day, 10 days and the offered load

5.2

a )

Number

of beds

𝑐

Lost

demand

probability

𝐵(𝑐, 𝑎)(%)

Mean

number

of

patients

(𝐿)

Bed

occupancy

𝜌 (%)

𝜌 =𝑎′

𝑐

100 5.5*10-118

2.5 2.5

105 4.6*10-126

2.5 2.38

110 3.1*10-134

2.5 2.27

115 1.6*10-142

2.5 2.17

120 6.9*10-151

2.5 2.08

125 2.4*10-159

2.5 2

130 6.9*10-168

2.5 1.92

135 1.6*10-176

2.5 1.85

140 3.1*10-185

2.5 1.78

145 5.1*10-194

2.5 1.72

150 7.1*10-203

2.5 1.67

From the table, we wish to have at most 100 beds in

a hospital, because the lost demand probabilities

approach zero rapidly.

The cost model (optimising the average cost per unit

time)

Here, beds correspond to the inventory

where idle beds are the on-hand inventory and

occupied beds are unfilled orders. A patient arrival

corresponds to a demand and the subsequent length

of stay of the patient corresponds to waiting time for

replenishment. Patients who are turned away (i.e.

lost demand) because there are no empty beds,

corresponds to unsatisfied demands.

We consider ℎ as a holding cost per day for

each empty bed, ℎ>0 and 𝜋 as a fixed penalty cost

incurred for each patient that is turned away (i.e. lost

demand), 𝜋>0. We also consider 𝑝 as a profit per

patient per day. Therefore, the total cost to the

customer is the sum of variable costs (treatment),

fixed costs (holding costs), profit and penalty cost

(of lost demand). We observe that, the average

demand that is lost per unit time is equal to

),( acB , the average idle-bed inventory is equal

to )],(1[ acBac ( by (3)) and therefore, the

average service provider revenue per day is given by

)],(1[)]},(1[{),()( acBpaacBachacBcr

(6)

Now, we want to find 𝑐 to maximise 𝑟(𝑐).

This is the optimal number of beds where we

balance the number of empty beds against the

number of delayed patients. We note that, for a

public health service, 𝑝 = 0. We are interested in

maximising revenue 𝑟(𝑐) or minimising cost which

is given by

)]},(1[{),()( acBachacBcg (7)

Here, the cost minimisation means the optimal

balance between holding cost ℎ and penalty cost 𝜋.

Now, we say that, we are indifferent

between the use of the consecutive 𝑐 and 𝑐 + 1

when 𝑔 𝑐 = 𝑔(𝑐 + 1). From this equation, we

deduce that this equation is equivalent to 1

1)],1(),([

hacBacB

(8)

This equation shows that, the optimal choice of 𝑐

depends only upon the parameters

and

h

. It

means that, for a given value of 𝑐, we may evaluate

the indifference by means of a graph of h

versus

. This graph is known as the indifference curve,

which describes equation (6). When the curves for

different values of 𝑐 are close together, we say that,

we are indifferent to the choice of h

for given

values of and . This means that, for such

choices of 𝑐, the optimal number of beds is not very

dependent on the ratio h

.

Here, we assume that the total cost per

patient per day is 200.Rs , where 100.Rs are

incurred with respect to the bed and 100.Rs with

respect to the treatment. We also assume that, the

holding cost 100.Rsh per day and the penalty

cost as o % of the total cost of turning away the

patient. (Because the lost demand probabilities are

near to zero.)

Table-2 The values of the average cost per unit time

)(. cgRs

Beds

𝑐

𝜋

ℎ= 20

𝑔1(𝑐)

𝜋

ℎ= 30

𝑔2(𝑐)

𝜋

ℎ= 40

𝑔3(𝑐)

100 9750 9750 9750

105 10250 10250 10250

110 10750 10750 10750

115 11250 11250 11250

120 11750 11750 11750

125 12250 12250 12250

130 12750 12750 12750

135 13250 13250 13250

140 13750 13750 13750

145 14250 14250 14250

150 14750 14750 14750

This table shows that, if we increase the

penalty cost to holding cost ratio h

three times




778 | P a g e

from 20 to 40, then the number of beds needed for

the minimal average cost is at most 100. Here, the

average cost remains constant as the ratio h

is

increased from 20 to 40. This means that the ratio

h

has no significant influence on the optimal

number of beds. Here, we note that we do not draw

the indifference curves. We consider the tabular

data.

Conclusions

From table-1, we can see that, for 100

beds, the lost demand probability is near to zero and

the bed occupancy is %5.2 . Also, the bed

occupancy is decreased as the number of beds

increase.

From table-2, we can see that, for 100

beds, the average cost per unit time is minimum and

it increases rapidly as the number of beds increase.

Hence, finally we conclude that, there

should be at most 100 beds in a hospital with

arrival rate 25 patients per day and service

rate 10 days per patient.

References 1) Cooper RB (1972). Introduction to

Queuing Theory. McMillan: New York.

2) Cox DR (1962). Renewal Theory.

Methuen: London.

3) Worthington DJ (1987). Queuing models

for hospital waiting lists. J Opl Res Soc 38:

413-422.

4) Worthington DJ (1991). Hospital waiting

list management models. J Opl Res Soc 42:

833-843.

5) Worthington D and Wall A (1999). Using

the discrete time modelling approach to

evaluate the time-dependent behaviour of

queuing systems. J Opl Res Soc 50: 777-

888.

6) Tijms HC (1986). Stochastic Modelling

and Analysis. A Computational Approach.

Wiley: Chichester.

7) Stevenson WJ (1996).

Production/Operations Management, 6th

edn. Irwin, McGraw-Hill, USA.

