Equilibrium balking strategies of customers in Markovian queues with two-stage working vacations

Applied Mathematics and Computation 248 (2014) 195–214

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Applied Mathematics and Computation journal homepage: www.elsevier.com/locate/amc

Equilibrium balking strategies of customers in Markovian queues with two-stage working vacations Wei Sun, Shiyong Li ⇑, Quan-Lin Li School of Economics and Management, Yanshan University, Qinhuangdao 066004, China

a r t i c l e

i n f o

Keywords: Markovian queues Two-stage working vacations Equilibrium balking strategy Information precision Social welfare

a b s t r a c t This paper studies the customers’ equilibrium balking behavior in some single-server Markovian queues with two-stage working vacations. That is, the server starts taking two successive working vacations when the system becomes empty, during which he provides low-rate service but maintains different service rates in the two-stage vacations. Based on different precision levels of system information, we discuss observable queues, partially observable queues and unobservable queues, respectively. For each type of queues, we get the customers’ equilibrium balking strategy and equilibrium social welfare per time unit, and numerically observe that their positive equilibrium strategy is unique. Especially, for partially observable queues, the customers’ equilibrium joining probabilities in vacation states are not necessarily smaller than that in busy state. Ó 2014 Elsevier Inc. All rights reserved.

1. Introduction For the work studying on customers’ equilibrium balking strategies in queues with classical vacation policies, Burnetas and Economou [1] first presented several Markovian queues with setup times and four precision levels of system information: fully observable, almost observable, almost unobservable and fully unobservable. Then Economou and Kanta [2] discussed an observable queue with server breakdowns. For the fully and partially observable queues, Liu et al. [10] introduced classical single vacation policy, whereas Wang and Zhang [21] and Li et al. [9] focused on server breakdowns and delayed repairs. Then Sun and Li [16] obtained customers’ equilibrium balking strategies with five types of decision criteria in some unobservable queues with multiple vacations. On the other hand, for discrete-time queueing systems, Ma et al. [12] investigated some Geo/Geo/1 queues with multiple vacations. Besides customers’ equilibrium balking strategies, there are also some work concerning socially optimal ones. Sun et al. [15,18] considered fully observable and unobservable queues with several types of setup/closedown policies: interruptible, skippable and insusceptible policies, respectively. Moreover, Guo and Hassin [4,5] elaborately studied fully observable and unobservable queues with homogeneous and heterogeneous customers under N-policy, respectively. Then Economou et al. [3] further discussed the unobservable and partially observable queues with general service and vacation times. Different from the work with various classical vacation policies above, recently, Sun and Li [17] and Zhang et al. [22] studied the customers’ equilibrium or socially optimal balking strategies in queues with multiple working vacations under different information levels. Moreover, Wang et al. [20] considered customers’ equilibrium strategies in some discrete-time queues with single working vacation. Actually, detailed description of working vacation policy is originally given by Servi ⇑ Corresponding author. E-mail addresses: [email protected] (W. Sun), [email protected] (S. Li), [email protected] (Q.-L. Li). http://dx.doi.org/10.1016/j.amc.2014.09.116 0096-3003/Ó 2014 Elsevier Inc. All rights reserved.

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W. Sun et al. / Applied Mathematics and Computation 248 (2014) 195–214

and Finn [14] and various system performance indices can be consulted in the survey given by Tian and Zhang [19]. In addition, besides [17,20,22], in the other related literature, the most commonly studied policies are also multiple or single working vacation policy, and service rate during a whole vacation period is always uniform. Using matrix analytic method, Li et al. [8] and Liu et al. [11] analyzed stationary queue length distribution and stochastic decomposition results of an Markovian queue and an M/G/1 queue with multiple exponentially distributed working vacations, respectively. Then Li and Tian [7] considered a GI/M/1 queue with single working vacation and derived various performance measures. However, in some service systems, the server takes neither single nor multiple working vacations, but takes n-stage working vacations, where n is a fixed number and n P 2. During a whole vacation period, the server provides lower-rate service by himself or by his replacements. Moreover, in view of system managers’ specific demands, the server needs to adjust his service rates (or his replacements may be changed) among different vacation states, so service rate during each vacation time can be distinct with each other. One case is that in order to avoid overstocking too many customers at the end of each vacation time, service rate can be arranged to increase progressively with n. Alternatively, it can also be arranged to decrease progressively with n if the manager wants to equably attract customers even at the earlier-stage vacation times. To explain the model, here we give a simple real-life application. For example, there are a chef (the owner) and several (n) assistants in a famous snack bar in a small town, and it can also provide takeout to the customers. During the chef’s short vacation period, the assistants, as reserve cooks, take over the duties of the chef in turn. In view of their different levels of cooking skill, they can adjust their order according to the queue length or other factors. In this paper, we assume the original case n ¼ 2. That is, we study some queues with two-stage working vacations, where the server provides low-rate service but maintains different service rates in the two successive working vacation times, respectively. The main objective of our work is to study the customers’ equilibrium balking strategies in some single-server Markovian queues with two-stage exponentially distributed working vacations. When customers arrive at the system, in the light of their acquired different precision levels of system information, they need to make decisions of whether to join the queue or not. Here we study three types of queueing systems: the observable queues, the partially observable queues and the unobservable queues. For the observable queues, we first derive the customers’ equilibrium balking thresholds both in server’s two-stage working vacation states and busy state. In equilibrium, the order of balking thresholds in the two-stage working vacations depends on the relation of service rates in the three states. Then in view of the relation of balking thresholds in the first-stage and second-stage working vacations, we get the stationary queue length distributions and equilibrium social welfare. For the partially observable queues, we derive the customers’ equilibrium joining probabilities and numerically find that their positive equilibrium mixed strategy is unique. Moreover, the customers’ equilibrium joining probabilities in vacation states are not necessarily lower than that in busy state, which depends on the values of the service rates. In addition, we also obtain the stationary queue length distribution and equilibrium social welfare. Finally, for the unobservable queues, we easily get the customers’ equilibrium mixed strategy and equilibrium social welfare based on the results given in the partially observable case. The paper is organized as follows: In Section 2, we describe the queueing models and different information precision levels. Sections 3–5 are devoted to the observable queues, the partially observable queues and the unobservable queues, respectively. For each type of queue, we derive the customers’ equilibrium balking strategies and equilibrium social welfare. In Section 6, we briefly conclude the paper. 2. Model description In this paper, we consider some single-server Markovian queueing systems with two-stage working vacations. Assume that customers’ potential arrival rate is k and the server’s regular service rate is lb . Once the system becomes empty, the server begins a first-stage working vacation V 1 , which follows an exponential distribution with parameter h. During this period, an arriving customer is served at a service rate lv 1 , which is lower than lb . As soon as the first-stage vacation finishes, the server continues to take a second-stage working vacation V 2 , which follows the same distribution with V 1 , no matter whether there are customers in the system or not, and then switches service rate from lv 1 to lv 2 , which is also lower than lb . After finishing the second-stage vacation, a regular busy period starts if there are still customers waiting in the queue. Otherwise, the server stays idle until one customer arrives. So we denominate this type of queues as M/M/1/(T-S) WV queues. Let ðNðtÞ; IðtÞÞ represent the system state at time t, where NðtÞ denotes the system occupancy, i.e., the number of customers in the system, and IðtÞ denotes the server state at time t, and

8 > < 0; IðtÞ ¼ 1; > : 2;

the server is busy or stays idle; the server is taking a first-stage vacation; the server is taking a second-stage vacation:

So the observable case means that arriving customers can observe system information of both NðtÞ and IðtÞ, and the partially observable case means they can observe IðtÞ but not NðtÞ, whereas the unobservable case means they can observe neither NðtÞ nor IðtÞ. Let us mark an arriving customer, and he can receive a reward R after service completion but has to bear a cost C for waiting a time unit. We adopt a linear cost function, so his expected net benefit, denoted by U, is U ¼ R  CE½W, where

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W. Sun et al. / Applied Mathematics and Computation 248 (2014) 195–214

