research article performance of an m/m/1 retrial queue...
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Research ArticlePerformance of an MM1 Retrial Queue with Working VacationInterruption and Classical Retrial Policy
Tao Li1 Liyuan Zhang2 and Shan Gao3
1School of Science Shandong University of Technology Zibo 255049 China2School of Business Shandong University of Technology Zibo 255049 China3School of Mathematics and Statistics Fuyang Normal College Fuyang 236037 China
Correspondence should be addressed to Tao Li liltaot126com
Received 8 December 2015 Revised 30 April 2016 Accepted 5 May 2016
Academic Editor Yi-Kuei Lin
Copyright copy 2016 Tao Li et al This is an open access article distributed under the Creative Commons Attribution License whichpermits unrestricted use distribution and reproduction in any medium provided the original work is properly cited
An MM1 retrial queue with working vacation interruption is considered Upon the arrival of a customer if the server is busy itwould join the orbit of infinite size The customers in the orbit will try for service one by one when the server is idle under theclassical retrial policy with retrial rate 119899120572 where 119899 is the size of the orbit During a working vacation period if there are customersin the system at a service completion instant the vacation will be interrupted Under the stable condition the probability generatingfunctions of the number of customers in the orbit are obtained Various system performance measures are also developed Finallysome numerical examples and cost optimization analysis are presented
1 Introduction
Queueing models with server vacations have been wellstudied in the past three decades and successfully appliedin manufacturing and production systems service systemsand communication systems Some vacation queues can befound in Tian and Zhang [1] On the basis of ordinaryvacation Servi and Finn [2] first introduced a class ofsemivacation policy known asworking vacation (WV)wherethe server provides service at a lower speed during thevacation period rather than stopping service completelyThe motivation of analyzing MM1WV queue is to modelapproximately a multiqueue system where each queue canbe served at one of two service rates From then onwardsseveral works [3ndash5] have appeared that analyzed the singleserver queue with working vacations In order to utilizethe server effectively Li and Tian [6] introduced vacationinterruption policy During the working vacation period ifat least one customer is present in the system at a servicecompletion epoch the server will interrupt the vacationand resume regular service Working vacation interruption
has become an important aspect Using the matrix ana-lytic method Li and Tian [7] analyzed a GIGeo1 queuewith working vacations and vacation interruption and Liet al [8] discussed the GIM1 queue Using the methodof a supplementary variable Zhang and Hou [9] investi-gated an MG1 queue with working vacations and vacationinterruption
Retrial queueing systems are described by the feature thatthe arriving customers who find the server busy join theretrial orbit to try again for their requests During the last twodecades retrial queues have been studied extensively becauseof their applications in telephone switching systems telecom-munication networks and computer systems Readers canrefer to Artalejo and Gomez-Corral [10] and Choudhury etal [11] Recently the retrial queueing systems with workingvacations have been investigated extensively Do [12] firststudied an MM1 retrial queue with working vacations Liet al [13] considered a discrete time GeoGeo1 retrial queuewith working vacations and vacation interruption and Liuand Song [14] introduced nonpersistent customers into theGeoGeo1 retrial queue with working vacations Tao et al
Hindawi Publishing CorporationAdvances in Operations ResearchVolume 2016 Article ID 4538031 9 pageshttpdxdoiorg10115520164538031
2 Advances in Operations Research
[15] discussed an MM1 retrial queue with collisions andworking vacation interruption under N-policy Using thematrix analytic method Do [12] Li et al [13] Liu and Song[14] and Tao et al [15] obtained the stationary probabilitydistribution and showed the conditional stochastic decompo-sition for the queue lengthUsing themethod of a supplemen-tary variable Aissani et al [16] and Jailaxmi et al [17] bothgeneralized the model of [12] to an MG1 queue Gao andWang [18] analyzed a Geo119883G1 retrial queue with generalretrial times and working vacation interruption and thecontinuous-time MG1 queue was investigated by Gao etal [19] Note that the retrial policy of the above papers iseither constant retrial policy or general retrial policy whereonly the customer at the head of the orbit can request aservice
Many of the queueing systems with repeated attemptsoperate under the classical retrial policy where each blockof customers generate a stream of repeated attempts inde-pendently of the rest of the customers in the orbit Intelephone systems the calls which will be reattempted canbe modeled by an infinite server queue where the timeuntil reattempt follows an exponential distribution Thusit needs to discuss a retrial queue with working vacationunder the classical retrial policy Ayyappan et al [20] firststudied such a queueing system In order to obtain thesteady state probability vector the orbit size is restricted to119872 such that any arriving customer finding the orbit full isconsidered lost In this paper we also analyze an MM1retrial queue with multiple working vacations under theclassical retrial policy and vacation interruption policy isconsidered Ayyappan et al [20] mainly focus on numericalcomputation and approximation and they apply the directtruncation method to find the steady state probability vectorWe place more emphasis on the analytic solutions andthe orbit size is always infinite in our model The noveltyof this investigation is that we use probability generatingfunction method to deal with the steady state joint distri-bution of the server state and the number of customersin the orbit and the ergodicity condition is obtained byFosterrsquos criterion Finally a cost minimization problem isdiscussed
This paper is organized as follows In Section 2 we estab-lish the model Using Fosterrsquos criterion the stable conditionis obtained In Section 3 we get the balance equations andderive the probability generating function of the number ofcustomers in the orbit Various performance measures of thismodel are discussed in Section 4 Section 5 presents somenumerical examples and cost optimization analysis FinallySection 6 concludes the paper
2 Model Formulation and Stable Condition
Let us consider an MM1 retrial queue with working vaca-tions and vacation interruption Customers arrive accordingto a Poisson process with rate 120582 Upon the arrival ofcustomers if the server is free arriving customers get serviceimmediately If the server is occupied customers are forced
to wait in the orbit of infinite size In this paper we adoptthe classical retrial policy with retrial rate 119899120572 where 119899 isthe orbit size Upon the arrival of retrials if the serveris busy retrials will go back to the orbit If the server isnot occupied arriving retrials get services immediately Theservice rate is120583when the system is not on vacationThe singleserver takes a working vacation when the system becomesempty Vacation time follows an exponential distributionwith parameter 120579 During the vacation period customers canbe served with rate 120578 At a service completion instant in thevacation period if there are customers in the system at thatmoment the vacation is interrupted and the server comesback to the normal working level The working vacation willcontinue if the system is empty At the end of each vacationthe server only takes another new vacation if the system isempty
Next we give an application example of this modelIn our daily life some social organizations can providetelephone psychological counseling Here we consider atelephone counseling system staffed with a psychologicalconsultant (called main server) and an assistant (calledsubstitute server) The assistant only provides service topeople (called customers) when the consultant does notwork Generally speaking there is a telephone operatorwho is responsible for establishing communications betweenservers and customers the operator also needs to note downthe callsrsquo information in a registration form (correspondingto the ldquoorbitrdquo) When a person makes a call and if theserver is free the operator notes down the information andthe customer is served immediately by the consultant orthe assistant If the server is busy on the other hand theoperator tells the customer to call again some time later(called retrial) and the customerrsquos information should alsobe noted down When a retrial customer makes a call againthe operator does not need to note down the informationagain Moreover if the customer in the registration formcompletes his service the telephone operator will make anote then he can know whether the customer completed hisservice or not When the operator finds that all customersin the registration form have completed their services (iethere is no customer in the orbit and the system is empty)the consultant will rest from his work (begin a vacation)During the consultantrsquos vacation period the assistant willprovide service If a service is completed by the assistantand if there are customers in the registration form who havenot received services (ie there are customers in the orbitand the system is nonempty) the consultant will come backfrom his vacation no matter whether his vacation ends ornot (vacation interruption happens) Furthermore when avacation ends and if all customers in the registration formhave completed their services the consultant begins anothervacation Otherwise the consultant takes over the assistantIn order to understand the customerrsquos need the consultantwill restart service no matter how long the assistant hasserved
Let 119876(119905) be the number of customers in the orbit at time119905 and let 119869(119905) be the state of server at time 119905 There are fourpossible states of the server as follows
Advances in Operations Research 3
119869 (119905) =
0 the server is in a working vacation period at time 119905 and the server is free
1 the server is in a working vacation period at time 119905 and the server is busy
2 the server is during a normal service period at time 119905 and the server is free
3 the server is during a normal service period at time 119905 and the server is busy
(1)
Note that when the orbit is empty it is impossible that theserver is free in a normal service period Thus 119876(119905) 119869(119905) isa Markov process with state space
Ω = (0 119895) 119895 = 0 1 3 cup (119896 119895) 119896 ge 1 119895 = 1 2 3 (2)
Remark 1 Note that in this model we consider workingvacation interruption policy so the state (119896 0) 119896 ge 1 doesnot exist
Define
119886119896= int
infin
0
(120582119909)119896
119896
119890minus120582119909
120583119890minus120583119909
119889119909
119887119896= int
infin
0
(120582119909)119896
119896
119890minus120582119909
119890minus120579119909
120578119890minus120578119909
119889119909
119888119896= int
infin
0
(120582119909)119896
119896
119890minus120582119909
120579119890minus120579119909
119890minus120578119909
119889119909
(3)
where 119886119896 119896 ge 0 is the probability that 119896 customers arrive
during a normal service time 119887119896 119896 ge 0 represents the
probability that 119896 customers arrive during a lower servicetime and the working vacation does not end 119888
119896 119896 ge 0
explains the probability that 119896 customers arrive before theworking vacation ends and no service is completed
Let 119905119899 119899 = 1 2 be the sequence of epochs at which
a normal service or a lower service completion occurs and119876119899= 119876(119905
+
119899) then the sequence of random variables 119876
119899 119899 =
1 2 forms an embedded Markov chain
Theorem 2 The embedded Markov chain 119876119899 119899 = 1 2
is ergodic if and only if 120588 = 120582120583 lt 1
Proof It is not difficult to see that 119876119899 119899 = 1 2 is an
irreducible and aperiodic Markov chain and the transitionprobability is given by
119875 =
(
(
(
(
11987500
11987501
11987502
11987503
sdot sdot sdot
11987510
11987511
11987512
11987513
sdot sdot sdot
0 11987521
11987522
11987523
sdot sdot sdot
0 0 11987532
11987533
sdot sdot sdot
d
)
)
)
)
(4)
where
119875119899119896
=
119887119896+
119896
sum
119895=0
119888119895119886119896minus119895
119899 = 0 119896 ge 0
119899120572
119899120572 + 120582
1198860 119899 ge 1 119896 = 119899 minus 1
119899120572
119899120572 + 120582
119886119896+1minus119899
+
120582
119899120572 + 120582
119886119896minus119899
119899 ge 1 119896 ge 119899
(5)
In order to discuss the ergodicity of the chain 119876119899 119899 =
1 2 we use Fosterrsquos criterion with the mean drift
119909119899= 119864 [119891 (119876
119896+1) minus 119891 (119876
119896) | 119876119896= 119899] (6)
where the test function 119891(119899) = 119899 From the transition matrixit can be easily proved that
119909119899=
120588
120579
120579 + 120578
+
120582
120579 + 120578
119899 = 0
120588 minus
119899120572
119899120572 + 120582
119899 ge 1
(7)
As 119899 rarr infin there exists 119909 = lim119899rarrinfin
119909119899= 120588 minus 1 Applying
Fosterrsquos criterion we can get that the embeddedMarkov chainis ergodic if 120588 lt 1 Besides since 119876
119899+1minus 119876119899ge minus1 applying
the criterion from Sennott et al [21] we can obtain that 120588 lt 1
is necessary for ergodicity
Since the arrival process is Poisson using PASTA prop-erty it can be shown from Burkersquos theorem (see [22 pp 187-188]) that the steady state probabilities of the Markov process119876(119905) 119869(119905) exist if and only if the stable condition120588 lt 1holdsNow we define the limiting probability
120587119896119895= 119875 119876 = 119896 119869 = 119895
= lim119905rarrinfin
119875 119876 (119905) = 119896 119869 (119905) = 119895 (119896 119895) isin Ω
(8)
3 Balance Equations and ProbabilityGenerating Functions
Figure 1 presents the state transition rate diagram of thesystem Under the stable condition 120588 lt 1 the set of balanceequations is given as follows
4 Advances in Operations Research
0 1 1 1 2 1 3 1
0 0 1 2 2 2 3 2
0 3 1 3 2 3 3 3
120579
120579120579120579120579
120582
120582 120582 120582 120582
120582
120582 120582 120582
120582120582120582
120578 120578 120578 120578
120583120583 120583 120583
120572 2120572 3120572 4120572
Figure 1 State transition rate diagram of the system
12058212058700= 12057812058701+ 12058312058703 (9)
(120582 + 120579 + 120578) 12058701= 12058212058700 (10)
(120582 + 120583) 12058703= 12057212058712+ 12057912058701 (11)
(120582 + 120579 + 120578) 1205871198991= 120582120587119899minus11
119899 ge 1 (12)
(120582 + 119899120572) 1205871198992= 1205781205871198991+ 1205831205871198993 119899 ge 1 (13)
(120582 + 120583) 1205871198993= 1205821205871198992+ 120582120587119899minus13
+ (119899 + 1) 120572120587119899+12
+ 1205791205871198991 119899 ge 1
(14)
Define the probability generating functions (pgfs)
1198661 (119911) =
infin
sum
119899=0
1205871198991119911119899
1198662 (119911) =
infin
sum
119899=1
1205871198992119911119899
1198663 (119911) =
infin
sum
119899=0
1205871198993119911119899
(15)
with 1198661015840119894(119911) = sum
infin
119899=1119899120587119899119894119911119899minus1
119894 = 1 2 3Multiplying by the appropriate power of 119911119899 in (10) and
(12) summing over 119899 and rearranging the terms we get
(120582 + 120579 + 120578 minus 120582119911)1198661 (119911) = 120582120587
00 (16)
Multiplying by the appropriate power of 119911119899 in (13) summingover 119899 and using (9) we obtain
1205821198662 (119911) + 120572119911119866
1015840
2(119911) = 120578119866
1 (119911) + 120583119866
3 (119911) minus 120578120587
01
minus 12058312058703
= 1205781198661 (119911) + 120583119866
3 (119911) minus 120582120587
00
(17)
In a similar way from (11) and (14) we derive
(120582 + 120583 minus 120582119911)1198663 (119911) = 120582119866
2 (119911) + 120572119866
1015840
2(119911) + 120579119866
1 (119911) (18)
From (16) we get
1198661 (119911) =
120582
120582 + 120579 + 120578 minus 120582119911
12058700 (19)
From (17) we obtain
1198663 (119911) =
120582
120583
1198662 (119911) +
120572119911
120583
1198661015840
2(119911) minus
120578
120583
1198661 (119911) +
120582
120583
12058700 (20)
Taking (20) into (18) using (19) after some computation wecan derive
1198661015840
2(119911) minus
1205822
(120583 minus 120582119911) 120572
1198662 (119911)
=
120582212058700
(120583 minus 120582119911) (1 minus 119911) 120572
+
12058212058700
(1 minus 119911) 120572
minus
12057812058212058700
(120582 + 120579 + 120578 minus 120582119911) (1 minus 119911) 120572
minus
(120579120583 + 120582120578) 12058212058700
(120582 + 120579 + 120578 minus 120582119911) (120583 minus 120582119911) (1 minus 119911) 120572
(21)
Solving the first-order differential equation we have
1198662 (119911) = (120583 minus 120582119911)
minus120582120572[
120582212058700
120572
1198701 (119911) +
12058212058700
120572
1198702 (119911)
minus
12058212057812058700
120572
1198703 (119911) minus
(120579120583 + 120582120578) 12058212058700
120572
1198704 (119911)]
(22)
where
1198701 (119911) = int
119911
0
(1 minus 119909)minus1(120583 minus 120582119909)
120582120572minus1119889119909
1198702 (119911) = int
119911
0
(1 minus 119909)minus1(120583 minus 120582119909)
120582120572119889119909
1198703 (119911)
= int
119911
0
(120582 + 120579 + 120578 minus 120582119909)minus1
(1 minus 119909)minus1(120583 minus 120582119909)
120582120572119889119909
1198704 (119911)
= int
119911
0
(120582 + 120579 + 120578 minus 120582119909)minus1
(1 minus 119909)minus1(120583 minus 120582119909)
120582120572minus1119889119909
(23)
From (17) and (18) substituting 11986610158402(119911) we get
1198663 (119911) =
120582 (1 minus 119911) 1198662 (119911) minus (120578 + 120579119911)119866
1 (119911) + 120582120587
00
(120583 minus 120582119911) (1 minus 119911)
(24)
Advances in Operations Research 5
From (19) (22) and (24) we can see that 1198661(119911) 119866
2(119911) and
1198663(119911) are expressed in terms of 120587
00 In the next section using
normalization condition we will obtain 12058700
4 Performance Measures
From (19) we have
1198661 (1) =
120582
120579 + 120578
12058700 (25)
1198661015840
1(1) = lim
119911rarr1
1198661015840
1(119911) = lim
119911rarr1
1205822
(120582 + 120579 + 120578 minus 120582119911)212058700
=
1205822
(120579 + 120578)212058700
(26)
From (22) we get
1198662 (1) = 119870120587
00 (27)
where
119870 = (120583 minus 120582)minus120582120572
[
1205822
120572
1198701 (1) +
120582
120572
1198702 (1) minus
120582120578
120572
1198703 (1)
minus
(120579120583 + 120582120578) 120582
120572
1198704 (1)]
(28)
From (24) using LrsquoHospital rule we obtain
1198663 (1) = lim119911rarr1
1198663 (119911)
= lim119911rarr1
minus1205821198662 (119911)+ 120582 (1 minus 119911) 119866
1015840
2(119911) minus 120579119866
1 (119911) minus (120578 + 120579119911)119866
1015840
1(119911)