8) Vinicchayakul, R (2000). Costing care in

geriatric medicine. MSc dissertation,

University of London.

9) McClean SI and Millard PH (1993).

Modelling in-patient bed usage behaviour

in a department of geriatric medicine.

Meth Inform Med 32: 79-81.

APPLICATION OF QUEUING THEORY TO CRICKET TEST MATCH

BHAVIN PATEL1 & PRAVIN BHATHAWALA

2

1Assistant Professor, Humanities Department, Sankalchand Patel College of Engineering, Visnagar, Gujarat, India

2Professor & Head (Retd), Department of Mathematics, VNSGU, Surat, Gujarat, India

ABSTRACT

In a cricket test match, two openers open the innings. That is, one pair is batting and the next batsman is waiting

in a dressing room (i.e. waiting in a queue). We can derive an arrival rate and service rate from observation. So, we can

apply the queuing theory to a cricket test match. As a result, we conclude that the result of the cricket test match is either a

win (by any one of the two teams) or a draw with respective probabilities.

KEYWORDS: Cricket Test Match, Queue, Queuing Theory, M/M/1 Queuing Model

HISTORY OF A TEST CRICKET

Test cricket is played between international cricket teams who are full members of the International Cricket

Council (ICC ). Test matches consist of two innings per team, having no limit in their number of overs. The duration of

tests has varied through test history, ranging from three days to timeless matches. Currently, the duration of tests is limited

to five days. The earliest match now recognised as a test was played between England and Australia in March 1877, since

then there have been over tests played by teams. The frequency of tests has steadily increased because of the

increase in the number of Test-playing countries.

INTRODUCTION

At the start of the cricket test match, two opening batsmen i.e. one pair of batsmen, open the innings. This pair of

batsmen is considered as a customer being served by a single server i.e. cricket pitch. The one down batsman i.e. next

batsman, is waiting in a dressing room i.e. this batsman is considered as a waiting customer in a queue.

From the innings of this two batsmen, we can derive the arrival rate and the service rate. We assume that, the

service rate is slightly higher than the arrival rate. That means that, the time for which one pair is batting, is the service

time for one pair of batsmen. And the time at which the incoming batsman occupies the pitch, is the arrival time for

incoming batsman ( i.e. the arrival time for a waiting batsman or customer ). So, the service rate is slightly higher than the

arrival rate. For this reason, we consider the utilization factor as . We observe that, the length of the queue is

customer. We also observe that, the average number of customers in the system is customers (i.e. pair is playing and

cricketer is waiting in a dressing room). We consider the following assumptions:

ASSUMPTIONS

We assume that, at the start of the innings the one down batsman has already padded up and is waiting for his

turn.

It is assumed that, no batsman is retired hurt or retired absent in the test match (or in each innings).

It is assumed that, there is no rain interruption during the test match.

International Journal of Mathematics and

Computer Applications Research (IJMCAR)

ISSN(P): 2249-6955; ISSN(E): 2249-8060

Vol. 3, Issue 5, Dec 2013, 71-76

© TJPRC Pvt. Ltd.

72 Bhavin Patel & Pravin Bhathawala

We assume that, the cricket stadium is a queuing system having components: arrival of cricketers, service

mechanism, waiting of cricketers and departures.

This queuing system is assumed to be in a steady-state.

We observe that, only one pair is served by a cricket pitch at a time. Other batsmen are waiting in a queue.

Thus, the queuing model that best illustrates this cricketing model, is queuing model.

QUEUING MODELS

There are two models for the single server case . The first model has no limit on the maximum number

in the system and the second model has finite limit on the maximum number. Both models assume an infinite-capacity

source. Arrivals occur at the rate customers per unit time and the service rate is customers per unit time.

Our cricket model is the second model that has finite limit on the maximum number and an infinite-capacity

source. In this model, the maximum number of customers is .

Letting , the probability of having customers in the system is

,

We find the probability of having customers in the system, using the identity

.

Assuming, , the geometric series will have the finite sum . Thus,

, provided .

Thus, the general formula for is given by

, .

The mathematical derivation of imposes the condition or . If , the geometric series will

not converge and the steady-state probabilities will not exist. Because unless the service rate is larger than the arrival rate,

queue length will continually increase and no steady-state can be reached.

STEADY-STATE MEASURES OF PERFORMANCE

The most commonly used measures of performance in a queuing situation are

Expected number of customers in system

Expected number of customers in queue

Expected waiting time in system

Expected waiting time in queue

Application of Queuing Theory to Cricket Test Match 73

Expected number of busy servers

The relationship between and (also and ) is known as Little’s formula. It is given as

.

It describes that, as the waiting time in system increases (or decreases), expected number of customers in system

increases (or decreases) and as the waiting time in queue increases (or decreases), expected number of customers in queue

increases (or decreases) and vice-versa. This is exactly the case in cricketing model.

CALCULATION OF PROBABILITIES

As we know the maximum number of customers (batsmen) in this model is and there is no limit in the number

of over they play, the test match will have a result (a win by any one of the two teams), if more than customers are

served. That means, if customers are served, the test match will have a result. In the context of cricketing model, this

means that, if wickets are fallen, the test match will have a result. Also, the test match will be drawn, if less than or

equal to customers are served. This means that, if less than or equal to wickets are fallen, the test match will be

drawn.

Thus, the probability of the test match resulting in a draw = P (less than or equal to 39 customers in the system)

=

=

=

The remaining probability will be of the test match resulting in a win or loss. That is, the probability of the test

match resulting a win by any one of the two teams = P (more than customers in the system) =

=

=

=

In other words, we can say that, test matches result in a draw. And test matches result in a

win (by any one of the two teams). That means that, out of test matches test matches result in a draw and

test matches result in a win (by any one of the two teams).