E½W represents his expected sojourn time in the system. Obviously, U ¼ 0 if he selects balking. We assume that the interarrival times, service times, and the two-stage working vacation times are mutually independent, and the service discipline is first in first out (FIFO). 3. The observable queues We first consider the customers’ equilibrium balking behavior in the observable case, i.e., arriving customers can observe both NðtÞ and IðtÞ at time t. Denote the sojourn time of the marked customer who joins the system in case he encounters the system state ðn; iÞ ði ¼ 0; 1; 2Þ as W i ðnÞ, and its mean as E½W i ðnÞ, and his expected net benefit after service completion as U i ðnÞ. So the customers’ equilibrium balking threshold at state i can be denoted by ne ðiÞ. Now we need to find the customers’ equilibrium integrated balking threshold strategy ðne ð0Þ; ne ð1Þ; ne ð2ÞÞ. For E½W 0 ðnÞ, it is obviously equal to n þ 1 service times at the regular service rate lb , that is,

E½W 0 ðnÞ ¼

nþ1

lb

ð3:1Þ

:

On the other hand, in order to derive E½W i ðnÞ ði ¼ 1; 2Þ, first we need to discuss three cases at state ðn; 1Þ: If the residual time of the first-stage working vacation, denoted by V 1R , is long enough for service completion of n þ 1 customers, his sojourn time W 1 ðnÞ of the marked customer equals n þ 1 service times at the service rate lv 1 ; If there are only j ð0 6 j 6 nÞ customers completing service during V 1R and other n þ 1  j customers completing service during the second-stage working vacation V 2 ; W 1 ðnÞ equals the sum of V 1R and n þ 1  j service times at the service rate lv 2 ; Otherwise, if there are j ð0 6 j 6 nÞ customers completing service during V 1R and i ð0 6 i 6 n  jÞ customers completing service during V 2 and other n þ 1  i  j customers completing service during the next busy period, W 1 ðnÞ equals the sum of V 1R and V 2 and n þ 1  i  j service times at the regular service rate lb . Hence, based on the Lemmas 1 and 2 given by Sun and Li [17], we get the Laplace–Stieltjes f  ðsÞ, as follows: transform (LST) of W 1 ðnÞ, denoted by W 1

n o f  ðsÞ ¼ P Sðnþ1Þ < V 1R W 1 v1

lv 1 þ h lv 1 þ h þ s



lv 1 þ h



h lv 1 þ h þ s

þ

¼

!jþ1 !jþ1

n n o X \ ðnþ1jÞ P SðvjÞ1 < V 1R < Sðvjþ1Þ < V2 Sv 2 1 j¼0

lv 2 þ h lv 2 þ h þ s

!nþ1j þ

lv 2 þ h þ s !nþ1

lv 2

j¼0



lb

lb

nþ1ij ¼

lb þ s

n X

lv 2 þ h þ s

nj n n X o X \ ðiÞ P SðvjÞ1 < V 1R < Sðvjþ1Þ Sv 2 < V 2 < Sðiþ1Þ v2 1 j¼0 i¼0

!iþ1 

lv 2 þ h

h h lv 1 þ h þ s lv 2 þ h þ s !nþ1

lv 1 lv 1 þ h þ s

þ

lv 1 þ h þ s

lv 1 þ h þ s

þ

!nþ1

lv 1 þ h

lv 1 ðlv 2 þ h þ sÞ lv 2 ðlv 1 þ h þ sÞ

nþ1 X nj n X

lb þ s

j¼0 i¼0

0

!nþ1

lv 1

lv 1 þ h þ s

!j !j

lv 1 lv 1 þ h þ s

lv 2 lv 2 þ h þ s

!nþ1

!i 

lb

ðiþjÞ

lb þ s !nþ1 1 A

hlv 2 lv 1 lv 2 @  ðh þ sÞðlv 1  lv 2 Þ lv 1 þ h þ s lv 2 þ h þ s 0 0 !nþ1 1   lv 1 lb h 2 lb lb nþ1 @ @ A  þ lb h þ ðlb  lv 2 Þs lb h þ ðlb  lv 1 Þs lb þ s lv 1 þ h þ s 0 !nþ1 !nþ1 11 þ

þ

lv 2

ðh þ sÞðlv 1  lv 2 Þ

@

lv 2 lv 2 þ h þ s



lv 1 lv 1 þ h þ s

AA:

ð3:2Þ

f  ð0Þ, that is, Then we can easily get E½W 1 ðnÞ ¼  W 1 0

E½W 1 ðnÞ ¼

nþ1

lb

2lb  lv 1  lv 2 l2v 1  2lb lv 1 þ lb lv 2 þ þ lb h lb hðlv 1  lv 2 Þ

lv 1 lv 1 þ h

!nþ1

lv 2 ðlb  lv 2 Þ lv 2 þ lb hðlv 1  lv 2 Þ lv 2 þ h

!nþ1 :

ð3:3Þ

Then for the system state ðn; 2Þ, we need to discuss two cases: If the residual time of the second-stage working vacation, denoted by V 2R , is long enough for service completion of n þ 1 customers, W 2 ðnÞ equals n þ 1 service times at the service rate lv 2 ; Otherwise, if there are only j ð0 6 j 6 nÞ customers completing service during V 2R and other n þ 1  j customers completing service during the next busy period, W 2 ðnÞ equals the sum of V 2R and n þ 1  j service times at the regular service f  ðsÞ, as follows: rate lb . Hence, we get the LST of W 2 ðnÞ, denoted by W 2

198

W. Sun et al. / Applied Mathematics and Computation 248 (2014) 195–214 ne(0) n (1) e ne(2)

20

ne

15

10

5

0

1.5

2

2.5

3

3.5

μ

4

4.5

5

b

Fig. 1. Equilibrium threshold strategies in an observable queue when R ¼ 5; C ¼ 1; lv 1 ¼ 0:8; lv 2 ¼ 1, h ¼ 0:5.

n o f  ðsÞ ¼ P Sðnþ1Þ < V 2R W v2 2

¼

¼

¼

!nþ1

lv 2

lv 2 þ h

!nþ1

lv 2 þ h lv 2 þ h þ s

þ

lv 2 þ h þ s !nþ1



j¼0

!nþ1

lv 2 þ h

n n o X þ P SðvjÞ2 < V 2R < Sðvjþ1Þ 2 n X j¼0

!

h lv 2 þ h

lv 2

!j

lv 2 þ h

!jþ1 

lv 2 þ h lv 2 þ h þ s lv 2 þ h

þ

lb

f 0 ð0Þ, that is, Then we get E½W 2 ðnÞ ¼  W 2

E½W 2 ðnÞ ¼

nþ1

lb

0

lb

nþ1j

lb þ s

!j

nþ1

n X lv 2 ðlb þ sÞ h lv 2 þ h þ s lb þ s lv 2 þ h þ s j¼0 lb ðlv 2 þ h þ sÞ 0 !nþ1 !nþ1 1  nþ1 lv 2 l lb h l v b 2 @ A: þ  lv 2 þ h þ s lb h þ ðlb  lv 2 Þs lb þ s lv 2 þ h þ s

lv 2

nþ1j

lb þ s

!jþ1 

lv 2 þ h þ s

lb

!nþ1 1 A:

lb  lv 2 @ lv 2 þ 1 h lb lv 2 þ h

ð3:4Þ

ð3:5Þ

Based on (3.1), (3.3) and (3.5), we can get the expressions of U i ðnÞ ¼ R  CE½W i ðnÞ ði ¼ 0; 1; 2Þ, respectively. In equilibrium, the marked customer’s expected net benefit after service completion, denoted by U ei ðnÞ, equals 0, i.e., U ei ðnÞ ¼ 0. So solving U ei ðnÞ ¼ 0, we can derive the unique feasible root ne ðiÞ.1 Hence the customers’ equilibrium integrated balking threshold strategy is ðne ð0Þ; ne ð1Þ; ne ð2ÞÞ ¼ ðbne ð0Þc; bne ð1Þc; bne ð2ÞcÞ. Fig. 1 shows that ne ðiÞ ði ¼ 0; 1; 2Þ all increase with lb , and the relation ne ð0Þ P ne ð2Þ P ne ð1Þ always maintains when lv 1 < lv 2 , which is obvious. On the other hand, in case lv 1 is much larger than lv 2 , the nth customer arriving in the firststage vacation is much more expected to finish his service during that vacation than the nth customer arriving in the second-stage vacation when lb ð> lv 1 Þ is relatively small. However, with the increase of lb , compared with customers arriving in the first-stage vacation who have risk of switching service rate from fast to low value, customers arriving in the secondstage vacation face a much more optimistic situation although the value of lv 2 is low, so the increasing rate of ne ð2Þ is much faster than that of ne ð1Þ. Therefore, Fig. 2 shows that initially ne ð1Þ P ne ð2Þ and then ne ð2Þ P ne ð1Þ with lb . Summarily, we can numerically observe that for all feasible lb ; ne ð0Þ P ne ð2Þ P ne ð1Þ when lv 1 < lv 2 or lv 1 does not exceed lv 2 too much. However, it is not true when lv 1 exceeds lv 2 too much and lv 1 is close to lb . Next we will derive the stationary queue length distribution in observable queues. In view of the possible relations of ne ð1Þ and ne ð2Þ, we discuss two cases: ne ð1Þ > ne ð2Þ and ne ð1Þ < ne ð2Þ.2 For the first case, the equilibrium transition diagram is depicted in Fig. 1. So state space of the Markovian process fNðtÞ; IðtÞg is