minus120582 (1 minus 119911) minus (120583 minus 120582119911)
=
1205821198662 (1) + 120579119866
1 (1) + (120578 + 120579)119866
1015840
1(1)
120583 minus 120582
=
120582119870 (120579 + 120578) + 120582120579 + 1205822
(120583 minus 120582) (120579 + 120578)
12058700
(29)
Since
12058700+ 1198661 (1) + 119866
2 (1) + 119866
3 (1) = 1 (30)
taking (25) (27) and (29) into the above equation we canderive
12058700= [1 +
120582
120579 + 120578
+ 119870 +
120582119870 (120579 + 120578) + 120582120579 + 1205822
(120583 minus 120582) (120579 + 120578)
]
minus1
= [
120582 (120583 minus 120578) + 120583 (1 + 119870) (120579 + 120578)
(120583 minus 120582) (120579 + 120578)
]
minus1
(31)
Thus 12058700
is found out From (19) (22) and (24) 1198661(119911)
1198662(119911) and 119866
3(119911) are completely computed And the pgf of
the number of customers in the orbit is given by
119866 (119911) = 12058700+ 1198661 (119911) + 119866
2 (119911) + 119866
3 (119911) (32)
The pgf of the number of customers in the system is given by
(119911) = 12058700+ 1199111198661 (119911) + 119866
2 (119911) + 119911119866
3 (119911) (33)
Let 119864[119871119895] denote the average number of customers in the
orbit when the serverrsquos state is 119895 119895 = 1 2 3Differentiating (19) with respect to 119911 we have
119864 [1198711] = lim119911rarr1
1198661015840
1(119911) =
1205822
(120579 + 120578)212058700 (34)
From (18) we can get
119864 [1198712] = lim119911rarr1
1198661015840
2(119911) =
120583
120572
1198663 (1) minus
120582
120572
1198662 (1) minus
120579
120572
1198661 (1) (35)
Differentiating (24) with respect to 119911 and using LrsquoHospitalrule twice the expression for 119864[119871
3] is obtained as
119864 [1198713] = lim119911rarr1
1198661015840
3(119911) =
(120583 minus 120582) (120579 + 120578)11986610158401015840
1(1) + 2 (120579120583 + 120582120578) 119864 [119871
1] + 2120582 (120583 minus 120582) 119864 [119871
2] + 2120582 [120579119866
1 (1) + 120582119866
2 (1)]
2 (120583 minus 120582)2
(36)
and differentiating (19) twice with respect to 119911 and taking 119911 =1 yield 11986610158401015840
1(1) as
11986610158401015840
1(1) =
21205823
(120579 + 120578)312058700 (37)
Therefore the average orbit length (119864[119871]) is given by
119864 [119871] = lim119911rarr1
1198661015840(119911) = 119864 [119871
1] + 119864 [119871
2] + 119864 [119871
3] (38)
And the average system length (119864[]) is derived as
119864 [] = lim119911rarr1
1198661015840(119911) = 119864 [119871] + 119866
1 (1) + 119866
3 (1) (39)
The probability that the server is busy is
119875119887= 119875 119869 = 1 + 119875 119869 = 3 = 119866
1 (1) + 119866
3 (1) (40)
6 Advances in Operations Research
The probability that the server is free is
119875119891= 119875 119869 = 0 + 119875 119869 = 2 = 120587
00+ 1198662 (1) = 1 minus 119875
119887 (41)
The probability that the server is in a working vacation periodis given by
119875119908= 119875 119869 = 0 + 119875 119869 = 1 = 120587
00+ 1198661 (1) (42)
The probability that the server is in a normal service periodis given by
119875119899= 119875 119869 = 2 + 119875 119869 = 3 = 119866
1 (2) + 119866
2 (3)
= 1 minus 119875119908
(43)
Let 119864[119882] (119864[]) be the expected waiting (sojourn) time of acustomer in the orbit (system) using Littlersquos formula
119864 [119882] =
119864 [119871]
120582
119864 [] =
119864 []
120582
(44)
The system busy period 119879 is defined as the period that startsat an epochwhen an arriving customer finds an empty systemand ends at the departure epoch at which the system is emptyUsing the theory of regenerative process we get
12058700=
119864 [11987900]
119864 [11987900] + 119864 [119879]
(45)
where 11987900
is the time length that the system is in the state(0 0) Since the interarrival time between two customersfollows exponential distribution with parameter 120582 we have119864[11987900] = 1120582 Thus 119864[119879] = 120582
minus1(120587minus1
00minus 1)
5 Numerical Results
In this section under the stable condition we present somenumerical examples to illustrate the effect of the varyingparameters on themeanorbit length119864[119871] the probability thatthe system is busy 1minus120587
00 and the probabilities of serverrsquos state
(119875119887 119875119891 119875119908 and 119875
119899) Moreover a cost minimization problem
is also discussed The various parameters of this model arearbitrarily chosen as 120582 = 2 120578 = 1 120583 = 5 120572 = 3 and120579 = 02 unless they are considered as variables or their valuesare mentioned in Figures 2 3 4 and 5
51 Sensitivity Analysis Figures 2 and 3 illustrate the effect of120579 on 119864[119871] and 1 minus 120587
00for different values of 120578 respectively
When 120578 lt 120583 we find that 119864[119871] and 1 minus 12058700
both decrease as120579 increases and 120579 has a noticeable effect on the performancemeasures which cannot be ignored An especial case is 120578 = 120583that is the service rate in the vacation period is equal to theservice rate in the normal period and the model we considerreduces to anMM1 retrial queue with classical retrial policyClearly 120579 has no effect on119864[119871] and 1minus120587
00 which agrees with
the intuitive expectation
0 1 2 3 4 5 6 7 8 9 10
08
1
12
14
16
18
2
Expe
cted
que
ue le
ngth
E[L]
120579
120578 = 1
120578 = 3120578 = 5
Figure 2 The effect of 120579 on 119864[119871] for different values of 120578
0 1 2 3 4 5 6 7 8 9 10055
06
065
07
075
08
085
09
120579
120578 = 1
120578 = 3120578 = 5
1minus12058700
Figure 3 The effect of 120579 on 1 minus 12058700
for different values of 120578
The effect of 120578 on 119864[119871] is presented in Figure 4 asexpected 119864[119871] decreases with increasing values of 120578 When120578 is large it can be observed that the effect of 120578 on 119864[119871] is notobvious as 120578 increases this is because we consider workingvacation interruption policy If the system is not empty at aservice completion instant in the vacation period the serverwill come back to the normal working level Moreover sinceconstant retrial policy means that only the customer at thehead of the orbit can request a service we can find that 119864[119871]of MM1 queue with classical retrial policy is smaller thanthat of MM1 queue with constant retrial policy As shownin Figure 5 119875
119887decreases with increasing values of 120578 and 119875
119891
has the opposite tendency The reason is that as 120578 increases
Advances in Operations Research 7
0 05 1 15 2 25 3 35 4 45 50
2
4
6
8
10
12
14Ex
pect
ed q
ueue
leng
th E[L]
MM1 with classical retrialMM1 with constant retrial
120578
Figure 4 The effect of 120578 on 119864[119871] for two different retrial policies
0 05 1 15 2 25 3 35 4 45 501
02
03
04
05
06
07
08
09
120578
PbPf
Figure 5 The probability of serverrsquos state with the change of 120578
themeanorbit length119864[119871] and themean service time 1120578bothdecrease
Figure 6 indicates that 119864[119871] decreases with increasingvalues of 120572 this is because the mean retrial time decreases as120572 increases From the instant when the server becomes idlean external potential primary customer and retrial customerscompete to access the server and the smaller the mean retrialtime is the bigger the probability that the server is busy iswhich leads to decreasing of 119864[119871] Furthermore under thesame condition decreasing of 119864[119871] results in the increasein the probability that the system is empty and Figure 7reveals that 119875
119908increases with increasing values of 120572 while 119875
119899
decreases as 120572 increases In Figure 6 compared with constantretrial policy we can also find that classical retrial policydecreases the waiting jobs effectively
2 25 3 35 4 45 5 55 615
2
25
3
35
4
45
5
55
6
Expe
cted
que
ue le
ngth
E[L]
MM1 with classical retrialMM1 with constant retrial
120572
Figure 6 The effect of 120572 on 119864[119871] for two different retrial policies
2 25 3 35 4 45 5 55 6035
04
045
05
055
06
065
07
PnPw
120572
Figure 7 The probability of serverrsquos state with the change of 120572
52 Cost Analysis In this subsection we establish a costfunction to search for the optimal service rate 120578 so as tominimize the expected operating cost function per unit time
Define the following cost elements
119862119871 cost per unit time for each customer present in the
orbit
119862120583 cost per unit time for service during a normal
service period
119862120578 cost per unit time for service in aworking vacation
period
119862120579 fixed cost per unit time during a vacation period
8 Advances in Operations Research
0 05 1 15 2 25 3 35 4 45 565
70
75
80
85
90
95
100
105
Expe
cted
ope
ratin
g co
st pe
r uni
t tim
e
120578
Figure 8 Effect of 120578 on the expected operating cost per unit time
Based on the definitions of each cost element listed abovethe expected operating cost function per unit time can begiven by
min120578
119891 (120578) = 119862119871119864 [119871] + 119862
120583120583 + 119862
120578120578 + 119862120579120579 (46)
Because the expected operating cost function is highly non-linear and complex it is not easy to get the derivative of itWeassume 119862
119871= 6 119862
120583= 10 119862
120578= 8 and 119862
120579= 4 and develop
approximations by MATLAB program to find the optimumvalue of 120578 say 120578lowast
From Figure 8 we can see that there is an optimal servicerate 120578 to minimize the cost Implementing the computersoftware MATLAB by the parabolic method and controllingthe error by 10
minus4 we find the solution 120578lowast
= 08584 with119891(120578lowast) = 690971
6 Conclusion
This paper analyzes a single server retrial queue with workingvacation interruption under the classical retrial policy Usingembedded Markov chain and Fosterrsquos criterion we get thecondition of stability The pgf of the number of customersin the orbit is obtained and some important performancemeasures are also discussed Moreover the effects of var-ious parameters on the system performance measures areexamined numerically Under the stable condition a costminimization problem is considered For future researchusing the same method one can investigate a similar modelwith batch arrival and without vacation interruption
Competing Interests
The authors declare that there are no competing interestsregarding the publication of this paper
Acknowledgments
This work is supported by the National Natural ScienceFoundation of China (nos 11301306 and 11401348) and theNatural Science Foundation of Anhui Higher EducationInstitutions (no KJ2014ZD21)
References
[1] N Tian and Z G ZhangVacation QueueingModels-Theory andApplications Springer New York NY USA 2006
[2] L D Servi and S G Finn ldquoMM1 queues with workingvacations (MM1WV)rdquo Performance Evaluation vol 50 no1 pp 41ndash52 2002
[3] Y Baba ldquoAnalysis of a GIM1 queue with multiple workingvacationsrdquo Operations Research Letters vol 33 no 2 pp 201ndash209 2005
[4] D-A Wu and H Takagi ldquoMG1 queue with multiple workingvacationsrdquo Performance Evaluation vol 63 no 7 pp 654ndash6812006
[5] J-H Li N-S Tian and W-Y Liu ldquoDiscrete-time GIGeo1queue with multiple working vacationsrdquoQueueing Systems vol56 no 1 pp 53ndash63 2007
[6] J Li andNTian ldquoTheMM1 queuewithworking vacations andvacation interruptionsrdquo Journal of Systems Science and SystemsEngineering vol 16 no 1 pp 121ndash127 2007
[7] J-H Li and N-S Tian ldquoThe discrete-time GIGeo1 queuewith working vacations and vacation interruptionrdquo AppliedMathematics and Computation vol 185 no 1 pp 1ndash10 2007
[8] J-H Li N-S Tian and Z-Y Ma ldquoPerformance analysis ofGIM1 queue with working vacations and vacation interrup-tionrdquo Applied Mathematical Modelling vol 32 no 12 pp 2715ndash2730 2008
[9] M Zhang and Z Hou ldquoPerformance analysis of MG1 queuewith working vacations and vacation interruptionrdquo Journal ofComputational and Applied Mathematics vol 234 no 10 pp2977ndash2985 2010
[10] J R Artalejo and A Gomez-Corral Retrial Queueing SystemsA Computational Approach Springer Berlin Germany 2008
[11] G Choudhury L Tadj and M Deka ldquoAn unreliable serverretrial queue with two phases of service and general retrialtimes under Bernoulli vacation schedulerdquo Quality Technologyamp Quantitative Management vol 12 no 4 pp 437ndash464 2015
[12] T V Do ldquoMM1 retrial queue with working vacationsrdquo ActaInformatica vol 47 no 1 pp 67ndash75 2010
[13] T Li Z Wang and Z Liu ldquoGeoGeo1 retrial queue with work-ing vacations and vacation interruptionrdquo Journal of AppliedMathematics and Computing vol 39 no 1-2 pp 131ndash143 2012
[14] Z Liu and Y Song ldquoGeoGeo1 retrial queue with non-persistent customers and working vacationsrdquo Journal of AppliedMathematics and Computing vol 42 no 1-2 pp 103ndash115 2013
[15] L Tao Z Liu andZWang ldquoMM1 retrial queuewith collisionsand working vacation interruption under N-policyrdquo RAIROOperations Research vol 46 no 4 pp 355ndash371 2012
[16] A Aissani S Taleb T Kernane G Saidi and D HamadoucheldquoAn MG1 retrial queue with working vacationrdquo in Advancesin Systems Science Proceedings of the International Conferenceon Systems Science 2013 (ICSS 2013) vol 240 of Advancesin Intelligent Systems and Computing pp 443ndash452 SpringerBerlin Germany 2014
Advances in Operations Research 9
[17] V Jailaxmi RArumuganathan andM S Kumar ldquoPerformanceanalysis of single server non-Markovian retrial queue withworking vacation and constant retrial policyrdquo RAIRO Opera-tions Research vol 48 no 3 pp 381ndash398 2014
[18] S Gao and J Wang ldquoDiscrete-time Geo119883G1 retrial queuewith general retrial times working vacations and vacationinterruptionrdquo Quality Technology amp Quantitative Managementvol 10 no 4 pp 495ndash512 2013
[19] S Gao J Wang and W W Li ldquoAn MG1 retrial queuewith general retrial times working vacations and vacationinterruptionrdquo Asia-Pacific Journal of Operational Research vol31 no 2 Article ID 1440006 25 pages 2014
[20] G Ayyappan A Ganapathy and G Sekar ldquoRetrial queueingsystemwith single working vacation under pre-emptive priorityservicerdquo International Journal of Computer Applications vol 2no 2 pp 28ndash35 2010
[21] L I Sennott P A Humblet and R L Tweedie ldquoMean drifts andthe nonergodicity of Markov chainsrdquo Operations Research vol31 no 4 pp 783ndash789 1983
[22] R B Cooper Introduction to QueueingTheory North-HollandNew York NY USA 1981
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical Problems in Engineering
Hindawi Publishing Corporationhttpwwwhindawicom
Differential EquationsInternational Journal of
Volume 2014
Applied MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Mathematical PhysicsAdvances in
Complex AnalysisJournal of
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OptimizationJournal of
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CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
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Operations ResearchAdvances in
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Decision SciencesAdvances in
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
2 Advances in Operations Research
[15] discussed an MM1 retrial queue with collisions andworking vacation interruption under N-policy Using thematrix analytic method Do [12] Li et al [13] Liu and Song[14] and Tao et al [15] obtained the stationary probabilitydistribution and showed the conditional stochastic decompo-sition for the queue lengthUsing themethod of a supplemen-tary variable Aissani et al [16] and Jailaxmi et al [17] bothgeneralized the model of [12] to an MG1 queue Gao andWang [18] analyzed a Geo119883G1 retrial queue with generalretrial times and working vacation interruption and thecontinuous-time MG1 queue was investigated by Gao etal [19] Note that the retrial policy of the above papers iseither constant retrial policy or general retrial policy whereonly the customer at the head of the orbit can request aservice
Many of the queueing systems with repeated attemptsoperate under the classical retrial policy where each blockof customers generate a stream of repeated attempts inde-pendently of the rest of the customers in the orbit Intelephone systems the calls which will be reattempted canbe modeled by an infinite server queue where the timeuntil reattempt follows an exponential distribution Thusit needs to discuss a retrial queue with working vacationunder the classical retrial policy Ayyappan et al [20] firststudied such a queueing system In order to obtain thesteady state probability vector the orbit size is restricted to119872 such that any arriving customer finding the orbit full isconsidered lost In this paper we also analyze an MM1retrial queue with multiple working vacations under theclassical retrial policy and vacation interruption policy isconsidered Ayyappan et al [20] mainly focus on numericalcomputation and approximation and they apply the directtruncation method to find the steady state probability vectorWe place more emphasis on the analytic solutions andthe orbit size is always infinite in our model The noveltyof this investigation is that we use probability generatingfunction method to deal with the steady state joint distri-bution of the server state and the number of customersin the orbit and the ergodicity condition is obtained byFosterrsquos criterion Finally a cost minimization problem isdiscussed
This paper is organized as follows In Section 2 we estab-lish the model Using Fosterrsquos criterion the stable conditionis obtained In Section 3 we get the balance equations andderive the probability generating function of the number ofcustomers in the orbit Various performance measures of thismodel are discussed in Section 4 Section 5 presents somenumerical examples and cost optimization analysis FinallySection 6 concludes the paper
2 Model Formulation and Stable Condition
Let us consider an MM1 retrial queue with working vaca-tions and vacation interruption Customers arrive accordingto a Poisson process with rate 120582 Upon the arrival ofcustomers if the server is free arriving customers get serviceimmediately If the server is occupied customers are forced
to wait in the orbit of infinite size In this paper we adoptthe classical retrial policy with retrial rate 119899120572 where 119899 isthe orbit size Upon the arrival of retrials if the serveris busy retrials will go back to the orbit If the server isnot occupied arriving retrials get services immediately Theservice rate is120583when the system is not on vacationThe singleserver takes a working vacation when the system becomesempty Vacation time follows an exponential distributionwith parameter 120579 During the vacation period customers canbe served with rate 120578 At a service completion instant in thevacation period if there are customers in the system at thatmoment the vacation is interrupted and the server comesback to the normal working level The working vacation willcontinue if the system is empty At the end of each vacationthe server only takes another new vacation if the system isempty