ACTUAL OBSERVATION

In some situation, we have observed that, there are only or or wickets had fallen during a cricket test

match, still the test match had resulted in a win or loss (a win by one team and a loss by other team). In this case, the

74 Bhavin Patel & Pravin Bhathawala

probabilities are varied but they are nearer to the probabilities, we have already obtained by calculation using queuing

theory

In other situation, the team having won the toss or batting first, sets a big total, in their 1st innings. The other team

in their 1st innings, gets following on by a captain of the opposite team. And starts their 2

nd innings immediately, after their

1st innings and got out inside the lead of the opposite team. In this case, the number of wickets that would be falling is

nearer to or , still the team having a big total, wins a cricket test match. For this case also, we get the nearer

probabilities.

In some other situation, the team having batted first, post a reasonable total around or in their 1st

innings. The other team in their 1st innings, sets a big total, having reasonable lead. The opposite team in their 2

nd innings

got out inside this lead. So, the team having a big total, wins a cricket test match. In this case also, the number of wickets

that would be falling is nearer to or . Again, we get the nearer probabilities.

Thus, in general, we can say that, the probability of drawn test matches is and the probability of

resulted test matches is . These probabilities can be verified from the actual test records as follows:

Actual Test Records (As On December 2012)

Table 1

Teams Matches Won Lost Tied Drawn Drawn %

England 930 331 268 0 331 35.59

Australia 748 351 195 2 200 26.74

West Indies 488 158 162 1 167 34.22

India 468 115 149 1 203 43.38

New Zealand 377 72 154 0 151 40.05

Pakistan 370 115 101 0 154 41.62

South Africa 372 132 126 0 114 30.65

Sri Lanka 219 65 79 0 75 34.24

Zimbabwe 87 9 52 0 26 29.88

Bangladesh 75 3 65 0 7 9.33

Total 4134 1351 1351 4 1428 34.54

Excluding the ICC World XI test against Australia.

Including tied test matches.

From this table, we observe that, of test matches have been drawn, which is almost equal to .

We see that, there is no much difference between these percentages. Thus, we can verify our results.

Table 2

Total Matches

Played

Number of Wins

( By Any One of

the Two Teams )

Number of Matches

Drawn

Actual Probability

of Drawn Matches

CONCLUSIONS

As a result of this research paper, we conclude that, the probability of a cricket test match, resulting in a draw is

. And the probability of a cricket test match, resulting in a win (by any one of the two teams) is .

Application of Queuing Theory to Cricket Test Match 75

In other words, out of cricket test matches, that would be played, almost test matches result in a draw and almost

test matches result in a win (by any one of the two teams)

The difference between the actual probability and the calculated probability is

which is negligible. So, we can say that, the calculated probability gives the idea of the actual probability of a drawn

cricket test match. Thus, queuing theory can be applied to a cricket test match to calculate the required probability.

The probability of a cricket test match, resulting in a draw is which is approximately equal to

i.e. it is approximately equal to . The probability of a cricket test match, resulting in a win (by any one of the

two teams) is which is approximately equal to i.e. it is approximately equal to . That means, the

sum of the probabilities of a drawn test match and a resulted test match is i.e.

We also conclude that, the utilization factor of the cricket pitch is . That is, the pitch is occupied by one

pair of batsmen for of the total time of the test match. Further, the probability of zero customers in the system is

. That means that, the pitch is not occupied by any pair of batsmen for of the total

time of the test match.

REFERENCES

1. H.A. Taha, Operations Research-An Introduction. 8th

Edition, ISBN 0131889230. Pearson Education, 2007.

2. J.D.C. Little, “A Proof for the Queuing Formula: ”, Operations Research, vol. 9(3), 1961, pp. 383-387,

doi:10.2307/167570.

3. Cooper RB (1972). Introduction to Queuing Theory. McMillan: New York.

4. Tijms HC (1986). Stochastic Modelling and Analysis. A Computational Approach. Wiley: Chichester.

5. K. Rust, “Using Little’s Law to Estimate Cycle Time and Cost”, Proceedings of the 2008 Winter Simulation

Conference, IEEE Press, Dec. 2008, doi:10.1109/WSC.2008.4736323.

6. Worthington D and Wall A (1999). Using the discrete time modelling approach to evaluate the time-dependent

behaviour of queuing systems. J Opl Res Soc 50: 777-888.

7. Stevenson WJ (1996). Production/Operations Management, 6th

edn. Irwin, McGraw-Hill, USA.

IOSR Journal of Mathematics (IOSR-JM)

e-ISSN: 2278-3008, p-ISSN:2319-7676. Volume 10, Issue 2 Ver. III (Mar-Apr. 2014), PP 33-35

www.iosrjournals.org

www.iosrjournals.org 33 | Page

An Application Of Queuing Theory To The TV Show Kaun

Banega Crorepati ( KBC )

Bhavin Patel1 and Pravin Bhathawala

2

1Assistant Professor, Humanities Department, Sankalchand Patel College of Engineering, Visnagar, Gujarat,

India; 2Professor of Mathematics,Auro University, Opp. ONGC, Hajira Road, Surat, Gujarat, India.

Abstract: The present paper explains the application of queuing theory to one episode of the TV show Kaun

Banega Crorepati during a week. By analysis of one episode, we can understand the general scenario of the

KBC. In one episode of this show, contestants or participants wait for their turn to go on the hot seat. At one

time only one contestant can seat on the hot seat and other contestants are waiting in a queue. So, we can apply

queuing theory to the TV show Kaun Banega Crorepati. We also seek to find the probability of one contestant to

be selected for the hot seat and the expectation of contestants on each day of a week.