Xob1 ¼ fðn; 0Þ : 0 6 n 6 ne ð0Þ þ 1g

[

fðn; jÞ : 0 6 n 6 ne ð1Þ þ 1;

j ¼ 1; 2g:

1 Under the model assumptions of FCFS service discipline and independence of arrival process and two-stage vacation times, a customer joining in vacation states only gives negative externality to his following customers. So U 1 ðnÞ is surely monotonously decreasing with n. 2 The case of ne ð1Þ ¼ ne ð2Þ can be regarded as the common special case of ne ð1Þ > ne ð2Þ and ne ð1Þ < ne ð2Þ.

W. Sun et al. / Applied Mathematics and Computation 248 (2014) 195–214 30

199

n (0) e n (1) e ne(2)

28 26 24

n

e

22 20 18 16 14 12 10 3

3.5

4

4.5

μ

5

5.5

6

b

Fig. 2. Equilibrium threshold strategies in an observable queue when R ¼ 5; C ¼ 1; lv 1 ¼ 3; lv 2 ¼ 1, h ¼ 0:5.

Denote the stationary queue length distribution as

pnj ¼ PfN ¼ n; I ¼ jg ¼ t!1 limPfNðtÞ ¼ n; IðtÞ ¼ jg; ðn; jÞ 2 Xob1 : Then the stationary transition probability equations can be written as

p00 k ¼ p02 h; pn0 ðk þ lb Þ ¼ pn10 k þ pnþ10 lb þ pn2 h; 1 6 n 6 ne ð1Þ þ 1 pn0 ðk þ lb Þ ¼ pn10 k þ pnþ10 lb ; ne ð1Þ þ 2 6 n 6 ne ð0Þ pne ð0Þþ10 lb ¼ pne ð0Þ0 k; p01 ðk þ hÞ ¼ p10 lb þ p11 lv 1 ; pn1 ðk þ h þ lv 1 Þ ¼ pn11 k þ pnþ11 lv 1 ; 1 6 n 6 ne ð1Þ pne ð1Þþ11 ðh þ lv 1 Þ ¼ pne ð1Þ1 k; p02 ðk þ hÞ ¼ p01 h þ p12 lv 2 ; pn2 ðk þ h þ lv 2 Þ ¼ pn12 k þ pnþ12 lv 2 þ pn1 h; 1 6 n 6 ne ð2Þ pne ð2Þþ12 ðh þ lv 2 Þ ¼ pne ð2Þ2 k þ pne ð2Þþ22 lv 2 þ pne ð2Þþ11 h; pn2 ðh þ lv 2 Þ ¼ pnþ12 lv 2 þ pn1 h; ne ð2Þ þ 2 6 n 6 ne ð1Þ pne ð1Þþ12 ðh þ lv 2 Þ ¼ pne ð1Þþ11 h:

ð3:6Þ ð3:7Þ ð3:8Þ ð3:9Þ ð3:10Þ ð3:11Þ ð3:12Þ ð3:13Þ ð3:14Þ ð3:15Þ ð3:16Þ ð3:17Þ

We first consider the probabilities fpn1 j 0 6 n 6 ne ð1Þ þ 1g in the first-stage vacation state. From (3.11), we notice that they are the solutions of the following homogeneous linear difference equation with constant coefficients

lv 1 xnþ1  ðk þ h þ lv 1 Þxn þ kxn1 ¼ 0; n ¼ 1; 2; . . . ; ne ð1Þ:

ð3:18Þ

So its corresponding characteristic equation is

lv 1 x2  ðk þ h þ lv 1 Þx þ k ¼ 0; which has two roots

x1;2 ¼

ðk þ h þ lv 1 Þ 

qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ðk þ h þ lv 1 Þ2  4klv 1 2lv 1

ð3:19Þ

:

n   The general solution of (3.18), denoted by xhom , is xhom ¼ A1 xn n n 1 þ B1 x2 (obviously x1 – x2 ), where A1 ; B1 are the coefficients to be determined. From (3.10) and (3.12), we obtain

8 > > < A1 ¼ xne ð1Þ ððhþlv 2

> > : B1 ¼

ne ð1Þ ððhþ

x2 1

lv 1 Þx2 kÞlb p10 ne ð1Þ ððhþ

Þx2 kÞðkþhlv x1 Þx1 1

n ð1Þ x1 e ððhþ v Þx1 kÞ b 10 1 ne ð1Þ   ððhþ v Þx2 kÞðkþh v 1 Þx1 kÞðkþh v 1 x2 Þx2 1

l

n ð1Þ x1 e ððhþ

lv 1 Þx1 kÞðkþhlv 1 x2 Þ

l

l

lp

l

 v 1 x1 Þ

l

; ð3:20Þ ;

200

W. Sun et al. / Applied Mathematics and Computation 248 (2014) 195–214

thus, n pn1 ¼ A1 xn n ¼ 0; 1; . . . ; ne ð1Þ þ 1; 1 þ B1 x2

ð3:21Þ

where x1 ; x2 and A1 ; B1 are given by (3.19) and (3.20), respectively. For the second-stage vacation state, we first consider the probabilities fpn2 jne ð2Þ þ 2 6 n 6 ne ð1Þ þ 1g. From (3.16), they are the solutions of the following nonhomogeneous linear difference equation n lv 2 xnþ1  ðh þ lv 2 Þxn ¼ pn1 h ¼ hðA1 xn n ¼ ne ð2Þ þ 2; . . . ; ne ð1Þ: 1 þ B1 x2 Þ;

ð3:22Þ

The general solution of the homogeneous version of (3.22) is xhom ¼ B3 ^ xn2 , where n

^x2 ¼

h þ lv 2

lv 2

ð3:23Þ

:

gen spec Hence the general solution of (3.22), denoted by xgen ¼ xhom þ xspec is a specific solution of (3.22). n , is given as xn n , where xn n   Because the nonhomogeneous part of (3.22) is geometric with parameter x1 and x2 , we consider a specific solution of the n form xspec ¼ C 2 xn n 1 þ D2 x2 . Substituting it into (3.22), we get

8 1 ; < C 2 ¼ l xhA v 2 1 ðhþlv 2 Þ hB1 : D2 ¼ : 

ð3:24Þ

lv 2 x2 ðhþlv 2 Þ

Therefore, n n ^n xgen n ¼ B3 x2 þ C 2 x1 þ D2 x2 ;

n ¼ ne ð2Þ þ 2; . . . ; ne ð1Þ þ 1;

ð3:25Þ

where B3 is to be determined. Taking into account (3.17), then obtain

B3 ¼

    n ð1Þþ1 n ð1Þþ1 n ð1Þþ1 n ð1Þþ1 h A 1 x1 e þ B1 x2 e þ D2 x2 e  ðh þ lv 2 Þ C 2 x1 e n ð1Þþ1

ðh þ lv 2 Þ^x2e

ð3:26Þ

;

thus, from (3.25), n pn2 ¼ B3 ^xn2 þ C 2 xn n ¼ ne ð2Þ þ 2; . . . ; ne ð1Þ þ 1 1 þ D2 x2 ;