Next we give an application example of this modelIn our daily life some social organizations can providetelephone psychological counseling Here we consider atelephone counseling system staffed with a psychologicalconsultant (called main server) and an assistant (calledsubstitute server) The assistant only provides service topeople (called customers) when the consultant does notwork Generally speaking there is a telephone operatorwho is responsible for establishing communications betweenservers and customers the operator also needs to note downthe callsrsquo information in a registration form (correspondingto the ldquoorbitrdquo) When a person makes a call and if theserver is free the operator notes down the information andthe customer is served immediately by the consultant orthe assistant If the server is busy on the other hand theoperator tells the customer to call again some time later(called retrial) and the customerrsquos information should alsobe noted down When a retrial customer makes a call againthe operator does not need to note down the informationagain Moreover if the customer in the registration formcompletes his service the telephone operator will make anote then he can know whether the customer completed hisservice or not When the operator finds that all customersin the registration form have completed their services (iethere is no customer in the orbit and the system is empty)the consultant will rest from his work (begin a vacation)During the consultantrsquos vacation period the assistant willprovide service If a service is completed by the assistantand if there are customers in the registration form who havenot received services (ie there are customers in the orbitand the system is nonempty) the consultant will come backfrom his vacation no matter whether his vacation ends ornot (vacation interruption happens) Furthermore when avacation ends and if all customers in the registration formhave completed their services the consultant begins anothervacation Otherwise the consultant takes over the assistantIn order to understand the customerrsquos need the consultantwill restart service no matter how long the assistant hasserved
Let 119876(119905) be the number of customers in the orbit at time119905 and let 119869(119905) be the state of server at time 119905 There are fourpossible states of the server as follows
Advances in Operations Research 3
119869 (119905) =
0 the server is in a working vacation period at time 119905 and the server is free
1 the server is in a working vacation period at time 119905 and the server is busy
2 the server is during a normal service period at time 119905 and the server is free
3 the server is during a normal service period at time 119905 and the server is busy
(1)
Note that when the orbit is empty it is impossible that theserver is free in a normal service period Thus 119876(119905) 119869(119905) isa Markov process with state space
Ω = (0 119895) 119895 = 0 1 3 cup (119896 119895) 119896 ge 1 119895 = 1 2 3 (2)
Remark 1 Note that in this model we consider workingvacation interruption policy so the state (119896 0) 119896 ge 1 doesnot exist
Define
119886119896= int
infin
0
(120582119909)119896
119896
119890minus120582119909
120583119890minus120583119909
119889119909
119887119896= int
infin
0
(120582119909)119896
119896
119890minus120582119909
119890minus120579119909
120578119890minus120578119909
119889119909
119888119896= int
infin
0
(120582119909)119896
119896
119890minus120582119909
120579119890minus120579119909
119890minus120578119909
119889119909
(3)
where 119886119896 119896 ge 0 is the probability that 119896 customers arrive
during a normal service time 119887119896 119896 ge 0 represents the
probability that 119896 customers arrive during a lower servicetime and the working vacation does not end 119888
119896 119896 ge 0
explains the probability that 119896 customers arrive before theworking vacation ends and no service is completed
Let 119905119899 119899 = 1 2 be the sequence of epochs at which
a normal service or a lower service completion occurs and119876119899= 119876(119905
+
119899) then the sequence of random variables 119876
119899 119899 =
1 2 forms an embedded Markov chain
Theorem 2 The embedded Markov chain 119876119899 119899 = 1 2
is ergodic if and only if 120588 = 120582120583 lt 1
Proof It is not difficult to see that 119876119899 119899 = 1 2 is an
irreducible and aperiodic Markov chain and the transitionprobability is given by
119875 =
(
(
(
(
11987500
11987501
11987502
11987503
sdot sdot sdot
11987510
11987511
11987512
11987513
sdot sdot sdot
0 11987521
11987522
11987523
sdot sdot sdot
0 0 11987532
11987533
sdot sdot sdot
d
)
)
)
)
(4)
where
119875119899119896
=
119887119896+
119896
sum
119895=0
119888119895119886119896minus119895
119899 = 0 119896 ge 0
119899120572
119899120572 + 120582
1198860 119899 ge 1 119896 = 119899 minus 1
119899120572
119899120572 + 120582
119886119896+1minus119899
+
120582
119899120572 + 120582
119886119896minus119899
119899 ge 1 119896 ge 119899
(5)
In order to discuss the ergodicity of the chain 119876119899 119899 =
1 2 we use Fosterrsquos criterion with the mean drift
119909119899= 119864 [119891 (119876
119896+1) minus 119891 (119876
119896) | 119876119896= 119899] (6)
where the test function 119891(119899) = 119899 From the transition matrixit can be easily proved that
119909119899=
120588
120579
120579 + 120578
+
120582
120579 + 120578
119899 = 0
120588 minus
119899120572
119899120572 + 120582
119899 ge 1
(7)
As 119899 rarr infin there exists 119909 = lim119899rarrinfin
119909119899= 120588 minus 1 Applying
Fosterrsquos criterion we can get that the embeddedMarkov chainis ergodic if 120588 lt 1 Besides since 119876
119899+1minus 119876119899ge minus1 applying
the criterion from Sennott et al [21] we can obtain that 120588 lt 1
is necessary for ergodicity
Since the arrival process is Poisson using PASTA prop-erty it can be shown from Burkersquos theorem (see [22 pp 187-188]) that the steady state probabilities of the Markov process119876(119905) 119869(119905) exist if and only if the stable condition120588 lt 1holdsNow we define the limiting probability
120587119896119895= 119875 119876 = 119896 119869 = 119895
= lim119905rarrinfin
119875 119876 (119905) = 119896 119869 (119905) = 119895 (119896 119895) isin Ω
(8)
3 Balance Equations and ProbabilityGenerating Functions
Figure 1 presents the state transition rate diagram of thesystem Under the stable condition 120588 lt 1 the set of balanceequations is given as follows
4 Advances in Operations Research
0 1 1 1 2 1 3 1
0 0 1 2 2 2 3 2
0 3 1 3 2 3 3 3
120579
120579120579120579120579
120582
120582 120582 120582 120582
120582
120582 120582 120582
120582120582120582
120578 120578 120578 120578
120583120583 120583 120583
120572 2120572 3120572 4120572
Figure 1 State transition rate diagram of the system
12058212058700= 12057812058701+ 12058312058703 (9)
(120582 + 120579 + 120578) 12058701= 12058212058700 (10)
(120582 + 120583) 12058703= 12057212058712+ 12057912058701 (11)
(120582 + 120579 + 120578) 1205871198991= 120582120587119899minus11
119899 ge 1 (12)
(120582 + 119899120572) 1205871198992= 1205781205871198991+ 1205831205871198993 119899 ge 1 (13)
(120582 + 120583) 1205871198993= 1205821205871198992+ 120582120587119899minus13
+ (119899 + 1) 120572120587119899+12
+ 1205791205871198991 119899 ge 1
(14)
Define the probability generating functions (pgfs)
1198661 (119911) =
infin
sum
119899=0
1205871198991119911119899
1198662 (119911) =
infin
sum
119899=1
1205871198992119911119899
1198663 (119911) =
infin
sum
119899=0
1205871198993119911119899
(15)
with 1198661015840119894(119911) = sum
infin
119899=1119899120587119899119894119911119899minus1
119894 = 1 2 3Multiplying by the appropriate power of 119911119899 in (10) and
(12) summing over 119899 and rearranging the terms we get
(120582 + 120579 + 120578 minus 120582119911)1198661 (119911) = 120582120587
00 (16)
Multiplying by the appropriate power of 119911119899 in (13) summingover 119899 and using (9) we obtain
1205821198662 (119911) + 120572119911119866
1015840
2(119911) = 120578119866
1 (119911) + 120583119866
3 (119911) minus 120578120587
01
minus 12058312058703
= 1205781198661 (119911) + 120583119866
3 (119911) minus 120582120587
00
(17)
In a similar way from (11) and (14) we derive
(120582 + 120583 minus 120582119911)1198663 (119911) = 120582119866
2 (119911) + 120572119866
1015840
2(119911) + 120579119866
1 (119911) (18)
From (16) we get
1198661 (119911) =
120582
120582 + 120579 + 120578 minus 120582119911
12058700 (19)
From (17) we obtain
1198663 (119911) =
120582
120583
1198662 (119911) +
120572119911
120583
1198661015840
2(119911) minus
120578
120583
1198661 (119911) +
120582
120583
12058700 (20)
Taking (20) into (18) using (19) after some computation wecan derive
1198661015840
2(119911) minus
1205822
(120583 minus 120582119911) 120572
1198662 (119911)
=
120582212058700
(120583 minus 120582119911) (1 minus 119911) 120572
+
12058212058700
(1 minus 119911) 120572
minus
12057812058212058700
(120582 + 120579 + 120578 minus 120582119911) (1 minus 119911) 120572
minus
(120579120583 + 120582120578) 12058212058700
(120582 + 120579 + 120578 minus 120582119911) (120583 minus 120582119911) (1 minus 119911) 120572
(21)
Solving the first-order differential equation we have
1198662 (119911) = (120583 minus 120582119911)
minus120582120572[
120582212058700
120572
1198701 (119911) +
12058212058700
120572
1198702 (119911)
minus
12058212057812058700
120572
1198703 (119911) minus
(120579120583 + 120582120578) 12058212058700
120572
1198704 (119911)]
(22)
where
1198701 (119911) = int
119911
0
(1 minus 119909)minus1(120583 minus 120582119909)
120582120572minus1119889119909
1198702 (119911) = int
119911
0
(1 minus 119909)minus1(120583 minus 120582119909)
120582120572119889119909
1198703 (119911)
= int
119911
0
(120582 + 120579 + 120578 minus 120582119909)minus1
(1 minus 119909)minus1(120583 minus 120582119909)
120582120572119889119909
1198704 (119911)
= int
119911
0
(120582 + 120579 + 120578 minus 120582119909)minus1
(1 minus 119909)minus1(120583 minus 120582119909)
120582120572minus1119889119909
(23)
From (17) and (18) substituting 11986610158402(119911) we get
1198663 (119911) =
120582 (1 minus 119911) 1198662 (119911) minus (120578 + 120579119911)119866
1 (119911) + 120582120587
00
(120583 minus 120582119911) (1 minus 119911)
(24)
Advances in Operations Research 5
From (19) (22) and (24) we can see that 1198661(119911) 119866
2(119911) and
1198663(119911) are expressed in terms of 120587
00 In the next section using
normalization condition we will obtain 12058700
4 Performance Measures
From (19) we have
1198661 (1) =
120582
120579 + 120578
12058700 (25)
1198661015840
1(1) = lim
119911rarr1
1198661015840
1(119911) = lim
119911rarr1
1205822
(120582 + 120579 + 120578 minus 120582119911)212058700
=
1205822
(120579 + 120578)212058700
(26)
From (22) we get
1198662 (1) = 119870120587
00 (27)
where
119870 = (120583 minus 120582)minus120582120572
[
1205822
120572
1198701 (1) +
120582
120572
1198702 (1) minus
120582120578
120572
1198703 (1)
minus
(120579120583 + 120582120578) 120582
120572
1198704 (1)]
(28)
From (24) using LrsquoHospital rule we obtain
1198663 (1) = lim119911rarr1
1198663 (119911)
= lim119911rarr1
minus1205821198662 (119911)+ 120582 (1 minus 119911) 119866
1015840
2(119911) minus 120579119866
1 (119911) minus (120578 + 120579119911)119866
1015840
1(119911)
minus120582 (1 minus 119911) minus (120583 minus 120582119911)
=
1205821198662 (1) + 120579119866
1 (1) + (120578 + 120579)119866
1015840
1(1)
120583 minus 120582
=
120582119870 (120579 + 120578) + 120582120579 + 1205822
(120583 minus 120582) (120579 + 120578)
12058700
(29)
Since
12058700+ 1198661 (1) + 119866
2 (1) + 119866
3 (1) = 1 (30)
taking (25) (27) and (29) into the above equation we canderive
12058700= [1 +
120582
120579 + 120578
+ 119870 +
120582119870 (120579 + 120578) + 120582120579 + 1205822
(120583 minus 120582) (120579 + 120578)
]
minus1
= [
120582 (120583 minus 120578) + 120583 (1 + 119870) (120579 + 120578)
(120583 minus 120582) (120579 + 120578)
]
minus1
(31)
Thus 12058700
is found out From (19) (22) and (24) 1198661(119911)
1198662(119911) and 119866
3(119911) are completely computed And the pgf of
the number of customers in the orbit is given by
119866 (119911) = 12058700+ 1198661 (119911) + 119866
2 (119911) + 119866
3 (119911) (32)
The pgf of the number of customers in the system is given by
(119911) = 12058700+ 1199111198661 (119911) + 119866
2 (119911) + 119911119866
3 (119911) (33)
Let 119864[119871119895] denote the average number of customers in the
orbit when the serverrsquos state is 119895 119895 = 1 2 3Differentiating (19) with respect to 119911 we have
119864 [1198711] = lim119911rarr1
1198661015840
1(119911) =
1205822
(120579 + 120578)212058700 (34)
From (18) we can get
119864 [1198712] = lim119911rarr1
1198661015840
2(119911) =
120583
120572
1198663 (1) minus
120582
120572
1198662 (1) minus
120579
120572
1198661 (1) (35)
Differentiating (24) with respect to 119911 and using LrsquoHospitalrule twice the expression for 119864[119871
3] is obtained as
119864 [1198713] = lim119911rarr1
1198661015840
3(119911) =
(120583 minus 120582) (120579 + 120578)11986610158401015840
1(1) + 2 (120579120583 + 120582120578) 119864 [119871
1] + 2120582 (120583 minus 120582) 119864 [119871
2] + 2120582 [120579119866
1 (1) + 120582119866
2 (1)]
2 (120583 minus 120582)2
(36)
and differentiating (19) twice with respect to 119911 and taking 119911 =1 yield 11986610158401015840
1(1) as
11986610158401015840
1(1) =
21205823
(120579 + 120578)312058700 (37)
Therefore the average orbit length (119864[119871]) is given by
119864 [119871] = lim119911rarr1
1198661015840(119911) = 119864 [119871
1] + 119864 [119871
2] + 119864 [119871
3] (38)
And the average system length (119864[]) is derived as
119864 [] = lim119911rarr1
1198661015840(119911) = 119864 [119871] + 119866
1 (1) + 119866
3 (1) (39)
The probability that the server is busy is
119875119887= 119875 119869 = 1 + 119875 119869 = 3 = 119866
1 (1) + 119866
3 (1) (40)
6 Advances in Operations Research
The probability that the server is free is
119875119891= 119875 119869 = 0 + 119875 119869 = 2 = 120587
00+ 1198662 (1) = 1 minus 119875
119887 (41)
The probability that the server is in a working vacation periodis given by
119875119908= 119875 119869 = 0 + 119875 119869 = 1 = 120587
00+ 1198661 (1) (42)
The probability that the server is in a normal service periodis given by
119875119899= 119875 119869 = 2 + 119875 119869 = 3 = 119866
1 (2) + 119866
2 (3)
= 1 minus 119875119908
(43)
Let 119864[119882] (119864[]) be the expected waiting (sojourn) time of acustomer in the orbit (system) using Littlersquos formula
119864 [119882] =
119864 [119871]
120582
119864 [] =
119864 []
120582
(44)
The system busy period 119879 is defined as the period that startsat an epochwhen an arriving customer finds an empty systemand ends at the departure epoch at which the system is emptyUsing the theory of regenerative process we get
12058700=
119864 [11987900]
119864 [11987900] + 119864 [119879]
(45)
where 11987900
is the time length that the system is in the state(0 0) Since the interarrival time between two customersfollows exponential distribution with parameter 120582 we have119864[11987900] = 1120582 Thus 119864[119879] = 120582
minus1(120587minus1
00minus 1)
5 Numerical Results
In this section under the stable condition we present somenumerical examples to illustrate the effect of the varyingparameters on themeanorbit length119864[119871] the probability thatthe system is busy 1minus120587
00 and the probabilities of serverrsquos state
(119875119887 119875119891 119875119908 and 119875
119899) Moreover a cost minimization problem
is also discussed The various parameters of this model arearbitrarily chosen as 120582 = 2 120578 = 1 120583 = 5 120572 = 3 and120579 = 02 unless they are considered as variables or their valuesare mentioned in Figures 2 3 4 and 5
51 Sensitivity Analysis Figures 2 and 3 illustrate the effect of120579 on 119864[119871] and 1 minus 120587
00for different values of 120578 respectively
When 120578 lt 120583 we find that 119864[119871] and 1 minus 12058700
both decrease as120579 increases and 120579 has a noticeable effect on the performancemeasures which cannot be ignored An especial case is 120578 = 120583that is the service rate in the vacation period is equal to theservice rate in the normal period and the model we considerreduces to anMM1 retrial queue with classical retrial policyClearly 120579 has no effect on119864[119871] and 1minus120587
00 which agrees with
the intuitive expectation
0 1 2 3 4 5 6 7 8 9 10
08
1
12
14
16
18
2
Expe
cted
que
ue le
ngth
E[L]
120579
120578 = 1
120578 = 3120578 = 5
Figure 2 The effect of 120579 on 119864[119871] for different values of 120578
0 1 2 3 4 5 6 7 8 9 10055
06
065
07
075
08
085
09
120579
120578 = 1
120578 = 3120578 = 5
1minus12058700
Figure 3 The effect of 120579 on 1 minus 12058700
for different values of 120578
The effect of 120578 on 119864[119871] is presented in Figure 4 asexpected 119864[119871] decreases with increasing values of 120578 When120578 is large it can be observed that the effect of 120578 on 119864[119871] is notobvious as 120578 increases this is because we consider workingvacation interruption policy If the system is not empty at aservice completion instant in the vacation period the serverwill come back to the normal working level Moreover sinceconstant retrial policy means that only the customer at thehead of the orbit can request a service we can find that 119864[119871]of MM1 queue with classical retrial policy is smaller thanthat of MM1 queue with constant retrial policy As shownin Figure 5 119875
119887decreases with increasing values of 120578 and 119875
119891
has the opposite tendency The reason is that as 120578 increases
Advances in Operations Research 7
0 05 1 15 2 25 3 35 4 45 50
2
4
6
8
10
12
14Ex
pect
ed q
ueue
leng
th E[L]
MM1 with classical retrialMM1 with constant retrial
120578
Figure 4 The effect of 120578 on 119864[119871] for two different retrial policies
0 05 1 15 2 25 3 35 4 45 501
02
03
04
05
06
07
08
09
120578
PbPf
Figure 5 The probability of serverrsquos state with the change of 120578
themeanorbit length119864[119871] and themean service time 1120578bothdecrease
Figure 6 indicates that 119864[119871] decreases with increasingvalues of 120572 this is because the mean retrial time decreases as120572 increases From the instant when the server becomes idlean external potential primary customer and retrial customerscompete to access the server and the smaller the mean retrialtime is the bigger the probability that the server is busy iswhich leads to decreasing of 119864[119871] Furthermore under thesame condition decreasing of 119864[119871] results in the increasein the probability that the system is empty and Figure 7reveals that 119875
119908increases with increasing values of 120572 while 119875
119899
decreases as 120572 increases In Figure 6 compared with constantretrial policy we can also find that classical retrial policydecreases the waiting jobs effectively
2 25 3 35 4 45 5 55 615
2
25
3
35
4
45
5
55
6
Expe
cted
que
ue le
ngth
E[L]
MM1 with classical retrialMM1 with constant retrial
120572
Figure 6 The effect of 120572 on 119864[119871] for two different retrial policies
2 25 3 35 4 45 5 55 6035
04
045
05
055
06
065
07
PnPw
120572
Figure 7 The probability of serverrsquos state with the change of 120572