Keywords: Queue; Queuing Theory; Kaun Banega Crorepati (KBC); Hot seat.

I. Introduction There are three episodes of KBC during each week from Friday to Sunday. In one episode of the TV

show Kaun Banega Crorepati, contestants or participants are allowed to participate in the show. Depending

on the Fastest Finger First (FFF) contest, one participant can occupy the hot seat in front of Amitabh Bachchan.

From the durarion of one participant on the hot seat, we can derive the service rate of the hot seat. If we assume

that, the time span of FFF contest is negligible, the arrival rate is same as the service rate.

The first contestant is considered as a customer being served by a single server i.e. the hot seat. The

other contestants are considered as the waiting customers in a queue. So, we can apply queuing theory to KBC.

We assume that, the arrival rate is same as the service rate for the hot seat. For this reason, we consider the

utilization factor for the hot seat as . Because of the single server, that is, the single hot seat, queuing

model can be useful for analysing the TV show KBC. M stands for Markovian arrival time and Markovian

service time distributions. The service time for one customer is random and therefore, the arrival time is also

random.

Queuing Model: In this model, the customers are served by the single server i.e. the hot seat. The

maximum number of customers in the system is . This model is the single server model with finite limit .

is the maximum number of customers in the system (maximum queue length ). In the context of KBC,

during each week. That means, there are customers to be served by the hot seat in the episodes of

one week. We assume that, the arrival rate is λ customers per unit time and the service rate is μ customers per

unit time.

During three episodes of KBC in one week, if contestants are served, then no more contestants are

allowed. That is, when the number of customers reaches , no more customers are allowed in the system. Thus,

we have

{

.

.

Setting

, the probability of customers in the system is

{

.

The value of , the probability of customers in the system, is determined by the equation,

∑ , which gives

( )

{

.

Hence, the probability of customers in the system is

{

( )

.

An Application Of Queuing Theory To The TV Show Kaun Banega Crorepati ( KBC )


The value of

need not be less than in this case, as the number of customers has limit . In the

context of KBC, we take .

The expected number of customers in the system is determined by

∑

( )

( )

* ( )

+

.

This means that, the expected number of customers ( contestants ) in the three episodes of KBC during

each week is,

.

The relationship between and ( also and ) is known as Little’s formula and is given as

and .

The parameter is the effective arrival rate at the system. When all arriving customers are allowed in

the system, it is equal to the arrival rate .If some customers can not be allowed in the system, then . In

this case, is the rate at which, the customers are lost i.e. can not be allowed in the system. So, an

arriving customer may enter the system or will be lost with rates or , which means that,

. A customer will be lost from the system, if customers are already in the system. This means that, the

proportion of customers that will be lost from the system is . Thus,

.

( ).

In the context of KBC, (

) (

) (

) .

.

Obviously, , since .

By definition, (

)=(

) (

).

.

Multiplying this formula by , we can determine relationship between and , using Little’s formula, that

is

.

In the context of KBC,

(

)

.

That is, the expected number of customers ( or contestants ) in the queue are . That means that,

contestants are waiting in the queue actually, in a week.

Sample Space: During three episodes of KBC, everytime onecontestant is chosen from the group of

contestants. We name these contestants as . Any one contestant among these is chosen

through the contest of FFF. In this event, the sample space is * +. In this event, the probability

of each contestant to be selected is same and is

( )

, .

Therefore, the expectation of contestants or the expected number of contestants in each week is

( ) ∑ ( ) (

) (

) (

)

( )

(

)

.

An Application Of Queuing Theory To The TV Show Kaun Banega Crorepati ( KBC )


II. Calculation of probabilities:

On a first day of a week i.e. on Friday, the probability of each contestant to be selected is

and

the expected number of contestants to be served by the hot seat is out of contestants during a week. We

assume that, on average contestants are served by the hot seat on day one.

On a second day of the week i.e. on Saturday, the probability of each contestant to be selected is

because there are contestants remaining on day two. Therefore, the expected number of contestants to

be served on second day of the week is

( ) ∑ ( ) (

) (

) (

)

( )

(

)

.

Again, we assume that, on average contestants are served by the hot seat on day two. On a third day

of the week i.e. on Sunday, the probability of each contestant to be selected is

because there are

contestants remaining on day three. Therefore, the expected number of contestants to be served on third day of

the week is

( ) ∑ ( ) (

) (

) (

)

( )

(

)

.

III. Conclusion

We conclude that, theprobability of each contestant on first day is

and the expectation is

contestants for a week of three days. The probability of each contestant on second day is


contestants for two days of a week. The probability of each contestant on third day is


contestants for one day of a week. So, we can say that, as the probability on each day increases, the

expectation also increases.

References: [1]. J.D.C. Little, “A Proof for the Queuing Formula: ”, Operations Research, vol. 9(3), 1961, pp. 383-387, doi:10.2307/167570.

[2]. Cooper RB (1972). Introduction to Queuing Theory. McMillan: New York. [3]. Tijms HC (1986). Stochastic Modelling and Analysis. A Computational Approach. Wiley: Chichester.

[4]. K. Rust, “Using Little’s Law to Estimate Cycle Time and Cost”, Proceedings of the 2008 Winter Simulation Conference, IEEE

Press, Dec. 2008, doi:10.1109/WSC.2008.4736323. [5]. Worthington D and Wall A (1999). Using the discrete time modelling approach to evaluate the time-dependent behaviour of

queuing systems. J Opl Res Soc50: 777-888.

[6]. H.A. Taha, Operations Research-An Introduction. 8th Edition, ISBN 0131889230. Pearson Education, 2007.