ð3:27Þ

can be obtained, where C 2 ; D2 and B3 are given by (3.24) and (3.26), respectively. Then we consider the probabilities fpn2 j0 6 n 6 ne ð2Þ þ 1g. From (3.14), they are the solutions of the following nonhomogeneous linear difference equation n lv 2 xnþ1  ðk þ h þ lv 2 Þxn þ kxn1 ¼ pn1 h ¼ hðA1 xn n ¼ 1; 2; . . . ; ne ð2Þ: 1 þ B1 x2 Þ;

The general solution of the homogeneous version of (3.28) is

~x1;2 ¼

qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ðk þ h þ lv 2 Þ  ðk þ h þ lv 2 Þ2  4klv 2 2lv 2

xhom n

¼

xn1 A2 ~

þ

xn2 , B2 ~

ð3:28Þ

where

ð3:29Þ

:

spec n So the general solution of (3.28) is xgen ¼ xhom þ xspec ¼ C 1 xn n n , where xn n 1 þ D1 x2 . Substituting it into (3.28), we get

8 hA1 x > < C 1 ¼ l x2 ðhþkþ1l Þx þk ; v2 1 v2 1 hB1 x2 > :D ¼ : 1

ð3:30Þ

lv 2 x2 ðhþkþlv Þx2 þk 2 2

Therefore, n n ~n ~n xgen n ¼ A2 x1 þ B2 x2 þ C 1 x1 þ D1 x2 ;

n ¼ 0; 1; . . . ; ne ð2Þ þ 1;

ð3:31Þ

where A2 ; B2 are to be determined. Taking into account (3.13), (3.15) and (3.27), then obtain         8 ne ð2Þ n ð2Þ n ð2Þþ1 n ð2Þþ1 > ~ k  ðlv 2 þ hÞx1 þ D1 x2 e k  ðlv 2 þ hÞx2 þ h A1 x1 e þ B1 x2 e > > A2 ¼ ðk þ hÞ  lv 2 x2 C 1 x1 > >      > > n ð2Þþ2 n ð2Þþ2 n ð2Þþ2 n ð2Þ > þlv 2 B3 ^x2e þ C 2 x1 e þ D2 x2 e ðlv 2 þ hÞ~x2  k hðA1 þ B1 Þ þ lv 2 ðC 1 x1 þ D1 x2 Þ  ðk þ hÞðC 1 þ D1 Þ  ~x2e > > > > >       1 > > n ð2Þ n ð2Þ > < ðlv 2 þ hÞ~x1  k ðk þ hÞ  lv 2 ~x2  ~x2e ðlv 2 þ hÞ~x2  k ðk þ hÞ  lv 2 ~x1 ;  ~x1e         n ð2Þ n ð2Þ n ð2Þþ1 n ð2Þþ1 > e > k  ðlv 2 þ hÞx1 þ D1 x2 e k  ðlv 2 þ hÞx2 þ h A1 x1 e þ B1 x2 e > B2 ¼ ðk þ hÞ  lv 2 ~x1 C 1 x1 > > >      > > n ð2Þþ2 n ð2Þþ2 n ð2Þþ2 n ð2Þ > > þlv 2 B3 ^x2e þ C 2 x1 e þ D2 x2 e ðlv 2 þ hÞ~x1  k hðA1 þ B1 Þ þ lv 2 ðC 1 x1 þ D1 x2 Þ  ðk þ hÞðC 1 þ D1 Þ  ~x1e > > > >       1 > > n ð2Þ n ð2Þ : ðlv þ hÞ~x2  k ðk þ hÞ  lv ~x1  ~x1e ðlv þ hÞ~x1  k ðk þ hÞ  lv ~x2 ;  ~x2e 2

2

2

2

ð3:32Þ

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W. Sun et al. / Applied Mathematics and Computation 248 (2014) 195–214

thus, from (3.31), n pn2 ¼ A2 ~xn1 þ B2 ~xn2 þ C 1 xn n ¼ 0; 1; . . . ; ne ð2Þ þ 1 1 þ D 1 x2 ;

ð3:33Þ

can be obtained, where C 1 ; D1 and A2 ; B2 are given by (3.30) and (3.32), respectively. Sequentially analyzing the server’s busy state, we first derive the probabilities fpn0 j0 6 n 6 ne ð2Þ þ 1g. From (3.7), they are the solutions of the following nonhomogeneous linear difference equation n lb xnþ1  ðk þ lb Þxn þ kxn1 ¼ pn2 h ¼ hðA2 ~xn1 þ B2 ~xn2 þ C 1 xn n ¼ 1; 2; . . . ; ne ð2Þ: 1 þ D1 x2 Þ;

So the general solution of (3.34) is tuting it into (3.34), we get

xgen n

xspec n ,

¼ xhom þ n

where xhom ¼ A4 þ B4 qn and n

xspec n

ð3:34Þ

n ¼ E1 ~ xn1 þ F 1 ~ xn1 þ C 3 xn 1 þ D3 x2 . Substi-

8 hA2 ~x1 E1 ¼ l ~x2 ðkþ ; > lb Þ~x1 þk > b 1 > > > ~ hB x > > < F 1 ¼ l ~x2 ðkþ2l 2Þ~x þk ; b 2

2

b

ð3:35Þ



hC x > C 3 ¼ l x2 ðkþ1l1 Þx þk ; > > b 1 b 1 > > > > hD1 x2 :D ¼ : 3 l x2 ðkþl Þx þk b 2

b

2

Therefore, n n n ~n ~n xgen n ¼ A4 þ B4 q þ E1 x1 þ F 1 x1 þ C 3 x1 þ D3 x2 ;

n ¼ 0; 1; . . . ; ne ð2Þ þ 1;

ð3:36Þ

where A4 ; B4 are to be determined. Taking into account (3.6), then obtain

8   ~ ~ > < A4 ¼ ðhðA2 þB2 þC1 þD1 ÞkðE1 þF 1 þC 3 þD3 ÞÞlb ðp10 ðE1 x1 þF 1 x2 þC3 x1 þD3 x2 ÞÞ ; kl b

ð3:37Þ

  > : B4 ¼ ðhðA2 þB2 þC 1 þD1 ÞkðE1 þF 1 þC 3 þD3 ÞÞkðp10 ðE1 ~x1 þF 1 ~x2 þC 3 x1 þD3 x2 ÞÞ ; kð1qÞ

thus, from (3.36), n pn0 ¼ A4 þ B4 qn þ E1 ~xn1 þ F 1 ~xn1 þ C 3 xn n ¼ 0; 1; . . . ; ne ð2Þ þ 1; 1 þ D 3 x2 ;

ð3:38Þ

can be obtained, where E1 ; F 1 ; C 3 ; D3 are given by (3.35) and A4 ; B4 are given by (3.37), respectively. Then we derive the probabilities fpn0 jne ð2Þ þ 2 6 n 6 ne ð1Þ þ 1g. They are the solutions of the following nonhomogeneous linear difference equation n lb xnþ1  ðk þ lb Þxn þ kxn1 ¼ pn2 h ¼ hðB3 ~xn2 þ C 2 xn n ¼ ne ð2Þ þ 2; . . . ; ne ð1Þ þ 1: 1 þ D2 x2 Þ;

So the general solution of (3.39) is into (3.39), we get

xgen n

¼

xhom n

þ

xspec n ,

where

xhom n

n

¼ A5 þ B5 q and

xspec n

8 hB3 ~x2 F ¼ ; > > > 2 lb ~x22 ðkþlb Þ~x2 þk > < hC 2 x1 C 4 ¼ l x2 ðkþl Þx þk ; b 1 b 1 > > > hD2 x2 > : D4 ¼ l x2 ðkþl Þx þk : b 2

b

¼

F2~ xn2

þ

C 4 xn 1

ð3:39Þ þ

D4 xn 2 .