52 Cost Analysis In this subsection we establish a costfunction to search for the optimal service rate 120578 so as tominimize the expected operating cost function per unit time
Define the following cost elements
119862119871 cost per unit time for each customer present in the
orbit
119862120583 cost per unit time for service during a normal
service period
119862120578 cost per unit time for service in aworking vacation
period
119862120579 fixed cost per unit time during a vacation period
8 Advances in Operations Research
0 05 1 15 2 25 3 35 4 45 565
70
75
80
85
90
95
100
105
Expe
cted
ope
ratin
g co
st pe
r uni
t tim
e
120578
Figure 8 Effect of 120578 on the expected operating cost per unit time
Based on the definitions of each cost element listed abovethe expected operating cost function per unit time can begiven by
min120578
119891 (120578) = 119862119871119864 [119871] + 119862
120583120583 + 119862
120578120578 + 119862120579120579 (46)
Because the expected operating cost function is highly non-linear and complex it is not easy to get the derivative of itWeassume 119862
119871= 6 119862
120583= 10 119862
120578= 8 and 119862
120579= 4 and develop
approximations by MATLAB program to find the optimumvalue of 120578 say 120578lowast
From Figure 8 we can see that there is an optimal servicerate 120578 to minimize the cost Implementing the computersoftware MATLAB by the parabolic method and controllingthe error by 10
minus4 we find the solution 120578lowast
= 08584 with119891(120578lowast) = 690971
6 Conclusion
This paper analyzes a single server retrial queue with workingvacation interruption under the classical retrial policy Usingembedded Markov chain and Fosterrsquos criterion we get thecondition of stability The pgf of the number of customersin the orbit is obtained and some important performancemeasures are also discussed Moreover the effects of var-ious parameters on the system performance measures areexamined numerically Under the stable condition a costminimization problem is considered For future researchusing the same method one can investigate a similar modelwith batch arrival and without vacation interruption
Competing Interests
The authors declare that there are no competing interestsregarding the publication of this paper
Acknowledgments
This work is supported by the National Natural ScienceFoundation of China (nos 11301306 and 11401348) and theNatural Science Foundation of Anhui Higher EducationInstitutions (no KJ2014ZD21)
References
[1] N Tian and Z G ZhangVacation QueueingModels-Theory andApplications Springer New York NY USA 2006
[2] L D Servi and S G Finn ldquoMM1 queues with workingvacations (MM1WV)rdquo Performance Evaluation vol 50 no1 pp 41ndash52 2002
[3] Y Baba ldquoAnalysis of a GIM1 queue with multiple workingvacationsrdquo Operations Research Letters vol 33 no 2 pp 201ndash209 2005
[4] D-A Wu and H Takagi ldquoMG1 queue with multiple workingvacationsrdquo Performance Evaluation vol 63 no 7 pp 654ndash6812006
[5] J-H Li N-S Tian and W-Y Liu ldquoDiscrete-time GIGeo1queue with multiple working vacationsrdquoQueueing Systems vol56 no 1 pp 53ndash63 2007
[6] J Li andNTian ldquoTheMM1 queuewithworking vacations andvacation interruptionsrdquo Journal of Systems Science and SystemsEngineering vol 16 no 1 pp 121ndash127 2007
[7] J-H Li and N-S Tian ldquoThe discrete-time GIGeo1 queuewith working vacations and vacation interruptionrdquo AppliedMathematics and Computation vol 185 no 1 pp 1ndash10 2007
[8] J-H Li N-S Tian and Z-Y Ma ldquoPerformance analysis ofGIM1 queue with working vacations and vacation interrup-tionrdquo Applied Mathematical Modelling vol 32 no 12 pp 2715ndash2730 2008
[9] M Zhang and Z Hou ldquoPerformance analysis of MG1 queuewith working vacations and vacation interruptionrdquo Journal ofComputational and Applied Mathematics vol 234 no 10 pp2977ndash2985 2010
[10] J R Artalejo and A Gomez-Corral Retrial Queueing SystemsA Computational Approach Springer Berlin Germany 2008
[11] G Choudhury L Tadj and M Deka ldquoAn unreliable serverretrial queue with two phases of service and general retrialtimes under Bernoulli vacation schedulerdquo Quality Technologyamp Quantitative Management vol 12 no 4 pp 437ndash464 2015
[12] T V Do ldquoMM1 retrial queue with working vacationsrdquo ActaInformatica vol 47 no 1 pp 67ndash75 2010
[13] T Li Z Wang and Z Liu ldquoGeoGeo1 retrial queue with work-ing vacations and vacation interruptionrdquo Journal of AppliedMathematics and Computing vol 39 no 1-2 pp 131ndash143 2012
[14] Z Liu and Y Song ldquoGeoGeo1 retrial queue with non-persistent customers and working vacationsrdquo Journal of AppliedMathematics and Computing vol 42 no 1-2 pp 103ndash115 2013
[15] L Tao Z Liu andZWang ldquoMM1 retrial queuewith collisionsand working vacation interruption under N-policyrdquo RAIROOperations Research vol 46 no 4 pp 355ndash371 2012
[16] A Aissani S Taleb T Kernane G Saidi and D HamadoucheldquoAn MG1 retrial queue with working vacationrdquo in Advancesin Systems Science Proceedings of the International Conferenceon Systems Science 2013 (ICSS 2013) vol 240 of Advancesin Intelligent Systems and Computing pp 443ndash452 SpringerBerlin Germany 2014
Advances in Operations Research 9
[17] V Jailaxmi RArumuganathan andM S Kumar ldquoPerformanceanalysis of single server non-Markovian retrial queue withworking vacation and constant retrial policyrdquo RAIRO Opera-tions Research vol 48 no 3 pp 381ndash398 2014
[18] S Gao and J Wang ldquoDiscrete-time Geo119883G1 retrial queuewith general retrial times working vacations and vacationinterruptionrdquo Quality Technology amp Quantitative Managementvol 10 no 4 pp 495ndash512 2013
[19] S Gao J Wang and W W Li ldquoAn MG1 retrial queuewith general retrial times working vacations and vacationinterruptionrdquo Asia-Pacific Journal of Operational Research vol31 no 2 Article ID 1440006 25 pages 2014
[20] G Ayyappan A Ganapathy and G Sekar ldquoRetrial queueingsystemwith single working vacation under pre-emptive priorityservicerdquo International Journal of Computer Applications vol 2no 2 pp 28ndash35 2010
[21] L I Sennott P A Humblet and R L Tweedie ldquoMean drifts andthe nonergodicity of Markov chainsrdquo Operations Research vol31 no 4 pp 783ndash789 1983
[22] R B Cooper Introduction to QueueingTheory North-HollandNew York NY USA 1981
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical Problems in Engineering
Hindawi Publishing Corporationhttpwwwhindawicom
Differential EquationsInternational Journal of
Volume 2014
Applied MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical PhysicsAdvances in
Complex AnalysisJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
OptimizationJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Operations ResearchAdvances in
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Decision SciencesAdvances in
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
Advances in Operations Research 3
119869 (119905) =
0 the server is in a working vacation period at time 119905 and the server is free
1 the server is in a working vacation period at time 119905 and the server is busy
2 the server is during a normal service period at time 119905 and the server is free
3 the server is during a normal service period at time 119905 and the server is busy
(1)
Note that when the orbit is empty it is impossible that theserver is free in a normal service period Thus 119876(119905) 119869(119905) isa Markov process with state space
Ω = (0 119895) 119895 = 0 1 3 cup (119896 119895) 119896 ge 1 119895 = 1 2 3 (2)
Remark 1 Note that in this model we consider workingvacation interruption policy so the state (119896 0) 119896 ge 1 doesnot exist
Define
119886119896= int
infin
0
(120582119909)119896
119896
119890minus120582119909
120583119890minus120583119909
119889119909
119887119896= int
infin
0
(120582119909)119896
119896
119890minus120582119909
119890minus120579119909
120578119890minus120578119909
119889119909
119888119896= int
infin
0
(120582119909)119896
119896
119890minus120582119909
120579119890minus120579119909
119890minus120578119909
119889119909
(3)
where 119886119896 119896 ge 0 is the probability that 119896 customers arrive
during a normal service time 119887119896 119896 ge 0 represents the
probability that 119896 customers arrive during a lower servicetime and the working vacation does not end 119888
119896 119896 ge 0
explains the probability that 119896 customers arrive before theworking vacation ends and no service is completed
Let 119905119899 119899 = 1 2 be the sequence of epochs at which
a normal service or a lower service completion occurs and119876119899= 119876(119905
+
119899) then the sequence of random variables 119876
119899 119899 =
1 2 forms an embedded Markov chain
Theorem 2 The embedded Markov chain 119876119899 119899 = 1 2
is ergodic if and only if 120588 = 120582120583 lt 1
Proof It is not difficult to see that 119876119899 119899 = 1 2 is an
irreducible and aperiodic Markov chain and the transitionprobability is given by
119875 =
(
(
(
(
11987500
11987501
11987502
11987503
sdot sdot sdot
11987510
11987511
11987512
11987513
sdot sdot sdot
0 11987521
11987522
11987523
sdot sdot sdot
0 0 11987532
11987533
sdot sdot sdot
d
)
)
)
)
(4)
where
119875119899119896
=
119887119896+
119896
sum
119895=0
119888119895119886119896minus119895
119899 = 0 119896 ge 0
119899120572
119899120572 + 120582
1198860 119899 ge 1 119896 = 119899 minus 1
119899120572
119899120572 + 120582
119886119896+1minus119899
+
120582
119899120572 + 120582
119886119896minus119899
119899 ge 1 119896 ge 119899
(5)
In order to discuss the ergodicity of the chain 119876119899 119899 =
1 2 we use Fosterrsquos criterion with the mean drift
119909119899= 119864 [119891 (119876
119896+1) minus 119891 (119876
119896) | 119876119896= 119899] (6)
where the test function 119891(119899) = 119899 From the transition matrixit can be easily proved that
119909119899=
120588
120579
120579 + 120578
+
120582
120579 + 120578
119899 = 0
120588 minus
119899120572
119899120572 + 120582
119899 ge 1
(7)
As 119899 rarr infin there exists 119909 = lim119899rarrinfin
119909119899= 120588 minus 1 Applying
Fosterrsquos criterion we can get that the embeddedMarkov chainis ergodic if 120588 lt 1 Besides since 119876
119899+1minus 119876119899ge minus1 applying
the criterion from Sennott et al [21] we can obtain that 120588 lt 1
is necessary for ergodicity
Since the arrival process is Poisson using PASTA prop-erty it can be shown from Burkersquos theorem (see [22 pp 187-188]) that the steady state probabilities of the Markov process119876(119905) 119869(119905) exist if and only if the stable condition120588 lt 1holdsNow we define the limiting probability
120587119896119895= 119875 119876 = 119896 119869 = 119895
= lim119905rarrinfin
119875 119876 (119905) = 119896 119869 (119905) = 119895 (119896 119895) isin Ω
(8)
3 Balance Equations and ProbabilityGenerating Functions
Figure 1 presents the state transition rate diagram of thesystem Under the stable condition 120588 lt 1 the set of balanceequations is given as follows
4 Advances in Operations Research
0 1 1 1 2 1 3 1
0 0 1 2 2 2 3 2
0 3 1 3 2 3 3 3
120579
120579120579120579120579
120582
120582 120582 120582 120582
120582
120582 120582 120582
120582120582120582
120578 120578 120578 120578
120583120583 120583 120583
120572 2120572 3120572 4120572
Figure 1 State transition rate diagram of the system
12058212058700= 12057812058701+ 12058312058703 (9)
(120582 + 120579 + 120578) 12058701= 12058212058700 (10)
(120582 + 120583) 12058703= 12057212058712+ 12057912058701 (11)
(120582 + 120579 + 120578) 1205871198991= 120582120587119899minus11
119899 ge 1 (12)
(120582 + 119899120572) 1205871198992= 1205781205871198991+ 1205831205871198993 119899 ge 1 (13)
(120582 + 120583) 1205871198993= 1205821205871198992+ 120582120587119899minus13
+ (119899 + 1) 120572120587119899+12
+ 1205791205871198991 119899 ge 1
(14)
Define the probability generating functions (pgfs)
1198661 (119911) =
infin
sum
119899=0
1205871198991119911119899
1198662 (119911) =
infin
sum
119899=1
1205871198992119911119899
1198663 (119911) =
infin
sum
119899=0
1205871198993119911119899
(15)
with 1198661015840119894(119911) = sum
infin
119899=1119899120587119899119894119911119899minus1
119894 = 1 2 3Multiplying by the appropriate power of 119911119899 in (10) and
(12) summing over 119899 and rearranging the terms we get
(120582 + 120579 + 120578 minus 120582119911)1198661 (119911) = 120582120587
00 (16)
Multiplying by the appropriate power of 119911119899 in (13) summingover 119899 and using (9) we obtain
1205821198662 (119911) + 120572119911119866
1015840
2(119911) = 120578119866
1 (119911) + 120583119866
3 (119911) minus 120578120587
01
minus 12058312058703
= 1205781198661 (119911) + 120583119866
3 (119911) minus 120582120587
00
(17)
In a similar way from (11) and (14) we derive
(120582 + 120583 minus 120582119911)1198663 (119911) = 120582119866
2 (119911) + 120572119866
1015840
2(119911) + 120579119866
1 (119911) (18)
From (16) we get
1198661 (119911) =
120582
120582 + 120579 + 120578 minus 120582119911
12058700 (19)
From (17) we obtain
1198663 (119911) =
120582
120583
1198662 (119911) +
120572119911
120583
1198661015840
2(119911) minus
120578
120583
1198661 (119911) +
120582
120583
12058700 (20)
Taking (20) into (18) using (19) after some computation wecan derive
1198661015840
2(119911) minus
1205822
(120583 minus 120582119911) 120572
1198662 (119911)
=
120582212058700
(120583 minus 120582119911) (1 minus 119911) 120572
+
12058212058700
(1 minus 119911) 120572
minus
12057812058212058700
(120582 + 120579 + 120578 minus 120582119911) (1 minus 119911) 120572
minus
(120579120583 + 120582120578) 12058212058700
(120582 + 120579 + 120578 minus 120582119911) (120583 minus 120582119911) (1 minus 119911) 120572
(21)
Solving the first-order differential equation we have
1198662 (119911) = (120583 minus 120582119911)
minus120582120572[
120582212058700
120572
1198701 (119911) +
12058212058700
120572
1198702 (119911)
minus
12058212057812058700
120572
1198703 (119911) minus
(120579120583 + 120582120578) 12058212058700
120572
1198704 (119911)]
(22)
where
1198701 (119911) = int
119911
0
(1 minus 119909)minus1(120583 minus 120582119909)
120582120572minus1119889119909
1198702 (119911) = int
119911
0
(1 minus 119909)minus1(120583 minus 120582119909)
120582120572119889119909
1198703 (119911)
= int
119911
0
(120582 + 120579 + 120578 minus 120582119909)minus1
(1 minus 119909)minus1(120583 minus 120582119909)
120582120572119889119909
1198704 (119911)
= int
119911
0
(120582 + 120579 + 120578 minus 120582119909)minus1
(1 minus 119909)minus1(120583 minus 120582119909)
120582120572minus1119889119909
(23)
From (17) and (18) substituting 11986610158402(119911) we get
1198663 (119911) =
120582 (1 minus 119911) 1198662 (119911) minus (120578 + 120579119911)119866
1 (119911) + 120582120587
00
(120583 minus 120582119911) (1 minus 119911)
(24)
Advances in Operations Research 5
From (19) (22) and (24) we can see that 1198661(119911) 119866
2(119911) and
1198663(119911) are expressed in terms of 120587
00 In the next section using
normalization condition we will obtain 12058700
4 Performance Measures
From (19) we have
1198661 (1) =
120582
120579 + 120578
12058700 (25)
1198661015840
1(1) = lim
119911rarr1
1198661015840
1(119911) = lim
119911rarr1
1205822
(120582 + 120579 + 120578 minus 120582119911)212058700
=
1205822
(120579 + 120578)212058700
(26)
From (22) we get
1198662 (1) = 119870120587
00 (27)
where
119870 = (120583 minus 120582)minus120582120572
[
1205822
120572
1198701 (1) +
120582
120572
1198702 (1) minus
120582120578
120572
1198703 (1)
minus
(120579120583 + 120582120578) 120582
120572
1198704 (1)]
(28)
From (24) using LrsquoHospital rule we obtain
1198663 (1) = lim119911rarr1
1198663 (119911)
= lim119911rarr1
minus1205821198662 (119911)+ 120582 (1 minus 119911) 119866
1015840
2(119911) minus 120579119866
1 (119911) minus (120578 + 120579119911)119866
1015840
1(119911)
minus120582 (1 minus 119911) minus (120583 minus 120582119911)
=
1205821198662 (1) + 120579119866
1 (1) + (120578 + 120579)119866
1015840
1(1)
120583 minus 120582
=
120582119870 (120579 + 120578) + 120582120579 + 1205822
(120583 minus 120582) (120579 + 120578)
12058700
(29)
Since
12058700+ 1198661 (1) + 119866
2 (1) + 119866
3 (1) = 1 (30)
taking (25) (27) and (29) into the above equation we canderive
12058700= [1 +
120582
120579 + 120578
+ 119870 +
120582119870 (120579 + 120578) + 120582120579 + 1205822
(120583 minus 120582) (120579 + 120578)
]
minus1
= [
120582 (120583 minus 120578) + 120583 (1 + 119870) (120579 + 120578)
(120583 minus 120582) (120579 + 120578)
]
minus1
(31)
Thus 12058700
is found out From (19) (22) and (24) 1198661(119911)
1198662(119911) and 119866
3(119911) are completely computed And the pgf of
the number of customers in the orbit is given by
119866 (119911) = 12058700+ 1198661 (119911) + 119866
2 (119911) + 119866
3 (119911) (32)
The pgf of the number of customers in the system is given by
(119911) = 12058700+ 1199111198661 (119911) + 119866
2 (119911) + 119911119866
3 (119911) (33)
Let 119864[119871119895] denote the average number of customers in the
orbit when the serverrsquos state is 119895 119895 = 1 2 3Differentiating (19) with respect to 119911 we have
119864 [1198711] = lim119911rarr1
1198661015840
1(119911) =
1205822
(120579 + 120578)212058700 (34)
From (18) we can get
119864 [1198712] = lim119911rarr1
1198661015840
2(119911) =
120583
120572
1198663 (1) minus
120582
120572
1198662 (1) minus
120579
120572
1198661 (1) (35)
Differentiating (24) with respect to 119911 and using LrsquoHospitalrule twice the expression for 119864[119871
3] is obtained as
119864 [1198713] = lim119911rarr1
1198661015840
3(119911) =
(120583 minus 120582) (120579 + 120578)11986610158401015840
1(1) + 2 (120579120583 + 120582120578) 119864 [119871
1] + 2120582 (120583 minus 120582) 119864 [119871
2] + 2120582 [120579119866
1 (1) + 120582119866
2 (1)]
2 (120583 minus 120582)2
(36)
and differentiating (19) twice with respect to 119911 and taking 119911 =1 yield 11986610158401015840
1(1) as
11986610158401015840
1(1) =
21205823
(120579 + 120578)312058700 (37)
Therefore the average orbit length (119864[119871]) is given by
119864 [119871] = lim119911rarr1
1198661015840(119911) = 119864 [119871
1] + 119864 [119871
2] + 119864 [119871
3] (38)
And the average system length (119864[]) is derived as
119864 [] = lim119911rarr1
1198661015840(119911) = 119864 [119871] + 119866
1 (1) + 119866
3 (1) (39)
The probability that the server is busy is
119875119887= 119875 119869 = 1 + 119875 119869 = 3 = 119866
1 (1) + 119866
3 (1) (40)
6 Advances in Operations Research
The probability that the server is free is
119875119891= 119875 119869 = 0 + 119875 119869 = 2 = 120587
00+ 1198662 (1) = 1 minus 119875
119887 (41)
The probability that the server is in a working vacation periodis given by
119875119908= 119875 119869 = 0 + 119875 119869 = 1 = 120587
00+ 1198661 (1) (42)
The probability that the server is in a normal service periodis given by
119875119899= 119875 119869 = 2 + 119875 119869 = 3 = 119866
1 (2) + 119866
2 (3)
= 1 minus 119875119908
(43)
Let 119864[119882] (119864[]) be the expected waiting (sojourn) time of acustomer in the orbit (system) using Littlersquos formula
119864 [119882] =
119864 [119871]
120582
119864 [] =
119864 []
120582