International Journal of Scientific and Innovative Mathematical Research (IJSIMR)

Volume 2, Issue 5, May 2014, PP 474-480

ISSN 2347-307X (Print) & ISSN 2347-3142 (Online)

www.arcjournals.org

©ARC Page | 474

Application of Queuing Theory to ICC One Day Internationals

Bhavin Patel

Assistant Professor, Humanities Department,

Sankalchand Patel College of Engineering,

Visnagar, Gujarat, India

[email protected]

Pravin Bhathawala

Professor of Mathematics, Auro University,

Opp. ONGC, Hajira Road,

Surat, Gujarat, India

[email protected]

Abstract: In the present paper, we seek to apply queuing theory model to one day internationals( ODIs ).

In ODIs, one pair of batsmen open the innings. This pair of batsmen is considered as a customer being

served by a cricket pitch i.e. a single server. The one down batsman is considered as a waiting customer.

The utilization factor of a server can be obtained by observing the batting of a pair of batsmen in ODIs. As

a result we determine the probabilities of a tied ODIs including no result ( NR ) and resulted ODIs.

Keywords: Cricket, ODIs, Batsman, Pair of batsmen, Queuing theory.

1. INTRODUCTION

History of ODI Cricket: One day international ( ODI ) cricket is played between international

cricket teams who are full members of the International Cricket Council ( ICC ) as well as top six

Associate and Affiliate members. Unlike Test matches, ODIs consist of one innings per team,

having a limit in the number of overs. The limit of overs is currently overs per innings,

although in the past this has been or overs. ODI cricket is List-A cricket, so statistics and

records set in ODI matches also count toward List-A records. The earliest match now recognised

as an ODI was played between England and Australia in January , since then there have

been over ODIs played by teams. The frequency of matches has steadily increased

because of the increase in the number of ODI-playing countries.

At the start of the 1st innings of ODI, one pair of batsmen open the innings. The one down

batsman is waiting in a queue i.e. in a dressing room. The cricket pitch is considered as a single

server, which serves the customers ( cricketers ). From the innings of a pair of batsmen, we can

find the service rate and arrival rate. And hence, the utilization factor of the cricket pitch. We can

consider the time for which one pair is batting as the service time. At the time of falling a wicket

from among the pair of batsmen, the new batsman arrives. So the arrival time is same as the

service time. That means that, the service rate and arrival rate are same. Thus, we can consider the

utilization factor of the cricket pitch as .

Assumptions

We assume that, at the start of the 1st innings, the one down batsman has already padded

up and is waiting for his turn.

It is assumed that, no batsman is retired hurt or retired absent, in each innings.

We assume that, there is no rain during ODI.

2. ODI Model

This model is the single server model which assumes a finite system limit . is the maximum

number of customers in the system. (Maximum queue length .). In the context of ODI

model in each innings. That means, there are customers to be served or wickets to

be taken in each innings. Arrivals occur at the rate customers per unit time and the service rate

is customers per unit time.

When the number of customers in the system reaches , no more arrivals are allowed in the

system. Thus, we have,

mailto:[email protected]

mailto:[email protected]

Bhavin Patel & Pravin Bhathawala

International Journal of Scientific and Innovative Mathematical Research (IJSIMR) Page 475

.

.

Letting, , the probability of customers in the system is

.

The value of , the probability of customers in the system, is determined by the equation,

, which gives

.

Hence, the probability of customers in the system is

.

The value of need not be less than in this case, as the number of customers has limit .

In the context of ODI model, we let .


.

This suggests that, the expected number of customers ( wickets ) in each innings is

.

The relationship between and ( also and ) is known as Little’s formula and is given

as

and .

The parameter is the effective arrival rate at the system. It is equal to the arrival rate , when

all arriving customers can join the system. If some customers can not join because the system is

full then .

In the context of ODI model, if , then the effective arrival rate , rather than the arrival

rate , is the rate that matters. The effective arrival rate can be computed by observing the

schematic diagram given below, where customers arrive from the source at the rate customers

per hour.



Relationship between , and

An arriving customer may enter the system or will be lost with rates or , which means

that,

.

A customer will be lost from the system, if customers are already in the system. This means

that, the proportion of customers that will not be able to enter the system is . Thus,

.

.

In the context of ODI model, .

.

Obviously, , since .

By definition, = .

.

Multiplying this formula by , we can determine relationship between and , using

Little’s formula, that is

.

In the context of ODI model,

.

That is, the expected number of customers ( wickets ) in the queue . That means that,

almost batsmen are waiting in the queue in each innings.

By definition, the difference between the average number in the system and the average

number in the queue must be equal to the average number of busy servers . Therefore, we

have,

.



Also, it follows that, .

In the context of ODI model, the average number of busy servers is

. That means that, busy server.

It follows that, . That means that, is the

utilization factor.

3. CALCULATION OF PROBABILITIES

Before the start of the ODI match, a coin is tossed to decide which team will bat first. We define

the events of batting in the 1st innings and batting in the 2

nd innings.

Let be the event that, the team bat in the 1st innings and be the event that, the team bat in

the 2nd

innings.

.

Also, we define the event that, the 1st innings is completed, irrespective of the number of wickets

falling in the innings and the events of team having 1st innings and 2

nd innings win.

Let be the event that, the 1st innings is completed, irrespective of the number of wickets falling

in the innings.

( )

.

If there are less than wickets falling in the 1st innings, then also is taken as , since it is

irrespective of the number of wickets falling.

Let be the event that, team having 1st innings win and be the event that, team having 2

nd

innings win.

Then, the probability of team having 2nd

innings win is

( )

=

.

The probability of team having 1st innings win is

.