Substituting it

ð3:40Þ

2

Therefore, n n n ~n xgen n ¼ A5 þ B5 q þ F 2 x2 þ C 4 x1 þ D4 x2 ;

n ¼ ne ð2Þ þ 2; . . . ; ne ð1Þ þ 1;

ð3:41Þ

where A5 ; B5 are to be determined. Taking into account (3.38), then obtain

8 > < A5 > : B5

n ð2Þþ1

¼ A4 þ

E1 ~x1e

n ð2Þþ1

¼ B4 þ

E1 ~x1e

n ð2Þþ1

ð~x1 qÞþ~x2e 1q

ð~x2 qÞðF 1 F 2 Þ

n ð2Þþ1 ð1~x1 Þþ~x2e ð1~x2 ÞðF 1 F 2 Þ

qne ð2Þþ1 ð1qÞ

þ þ

ne ð2Þþ1  n ð2Þþ1  ðx1 qÞðC 3 C 4 Þþx2 e ðx2 qÞðD3 D4 Þ

x1

1q

ne ð2Þþ1 n ð2Þþ1 ð1x1 ÞðC 3 C 4 Þþx2 e ð1x2 ÞðD3 D4 Þ qne ð2Þþ1 ð1qÞ

x1

; ð3:42Þ

;

thus, from (3.41), n pn0 ¼ A5 þ B5 qn þ F 2 ~xn2 þ C 4 xn n ¼ ne ð2Þ þ 2; . . . ; ne ð1Þ þ 1 1 þ D 4 x2 ;

ð3:43Þ

can be obtained, where F 2 ; C 4 ; D4 are given by (3.40) and A5 ; B5 are given by (3.42), respectively. Finally, we derive the probabilities fpn0 jne ð1Þ þ 2 6 n 6 ne ð0Þ þ 1g. From (3.8), we get xhom ¼ A6 þ B6 qn , where A6 ; B6 are to n be determined. Taking into account (3.9) and (3.43), we obtain A6 ¼ 0 and n ð1Þþ2

B6 ¼

A5 þ B5 qne ð1Þþ2 þ F 2 ~x2e

q

ne ð1Þþ2

þ C 4 x1

ne ð1Þþ2

ne ð1Þþ2

þ D 4 x2

:

ð3:44Þ

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W. Sun et al. / Applied Mathematics and Computation 248 (2014) 195–214

Therefore,

pn0 ¼ B6 qn ; n ¼ ne ð1Þ þ 2; . . . ; ne ð0Þ þ 1;

ð3:45Þ

where B6 are given by (3.44). In summary, we find that the obtained stationary state probabilities fpnj jðn; jÞ 2 Xob1 g are all related to p10 . Using the normalization condition, we get the results in the following theorem. Theorem 3.1. For the observable Markovian queue with two-stage working vacations and state space Xob1 ¼ S fðn; 0Þ : 0 6 n 6 ne ð0Þ þ 1g fðn; jÞ : 0 6 n 6 ne ð1Þ þ 1; j ¼ 1; 2g, where ðne ð0Þ; ne ð1Þ; ne ð2ÞÞ are the customers’ equilibrium threshold strategies, the stationary queue length distribution fpnj jðn; jÞ 2 Xob1 g is: n pn1 ¼ A1 xn n ¼ 0; 1; . . . ; ne ð1Þ þ 1 1 þ B1 x2 ;

pn2 ¼

A2 ~xn1

þ

B2 ~xn2

þ

C 1 xn 1

þ

D1 xn 2 ;

ð3:46Þ

n ¼ 0; 1; . . . ; ne ð2Þ þ 1

ð3:47Þ

n pn2 ¼ B3 ^xn2 þ C 2 xn n ¼ ne ð2Þ þ 2; . . . ; ne ð1Þ þ 1 1 þ D2 x2 ; n

pn0 ¼ A4 þ B4 q þ

E1 ~xn1

þ

F 1 ~xn1

þ

C 3 xn 1

þ

D3 xn 2 ;

ð3:48Þ

n ¼ 0; 1; . . . ; ne ð2Þ þ 1

ð3:49Þ

n pn0 ¼ A5 þ B5 qn þ F 2 ~xn2 þ C 4 xn n ¼ ne ð2Þ þ 2; . . . ; ne ð1Þ þ 1 1 þ D4 x2 ;

ð3:50Þ

pn0 ¼ B6 qn ; n ¼ ne ð1Þ þ 2; . . . ; ne ð0Þ þ 1

ð3:51Þ

where q ¼ k=lb – 1 and ¼ 1; 2; 4; 5Þ; Bj ð1 6 j  6Þ; C k ; Dk ð1 6 k  4Þ; E1 ; F 1 ; F 2 are given by (3.19), (3.20), (3.23), (3.24), (3.26), (3.29), (3.30), (3.32), (3.35), (3.37), (3.40), (3.42), (3.44), respectively, and p10 can be solved by the normalization equation: x1;2 ; ~ x1;2 ; ^ x2 ; Ai ði

X ðn;jÞ2Xob

pnj ¼ 1:

ð3:52Þ

1

Based on Fig. 3 and Theorem 3.1, an arriving customer will select balking when he finds the system is at state ðne ð0Þ þ 1; 0Þ; ðne ð1Þ þ 1; 1Þ or fðn; 2Þ : ne ð2Þ þ 1 6 n 6 ne ð1Þ þ 1g. So the equilibrium social welfare per time unit, denoted by U s ðne ð0Þ; ne ð1Þ; ne ð2ÞÞ, equals

U s ðne ð0Þ; ne ð1Þ; ne ð2ÞÞ ¼ kR 1  pne ð0Þþ10  pne ð1Þþ11 

neX ð1Þþ1

!

pn2  C

neX ð0Þþ1

neX ð1Þþ1

n¼1

n¼1

n¼ne ð2Þþ1

npn0 þ

! nðpn1 þ pn2 Þ :

ð3:53Þ

As for the second case ne ð1Þ < ne ð2Þ, the equilibrium transition diagram is depicted in Fig. 4. So the state space of the Markovian process fNðtÞ; IðtÞg is

Xob2 ¼ fðn; jÞ : 0 6 n 6 nð jÞ þ 1;

1,1

0,1

b

1,2 v2

0,0

...

ne(2), 1

v1

v1

0,2

2,1

j ¼ 0; 1; 2g:

...

ne(2), 2

v2

ne(1), 1

...

ne(2) +1,2

ne(2), 0

...

ne(1), 2

ne(1) +1,2 v2

ne(2) +1,0 b

ne(1) +1,1 v1

v2

2,0 b

...

v1

2,2

1,0

ne(2) +1,1

...

ne(1), 0

ne(1) +1,0

...

b

Fig. 3. Equilibrium transition rate diagram for observable queues in case ne ð1Þ > ne ð2Þ.

ne(0), 0

ne(0) +1,0 b

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W. Sun et al. / Applied Mathematics and Computation 248 (2014) 195–214

1,1

0,1

b

1,2 v2

0,0

ne(1), 1

...

ne(1) +1,1

v1

v1

0,2

2,1

v1

ne(1), 2

...

2,2

ne(1) +1,2

v2

1,0

ne(2), 2

...

v2

v2

2,0

ne(1), 0

...

ne(1) +1,0

b

ne(2) +1,2

ne(2), 0

...

b

ne(2) +1,0

...

ne(0), 0

b

ne(0) +1,0 b

Fig. 4. Equilibrium transition rate diagram for observable queues in case ne ð1Þ < ne ð2Þ.

16 14

U (n (0),n (1),n (2)) s e e e

12 10 8 6 4 2 0

0

0.5

1

1.5

2

2.5

3

3.5

4

λ Fig. 5. Equilibrium social welfare for case ne ð1Þ > ne ð2Þ when R ¼ 10; C ¼ 1; lb ¼ 3; lv 1 ¼ 2:5; lv 2 ¼ 1, h ¼ 0:2.