(44)
The system busy period 119879 is defined as the period that startsat an epochwhen an arriving customer finds an empty systemand ends at the departure epoch at which the system is emptyUsing the theory of regenerative process we get
12058700=
119864 [11987900]
119864 [11987900] + 119864 [119879]
(45)
where 11987900
is the time length that the system is in the state(0 0) Since the interarrival time between two customersfollows exponential distribution with parameter 120582 we have119864[11987900] = 1120582 Thus 119864[119879] = 120582
minus1(120587minus1
00minus 1)
5 Numerical Results
In this section under the stable condition we present somenumerical examples to illustrate the effect of the varyingparameters on themeanorbit length119864[119871] the probability thatthe system is busy 1minus120587
00 and the probabilities of serverrsquos state
(119875119887 119875119891 119875119908 and 119875
119899) Moreover a cost minimization problem
is also discussed The various parameters of this model arearbitrarily chosen as 120582 = 2 120578 = 1 120583 = 5 120572 = 3 and120579 = 02 unless they are considered as variables or their valuesare mentioned in Figures 2 3 4 and 5
51 Sensitivity Analysis Figures 2 and 3 illustrate the effect of120579 on 119864[119871] and 1 minus 120587
00for different values of 120578 respectively
When 120578 lt 120583 we find that 119864[119871] and 1 minus 12058700
both decrease as120579 increases and 120579 has a noticeable effect on the performancemeasures which cannot be ignored An especial case is 120578 = 120583that is the service rate in the vacation period is equal to theservice rate in the normal period and the model we considerreduces to anMM1 retrial queue with classical retrial policyClearly 120579 has no effect on119864[119871] and 1minus120587
00 which agrees with
the intuitive expectation
0 1 2 3 4 5 6 7 8 9 10
08
1
12
14
16
18
2
Expe
cted
que
ue le
ngth
E[L]
120579
120578 = 1
120578 = 3120578 = 5
Figure 2 The effect of 120579 on 119864[119871] for different values of 120578
0 1 2 3 4 5 6 7 8 9 10055
06
065
07
075
08
085
09
120579
120578 = 1
120578 = 3120578 = 5
1minus12058700
Figure 3 The effect of 120579 on 1 minus 12058700
for different values of 120578
The effect of 120578 on 119864[119871] is presented in Figure 4 asexpected 119864[119871] decreases with increasing values of 120578 When120578 is large it can be observed that the effect of 120578 on 119864[119871] is notobvious as 120578 increases this is because we consider workingvacation interruption policy If the system is not empty at aservice completion instant in the vacation period the serverwill come back to the normal working level Moreover sinceconstant retrial policy means that only the customer at thehead of the orbit can request a service we can find that 119864[119871]of MM1 queue with classical retrial policy is smaller thanthat of MM1 queue with constant retrial policy As shownin Figure 5 119875
119887decreases with increasing values of 120578 and 119875
119891
has the opposite tendency The reason is that as 120578 increases
Advances in Operations Research 7
0 05 1 15 2 25 3 35 4 45 50
2
4
6
8
10
12
14Ex
pect
ed q
ueue
leng
th E[L]
MM1 with classical retrialMM1 with constant retrial
120578
Figure 4 The effect of 120578 on 119864[119871] for two different retrial policies
0 05 1 15 2 25 3 35 4 45 501
02
03
04
05
06
07
08
09
120578
PbPf
Figure 5 The probability of serverrsquos state with the change of 120578
themeanorbit length119864[119871] and themean service time 1120578bothdecrease
Figure 6 indicates that 119864[119871] decreases with increasingvalues of 120572 this is because the mean retrial time decreases as120572 increases From the instant when the server becomes idlean external potential primary customer and retrial customerscompete to access the server and the smaller the mean retrialtime is the bigger the probability that the server is busy iswhich leads to decreasing of 119864[119871] Furthermore under thesame condition decreasing of 119864[119871] results in the increasein the probability that the system is empty and Figure 7reveals that 119875
119908increases with increasing values of 120572 while 119875
119899
decreases as 120572 increases In Figure 6 compared with constantretrial policy we can also find that classical retrial policydecreases the waiting jobs effectively
2 25 3 35 4 45 5 55 615
2
25
3
35
4
45
5
55
6
Expe
cted
que
ue le
ngth
E[L]
MM1 with classical retrialMM1 with constant retrial
120572
Figure 6 The effect of 120572 on 119864[119871] for two different retrial policies
2 25 3 35 4 45 5 55 6035
04
045
05
055
06
065
07
PnPw
120572
Figure 7 The probability of serverrsquos state with the change of 120572
52 Cost Analysis In this subsection we establish a costfunction to search for the optimal service rate 120578 so as tominimize the expected operating cost function per unit time
Define the following cost elements
119862119871 cost per unit time for each customer present in the
orbit
119862120583 cost per unit time for service during a normal
service period
119862120578 cost per unit time for service in aworking vacation
period
119862120579 fixed cost per unit time during a vacation period
8 Advances in Operations Research
0 05 1 15 2 25 3 35 4 45 565
70
75
80
85
90
95
100
105
Expe
cted
ope
ratin
g co
st pe
r uni
t tim
e
120578
Figure 8 Effect of 120578 on the expected operating cost per unit time
Based on the definitions of each cost element listed abovethe expected operating cost function per unit time can begiven by
min120578
119891 (120578) = 119862119871119864 [119871] + 119862
120583120583 + 119862
120578120578 + 119862120579120579 (46)
Because the expected operating cost function is highly non-linear and complex it is not easy to get the derivative of itWeassume 119862
119871= 6 119862
120583= 10 119862
120578= 8 and 119862
120579= 4 and develop
approximations by MATLAB program to find the optimumvalue of 120578 say 120578lowast
From Figure 8 we can see that there is an optimal servicerate 120578 to minimize the cost Implementing the computersoftware MATLAB by the parabolic method and controllingthe error by 10
minus4 we find the solution 120578lowast
= 08584 with119891(120578lowast) = 690971
6 Conclusion
This paper analyzes a single server retrial queue with workingvacation interruption under the classical retrial policy Usingembedded Markov chain and Fosterrsquos criterion we get thecondition of stability The pgf of the number of customersin the orbit is obtained and some important performancemeasures are also discussed Moreover the effects of var-ious parameters on the system performance measures areexamined numerically Under the stable condition a costminimization problem is considered For future researchusing the same method one can investigate a similar modelwith batch arrival and without vacation interruption
Competing Interests
The authors declare that there are no competing interestsregarding the publication of this paper
Acknowledgments
This work is supported by the National Natural ScienceFoundation of China (nos 11301306 and 11401348) and theNatural Science Foundation of Anhui Higher EducationInstitutions (no KJ2014ZD21)
References
[1] N Tian and Z G ZhangVacation QueueingModels-Theory andApplications Springer New York NY USA 2006
[2] L D Servi and S G Finn ldquoMM1 queues with workingvacations (MM1WV)rdquo Performance Evaluation vol 50 no1 pp 41ndash52 2002
[3] Y Baba ldquoAnalysis of a GIM1 queue with multiple workingvacationsrdquo Operations Research Letters vol 33 no 2 pp 201ndash209 2005
[4] D-A Wu and H Takagi ldquoMG1 queue with multiple workingvacationsrdquo Performance Evaluation vol 63 no 7 pp 654ndash6812006
[5] J-H Li N-S Tian and W-Y Liu ldquoDiscrete-time GIGeo1queue with multiple working vacationsrdquoQueueing Systems vol56 no 1 pp 53ndash63 2007
[6] J Li andNTian ldquoTheMM1 queuewithworking vacations andvacation interruptionsrdquo Journal of Systems Science and SystemsEngineering vol 16 no 1 pp 121ndash127 2007
[7] J-H Li and N-S Tian ldquoThe discrete-time GIGeo1 queuewith working vacations and vacation interruptionrdquo AppliedMathematics and Computation vol 185 no 1 pp 1ndash10 2007
[8] J-H Li N-S Tian and Z-Y Ma ldquoPerformance analysis ofGIM1 queue with working vacations and vacation interrup-tionrdquo Applied Mathematical Modelling vol 32 no 12 pp 2715ndash2730 2008
[9] M Zhang and Z Hou ldquoPerformance analysis of MG1 queuewith working vacations and vacation interruptionrdquo Journal ofComputational and Applied Mathematics vol 234 no 10 pp2977ndash2985 2010
[10] J R Artalejo and A Gomez-Corral Retrial Queueing SystemsA Computational Approach Springer Berlin Germany 2008
[11] G Choudhury L Tadj and M Deka ldquoAn unreliable serverretrial queue with two phases of service and general retrialtimes under Bernoulli vacation schedulerdquo Quality Technologyamp Quantitative Management vol 12 no 4 pp 437ndash464 2015
[12] T V Do ldquoMM1 retrial queue with working vacationsrdquo ActaInformatica vol 47 no 1 pp 67ndash75 2010
[13] T Li Z Wang and Z Liu ldquoGeoGeo1 retrial queue with work-ing vacations and vacation interruptionrdquo Journal of AppliedMathematics and Computing vol 39 no 1-2 pp 131ndash143 2012
[14] Z Liu and Y Song ldquoGeoGeo1 retrial queue with non-persistent customers and working vacationsrdquo Journal of AppliedMathematics and Computing vol 42 no 1-2 pp 103ndash115 2013
[15] L Tao Z Liu andZWang ldquoMM1 retrial queuewith collisionsand working vacation interruption under N-policyrdquo RAIROOperations Research vol 46 no 4 pp 355ndash371 2012
[16] A Aissani S Taleb T Kernane G Saidi and D HamadoucheldquoAn MG1 retrial queue with working vacationrdquo in Advancesin Systems Science Proceedings of the International Conferenceon Systems Science 2013 (ICSS 2013) vol 240 of Advancesin Intelligent Systems and Computing pp 443ndash452 SpringerBerlin Germany 2014
Advances in Operations Research 9
[17] V Jailaxmi RArumuganathan andM S Kumar ldquoPerformanceanalysis of single server non-Markovian retrial queue withworking vacation and constant retrial policyrdquo RAIRO Opera-tions Research vol 48 no 3 pp 381ndash398 2014
[18] S Gao and J Wang ldquoDiscrete-time Geo119883G1 retrial queuewith general retrial times working vacations and vacationinterruptionrdquo Quality Technology amp Quantitative Managementvol 10 no 4 pp 495ndash512 2013
[19] S Gao J Wang and W W Li ldquoAn MG1 retrial queuewith general retrial times working vacations and vacationinterruptionrdquo Asia-Pacific Journal of Operational Research vol31 no 2 Article ID 1440006 25 pages 2014
[20] G Ayyappan A Ganapathy and G Sekar ldquoRetrial queueingsystemwith single working vacation under pre-emptive priorityservicerdquo International Journal of Computer Applications vol 2no 2 pp 28ndash35 2010
[21] L I Sennott P A Humblet and R L Tweedie ldquoMean drifts andthe nonergodicity of Markov chainsrdquo Operations Research vol31 no 4 pp 783ndash789 1983
[22] R B Cooper Introduction to QueueingTheory North-HollandNew York NY USA 1981
Submit your manuscripts athttpwwwhindawicom
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Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical Problems in Engineering
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Differential EquationsInternational Journal of
Volume 2014
Applied MathematicsJournal of
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Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical PhysicsAdvances in
Complex AnalysisJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
OptimizationJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Operations ResearchAdvances in
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Decision SciencesAdvances in
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
4 Advances in Operations Research
0 1 1 1 2 1 3 1
0 0 1 2 2 2 3 2
0 3 1 3 2 3 3 3
120579
120579120579120579120579
120582
120582 120582 120582 120582
120582
120582 120582 120582
120582120582120582
120578 120578 120578 120578
120583120583 120583 120583
120572 2120572 3120572 4120572
Figure 1 State transition rate diagram of the system
12058212058700= 12057812058701+ 12058312058703 (9)
(120582 + 120579 + 120578) 12058701= 12058212058700 (10)
(120582 + 120583) 12058703= 12057212058712+ 12057912058701 (11)
(120582 + 120579 + 120578) 1205871198991= 120582120587119899minus11
119899 ge 1 (12)
(120582 + 119899120572) 1205871198992= 1205781205871198991+ 1205831205871198993 119899 ge 1 (13)
(120582 + 120583) 1205871198993= 1205821205871198992+ 120582120587119899minus13
+ (119899 + 1) 120572120587119899+12
+ 1205791205871198991 119899 ge 1
(14)
Define the probability generating functions (pgfs)
1198661 (119911) =
infin
sum
119899=0
1205871198991119911119899
1198662 (119911) =
infin
sum
119899=1
1205871198992119911119899
1198663 (119911) =
infin
sum
119899=0
1205871198993119911119899
(15)
with 1198661015840119894(119911) = sum
infin
119899=1119899120587119899119894119911119899minus1
119894 = 1 2 3Multiplying by the appropriate power of 119911119899 in (10) and
(12) summing over 119899 and rearranging the terms we get
(120582 + 120579 + 120578 minus 120582119911)1198661 (119911) = 120582120587
00 (16)
Multiplying by the appropriate power of 119911119899 in (13) summingover 119899 and using (9) we obtain
1205821198662 (119911) + 120572119911119866
1015840
2(119911) = 120578119866
1 (119911) + 120583119866
3 (119911) minus 120578120587
01
minus 12058312058703
= 1205781198661 (119911) + 120583119866
3 (119911) minus 120582120587
00
(17)
In a similar way from (11) and (14) we derive
(120582 + 120583 minus 120582119911)1198663 (119911) = 120582119866
2 (119911) + 120572119866
1015840
2(119911) + 120579119866
1 (119911) (18)
From (16) we get
1198661 (119911) =
120582
120582 + 120579 + 120578 minus 120582119911
12058700 (19)
From (17) we obtain
1198663 (119911) =
120582
120583
1198662 (119911) +
120572119911
120583
1198661015840
2(119911) minus
120578
120583
1198661 (119911) +
120582
120583
12058700 (20)
Taking (20) into (18) using (19) after some computation wecan derive
1198661015840
2(119911) minus
1205822
(120583 minus 120582119911) 120572
1198662 (119911)
=
120582212058700
(120583 minus 120582119911) (1 minus 119911) 120572
+
12058212058700
(1 minus 119911) 120572
minus
12057812058212058700
(120582 + 120579 + 120578 minus 120582119911) (1 minus 119911) 120572
minus
(120579120583 + 120582120578) 12058212058700
(120582 + 120579 + 120578 minus 120582119911) (120583 minus 120582119911) (1 minus 119911) 120572
(21)
Solving the first-order differential equation we have
1198662 (119911) = (120583 minus 120582119911)
minus120582120572[
120582212058700
120572
1198701 (119911) +
12058212058700
120572
1198702 (119911)
minus
12058212057812058700
120572
1198703 (119911) minus
(120579120583 + 120582120578) 12058212058700
120572
1198704 (119911)]
(22)
where
1198701 (119911) = int
119911
0
(1 minus 119909)minus1(120583 minus 120582119909)
120582120572minus1119889119909
1198702 (119911) = int
119911
0
(1 minus 119909)minus1(120583 minus 120582119909)
120582120572119889119909
1198703 (119911)
= int
119911
0
(120582 + 120579 + 120578 minus 120582119909)minus1
(1 minus 119909)minus1(120583 minus 120582119909)
120582120572119889119909
1198704 (119911)
= int
119911
0
(120582 + 120579 + 120578 minus 120582119909)minus1
(1 minus 119909)minus1(120583 minus 120582119909)
120582120572minus1119889119909
(23)
From (17) and (18) substituting 11986610158402(119911) we get
1198663 (119911) =
120582 (1 minus 119911) 1198662 (119911) minus (120578 + 120579119911)119866
1 (119911) + 120582120587
00
(120583 minus 120582119911) (1 minus 119911)
(24)
Advances in Operations Research 5
From (19) (22) and (24) we can see that 1198661(119911) 119866
2(119911) and
1198663(119911) are expressed in terms of 120587
00 In the next section using
normalization condition we will obtain 12058700
4 Performance Measures
From (19) we have
1198661 (1) =
120582
120579 + 120578
12058700 (25)
1198661015840
1(1) = lim
119911rarr1
1198661015840
1(119911) = lim
119911rarr1
1205822
(120582 + 120579 + 120578 minus 120582119911)212058700
=
1205822
(120579 + 120578)212058700
(26)
From (22) we get
1198662 (1) = 119870120587
00 (27)
where
119870 = (120583 minus 120582)minus120582120572
[
1205822
120572
1198701 (1) +
120582
120572
1198702 (1) minus
120582120578
120572
1198703 (1)
minus
(120579120583 + 120582120578) 120582
120572
1198704 (1)]
(28)
From (24) using LrsquoHospital rule we obtain
1198663 (1) = lim119911rarr1
1198663 (119911)
= lim119911rarr1
minus1205821198662 (119911)+ 120582 (1 minus 119911) 119866
1015840
2(119911) minus 120579119866
1 (119911) minus (120578 + 120579119911)119866
1015840
1(119911)
minus120582 (1 minus 119911) minus (120583 minus 120582119911)
=
1205821198662 (1) + 120579119866
1 (1) + (120578 + 120579)119866
1015840
1(1)
120583 minus 120582
=
120582119870 (120579 + 120578) + 120582120579 + 1205822
(120583 minus 120582) (120579 + 120578)
12058700
(29)
Since
12058700+ 1198661 (1) + 119866
2 (1) + 119866
3 (1) = 1 (30)
taking (25) (27) and (29) into the above equation we canderive
12058700= [1 +
120582
120579 + 120578
+ 119870 +
120582119870 (120579 + 120578) + 120582120579 + 1205822
(120583 minus 120582) (120579 + 120578)
]
minus1
= [
120582 (120583 minus 120578) + 120583 (1 + 119870) (120579 + 120578)
(120583 minus 120582) (120579 + 120578)
]
minus1
(31)
Thus 12058700
is found out From (19) (22) and (24) 1198661(119911)
1198662(119911) and 119866
3(119911) are completely computed And the pgf of
the number of customers in the orbit is given by
119866 (119911) = 12058700+ 1198661 (119911) + 119866
2 (119911) + 119866
3 (119911) (32)
The pgf of the number of customers in the system is given by
(119911) = 12058700+ 1199111198661 (119911) + 119866
2 (119911) + 119911119866
3 (119911) (33)
Let 119864[119871119895] denote the average number of customers in the
orbit when the serverrsquos state is 119895 119895 = 1 2 3Differentiating (19) with respect to 119911 we have
119864 [1198711] = lim119911rarr1
1198661015840
1(119911) =
1205822
(120579 + 120578)212058700 (34)
From (18) we can get
119864 [1198712] = lim119911rarr1
1198661015840
2(119911) =
120583
120572
1198663 (1) minus
120582
120572
1198662 (1) minus
120579
120572
1198661 (1) (35)
Differentiating (24) with respect to 119911 and using LrsquoHospitalrule twice the expression for 119864[119871
3] is obtained as
119864 [1198713] = lim119911rarr1
1198661015840
3(119911) =
(120583 minus 120582) (120579 + 120578)11986610158401015840
1(1) + 2 (120579120583 + 120582120578) 119864 [119871
1] + 2120582 (120583 minus 120582) 119864 [119871
2] + 2120582 [120579119866
1 (1) + 120582119866
2 (1)]
2 (120583 minus 120582)2
(36)
and differentiating (19) twice with respect to 119911 and taking 119911 =1 yield 11986610158401015840
1(1) as
11986610158401015840
1(1) =
21205823
(120579 + 120578)312058700 (37)
Therefore the average orbit length (119864[119871]) is given by
119864 [119871] = lim119911rarr1
1198661015840(119911) = 119864 [119871
1] + 119864 [119871
2] + 119864 [119871
3] (38)
And the average system length (119864[]) is derived as
119864 [] = lim119911rarr1
1198661015840(119911) = 119864 [119871] + 119866
1 (1) + 119866
3 (1) (39)
The probability that the server is busy is
119875119887= 119875 119869 = 1 + 119875 119869 = 3 = 119866