Now, we define the event that, the ODI match is tied or have no result ( NR ).

Let be the event that, the ODI match is resulting in a tie or having no result ( NR ). Then, the

probability of tied ODI match is



.

We observe that, the total probability is . That is, the sum of probabilities of team having 1st

innings win, team having 2nd

innings win and a tied match including no result (NR), is . Hence,

the probability of winning an ODI match by any one of the two teams is

.

In other words, we can say that, ODIs result in a tie or having no result (NR). And

ODIs result in a win ( by any one of the two teams ). That means that, out of ODIs

ODIs result in a tie or having no result ( NR ) and ODIs result in a win ( by any one of

the two teams ).

Also, the probability of zero batsmen on the pitch in single innings is , that is, the

proportion of zero batsmen in single innings is .

Actual ODI Records: ( As on January-2013 )

Teams Matches Won Lost Tied NR Tied+NR

%

England 595 290 278 7 20 4.54

Australia 811 500 275 9 27 4.44

West Indies 690 356 304 6 24 4.35

India 817 405 371 6 35 5.02

New Zealand 635 272 325 5 33 5.98

Pakistan 784 421 340 6 17 2.93

South Africa 482 299 165 5 13 3.73

Sri Lanka 681 318 330 4 29 4.85

Zimbabwe 410 107 289 5 9 3.41

Bangladesh 267 75 190 0 2 0.75

Kenya 148 41 102 0 5 3.38

Ireland 74 34 35 1 4 6.76

Total 6394 3061 3061 54 218 4.25

Excluding the records of teams, whose span for ODIs is either one, two, four or few years

or having no tied matches, like USA, UAE, Namibia, Honk Kong, Africa XI or Asia XI.

(Excluding the records of teams having very short duration of international cricket.)

Including the matches which are abandoned without a ball being bowled.

From this table, we observe that, of ODIs have been tied or having no result (NR), which

is almost equal to . We see that, there is no much difference between these percentages.

Thus, we can verify our results.



Table-1

Total ODIs

Played

Number of Wins

( by any one of the two

teams )

Number of Tied ODIs

including NR

Actual Probability of

Tied+NR

ODIs

At the start of the ODI match, we have no knowledge of a tied match or an ODI match having no

result. That’s why we consider two events that the team batting first win and the team batting

second win, with their probabilities and respectively. The probability of team batting first

and team batting second is each.

The expected value of a variable is given by . Therefore, the expected

probability of the two events is

.

We now verify this probability by actual record of ODI matches as follows:

Actual ODI Records: (As on 7 June-2013)

We consider the top ten ODI teams in the world only.

Teams Matches Won Lost Tied NR Won %

Australia 453 282 152 4 15 62.25

Bangladesh 134 31 103 0 0 23.13

England 288 122 155 3 8 42.36

India 390 177 191 4 18 45.38

New Zealand 323 133 177 2 11 41.18

Pakistan 422 226 177 4 15 53.55

South Africa 236 146 82 2 6 61.86

Sri Lanka 337 159 162 2 14 47.18

West Indies 311 151 142 3 15 48.55

Zimbabwe 210 57 148 3 2 27.14

Total 3104 1484 1489 27 104 47.81

From this table, we observe that, of ODIs have been won by team batting first, if we

consider the top ten ODI teams in the world. The expected percentage is . We see that,

there is no much difference between these percentages. Thus, we can verify our results.

Table-2

Total ODIs

Played

Number of Wins

by team batting

first

Number of Wins by

team batting second

Actual Probability

of team batting first

Actual Probability

of team batting

second

3104 1484 1489 0.4781 0.4797



From the above table, we can see that the expected probability match with the actual probabilities.

4. CONCLUSION

As a result, we conclude that, the probability of the ODI match resulting in a win by any one of

the two teams, is . That means that, out of ODI matches, ODIs result in a win by any one

of the two teams. The probability of the ODI match resulting in a tie or having no result (NR), is

. That means that, out of ODI matches, ODI results in a tie or have no result (NR). In other

words, we can say that, ODIs result in a tie or having no result (NR). And ODIs

result in a win ( by any one of the two teams ).

Also, the expected probability of the team batting first win or the team batting second win is

. That means that, the expected percentage of the team batting first win or the team batting

second win is .

Also, the probability of zero batsmen on the pitch in single innings is . That means that, out of

innings, in innings there is no batsman to be served by a pitch.

REFERENCES

[1] Rust K., “Using Little’s Law to Estimate Cycle Time and Cost”, Proceedings of the 2008

Winter Simulation Conference, IEEE Press, Dec. 2008, doi:10.1109/WSC.2008.4736323.

[2] Taha H.A., Operations Research-An Introduction. 8th Edition, ISBN 0131889230. Pearson

Education, 2007.

[3] Worthington D and Wall A (1999). Using the discrete time modelling approach to evaluate

the time-dependent behaviour of queuing systems. J Opl Res Soc 50: 777-888.

[4] Stevenson WJ (1996). Production/Operations Management, 6th edn. Irwin, McGraw-Hill,

USA.

[5] Tijms HC (1986). Stochastic Modelling and Analysis. A Computational Approach. Wiley:

Chichester.

[6] Cooper RB (1972). Introduction to Queuing Theory. McMillan: New York.

[7] Little J.D.C., “A Proof for the Queuing Formula: ”, Operations Research, vol. 9(3),

1961, pp. 383-387, doi:10.2307/167570.