Then the stationary transition probability equations can be written as

p00 k ¼ p02 h; pn0 ðk þ lb Þ ¼ pn10 k þ pnþ10 lb þ pn2 h; 1 6 n 6 ne ð2Þ þ 1 pn0 ðk þ lb Þ ¼ pn10 k þ pnþ10 lb ; ne ð2Þ þ 2 6 n 6 ne ð0Þ pne ð0Þþ10 lb ¼ pne ð0Þ0 k; p01 ðk þ hÞ ¼ p10 lb þ p11 lv 1 ; pn1 ðk þ h þ lv 1 Þ ¼ pn11 k þ pnþ11 lv 1 ; 1 6 n 6 ne ð1Þ pne ð1Þþ11 ðh þ lv 1 Þ ¼ pne ð1Þ1 k; p02 ðk þ hÞ ¼ p01 h þ p12 lv 2 ; pn2 ðk þ h þ lv 2 Þ ¼ pn12 k þ pnþ12 lv 2 þ pn1 h; 1 6 n 6 ne ð1Þ þ 1 pn2 ðk þ h þ lv 2 Þ ¼ pn12 k þ pnþ12 lv 2 ; ne ð1Þ þ 2 6 n 6 ne ð2Þ pne ð2Þþ12 lv 2 ¼ pne ð2Þ2 k:

ð3:54Þ ð3:55Þ ð3:56Þ ð3:57Þ ð3:58Þ ð3:59Þ ð3:60Þ ð3:61Þ ð3:62Þ ð3:63Þ ð3:64Þ

Using the similar analysis method with that in the first case, we can also get the stationary queue length distribution fpnj jðn; jÞ 2 Xob2 g for the second case.3 Based on Fig. 4, the customers’ balking states are ðne ð0Þ þ 1; 0Þ; ðne ð1Þ þ 1; 1Þ or ðne ð2Þ þ 1; 2Þ. So the equilibrium social welfare per time unit equals 3 In order to avoid repetition, here we omit the analysis process of deriving the stationary queue length distribution for the case ne ð1Þ < ne ð2Þ, which is similar with that in the case ne ð1Þ > ne ð2Þ.

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W. Sun et al. / Applied Mathematics and Computation 248 (2014) 195–214 14

Us(ne(0),ne(1),ne(2))

12 10 8 6 4 2 0

0

0.5

1

1.5

2

2.5

3

3.5

4

λ Fig. 6. Equilibrium social welfare for case ne ð1Þ < ne ð2Þ when R ¼ 10; C ¼ 1; lb ¼ 3; lv 1 ¼ 1:2; lv 2 ¼ 1, h ¼ 0:2.

  U s ðne ð0Þ; ne ð1Þ; ne ð2ÞÞ ¼ kR 1  pne ð0Þþ10  pne ð1Þþ11  C

neX ð0Þþ1

npn0 þ

n¼1

neX ð1Þþ1

neX ð2Þþ1

n¼1

n¼1

npn1 þ

!

npn2 :

ð3:65Þ

Figs. 5 and 6 show the tendency of the equilibrium social welfare with k for the cases ne ð1Þ > ne ð2Þ and ne ð1Þ < ne ð2Þ, respectively. We observe that U s ðne ð0Þ; ne ð1Þ; ne ð2ÞÞ first gradually increases and achieves a maximum, then decreases as k continues increasing no matter whichever case. The reason for this behavior is that the system is rarely crowded and joining customers have no need to wait long time for service when k is small, which makes the social welfare improve. However, customers face longer waiting delay and more waiting cost if they decide to join as k keeps increasing, which has a negative effect on the equilibrium social welfare. 4. The partially observable queues Next we consider the partially observable case, that is, arriving customers only know the server’s state IðtÞ but cannot observe the system occupancy NðtÞ at time t. So the customers’ decision problem can be expressed by a set of joining probabilities ðq0 ; q1 ; q2 Þð0 6 qi  1; i ¼ 0; 1; 2Þ, and their equilibrium mixed strategy is denoted by ðqe ð0Þ; qe ð1Þ; qe ð2ÞÞ. In order to derive ðqe ð0Þ; qe ð1Þ; qe ð2ÞÞ, we first try to get the stationary queue length distribution. The transition diagram is depicted in Fig. 7. Based on Fig. 7, we can obviously observe that fNðtÞ; IðtÞg is a quasi-birth-and-death (QBD) process with the state space

Xpo ¼ fðk; jÞ : k P 0;

j ¼ 0; 1; 2g:

q1

1,1

0,1 v1

q2

0,2 b

0,0

q1

q1

2,1

...

n,1

v1

v1

q2

q2

1,2

2,2

...

n,2

v2

v2

v2

q0

q0

q0

1,0

2,0 b

...

n,0

n+1, 1

...

n+1, 2

...

n+1, 0

...

b

Fig. 7. Transition rate diagram for partially observable queues.

W. Sun et al. / Applied Mathematics and Computation 248 (2014) 195–214

205

If q ¼ kq0 =lb < 1, let ðN; IÞ be the stationary limit of the QBD process fNðtÞ; IðtÞg. Denote the stationary queue length distribution as

p ¼ ðp0 ; p1 ; p2 ; . . .Þ; pk ¼ ðpk0 ; pk1 ; pk2 Þ; k P 0; pkj ¼ PfN ¼ k; I ¼ jg ¼ t!1 lim PfNðtÞ ¼ k; IðtÞ ¼ jg; ðk; jÞ 2 Xpo and the infinitesimal generator of the process as Q , then we get the stationary transition probability equations pQ ¼ 0 as follows

p00 kq0 ¼ p02 h; pk0 ðkq0 þ lb Þ ¼ pk10 kq0 þ pk2 h þ pkþ10 lb ; k P 1; p01 ðkq1 þ hÞ ¼ p10 lb þ p11 lv 1 ;

ð4:3Þ

pk1 ðkq1 þ h þ lv 1 Þ ¼ pk11 kq1 þ pkþ11 lv 1 ; k P 1;

ð4:4Þ

p02 ðkq2 þ hÞ ¼ p01 h þ p12 lv 2 ;

ð4:5Þ

pk2 ðkq2 þ h þ lv 2 Þ ¼ pk12 kq2 þ pkþ12 lv 2 þ pk1 h; k P 1:

ð4:6Þ

ð4:1Þ ð4:2Þ

So using the lexicographical ordering for the system states, Q can be written in the block-diagonal form as follows:

0

1

A00

C

B B10 B B Q ¼B B B @

A

C

B

A

C

B

A C .. .. .. . . .

C C C C; C C A

ð4:7Þ

where

0

B A00 ¼ @ 0

0

ðh þ kq1 Þ

h

0

ðh þ kq2 Þ

h 0

lb

B B¼@ 0 0 0 B A¼@

0

0

1

0 C A;

lv 1 0

1

0

kq0

lv 2

0

kq0

B C¼@ 0 0

0

C A;

B10

0

0

kq1 0

0 B ¼ @0

lb lv 1

0

0

1 0 0 C A;

lv 2

1

C 0 A; kq2

ðkq0 þ lb Þ

0

0

1

0

ðkq1 þ h þ lv 1 Þ

h

h

0

ðkq2 þ h þ lv 2 Þ

C A:

Now we need to solve the following matrix quadratic equation

R2 B þ RA þ C ¼ 0

ð4:8Þ

for its minimal non-negative solution which is called the rate matrix R. Based on the special structure of A; B; C, we conclude that R has the similar form. So we assume

0

r 11 B R ¼ @ r 21 r 31

0 r 22

1 0 C r 23 A;

0

r 33

ð4:9Þ

Substituting it into (4.8), we get the set of equations

8 2 r 11 lb  r11 ðkq0 þ lb Þ þ kq0 ¼ 0; > > > > > > > ðr21 r11 þ r 22 r 21 þ r 23 r 31 Þlb  r 21 ðkq0 þ lb Þ þ r23 h ¼ 0; > > > > > r 2 l  r ðh þ kq þ l Þ þ kq ¼ 0; < 22

v1

22

1

v1

1

> ðr22 r23 þ r 23 r 33 Þlv 2 þ r22 h  r 23 ðh þ kq2 þ lv 2 Þ ¼ 0; > > > > > > > ðr31 r11 þ r 33 r 31 Þlb  r31 ðkq0 þ lb Þ þ r 33 h ¼ 0 > > > > : r 2 l  r ðh þ kq þ l Þ þ kq ¼ 0: 33 2 2 33 v 2 v2

ð4:10Þ

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W. Sun et al. / Applied Mathematics and Computation 248 (2014) 195–214