1 (1) + 119866
3 (1) (40)
6 Advances in Operations Research
The probability that the server is free is
119875119891= 119875 119869 = 0 + 119875 119869 = 2 = 120587
00+ 1198662 (1) = 1 minus 119875
119887 (41)
The probability that the server is in a working vacation periodis given by
119875119908= 119875 119869 = 0 + 119875 119869 = 1 = 120587
00+ 1198661 (1) (42)
The probability that the server is in a normal service periodis given by
119875119899= 119875 119869 = 2 + 119875 119869 = 3 = 119866
1 (2) + 119866
2 (3)
= 1 minus 119875119908
(43)
Let 119864[119882] (119864[]) be the expected waiting (sojourn) time of acustomer in the orbit (system) using Littlersquos formula
119864 [119882] =
119864 [119871]
120582
119864 [] =
119864 []
120582
(44)
The system busy period 119879 is defined as the period that startsat an epochwhen an arriving customer finds an empty systemand ends at the departure epoch at which the system is emptyUsing the theory of regenerative process we get
12058700=
119864 [11987900]
119864 [11987900] + 119864 [119879]
(45)
where 11987900
is the time length that the system is in the state(0 0) Since the interarrival time between two customersfollows exponential distribution with parameter 120582 we have119864[11987900] = 1120582 Thus 119864[119879] = 120582
minus1(120587minus1
00minus 1)
5 Numerical Results
In this section under the stable condition we present somenumerical examples to illustrate the effect of the varyingparameters on themeanorbit length119864[119871] the probability thatthe system is busy 1minus120587
00 and the probabilities of serverrsquos state
(119875119887 119875119891 119875119908 and 119875
119899) Moreover a cost minimization problem
is also discussed The various parameters of this model arearbitrarily chosen as 120582 = 2 120578 = 1 120583 = 5 120572 = 3 and120579 = 02 unless they are considered as variables or their valuesare mentioned in Figures 2 3 4 and 5
51 Sensitivity Analysis Figures 2 and 3 illustrate the effect of120579 on 119864[119871] and 1 minus 120587
00for different values of 120578 respectively
When 120578 lt 120583 we find that 119864[119871] and 1 minus 12058700
both decrease as120579 increases and 120579 has a noticeable effect on the performancemeasures which cannot be ignored An especial case is 120578 = 120583that is the service rate in the vacation period is equal to theservice rate in the normal period and the model we considerreduces to anMM1 retrial queue with classical retrial policyClearly 120579 has no effect on119864[119871] and 1minus120587
00 which agrees with
the intuitive expectation
0 1 2 3 4 5 6 7 8 9 10
08
1
12
14
16
18
2
Expe
cted
que
ue le
ngth
E[L]
120579
120578 = 1
120578 = 3120578 = 5
Figure 2 The effect of 120579 on 119864[119871] for different values of 120578
0 1 2 3 4 5 6 7 8 9 10055
06
065
07
075
08
085
09
120579
120578 = 1
120578 = 3120578 = 5
1minus12058700
Figure 3 The effect of 120579 on 1 minus 12058700
for different values of 120578
The effect of 120578 on 119864[119871] is presented in Figure 4 asexpected 119864[119871] decreases with increasing values of 120578 When120578 is large it can be observed that the effect of 120578 on 119864[119871] is notobvious as 120578 increases this is because we consider workingvacation interruption policy If the system is not empty at aservice completion instant in the vacation period the serverwill come back to the normal working level Moreover sinceconstant retrial policy means that only the customer at thehead of the orbit can request a service we can find that 119864[119871]of MM1 queue with classical retrial policy is smaller thanthat of MM1 queue with constant retrial policy As shownin Figure 5 119875
119887decreases with increasing values of 120578 and 119875
119891
has the opposite tendency The reason is that as 120578 increases
Advances in Operations Research 7
0 05 1 15 2 25 3 35 4 45 50
2
4
6
8
10
12
14Ex
pect
ed q
ueue
leng
th E[L]
MM1 with classical retrialMM1 with constant retrial
120578
Figure 4 The effect of 120578 on 119864[119871] for two different retrial policies
0 05 1 15 2 25 3 35 4 45 501
02
03
04
05
06
07
08
09
120578
PbPf
Figure 5 The probability of serverrsquos state with the change of 120578
themeanorbit length119864[119871] and themean service time 1120578bothdecrease
Figure 6 indicates that 119864[119871] decreases with increasingvalues of 120572 this is because the mean retrial time decreases as120572 increases From the instant when the server becomes idlean external potential primary customer and retrial customerscompete to access the server and the smaller the mean retrialtime is the bigger the probability that the server is busy iswhich leads to decreasing of 119864[119871] Furthermore under thesame condition decreasing of 119864[119871] results in the increasein the probability that the system is empty and Figure 7reveals that 119875
119908increases with increasing values of 120572 while 119875
119899
decreases as 120572 increases In Figure 6 compared with constantretrial policy we can also find that classical retrial policydecreases the waiting jobs effectively
2 25 3 35 4 45 5 55 615
2
25
3
35
4
45
5
55
6
Expe
cted
que
ue le
ngth
E[L]
MM1 with classical retrialMM1 with constant retrial
120572
Figure 6 The effect of 120572 on 119864[119871] for two different retrial policies
2 25 3 35 4 45 5 55 6035
04
045
05
055
06
065
07
PnPw
120572
Figure 7 The probability of serverrsquos state with the change of 120572
52 Cost Analysis In this subsection we establish a costfunction to search for the optimal service rate 120578 so as tominimize the expected operating cost function per unit time
Define the following cost elements
119862119871 cost per unit time for each customer present in the
orbit
119862120583 cost per unit time for service during a normal
service period
119862120578 cost per unit time for service in aworking vacation
period
119862120579 fixed cost per unit time during a vacation period
8 Advances in Operations Research
0 05 1 15 2 25 3 35 4 45 565
70
75
80
85
90
95
100
105
Expe
cted
ope
ratin
g co
st pe
r uni
t tim
e
120578
Figure 8 Effect of 120578 on the expected operating cost per unit time
Based on the definitions of each cost element listed abovethe expected operating cost function per unit time can begiven by
min120578
119891 (120578) = 119862119871119864 [119871] + 119862
120583120583 + 119862
120578120578 + 119862120579120579 (46)
Because the expected operating cost function is highly non-linear and complex it is not easy to get the derivative of itWeassume 119862
119871= 6 119862
120583= 10 119862
120578= 8 and 119862
120579= 4 and develop
approximations by MATLAB program to find the optimumvalue of 120578 say 120578lowast
From Figure 8 we can see that there is an optimal servicerate 120578 to minimize the cost Implementing the computersoftware MATLAB by the parabolic method and controllingthe error by 10
minus4 we find the solution 120578lowast
= 08584 with119891(120578lowast) = 690971
6 Conclusion
This paper analyzes a single server retrial queue with workingvacation interruption under the classical retrial policy Usingembedded Markov chain and Fosterrsquos criterion we get thecondition of stability The pgf of the number of customersin the orbit is obtained and some important performancemeasures are also discussed Moreover the effects of var-ious parameters on the system performance measures areexamined numerically Under the stable condition a costminimization problem is considered For future researchusing the same method one can investigate a similar modelwith batch arrival and without vacation interruption
Competing Interests
The authors declare that there are no competing interestsregarding the publication of this paper
Acknowledgments
This work is supported by the National Natural ScienceFoundation of China (nos 11301306 and 11401348) and theNatural Science Foundation of Anhui Higher EducationInstitutions (no KJ2014ZD21)
References
[1] N Tian and Z G ZhangVacation QueueingModels-Theory andApplications Springer New York NY USA 2006
[2] L D Servi and S G Finn ldquoMM1 queues with workingvacations (MM1WV)rdquo Performance Evaluation vol 50 no1 pp 41ndash52 2002
[3] Y Baba ldquoAnalysis of a GIM1 queue with multiple workingvacationsrdquo Operations Research Letters vol 33 no 2 pp 201ndash209 2005
[4] D-A Wu and H Takagi ldquoMG1 queue with multiple workingvacationsrdquo Performance Evaluation vol 63 no 7 pp 654ndash6812006
[5] J-H Li N-S Tian and W-Y Liu ldquoDiscrete-time GIGeo1queue with multiple working vacationsrdquoQueueing Systems vol56 no 1 pp 53ndash63 2007
[6] J Li andNTian ldquoTheMM1 queuewithworking vacations andvacation interruptionsrdquo Journal of Systems Science and SystemsEngineering vol 16 no 1 pp 121ndash127 2007
[7] J-H Li and N-S Tian ldquoThe discrete-time GIGeo1 queuewith working vacations and vacation interruptionrdquo AppliedMathematics and Computation vol 185 no 1 pp 1ndash10 2007
[8] J-H Li N-S Tian and Z-Y Ma ldquoPerformance analysis ofGIM1 queue with working vacations and vacation interrup-tionrdquo Applied Mathematical Modelling vol 32 no 12 pp 2715ndash2730 2008
[9] M Zhang and Z Hou ldquoPerformance analysis of MG1 queuewith working vacations and vacation interruptionrdquo Journal ofComputational and Applied Mathematics vol 234 no 10 pp2977ndash2985 2010
[10] J R Artalejo and A Gomez-Corral Retrial Queueing SystemsA Computational Approach Springer Berlin Germany 2008
[11] G Choudhury L Tadj and M Deka ldquoAn unreliable serverretrial queue with two phases of service and general retrialtimes under Bernoulli vacation schedulerdquo Quality Technologyamp Quantitative Management vol 12 no 4 pp 437ndash464 2015
[12] T V Do ldquoMM1 retrial queue with working vacationsrdquo ActaInformatica vol 47 no 1 pp 67ndash75 2010
[13] T Li Z Wang and Z Liu ldquoGeoGeo1 retrial queue with work-ing vacations and vacation interruptionrdquo Journal of AppliedMathematics and Computing vol 39 no 1-2 pp 131ndash143 2012
[14] Z Liu and Y Song ldquoGeoGeo1 retrial queue with non-persistent customers and working vacationsrdquo Journal of AppliedMathematics and Computing vol 42 no 1-2 pp 103ndash115 2013
[15] L Tao Z Liu andZWang ldquoMM1 retrial queuewith collisionsand working vacation interruption under N-policyrdquo RAIROOperations Research vol 46 no 4 pp 355ndash371 2012
[16] A Aissani S Taleb T Kernane G Saidi and D HamadoucheldquoAn MG1 retrial queue with working vacationrdquo in Advancesin Systems Science Proceedings of the International Conferenceon Systems Science 2013 (ICSS 2013) vol 240 of Advancesin Intelligent Systems and Computing pp 443ndash452 SpringerBerlin Germany 2014
Advances in Operations Research 9
[17] V Jailaxmi RArumuganathan andM S Kumar ldquoPerformanceanalysis of single server non-Markovian retrial queue withworking vacation and constant retrial policyrdquo RAIRO Opera-tions Research vol 48 no 3 pp 381ndash398 2014
[18] S Gao and J Wang ldquoDiscrete-time Geo119883G1 retrial queuewith general retrial times working vacations and vacationinterruptionrdquo Quality Technology amp Quantitative Managementvol 10 no 4 pp 495ndash512 2013
[19] S Gao J Wang and W W Li ldquoAn MG1 retrial queuewith general retrial times working vacations and vacationinterruptionrdquo Asia-Pacific Journal of Operational Research vol31 no 2 Article ID 1440006 25 pages 2014
[20] G Ayyappan A Ganapathy and G Sekar ldquoRetrial queueingsystemwith single working vacation under pre-emptive priorityservicerdquo International Journal of Computer Applications vol 2no 2 pp 28ndash35 2010
[21] L I Sennott P A Humblet and R L Tweedie ldquoMean drifts andthe nonergodicity of Markov chainsrdquo Operations Research vol31 no 4 pp 783ndash789 1983
[22] R B Cooper Introduction to QueueingTheory North-HollandNew York NY USA 1981
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical Problems in Engineering
Hindawi Publishing Corporationhttpwwwhindawicom
Differential EquationsInternational Journal of
Volume 2014
Applied MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical PhysicsAdvances in
Complex AnalysisJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
OptimizationJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Operations ResearchAdvances in
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Decision SciencesAdvances in
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
Advances in Operations Research 5
From (19) (22) and (24) we can see that 1198661(119911) 119866
2(119911) and
1198663(119911) are expressed in terms of 120587
00 In the next section using
normalization condition we will obtain 12058700
4 Performance Measures
From (19) we have
1198661 (1) =
120582
120579 + 120578
12058700 (25)
1198661015840
1(1) = lim
119911rarr1
1198661015840
1(119911) = lim
119911rarr1
1205822
(120582 + 120579 + 120578 minus 120582119911)212058700
=
1205822
(120579 + 120578)212058700
(26)
From (22) we get
1198662 (1) = 119870120587
00 (27)
where
119870 = (120583 minus 120582)minus120582120572
[
1205822
120572
1198701 (1) +
120582
120572
1198702 (1) minus
120582120578
120572
1198703 (1)
minus
(120579120583 + 120582120578) 120582
120572
1198704 (1)]
(28)
From (24) using LrsquoHospital rule we obtain
1198663 (1) = lim119911rarr1
1198663 (119911)
= lim119911rarr1
minus1205821198662 (119911)+ 120582 (1 minus 119911) 119866
1015840
2(119911) minus 120579119866
1 (119911) minus (120578 + 120579119911)119866
1015840
1(119911)
minus120582 (1 minus 119911) minus (120583 minus 120582119911)
=
1205821198662 (1) + 120579119866
1 (1) + (120578 + 120579)119866
1015840
1(1)
120583 minus 120582
=
120582119870 (120579 + 120578) + 120582120579 + 1205822
(120583 minus 120582) (120579 + 120578)
12058700
(29)
Since
12058700+ 1198661 (1) + 119866
2 (1) + 119866
3 (1) = 1 (30)
taking (25) (27) and (29) into the above equation we canderive
12058700= [1 +
120582
120579 + 120578
+ 119870 +
120582119870 (120579 + 120578) + 120582120579 + 1205822
(120583 minus 120582) (120579 + 120578)
]
minus1
= [
120582 (120583 minus 120578) + 120583 (1 + 119870) (120579 + 120578)
(120583 minus 120582) (120579 + 120578)
]
minus1
(31)
Thus 12058700
is found out From (19) (22) and (24) 1198661(119911)
1198662(119911) and 119866
3(119911) are completely computed And the pgf of
the number of customers in the orbit is given by
119866 (119911) = 12058700+ 1198661 (119911) + 119866
2 (119911) + 119866
3 (119911) (32)
The pgf of the number of customers in the system is given by
(119911) = 12058700+ 1199111198661 (119911) + 119866
2 (119911) + 119911119866
3 (119911) (33)
Let 119864[119871119895] denote the average number of customers in the
orbit when the serverrsquos state is 119895 119895 = 1 2 3Differentiating (19) with respect to 119911 we have
119864 [1198711] = lim119911rarr1
1198661015840
1(119911) =
1205822
(120579 + 120578)212058700 (34)
From (18) we can get
119864 [1198712] = lim119911rarr1
1198661015840
2(119911) =
120583
120572
1198663 (1) minus
120582
120572
1198662 (1) minus
120579
120572
1198661 (1) (35)
Differentiating (24) with respect to 119911 and using LrsquoHospitalrule twice the expression for 119864[119871
3] is obtained as
119864 [1198713] = lim119911rarr1
1198661015840
3(119911) =
(120583 minus 120582) (120579 + 120578)11986610158401015840
1(1) + 2 (120579120583 + 120582120578) 119864 [119871
1] + 2120582 (120583 minus 120582) 119864 [119871
2] + 2120582 [120579119866
1 (1) + 120582119866
2 (1)]
2 (120583 minus 120582)2
(36)
and differentiating (19) twice with respect to 119911 and taking 119911 =1 yield 11986610158401015840
1(1) as
11986610158401015840
1(1) =
21205823
(120579 + 120578)312058700 (37)
Therefore the average orbit length (119864[119871]) is given by
119864 [119871] = lim119911rarr1
1198661015840(119911) = 119864 [119871
1] + 119864 [119871
2] + 119864 [119871
3] (38)
And the average system length (119864[]) is derived as
119864 [] = lim119911rarr1
1198661015840(119911) = 119864 [119871] + 119866
1 (1) + 119866
3 (1) (39)
The probability that the server is busy is
119875119887= 119875 119869 = 1 + 119875 119869 = 3 = 119866
1 (1) + 119866
3 (1) (40)
6 Advances in Operations Research
The probability that the server is free is
119875119891= 119875 119869 = 0 + 119875 119869 = 2 = 120587
00+ 1198662 (1) = 1 minus 119875
119887 (41)
The probability that the server is in a working vacation periodis given by
119875119908= 119875 119869 = 0 + 119875 119869 = 1 = 120587
00+ 1198661 (1) (42)
The probability that the server is in a normal service periodis given by
119875119899= 119875 119869 = 2 + 119875 119869 = 3 = 119866
1 (2) + 119866
2 (3)
= 1 minus 119875119908
(43)
Let 119864[119882] (119864[]) be the expected waiting (sojourn) time of acustomer in the orbit (system) using Littlersquos formula
119864 [119882] =
119864 [119871]
120582
119864 [] =
119864 []
120582
(44)
The system busy period 119879 is defined as the period that startsat an epochwhen an arriving customer finds an empty systemand ends at the departure epoch at which the system is emptyUsing the theory of regenerative process we get
12058700=
119864 [11987900]
119864 [11987900] + 119864 [119879]
(45)
where 11987900
is the time length that the system is in the state(0 0) Since the interarrival time between two customersfollows exponential distribution with parameter 120582 we have119864[11987900] = 1120582 Thus 119864[119879] = 120582
minus1(120587minus1
00minus 1)
5 Numerical Results
In this section under the stable condition we present somenumerical examples to illustrate the effect of the varyingparameters on themeanorbit length119864[119871] the probability thatthe system is busy 1minus120587
00 and the probabilities of serverrsquos state
(119875119887 119875119891 119875119908 and 119875
119899) Moreover a cost minimization problem
is also discussed The various parameters of this model arearbitrarily chosen as 120582 = 2 120578 = 1 120583 = 5 120572 = 3 and120579 = 02 unless they are considered as variables or their valuesare mentioned in Figures 2 3 4 and 5
51 Sensitivity Analysis Figures 2 and 3 illustrate the effect of120579 on 119864[119871] and 1 minus 120587
00for different values of 120578 respectively
When 120578 lt 120583 we find that 119864[119871] and 1 minus 12058700
both decrease as120579 increases and 120579 has a noticeable effect on the performancemeasures which cannot be ignored An especial case is 120578 = 120583that is the service rate in the vacation period is equal to theservice rate in the normal period and the model we considerreduces to anMM1 retrial queue with classical retrial policyClearly 120579 has no effect on119864[119871] and 1minus120587
00 which agrees with
the intuitive expectation
0 1 2 3 4 5 6 7 8 9 10
08
1
12
14
16
18
2
Expe
cted
que
ue le
ngth
E[L]
120579
120578 = 1
120578 = 3120578 = 5
Figure 2 The effect of 120579 on 119864[119871] for different values of 120578
0 1 2 3 4 5 6 7 8 9 10055
06
065
07
075
08
085
09
120579
120578 = 1
120578 = 3120578 = 5
1minus12058700
Figure 3 The effect of 120579 on 1 minus 12058700
for different values of 120578