AUTHORS’ BIOGRAPHY

Prof. Bhavin Patel is working as an assistant professor in Mathematics in

Sankalchand Patel College of Engineering, Visnagar since June-2005. He has

joined this institute as a lecturer in Mathematics on 21st June-2005. He is now

working as HOD of Applied Science & Humanities department. He completed

his M.Sc. in the year 2003 with Pure Mathematics from North Gujarat

University, Patan. He also completed M.Phil. in Mathematics in the year 2005

from Gujarat University, Ahmedabad. He is pursuing Ph.D. in Mathematics in

the subject of Operations Research from Kadi Sarva Vishwavidyalaya,

Gandhinagar.

Prof (Dr.) Pravin Bhathawala, Professor & Program Director, School of IT,

Auro university, Surat. He has 35 yrs of research & teaching experience and

has published 50 research papers in national and International journals. He has

produce around 35 Ph.D. and 20 M.Phil. His area of Research is Industrial and

Applied Mathematics together with Bio and Bio Medical Mathematics.

INTERNATIONAL JOURNAL OF MATHEMATICAL SCIENCES AND APPLICATIONSVolume 5 • Number 1 • January-June 2015

ICC WORLD TWENTY20-2014 ( WORLD CUP-2014 )-A CASE STUDY

Bhavin Patel1 and Pravin Bhathawala2

1Assistant Professor, Humanities Department, Sankalchand Patel College of Engineering, Visnagar, Gujarat India2Professor of Mathematics, Auro University, Opp. ONGC, Hajira Road, Surat, Gujarat,India

Abstract: The present paper describes the phenomena of the Twenty20 matches. In Twenty20 match, two openingbatsmen open the innings. From the time duration for which they are playing, we can find the service rate and thearrival rate for the pitch which serves the batsmen. Other batsmen are waiting for their turn. So, we apply queuingmodel M/M/1 to Twenty20 matches and determine the probability of win by first innings and second innings inTwenty20 match.

Keywords: Twenty20 match, queue, queuing model, game of cricket

HISTORY OF TWENTY-20 CRICKET:

Twenty 20 cricket, often abbreviated to T20, is a form of cricket originally introduced in England by the Englandand Wales Cricket Board (ECB) in 2003. A Twenty 20 game of cricket involves two teams, each has a singleinnings, batting for a maximum of 20 overs.

A Twenty20 game of cricket is completed in about three hours, with each innings lasting around 75 – 90minutes (with a 10–20 minutes interval ) thus bringing the popularity to the game of cricket. It was introduced toattract the spectators at the ground and viewers on tele-vision and as such it has been very successful game ofcricket. Since its inception the game has spread around the cricket world. It is played between the internationalteams that play game of cricket. On most international tours there is at least one Twenty20 match. The inauguralICC World Twenty 20 was played in South-Africa in 2007 with India winning by five runs against Pakistan in thefinal. Currently, Sri-lanka are the reigning champions after winning the ICC World Twenty 20-2014.

INTRODUCTION

As in the other two formats, Tests and ODIs, in Twenty20 match, two opening batsmen open the innings in eachinnings. This pair of batsmen is served by a single server i.e. cricket pitch. During the time period for which thesetwo are playing, the one down batsman is waiting in a dressing room i.e. waiting in a queue. By same way, otherbatsmen are also waiting for their turn. From the duration of time for opening batsmen, we can derive the servicerate and the arrival rate for the cricket pitch. The time for which one pair of batsmen is batting, is taken as theservice time. As soon as, the wicket falls, the new batsman arrives at the pitch. So, the arrival time of the waitingcustomers (cricketers) is same as the service time. The customers are being served by a single server i.e. the battingpitch. The service time is random, so the arrival time is also random. Thus, we can apply the M / M /1 queuingmodel to the Twenty 20 match. The utilization factor for the pitch is taken as � = 1.

M / M /1QUEUING MODEL

In this model, there is a limit on the number in the system (maximum queue length = N – 1). Example include a onelane drive-in window in a fast-food restaurant and an innings in a Twenty20 match in cricket.

8 BHAVIN PATEL AND PRAVIN BHATHAWALA

In these two examples, when the number of customers in the system is reached to N, no more arrivals areallowed in the system. Thus, we have,

, 0,1, 2, ..., 1

0, , 1, ... .... .n

n N

n N N

� � ��

, 0,1, 2, ... ....n n� � � �

Using ,�

� �� the probability of n customers in the system is

0 ,

0,

n

n

p n Np

n N

��

��

The value of is determined from the equation 0

1,nnp

�

�� which gives

20 (1 . ) 1Np ��

or

1

0

1, 1

1

1, 1

1

N

p

N

�

��

Thus,

1

(1 ), 1

1 , 0,1, ... .,1

, 11

n

N

np n N

N

�

� ��

� � ��

The value of �

� �� need not be less than 1 in the case of Twenty20 match, because arrivals at the system are

controlled by the system limit N. This means that �eff is the rate which matters in this case. Because customers willbe lost when there are N in the system,

lost Np� � �

� (1 ).eff lost N Np p� � � ��

In the case of Twenty20 match, .eff� � � .

Since 1, eff

��

�


ICC WORLD TWENTY-20, 2014 (WORLD CUP-2014)- A CASE STUDY 9

0N

s n nL np��

2 30 ( 2 3 ... )Np N��

0 (1 2 3 ... )p N� � � � �

1 ( 1)

1 2

N N

N

��

� .2S

NL �

EXPECTATION IN TWENTY20 MATCH:

In Twenty20 match, utilization factor � = 1 and the number of customers (wickets) falling is N =1 in each innings.Therefore, the expected number of customers (wickets) in each innings or expectation of customers (wickets) is10

5.2� That means in each innings, on average 5 wickets are falling. It may vary nearer 5 to i.e. 4 or 6 wickets in

each innings.