Solving (4.10), we get

8 kq0 > > > r 11 ¼ lb ; > > > > 2lv h2 r  > q1 > 1     ; > r 21 ¼  > >  r  2l r  2l r  þl  > l ðhþkq  l Þ r > 1 b v1 v 2 q1 v 1 q2 v 2 þh 2lv 1 r q2 lv 1 lv 2 q1 q2 > > > > > r > q > > r 22 ¼ 2l1 ; > v1 > <

ð4:11Þ

hr

q > r 23 ¼ l rþ 1l r ; > v 1 q2 v 2 q1 > > > >   > > > > h 2lv kðq1 r  þq2 r  Þ4l2v kq2 ðhþkq1 Þr r q q q q > 1 2 1 >  1   1 2  ; > r 31 ¼  > > > l ðhþkq1 lv Þ r  r 2lv r  2lv r  þlv þh 2lv r  lv lv > b q q q q q > 1 2 1 1 2 2 1 2 1 2 1 2 > > > > r > q2 > : r 33 ¼ 2l ;

v2

where

rq1 ¼ h þ kq1 þ lv 1 

qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ðh þ kq1 þ lv 1 Þ2  4kq1 lv 1 ;

rq2 ¼ h þ kq2 þ lv 2 

qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ðh þ kq2 þ lv 2 Þ2  4kq2 lv 2 ;

rþq2 ¼ h þ kq2 þ lv 2 þ

qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ðh þ kq2 þ lv 2 Þ2  4kq2 lv 2 :

Using the matrix analysis method (see [6,13]), we have

pk ¼ ðpk0 ; pk1 ; pk2 Þ ¼ ðp00 ; p01 ; p02 ÞRk ; k P 0:

ð4:12Þ

and ðp00 ; p01 ; p02 Þ satisfies

ðp00 ; p01 ; p02 ÞðRB10 þ A00 Þ ¼ 0;

ð4:13Þ

and the normalization condition is

ðp00 ; p01 ; p02 ÞðI  RÞ1 e ¼ 1;

ð4:14Þ

where I is identity matrix, e is unit vector, and

0

r 11 lb

0

1

r 21 lb þ r 22 lv 1  h  kq1

h þ r 23 lv 2

r 31 lb

r 33 lv 2  h  kq2

C A:

kq0

B RB10 þ A00 ¼ @ 0 h

ð4:15Þ

Substituting (4.15) into (4.13), we obtain

8 > < kq0 p00 þ hp02 ¼ 0; r 11 lb p00 þ ðr 21 lb þ r 22 lv 1  h  kq1 Þp01 þ r 31 lb p02 ¼ 0; > : ðh þ r l Þp þ ðr l  h  kq Þp ¼ 0: 23

v2

01

33

v2

2

ð4:16Þ

02

Solving (4.16), we get

  1 k q0 h þ k q2  r 33 lv 2 kq 0A @   : ðp00 ; p01 ; p02 Þ ¼ p00 1; ; h h h þ r23 lv 2 0

Furthermore,

0

r k11 B k1 k2 k2i B X X X j k1j Br r i11 r j22 r k2ji B 21 r 11 r 22 þ r 23 r 31 33 k R ¼B j¼0 i¼0 j¼0 B B k1 X B k1j @ r 31 r j11 r 33 j¼0

0 rk22 0

0

r23

1

C k1 C X k1j C r j22 r 33 C C; j¼0 C C C k A r 33

k P 1;

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W. Sun et al. / Applied Mathematics and Computation 248 (2014) 195–214

Substituting ðp00 ; p01 ; p02 Þ and Rk into (4.12), we can obtain the stationary queue length distribution as follows:  8  k  kq0 ðhþkq2 r33 lv Þp00  r  k  kq0 r 31 p00 p00 r 31 r 31 k k 2 21 > r11  r k22 þ ðr11 rr2223Þðr rk  ðr11 rr2223Þðr rk þ ðr11 rr3323Þðrr3122 r33 Þ rk33 ; k P 0; > hðhþr23 lv Þ r 11 r22 > pk0 ¼ hðhþr23 lv 2 Þ r11 þ hðr11 r33 Þ r11  r33 þ 11 r 33 Þ 11 22 r33 Þ 22 > 2 > > >   > > < k q hþk q r l p pk1 ¼ 0  2 33 v2 00 rk22 ; k P 0; h hþr 23 lv > 2 > > > >   > > > > p ¼ kq0 p00 ðhþkq2 r33 lv 2 Þr23 r k  r k  þ rk ; k P 0; : k2 33 33 h ðhþr23 l Þðr 22 r 33 Þ 22 v2

ð4:17Þ where p00 can be determined by (4.14). According to (4.17), the probability that the system is in state i ði ¼ 0; 1; 2Þ, denoted by pi , can be derived as follows:

p0 ¼ PfI ¼ 0g ¼

p1 ¼ PfI ¼ 1g ¼

p2 ¼ PfI ¼ 2g ¼

1 X

h  r 23 lv 2

k¼0

ð1  r 11 Þðh þ r23 lv 2 Þ

pk0 ¼ p00

ð1  r 11 Þð1  r 22 Þðh þ r 23 lv 2 Þh ! kq0 r 23 r 31 ðh  r 33 lv 2 þ kq2 Þ ; þ ð1  r 11 Þð1  r 22 Þð1  r33 Þðh þ r 23 lv 2 Þh

1 X

kq0 ðh þ kq2  r33 lv 2 Þp00

k¼0

hð1  r 22 Þðh þ r23 lv 2 Þ

pk1 ¼

1 X

pk2

k¼0

kq0 r 21 ðh þ r 33 lv 2  kq2 Þ



þ

kq0 r31 ð1  r 11 Þð1  r 33 Þh ð4:18Þ

ð4:19Þ

;

  0 1  r 23 h þ kq2  r33 lv 2  1 kq0 p00 @ 1 1 A:  ¼ þ  1  r 33 h h þ r 23 l ðr 22  r33 Þ 1  r 22 1  r 33

ð4:20Þ

v2

So we can get the conditional expected queue length in state i ði ¼ 0; 1; 2Þ, denoted by E½Li , as

P1 kpk0 E½L0  ¼ Pk¼1 ¼ 1 k¼0 pk0

þ





ðh þ kq2  r 33 lv 2 Þð1  r11 r22 Þr 21 ðh þ r 23 lv 2 Þð1  r11 Þ2 ð1  r 22 Þ2

þ

r 31 ð1  r 11 r 33 Þ ð1  r 11 Þ2 ð1  r 33 Þ2

r 31 r23 ðh þ kq2  r 33 lv 2 Þ r 21 ð2  ðr11 þ r 22 ÞÞ r 33 ð2  ðr11 þ r 33 ÞÞ  ðh þ r 23 lv 2 Þðr 22  r 33 Þ ð1  r 11 Þ2 ð1  r 22 Þ2 ð1  r 11 Þ2 ð1  r 33 Þ2 ðh þ kq2  r 33 lv 2 Þr21 ðh þ r 23 lv 2 Þð1  r 11 Þð1  r 22 Þ

þ

hr 11

þ

kq0 ð1  r 11 Þ2 !!

r31 h þ ð1  r 11 Þð1  r 33 Þ kq0 ð1  r 11 Þ

!1 1 þ r 33  ðr 11 þ r22 Þ ðh þ r23 lv 2 Þðr 11  r33 Þðr 22  r33 Þ ð1  r11 Þð1  r 22 Þ r31 r23 ðh þ kq2  r33 lv 2 Þ

P1 kpk1 p00 kq0 ðh þ kq2  r 33 lv 2 Þr 22 E½L1  ¼ Pk¼1 ¼ 1 hðh þ r 23 lv 2 Þð1  r 22 Þ2 k¼0 pk1

p00 kq0 ðh þ kq2  r33 lv 2 Þ hðh þ r23 lv 2 Þð1  r22 Þ

ð4:21Þ !1 ¼

r 22 ; 1  r22

ð4:22Þ

    0 1 P1 r 22 r 23 h þ k q2  r 33 lv 2 r23 r 33 h þ k q2  r 33 lv 2 kq0 p00 @ r33 k¼1 kpk2 A   E½L2  ¼ P1 ¼ þ  h ð1  r33 Þ2 k¼0 pk2 h þ r 23 lv ðr22  r 33 Þð1  r 22 Þ2 h þ r 23 lv ðr 22  r33 Þð1  r 33 Þ2 2