The effect of 120578 on 119864[119871] is presented in Figure 4 asexpected 119864[119871] decreases with increasing values of 120578 When120578 is large it can be observed that the effect of 120578 on 119864[119871] is notobvious as 120578 increases this is because we consider workingvacation interruption policy If the system is not empty at aservice completion instant in the vacation period the serverwill come back to the normal working level Moreover sinceconstant retrial policy means that only the customer at thehead of the orbit can request a service we can find that 119864[119871]of MM1 queue with classical retrial policy is smaller thanthat of MM1 queue with constant retrial policy As shownin Figure 5 119875
119887decreases with increasing values of 120578 and 119875
119891
has the opposite tendency The reason is that as 120578 increases
Advances in Operations Research 7
0 05 1 15 2 25 3 35 4 45 50
2
4
6
8
10
12
14Ex
pect
ed q
ueue
leng
th E[L]
MM1 with classical retrialMM1 with constant retrial
120578
Figure 4 The effect of 120578 on 119864[119871] for two different retrial policies
0 05 1 15 2 25 3 35 4 45 501
02
03
04
05
06
07
08
09
120578
PbPf
Figure 5 The probability of serverrsquos state with the change of 120578
themeanorbit length119864[119871] and themean service time 1120578bothdecrease
Figure 6 indicates that 119864[119871] decreases with increasingvalues of 120572 this is because the mean retrial time decreases as120572 increases From the instant when the server becomes idlean external potential primary customer and retrial customerscompete to access the server and the smaller the mean retrialtime is the bigger the probability that the server is busy iswhich leads to decreasing of 119864[119871] Furthermore under thesame condition decreasing of 119864[119871] results in the increasein the probability that the system is empty and Figure 7reveals that 119875
119908increases with increasing values of 120572 while 119875
119899
decreases as 120572 increases In Figure 6 compared with constantretrial policy we can also find that classical retrial policydecreases the waiting jobs effectively
2 25 3 35 4 45 5 55 615
2
25
3
35
4
45
5
55
6
Expe
cted
que
ue le
ngth
E[L]
MM1 with classical retrialMM1 with constant retrial
120572
Figure 6 The effect of 120572 on 119864[119871] for two different retrial policies
2 25 3 35 4 45 5 55 6035
04
045
05
055
06
065
07
PnPw
120572
Figure 7 The probability of serverrsquos state with the change of 120572
52 Cost Analysis In this subsection we establish a costfunction to search for the optimal service rate 120578 so as tominimize the expected operating cost function per unit time
Define the following cost elements
119862119871 cost per unit time for each customer present in the
orbit
119862120583 cost per unit time for service during a normal
service period
119862120578 cost per unit time for service in aworking vacation
period
119862120579 fixed cost per unit time during a vacation period
8 Advances in Operations Research
0 05 1 15 2 25 3 35 4 45 565
70
75
80
85
90
95
100
105
Expe
cted
ope
ratin
g co
st pe
r uni
t tim
e
120578
Figure 8 Effect of 120578 on the expected operating cost per unit time
Based on the definitions of each cost element listed abovethe expected operating cost function per unit time can begiven by
min120578
119891 (120578) = 119862119871119864 [119871] + 119862
120583120583 + 119862
120578120578 + 119862120579120579 (46)
Because the expected operating cost function is highly non-linear and complex it is not easy to get the derivative of itWeassume 119862
119871= 6 119862
120583= 10 119862
120578= 8 and 119862
120579= 4 and develop
approximations by MATLAB program to find the optimumvalue of 120578 say 120578lowast
From Figure 8 we can see that there is an optimal servicerate 120578 to minimize the cost Implementing the computersoftware MATLAB by the parabolic method and controllingthe error by 10
minus4 we find the solution 120578lowast
= 08584 with119891(120578lowast) = 690971
6 Conclusion
This paper analyzes a single server retrial queue with workingvacation interruption under the classical retrial policy Usingembedded Markov chain and Fosterrsquos criterion we get thecondition of stability The pgf of the number of customersin the orbit is obtained and some important performancemeasures are also discussed Moreover the effects of var-ious parameters on the system performance measures areexamined numerically Under the stable condition a costminimization problem is considered For future researchusing the same method one can investigate a similar modelwith batch arrival and without vacation interruption
Competing Interests
The authors declare that there are no competing interestsregarding the publication of this paper
Acknowledgments
This work is supported by the National Natural ScienceFoundation of China (nos 11301306 and 11401348) and theNatural Science Foundation of Anhui Higher EducationInstitutions (no KJ2014ZD21)
References
[1] N Tian and Z G ZhangVacation QueueingModels-Theory andApplications Springer New York NY USA 2006
[2] L D Servi and S G Finn ldquoMM1 queues with workingvacations (MM1WV)rdquo Performance Evaluation vol 50 no1 pp 41ndash52 2002
[3] Y Baba ldquoAnalysis of a GIM1 queue with multiple workingvacationsrdquo Operations Research Letters vol 33 no 2 pp 201ndash209 2005
[4] D-A Wu and H Takagi ldquoMG1 queue with multiple workingvacationsrdquo Performance Evaluation vol 63 no 7 pp 654ndash6812006
[5] J-H Li N-S Tian and W-Y Liu ldquoDiscrete-time GIGeo1queue with multiple working vacationsrdquoQueueing Systems vol56 no 1 pp 53ndash63 2007
[6] J Li andNTian ldquoTheMM1 queuewithworking vacations andvacation interruptionsrdquo Journal of Systems Science and SystemsEngineering vol 16 no 1 pp 121ndash127 2007
[7] J-H Li and N-S Tian ldquoThe discrete-time GIGeo1 queuewith working vacations and vacation interruptionrdquo AppliedMathematics and Computation vol 185 no 1 pp 1ndash10 2007
[8] J-H Li N-S Tian and Z-Y Ma ldquoPerformance analysis ofGIM1 queue with working vacations and vacation interrup-tionrdquo Applied Mathematical Modelling vol 32 no 12 pp 2715ndash2730 2008
[9] M Zhang and Z Hou ldquoPerformance analysis of MG1 queuewith working vacations and vacation interruptionrdquo Journal ofComputational and Applied Mathematics vol 234 no 10 pp2977ndash2985 2010
[10] J R Artalejo and A Gomez-Corral Retrial Queueing SystemsA Computational Approach Springer Berlin Germany 2008
[11] G Choudhury L Tadj and M Deka ldquoAn unreliable serverretrial queue with two phases of service and general retrialtimes under Bernoulli vacation schedulerdquo Quality Technologyamp Quantitative Management vol 12 no 4 pp 437ndash464 2015
[12] T V Do ldquoMM1 retrial queue with working vacationsrdquo ActaInformatica vol 47 no 1 pp 67ndash75 2010
[13] T Li Z Wang and Z Liu ldquoGeoGeo1 retrial queue with work-ing vacations and vacation interruptionrdquo Journal of AppliedMathematics and Computing vol 39 no 1-2 pp 131ndash143 2012
[14] Z Liu and Y Song ldquoGeoGeo1 retrial queue with non-persistent customers and working vacationsrdquo Journal of AppliedMathematics and Computing vol 42 no 1-2 pp 103ndash115 2013
[15] L Tao Z Liu andZWang ldquoMM1 retrial queuewith collisionsand working vacation interruption under N-policyrdquo RAIROOperations Research vol 46 no 4 pp 355ndash371 2012
[16] A Aissani S Taleb T Kernane G Saidi and D HamadoucheldquoAn MG1 retrial queue with working vacationrdquo in Advancesin Systems Science Proceedings of the International Conferenceon Systems Science 2013 (ICSS 2013) vol 240 of Advancesin Intelligent Systems and Computing pp 443ndash452 SpringerBerlin Germany 2014
Advances in Operations Research 9
[17] V Jailaxmi RArumuganathan andM S Kumar ldquoPerformanceanalysis of single server non-Markovian retrial queue withworking vacation and constant retrial policyrdquo RAIRO Opera-tions Research vol 48 no 3 pp 381ndash398 2014
[18] S Gao and J Wang ldquoDiscrete-time Geo119883G1 retrial queuewith general retrial times working vacations and vacationinterruptionrdquo Quality Technology amp Quantitative Managementvol 10 no 4 pp 495ndash512 2013
[19] S Gao J Wang and W W Li ldquoAn MG1 retrial queuewith general retrial times working vacations and vacationinterruptionrdquo Asia-Pacific Journal of Operational Research vol31 no 2 Article ID 1440006 25 pages 2014
[20] G Ayyappan A Ganapathy and G Sekar ldquoRetrial queueingsystemwith single working vacation under pre-emptive priorityservicerdquo International Journal of Computer Applications vol 2no 2 pp 28ndash35 2010
[21] L I Sennott P A Humblet and R L Tweedie ldquoMean drifts andthe nonergodicity of Markov chainsrdquo Operations Research vol31 no 4 pp 783ndash789 1983
[22] R B Cooper Introduction to QueueingTheory North-HollandNew York NY USA 1981
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical Problems in Engineering
Hindawi Publishing Corporationhttpwwwhindawicom
Differential EquationsInternational Journal of
Volume 2014
Applied MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical PhysicsAdvances in
Complex AnalysisJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
OptimizationJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Operations ResearchAdvances in
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Decision SciencesAdvances in
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
6 Advances in Operations Research
The probability that the server is free is
119875119891= 119875 119869 = 0 + 119875 119869 = 2 = 120587
00+ 1198662 (1) = 1 minus 119875
119887 (41)
The probability that the server is in a working vacation periodis given by
119875119908= 119875 119869 = 0 + 119875 119869 = 1 = 120587
00+ 1198661 (1) (42)
The probability that the server is in a normal service periodis given by
119875119899= 119875 119869 = 2 + 119875 119869 = 3 = 119866
1 (2) + 119866
2 (3)
= 1 minus 119875119908
(43)
Let 119864[119882] (119864[]) be the expected waiting (sojourn) time of acustomer in the orbit (system) using Littlersquos formula
119864 [119882] =
119864 [119871]
120582
119864 [] =
119864 []
120582
(44)
The system busy period 119879 is defined as the period that startsat an epochwhen an arriving customer finds an empty systemand ends at the departure epoch at which the system is emptyUsing the theory of regenerative process we get
12058700=
119864 [11987900]
119864 [11987900] + 119864 [119879]
(45)
where 11987900
is the time length that the system is in the state(0 0) Since the interarrival time between two customersfollows exponential distribution with parameter 120582 we have119864[11987900] = 1120582 Thus 119864[119879] = 120582
minus1(120587minus1
00minus 1)
5 Numerical Results
In this section under the stable condition we present somenumerical examples to illustrate the effect of the varyingparameters on themeanorbit length119864[119871] the probability thatthe system is busy 1minus120587
00 and the probabilities of serverrsquos state
(119875119887 119875119891 119875119908 and 119875
119899) Moreover a cost minimization problem
is also discussed The various parameters of this model arearbitrarily chosen as 120582 = 2 120578 = 1 120583 = 5 120572 = 3 and120579 = 02 unless they are considered as variables or their valuesare mentioned in Figures 2 3 4 and 5
51 Sensitivity Analysis Figures 2 and 3 illustrate the effect of120579 on 119864[119871] and 1 minus 120587
00for different values of 120578 respectively
When 120578 lt 120583 we find that 119864[119871] and 1 minus 12058700
both decrease as120579 increases and 120579 has a noticeable effect on the performancemeasures which cannot be ignored An especial case is 120578 = 120583that is the service rate in the vacation period is equal to theservice rate in the normal period and the model we considerreduces to anMM1 retrial queue with classical retrial policyClearly 120579 has no effect on119864[119871] and 1minus120587
00 which agrees with
the intuitive expectation
0 1 2 3 4 5 6 7 8 9 10
08
1
12
14
16
18
2
Expe
cted
que
ue le
ngth
E[L]
120579
120578 = 1
120578 = 3120578 = 5
Figure 2 The effect of 120579 on 119864[119871] for different values of 120578
0 1 2 3 4 5 6 7 8 9 10055
06
065
07
075
08
085
09
120579
120578 = 1
120578 = 3120578 = 5
1minus12058700
Figure 3 The effect of 120579 on 1 minus 12058700
for different values of 120578
The effect of 120578 on 119864[119871] is presented in Figure 4 asexpected 119864[119871] decreases with increasing values of 120578 When120578 is large it can be observed that the effect of 120578 on 119864[119871] is notobvious as 120578 increases this is because we consider workingvacation interruption policy If the system is not empty at aservice completion instant in the vacation period the serverwill come back to the normal working level Moreover sinceconstant retrial policy means that only the customer at thehead of the orbit can request a service we can find that 119864[119871]of MM1 queue with classical retrial policy is smaller thanthat of MM1 queue with constant retrial policy As shownin Figure 5 119875
119887decreases with increasing values of 120578 and 119875
119891
has the opposite tendency The reason is that as 120578 increases
Advances in Operations Research 7
0 05 1 15 2 25 3 35 4 45 50
2
4
6
8
10
12
14Ex
pect
ed q
ueue
leng
th E[L]
MM1 with classical retrialMM1 with constant retrial
120578
Figure 4 The effect of 120578 on 119864[119871] for two different retrial policies
0 05 1 15 2 25 3 35 4 45 501
02
03
04
05
06
07
08
09
120578
PbPf
Figure 5 The probability of serverrsquos state with the change of 120578
themeanorbit length119864[119871] and themean service time 1120578bothdecrease
Figure 6 indicates that 119864[119871] decreases with increasingvalues of 120572 this is because the mean retrial time decreases as120572 increases From the instant when the server becomes idlean external potential primary customer and retrial customerscompete to access the server and the smaller the mean retrialtime is the bigger the probability that the server is busy iswhich leads to decreasing of 119864[119871] Furthermore under thesame condition decreasing of 119864[119871] results in the increasein the probability that the system is empty and Figure 7reveals that 119875
119908increases with increasing values of 120572 while 119875
119899
decreases as 120572 increases In Figure 6 compared with constantretrial policy we can also find that classical retrial policydecreases the waiting jobs effectively
2 25 3 35 4 45 5 55 615
2
25
3
35
4
45
5
55
6
Expe
cted
que
ue le
ngth
E[L]
MM1 with classical retrialMM1 with constant retrial
120572
Figure 6 The effect of 120572 on 119864[119871] for two different retrial policies
2 25 3 35 4 45 5 55 6035
04
045
05
055
06
065
07
PnPw
120572
Figure 7 The probability of serverrsquos state with the change of 120572
52 Cost Analysis In this subsection we establish a costfunction to search for the optimal service rate 120578 so as tominimize the expected operating cost function per unit time
Define the following cost elements
119862119871 cost per unit time for each customer present in the
orbit
119862120583 cost per unit time for service during a normal
service period
119862120578 cost per unit time for service in aworking vacation
period
119862120579 fixed cost per unit time during a vacation period
8 Advances in Operations Research
0 05 1 15 2 25 3 35 4 45 565
70
75
80
85
90
95
100
105
Expe
cted
ope
ratin
g co
st pe
r uni
t tim
e
120578
Figure 8 Effect of 120578 on the expected operating cost per unit time
Based on the definitions of each cost element listed abovethe expected operating cost function per unit time can begiven by
min120578
119891 (120578) = 119862119871119864 [119871] + 119862
120583120583 + 119862
120578120578 + 119862120579120579 (46)
Because the expected operating cost function is highly non-linear and complex it is not easy to get the derivative of itWeassume 119862
119871= 6 119862
120583= 10 119862
120578= 8 and 119862
120579= 4 and develop
approximations by MATLAB program to find the optimumvalue of 120578 say 120578lowast
From Figure 8 we can see that there is an optimal servicerate 120578 to minimize the cost Implementing the computersoftware MATLAB by the parabolic method and controllingthe error by 10
minus4 we find the solution 120578lowast
= 08584 with119891(120578lowast) = 690971
6 Conclusion
This paper analyzes a single server retrial queue with workingvacation interruption under the classical retrial policy Usingembedded Markov chain and Fosterrsquos criterion we get thecondition of stability The pgf of the number of customersin the orbit is obtained and some important performancemeasures are also discussed Moreover the effects of var-ious parameters on the system performance measures areexamined numerically Under the stable condition a costminimization problem is considered For future researchusing the same method one can investigate a similar modelwith batch arrival and without vacation interruption
Competing Interests
The authors declare that there are no competing interestsregarding the publication of this paper
Acknowledgments
This work is supported by the National Natural ScienceFoundation of China (nos 11301306 and 11401348) and theNatural Science Foundation of Anhui Higher EducationInstitutions (no KJ2014ZD21)
References
[1] N Tian and Z G ZhangVacation QueueingModels-Theory andApplications Springer New York NY USA 2006
[2] L D Servi and S G Finn ldquoMM1 queues with workingvacations (MM1WV)rdquo Performance Evaluation vol 50 no1 pp 41ndash52 2002
[3] Y Baba ldquoAnalysis of a GIM1 queue with multiple workingvacationsrdquo Operations Research Letters vol 33 no 2 pp 201ndash209 2005
[4] D-A Wu and H Takagi ldquoMG1 queue with multiple workingvacationsrdquo Performance Evaluation vol 63 no 7 pp 654ndash6812006
[5] J-H Li N-S Tian and W-Y Liu ldquoDiscrete-time GIGeo1queue with multiple working vacationsrdquoQueueing Systems vol56 no 1 pp 53ndash63 2007
[6] J Li andNTian ldquoTheMM1 queuewithworking vacations andvacation interruptionsrdquo Journal of Systems Science and SystemsEngineering vol 16 no 1 pp 121ndash127 2007
[7] J-H Li and N-S Tian ldquoThe discrete-time GIGeo1 queuewith working vacations and vacation interruptionrdquo AppliedMathematics and Computation vol 185 no 1 pp 1ndash10 2007
[8] J-H Li N-S Tian and Z-Y Ma ldquoPerformance analysis ofGIM1 queue with working vacations and vacation interrup-tionrdquo Applied Mathematical Modelling vol 32 no 12 pp 2715ndash2730 2008
[9] M Zhang and Z Hou ldquoPerformance analysis of MG1 queuewith working vacations and vacation interruptionrdquo Journal ofComputational and Applied Mathematics vol 234 no 10 pp2977ndash2985 2010
[10] J R Artalejo and A Gomez-Corral Retrial Queueing SystemsA Computational Approach Springer Berlin Germany 2008
[11] G Choudhury L Tadj and M Deka ldquoAn unreliable serverretrial queue with two phases of service and general retrialtimes under Bernoulli vacation schedulerdquo Quality Technologyamp Quantitative Management vol 12 no 4 pp 437ndash464 2015
[12] T V Do ldquoMM1 retrial queue with working vacationsrdquo ActaInformatica vol 47 no 1 pp 67ndash75 2010
[13] T Li Z Wang and Z Liu ldquoGeoGeo1 retrial queue with work-ing vacations and vacation interruptionrdquo Journal of AppliedMathematics and Computing vol 39 no 1-2 pp 131ndash143 2012
[14] Z Liu and Y Song ldquoGeoGeo1 retrial queue with non-persistent customers and working vacationsrdquo Journal of AppliedMathematics and Computing vol 42 no 1-2 pp 103ndash115 2013