We applied queuing model to Twenty20 cricket in particular to the ICC World Twenty20-2014. As a result,firstly we got the expected number of wickets during World Cup-2014. The actual record of the World Cup-2014is as under:

Match No. Of Wkts No. Of Wkts Average Match No. of Wkts No. Of Wkts AverageSr. No. in 1st Inn. in 2nd Inn. No. Of Wkts Sr. No. in 1st Inn. in 2nd Inn. No. Of Wkts

1 7 3 5 13 4 4 4

2 7 8 7.5 14 5 7 6

3 6 1 3.5 15 5 7 6

4 5 10 7.5 16 7 10 8.5

5 7 3 5 17 5 10 7.5

6 6 8 7 18 10 10 10

7 10 1 5.5 19 5 3 4

8 7 10 8.5 20 6 10 8

9 9 10 9.5 21 6 4 5

10 4 4 4 22 4 4 4

11 8 4 6 23 4 4 4

12 7 2 4.5 Total: 23 144 137 140.5

• No. = number Wkts = Wickets Inn. = Innings

Actual Expectation is 140.5

6.10 623

� � wickets.

This data shows that, during ICC World Twenty20-2014, the average number of wickets that has been fallen is 6.This average is close to the ideal average 5. This is the real application of queuing models in the real world.

10 BHAVIN PATEL AND PRAVIN BHATHAWALA

CALCULATION OF PROBABILITIES IN TWENTY20 MATCH

Secondly, we seek to find the probability that the team batting first will win a Twenty20 match and the probabilitythat the team batting second will win. We neglect the tied Twenty20 matches, unlike ODIs, because in that case,the super over comes in the frame and gives the result of the match. For this, we define the events of team battingfirst and team batting second.

At the start of the game, the coin is tossed to decide which team will bat first. Let B1 be the event that the teambatting first and B2 be the event that the team batting second.

� 1 2

1( ) ( ) .

2P B P B� �

Let I be the event that the 1st innings is completed, irrespective of the number of wickets falling in the innings.

( ) ( 10 1 )P I P n in the st innings� � �

1 1 1(11 )

11 11 11times� � � ��

� P(I) = 1. .

Now, Let I1 be the event that the team batting first win and I2 be the event that the team batting second win.Then, the probability of I1 is determined by

1( )P I = P(B1)P(I) + P(B2) P(n = 10 in the 2nd innings)

= 1 1 1 1 1 12

(1)2 2 11 2 22 22

� ��

The probability of I2 is determined by

P(I2) = 2( ) ( 9 2 )ndP B P n in the innings�

= 1 1 1 1

... (10 times)2 11 11 11� ��

= 1 10 10

.2 11 22� � ��

This means that, out of Twenty20 matches 12 matches are won by team batting first and 10 matches are wonby team batting second. This can be verified from the data of the ICC World Twenty20-2014. The actual recordis as under:

Match Sr. No. 1st Inn. Win 2nd Inn. Win Match Sr. No. 1st Inn. Win 2nd Inn. Win

1 0 L 1 W 13 0 L 1 W

2 1 W 0 L 14 1 W 0 L

3 1 W 0 L 15 1 W 0 L

4 1 W 0 L 16 1 W 0 L

5 0 L 1 W 17 1 W 0 L

6 1 W 0 L 18 1 W 0 L

7 0 L 1 W 19 0 L 1 W

Table Cont’d

ICC WORLD TWENTY-20, 2014 (WORLD CUP-2014)- A CASE STUDY 11

The above table shows that during the ICC World Twenty20-2014, out of 23 Twenty20 matches 13 matcheshave been won by team batting first and 10 have been won by team batting second. We observe that the theoreticalprobabilities match with the actual probabilities.

CONCLUSION

We conclude that, in T20 game of cricket, out of 22 Twenty20 matches12 matches are won by team batting first

and 10 matches are won by team batting second. That is, the probability that team batting first win is 12

0.5522

�

and the probability that team batting second win is 10

0.4522

� .

In other words, it means that out of 100 Twenty20 matches 55 matches are won by team batting first andmatches are won by team batting second.

During ICC World Twenty20-2014, the probability that team batting first win is 13

0.5723

� and the probability

that team batting second win is 13

0.4323

� . This shows that the theoretical probabilities match with the actual

probabilities.

Moreover, during ICC World Twenty20-2014, the actual expectation of wickets falling is wickets.

REFERENCES

[1] H.A. Taha, “Operations Research-An Introduction”. 8th Edition, ISBN 0131889230. Pearson Education,2007.

[2] J.D.C. Little, “A Proof for the Queuing Formula: “, Operations Research, 9(3), 1961, pp. 383-387, doi:10.2307/167570.

[3] Cooper RB (1972). “Introduction to Queuing Theory”. McMillan: New York.

[4] Tijms HC (1986). “Stochastic Modelling and Analysis”. A Computational Approach. Wiley: Chichester.

[5] K. Rust, “Using Little’s Law to Estimate Cycle Time and Cost”, Proceedings of the 2008 Winter SimulationConference, IEEE Press, Dec. 2008, doi:10.1109/WSC.2008.4736323.

[6] Worthington D and Wall A (1999). Using the discrete time modelling approach to evaluate the time-dependentbehaviour of queuing systems. J Opl Res Soc 50: 777-888.

[7] Stevenson WJ (1996). Production/Operations Management, 6th edn. Irwin, McGraw-Hill, USA.

8 1 W 0 L 20 1 W 0 L

9 1 W 0 L 21 1 W 0 L

10 0 L 1 W 22 0 L 1 W

11 0 L 1 W 23 0 L 1 W

12 0 L 1 W Total:23 13 W 10 W

• 0 means Lost(L) and 1 means Won(W)

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