[15] L Tao Z Liu andZWang ldquoMM1 retrial queuewith collisionsand working vacation interruption under N-policyrdquo RAIROOperations Research vol 46 no 4 pp 355ndash371 2012
[16] A Aissani S Taleb T Kernane G Saidi and D HamadoucheldquoAn MG1 retrial queue with working vacationrdquo in Advancesin Systems Science Proceedings of the International Conferenceon Systems Science 2013 (ICSS 2013) vol 240 of Advancesin Intelligent Systems and Computing pp 443ndash452 SpringerBerlin Germany 2014
Advances in Operations Research 9
[17] V Jailaxmi RArumuganathan andM S Kumar ldquoPerformanceanalysis of single server non-Markovian retrial queue withworking vacation and constant retrial policyrdquo RAIRO Opera-tions Research vol 48 no 3 pp 381ndash398 2014
[18] S Gao and J Wang ldquoDiscrete-time Geo119883G1 retrial queuewith general retrial times working vacations and vacationinterruptionrdquo Quality Technology amp Quantitative Managementvol 10 no 4 pp 495ndash512 2013
[19] S Gao J Wang and W W Li ldquoAn MG1 retrial queuewith general retrial times working vacations and vacationinterruptionrdquo Asia-Pacific Journal of Operational Research vol31 no 2 Article ID 1440006 25 pages 2014
[20] G Ayyappan A Ganapathy and G Sekar ldquoRetrial queueingsystemwith single working vacation under pre-emptive priorityservicerdquo International Journal of Computer Applications vol 2no 2 pp 28ndash35 2010
[21] L I Sennott P A Humblet and R L Tweedie ldquoMean drifts andthe nonergodicity of Markov chainsrdquo Operations Research vol31 no 4 pp 783ndash789 1983
[22] R B Cooper Introduction to QueueingTheory North-HollandNew York NY USA 1981
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical Problems in Engineering
Hindawi Publishing Corporationhttpwwwhindawicom
Differential EquationsInternational Journal of
Volume 2014
Applied MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical PhysicsAdvances in
Complex AnalysisJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
OptimizationJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Operations ResearchAdvances in
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Decision SciencesAdvances in
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
Advances in Operations Research 7
0 05 1 15 2 25 3 35 4 45 50
2
4
6
8
10
12
14Ex
pect
ed q
ueue
leng
th E[L]
MM1 with classical retrialMM1 with constant retrial
120578
Figure 4 The effect of 120578 on 119864[119871] for two different retrial policies
0 05 1 15 2 25 3 35 4 45 501
02
03
04
05
06
07
08
09
120578
PbPf
Figure 5 The probability of serverrsquos state with the change of 120578
themeanorbit length119864[119871] and themean service time 1120578bothdecrease
Figure 6 indicates that 119864[119871] decreases with increasingvalues of 120572 this is because the mean retrial time decreases as120572 increases From the instant when the server becomes idlean external potential primary customer and retrial customerscompete to access the server and the smaller the mean retrialtime is the bigger the probability that the server is busy iswhich leads to decreasing of 119864[119871] Furthermore under thesame condition decreasing of 119864[119871] results in the increasein the probability that the system is empty and Figure 7reveals that 119875
119908increases with increasing values of 120572 while 119875
119899
decreases as 120572 increases In Figure 6 compared with constantretrial policy we can also find that classical retrial policydecreases the waiting jobs effectively
2 25 3 35 4 45 5 55 615
2
25
3
35
4
45
5
55
6
Expe
cted
que
ue le
ngth
E[L]
MM1 with classical retrialMM1 with constant retrial
120572
Figure 6 The effect of 120572 on 119864[119871] for two different retrial policies
2 25 3 35 4 45 5 55 6035
04
045
05
055
06
065
07
PnPw
120572
Figure 7 The probability of serverrsquos state with the change of 120572
52 Cost Analysis In this subsection we establish a costfunction to search for the optimal service rate 120578 so as tominimize the expected operating cost function per unit time
Define the following cost elements
119862119871 cost per unit time for each customer present in the
orbit
119862120583 cost per unit time for service during a normal
service period
119862120578 cost per unit time for service in aworking vacation
period
119862120579 fixed cost per unit time during a vacation period
8 Advances in Operations Research
0 05 1 15 2 25 3 35 4 45 565
70
75
80
85
90
95
100
105
Expe
cted
ope
ratin
g co
st pe
r uni
t tim
e
120578
Figure 8 Effect of 120578 on the expected operating cost per unit time
Based on the definitions of each cost element listed abovethe expected operating cost function per unit time can begiven by
min120578
119891 (120578) = 119862119871119864 [119871] + 119862
120583120583 + 119862
120578120578 + 119862120579120579 (46)
Because the expected operating cost function is highly non-linear and complex it is not easy to get the derivative of itWeassume 119862
119871= 6 119862
120583= 10 119862
120578= 8 and 119862
120579= 4 and develop
approximations by MATLAB program to find the optimumvalue of 120578 say 120578lowast
From Figure 8 we can see that there is an optimal servicerate 120578 to minimize the cost Implementing the computersoftware MATLAB by the parabolic method and controllingthe error by 10
minus4 we find the solution 120578lowast
= 08584 with119891(120578lowast) = 690971
6 Conclusion
This paper analyzes a single server retrial queue with workingvacation interruption under the classical retrial policy Usingembedded Markov chain and Fosterrsquos criterion we get thecondition of stability The pgf of the number of customersin the orbit is obtained and some important performancemeasures are also discussed Moreover the effects of var-ious parameters on the system performance measures areexamined numerically Under the stable condition a costminimization problem is considered For future researchusing the same method one can investigate a similar modelwith batch arrival and without vacation interruption
Competing Interests
The authors declare that there are no competing interestsregarding the publication of this paper
Acknowledgments
This work is supported by the National Natural ScienceFoundation of China (nos 11301306 and 11401348) and theNatural Science Foundation of Anhui Higher EducationInstitutions (no KJ2014ZD21)
References
[1] N Tian and Z G ZhangVacation QueueingModels-Theory andApplications Springer New York NY USA 2006
[2] L D Servi and S G Finn ldquoMM1 queues with workingvacations (MM1WV)rdquo Performance Evaluation vol 50 no1 pp 41ndash52 2002
[3] Y Baba ldquoAnalysis of a GIM1 queue with multiple workingvacationsrdquo Operations Research Letters vol 33 no 2 pp 201ndash209 2005
[4] D-A Wu and H Takagi ldquoMG1 queue with multiple workingvacationsrdquo Performance Evaluation vol 63 no 7 pp 654ndash6812006
[5] J-H Li N-S Tian and W-Y Liu ldquoDiscrete-time GIGeo1queue with multiple working vacationsrdquoQueueing Systems vol56 no 1 pp 53ndash63 2007
[6] J Li andNTian ldquoTheMM1 queuewithworking vacations andvacation interruptionsrdquo Journal of Systems Science and SystemsEngineering vol 16 no 1 pp 121ndash127 2007
[7] J-H Li and N-S Tian ldquoThe discrete-time GIGeo1 queuewith working vacations and vacation interruptionrdquo AppliedMathematics and Computation vol 185 no 1 pp 1ndash10 2007
[8] J-H Li N-S Tian and Z-Y Ma ldquoPerformance analysis ofGIM1 queue with working vacations and vacation interrup-tionrdquo Applied Mathematical Modelling vol 32 no 12 pp 2715ndash2730 2008
[9] M Zhang and Z Hou ldquoPerformance analysis of MG1 queuewith working vacations and vacation interruptionrdquo Journal ofComputational and Applied Mathematics vol 234 no 10 pp2977ndash2985 2010
[10] J R Artalejo and A Gomez-Corral Retrial Queueing SystemsA Computational Approach Springer Berlin Germany 2008
[11] G Choudhury L Tadj and M Deka ldquoAn unreliable serverretrial queue with two phases of service and general retrialtimes under Bernoulli vacation schedulerdquo Quality Technologyamp Quantitative Management vol 12 no 4 pp 437ndash464 2015
[12] T V Do ldquoMM1 retrial queue with working vacationsrdquo ActaInformatica vol 47 no 1 pp 67ndash75 2010
[13] T Li Z Wang and Z Liu ldquoGeoGeo1 retrial queue with work-ing vacations and vacation interruptionrdquo Journal of AppliedMathematics and Computing vol 39 no 1-2 pp 131ndash143 2012
[14] Z Liu and Y Song ldquoGeoGeo1 retrial queue with non-persistent customers and working vacationsrdquo Journal of AppliedMathematics and Computing vol 42 no 1-2 pp 103ndash115 2013
[15] L Tao Z Liu andZWang ldquoMM1 retrial queuewith collisionsand working vacation interruption under N-policyrdquo RAIROOperations Research vol 46 no 4 pp 355ndash371 2012
[16] A Aissani S Taleb T Kernane G Saidi and D HamadoucheldquoAn MG1 retrial queue with working vacationrdquo in Advancesin Systems Science Proceedings of the International Conferenceon Systems Science 2013 (ICSS 2013) vol 240 of Advancesin Intelligent Systems and Computing pp 443ndash452 SpringerBerlin Germany 2014
Advances in Operations Research 9
[17] V Jailaxmi RArumuganathan andM S Kumar ldquoPerformanceanalysis of single server non-Markovian retrial queue withworking vacation and constant retrial policyrdquo RAIRO Opera-tions Research vol 48 no 3 pp 381ndash398 2014
[18] S Gao and J Wang ldquoDiscrete-time Geo119883G1 retrial queuewith general retrial times working vacations and vacationinterruptionrdquo Quality Technology amp Quantitative Managementvol 10 no 4 pp 495ndash512 2013
[19] S Gao J Wang and W W Li ldquoAn MG1 retrial queuewith general retrial times working vacations and vacationinterruptionrdquo Asia-Pacific Journal of Operational Research vol31 no 2 Article ID 1440006 25 pages 2014
[20] G Ayyappan A Ganapathy and G Sekar ldquoRetrial queueingsystemwith single working vacation under pre-emptive priorityservicerdquo International Journal of Computer Applications vol 2no 2 pp 28ndash35 2010
[21] L I Sennott P A Humblet and R L Tweedie ldquoMean drifts andthe nonergodicity of Markov chainsrdquo Operations Research vol31 no 4 pp 783ndash789 1983
[22] R B Cooper Introduction to QueueingTheory North-HollandNew York NY USA 1981
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical Problems in Engineering
Hindawi Publishing Corporationhttpwwwhindawicom
Differential EquationsInternational Journal of
Volume 2014
Applied MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical PhysicsAdvances in
Complex AnalysisJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
OptimizationJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Operations ResearchAdvances in
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Decision SciencesAdvances in
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
8 Advances in Operations Research
0 05 1 15 2 25 3 35 4 45 565
70
75
80
85
90
95
100
105
Expe
cted
ope
ratin
g co
st pe
r uni
t tim
e
120578
Figure 8 Effect of 120578 on the expected operating cost per unit time
Based on the definitions of each cost element listed abovethe expected operating cost function per unit time can begiven by
min120578
119891 (120578) = 119862119871119864 [119871] + 119862
120583120583 + 119862
120578120578 + 119862120579120579 (46)
Because the expected operating cost function is highly non-linear and complex it is not easy to get the derivative of itWeassume 119862
119871= 6 119862
120583= 10 119862
120578= 8 and 119862
120579= 4 and develop
approximations by MATLAB program to find the optimumvalue of 120578 say 120578lowast
From Figure 8 we can see that there is an optimal servicerate 120578 to minimize the cost Implementing the computersoftware MATLAB by the parabolic method and controllingthe error by 10
minus4 we find the solution 120578lowast
= 08584 with119891(120578lowast) = 690971
6 Conclusion
This paper analyzes a single server retrial queue with workingvacation interruption under the classical retrial policy Usingembedded Markov chain and Fosterrsquos criterion we get thecondition of stability The pgf of the number of customersin the orbit is obtained and some important performancemeasures are also discussed Moreover the effects of var-ious parameters on the system performance measures areexamined numerically Under the stable condition a costminimization problem is considered For future researchusing the same method one can investigate a similar modelwith batch arrival and without vacation interruption
Competing Interests
The authors declare that there are no competing interestsregarding the publication of this paper
Acknowledgments
This work is supported by the National Natural ScienceFoundation of China (nos 11301306 and 11401348) and theNatural Science Foundation of Anhui Higher EducationInstitutions (no KJ2014ZD21)
References
[1] N Tian and Z G ZhangVacation QueueingModels-Theory andApplications Springer New York NY USA 2006
[2] L D Servi and S G Finn ldquoMM1 queues with workingvacations (MM1WV)rdquo Performance Evaluation vol 50 no1 pp 41ndash52 2002
[3] Y Baba ldquoAnalysis of a GIM1 queue with multiple workingvacationsrdquo Operations Research Letters vol 33 no 2 pp 201ndash209 2005
[4] D-A Wu and H Takagi ldquoMG1 queue with multiple workingvacationsrdquo Performance Evaluation vol 63 no 7 pp 654ndash6812006
[5] J-H Li N-S Tian and W-Y Liu ldquoDiscrete-time GIGeo1queue with multiple working vacationsrdquoQueueing Systems vol56 no 1 pp 53ndash63 2007
[6] J Li andNTian ldquoTheMM1 queuewithworking vacations andvacation interruptionsrdquo Journal of Systems Science and SystemsEngineering vol 16 no 1 pp 121ndash127 2007
[7] J-H Li and N-S Tian ldquoThe discrete-time GIGeo1 queuewith working vacations and vacation interruptionrdquo AppliedMathematics and Computation vol 185 no 1 pp 1ndash10 2007
[8] J-H Li N-S Tian and Z-Y Ma ldquoPerformance analysis ofGIM1 queue with working vacations and vacation interrup-tionrdquo Applied Mathematical Modelling vol 32 no 12 pp 2715ndash2730 2008
[9] M Zhang and Z Hou ldquoPerformance analysis of MG1 queuewith working vacations and vacation interruptionrdquo Journal ofComputational and Applied Mathematics vol 234 no 10 pp2977ndash2985 2010
[10] J R Artalejo and A Gomez-Corral Retrial Queueing SystemsA Computational Approach Springer Berlin Germany 2008
[11] G Choudhury L Tadj and M Deka ldquoAn unreliable serverretrial queue with two phases of service and general retrialtimes under Bernoulli vacation schedulerdquo Quality Technologyamp Quantitative Management vol 12 no 4 pp 437ndash464 2015
[12] T V Do ldquoMM1 retrial queue with working vacationsrdquo ActaInformatica vol 47 no 1 pp 67ndash75 2010
[13] T Li Z Wang and Z Liu ldquoGeoGeo1 retrial queue with work-ing vacations and vacation interruptionrdquo Journal of AppliedMathematics and Computing vol 39 no 1-2 pp 131ndash143 2012
[14] Z Liu and Y Song ldquoGeoGeo1 retrial queue with non-persistent customers and working vacationsrdquo Journal of AppliedMathematics and Computing vol 42 no 1-2 pp 103ndash115 2013
[15] L Tao Z Liu andZWang ldquoMM1 retrial queuewith collisionsand working vacation interruption under N-policyrdquo RAIROOperations Research vol 46 no 4 pp 355ndash371 2012
[16] A Aissani S Taleb T Kernane G Saidi and D HamadoucheldquoAn MG1 retrial queue with working vacationrdquo in Advancesin Systems Science Proceedings of the International Conferenceon Systems Science 2013 (ICSS 2013) vol 240 of Advancesin Intelligent Systems and Computing pp 443ndash452 SpringerBerlin Germany 2014
Advances in Operations Research 9
[17] V Jailaxmi RArumuganathan andM S Kumar ldquoPerformanceanalysis of single server non-Markovian retrial queue withworking vacation and constant retrial policyrdquo RAIRO Opera-tions Research vol 48 no 3 pp 381ndash398 2014
[18] S Gao and J Wang ldquoDiscrete-time Geo119883G1 retrial queuewith general retrial times working vacations and vacationinterruptionrdquo Quality Technology amp Quantitative Managementvol 10 no 4 pp 495ndash512 2013
[19] S Gao J Wang and W W Li ldquoAn MG1 retrial queuewith general retrial times working vacations and vacationinterruptionrdquo Asia-Pacific Journal of Operational Research vol31 no 2 Article ID 1440006 25 pages 2014
[20] G Ayyappan A Ganapathy and G Sekar ldquoRetrial queueingsystemwith single working vacation under pre-emptive priorityservicerdquo International Journal of Computer Applications vol 2no 2 pp 28ndash35 2010
[21] L I Sennott P A Humblet and R L Tweedie ldquoMean drifts andthe nonergodicity of Markov chainsrdquo Operations Research vol31 no 4 pp 783ndash789 1983
[22] R B Cooper Introduction to QueueingTheory North-HollandNew York NY USA 1981
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical Problems in Engineering
Hindawi Publishing Corporationhttpwwwhindawicom
Differential EquationsInternational Journal of
Volume 2014
Applied MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical PhysicsAdvances in
Complex AnalysisJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
OptimizationJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Operations ResearchAdvances in
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Decision SciencesAdvances in
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
Advances in Operations Research 9
[17] V Jailaxmi RArumuganathan andM S Kumar ldquoPerformanceanalysis of single server non-Markovian retrial queue withworking vacation and constant retrial policyrdquo RAIRO Opera-tions Research vol 48 no 3 pp 381ndash398 2014
[18] S Gao and J Wang ldquoDiscrete-time Geo119883G1 retrial queuewith general retrial times working vacations and vacationinterruptionrdquo Quality Technology amp Quantitative Managementvol 10 no 4 pp 495ndash512 2013
[19] S Gao J Wang and W W Li ldquoAn MG1 retrial queuewith general retrial times working vacations and vacationinterruptionrdquo Asia-Pacific Journal of Operational Research vol31 no 2 Article ID 1440006 25 pages 2014
[20] G Ayyappan A Ganapathy and G Sekar ldquoRetrial queueingsystemwith single working vacation under pre-emptive priorityservicerdquo International Journal of Computer Applications vol 2no 2 pp 28ndash35 2010
[21] L I Sennott P A Humblet and R L Tweedie ldquoMean drifts andthe nonergodicity of Markov chainsrdquo Operations Research vol31 no 4 pp 783ndash789 1983
[22] R B Cooper Introduction to QueueingTheory North-HollandNew York NY USA 1981
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical Problems in Engineering
Hindawi Publishing Corporationhttpwwwhindawicom
Differential EquationsInternational Journal of
Volume 2014
Applied MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical PhysicsAdvances in
Complex AnalysisJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
OptimizationJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Operations ResearchAdvances in
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Decision SciencesAdvances in
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical Problems in Engineering
Hindawi Publishing Corporationhttpwwwhindawicom
Differential EquationsInternational Journal of
Volume 2014
Applied MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical PhysicsAdvances in
Complex AnalysisJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
OptimizationJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Operations ResearchAdvances in
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Decision SciencesAdvances in
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of