2

    0 0 111 r r 23 h þ k q2  r 33 lv 2 23 h þ k q2  r 33 lv 2 kq p 1 0 00 @ @ AA      þ  1  r 33 ðr22  r 33 Þ h þ r 23 l ð1  r22 Þ ðr22  r 33 Þ h þ r 23 l ð1  r33 Þ h v2 v2   r 23 h þ kq2  r33 lv 2 r 33 1  : ¼ þ 1  r 33 1  r22 ð1  r22 Þðh þ r 23 l Þ þ r23 h þ kq  r 33 l 2 v2 v2

ð4:23Þ

Replacing n with E½L0  in (3.1), we can easily get the conditional expected sojourn time of the customers in busy state, denoted by E½W 0 , as

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W. Sun et al. / Applied Mathematics and Computation 248 (2014) 195–214

E½W 0  ¼

E½L0  þ 1

lb

¼

ðh þ kq2  r 33 lv 2 Þð2  ðr11 þ r 22 ÞÞr21 ðh þ r 23 lv 2 Þð1  r 11 Þ2 ð1  r 22 Þ2

þ

r 31 ð2  ðr 11 þ r 33 ÞÞ ð1  r 11 Þ2 ð1  r 33 Þ2

þ

h kq0 ð1  r 11 Þ2

! r 31 r 23 ðh þ kq2  r 33 lv 2 Þ r 11  r 33 r21 ð2  ðr 11 þ r 22 ÞÞ r 33 ð2  ðr11 þ r 33 ÞÞ :  ðh þ r 23 lv 2 Þðr11  r 33 Þðr22  r 33 Þ ð1  r 11 Þ2 ð1  r 22 Þ2 ð1  r 33 Þ2 !! lb ðh þ kq2  r33 lv 2 Þr21 1 þ r33  ðr 11 þ r22 Þ lb r31 lb h  þ þ ð1  r11 Þð1  r 22 Þ ðh þ r23 lv 2 Þð1  r 11 Þð1  r 22 Þ ð1  r11 Þð1  r 33 Þ kq0 ð1  r 11 Þ

þ

lb r31 r23 ðh þ kq2  r33 lv 2 Þ 1 þ r33  ðr11 þ r22 Þ  ðh þ r 23 lv 2 Þðr11  r 33 Þðr 22  r 33 Þ ð1  r 11 Þð1  r 22 Þ

!1 ð4:24Þ

:

As for the conditional expected sojourn time in the first-stage working vacation state, denoted by E½W 1 , we can similarly discuss the same three cases at state ðk; 1Þ ðk P 0Þ with those in the observable queues. So we first get the LS of the customers’ sojourn time, denoted by W 1 ðsÞ, as follows:

W 1 ðsÞ

¼

1 X

0

pk1 @

!kþ1

lv 1

þ

@

lv 1

!kþ1

!kþ1 1 A  lv 2 þ h þ s !kþ1 1

lv 2

ðh þ sÞðlv 1  lv 2 Þ lv 1 þ h þ s 0   lv 1 lb h 2 lb lb kþ1 @ @  þ lb h þ ðlb  lv 2 Þs lb h þ ðlb  lv 1 Þs lb þ s lv 1 þ h þ s 0 !kþ1 !kþ1 111 k¼0

lv 1 þ h þ s

0

hlv 2

0

l

l

A

l

v2 v2 v1 @ AAA  ðh þ sÞðlv 1  lv 2 Þ lv 2 þ h þ s lv 1 þ h þ s   0 1 k q0 h þ k q2  r 33 lv 2 p00 hlv 2 h2 lb @  þ ¼  þ lv 1 A:   lv 2 ð1  r22 Þ þ h þ s h h þ r 23 lv 2 ðlv 1 ð1  r 22 Þ þ h þ sÞ lb ð1  r 22 Þ þ s lv 2 ð1  r 22 Þ þ h þ s

þ

ð4:25Þ Therefore, the LST of the conditional sojourn time in the first-stage working vacation state is

f  ðsÞ ¼ 1 W  ðsÞ W 1 p1 1

0 1 2 hlv 2 1  r 22 h l b @  þ þ lv 1 A: ¼ lv 1 ð1  r22 Þ þ h þ s lb ð1  r22 Þ þ s lv ð1  r22 Þ þ h þ s lv 2 ð1  r22 Þ þ h þ s 2

ð4:26Þ

f 0 ð0Þ, that is, Then we get E½W 1  ¼  W 1

E½W 1  ¼

ðlb ð1  r 22 Þ þ hÞ2  lb ð1  r 22 Þ2 ðlb  lv 2 Þ : ðh þ lv 1 ð1  r22 ÞÞðh þ lv 2 ð1  r 22 ÞÞð1  r 22 Þlb

ð4:27Þ

On the other hand, for the conditional expected sojourn time in the second-stage working vacation state, denoted by E½W 2 , we can also discuss the same two cases at sate ðk; 2Þ ðk P 0Þ with those in the observable queues. So the LS of the customers’ sojourn time, denoted by W 2 ðsÞ, is

W 2 ðsÞ

¼

1 X

pk2 @

k¼0

¼

0

!kþ1

lv 2

lv 2 þ h þ s

0 

lb h lb @ þ lb h þ ðlb  lv 2 Þs lb þ s

kþ1

!kþ1 11 AA  lv 2 þ h þ s

lv 2

lb ðlv 2 ð1  r33 Þ þ hÞ þ lv 2 s l2v 2 ðh þ kq2  r33 lv 2 Þr23 kq0 p00  ðlv 2 ð1  r 33 Þ þ h þ sÞðlb ð1  r33 Þ þ sÞ ðh þ r 23 lv 2 Þðlv 2 s  lb ðh þ sÞÞ h  

ðlb  lv 2 Þs

l2b ðh þ kq2  r33 lv 2 Þr23 ðð1  r 22 Þlv 2 þ h þ sÞðð1  r33 Þlv 2 þ h þ sÞ ðh þ r 23 lv 2 Þðlv 2 s  lb ðh þ sÞÞ lb h ðð1  r 22 Þlb þ sÞðð1  r 33 Þlb þ sÞ



 :

So the LST of the conditional sojourn time in the second-stage working vacation state is

ð4:28Þ

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W. Sun et al. / Applied Mathematics and Computation 248 (2014) 195–214

lb ðlv 2 ð1  r33 Þ þ hÞ þ lv 2 s l2v 2 ðh þ kq2  r33 lv 2 Þr23  ðlv 2 ð1  r 33 Þ þ h þ sÞðlb ð1  r33 Þ þ sÞ ðh þ r23 lv 2 Þðlv 2 s  lb ðh þ sÞÞ ðlb  lv 2 Þs l2b ðh þ kq2  r33 lv 2 Þr23   ðð1  r 22 Þlv 2 þ h þ sÞðð1  r33 Þlv 2 þ h þ sÞ ðh þ r 23 lv 2 Þðlv 2 s  lb ðh þ sÞÞ

f  ðsÞ ¼ 1 W  ðsÞ ¼ W 2 p2 2

  h þ r 23 lv 2 ð1  r22 Þð1  r 33 Þ    :  ðð1  r 22 Þlb þ sÞðð1  r 33 Þlb þ sÞ h þ r 23 l ð1  r 22 Þ þ r23 h þ kq  r 33 l 2 v2 v2 

lb h

ð4:29Þ

f  ð0Þ, that is, Then we get E½W 2  ¼  W 2 0

E½W 2  ¼

lv 2 ðh þ r23 lv 2 Þð1  r22 Þð1  r33 Þ hðh þ r23 lv 2 Þð1  r22 Þðð1  r33 Þðlb þ lv 2 Þ þ hÞ þ ðlv 2 ð1  r33 Þ þ hÞ2 ðlv 2 ð1  r 33 Þ þ hÞ2 ðlb ð1  r 33 ÞÞ lv ðh þ kq2  r33 lv 2 Þr23 ðlb  lv 2 Þð1  r22 Þð1  r33 Þ ðh þ kq2  r33 lv 2 Þr23 ðlb  lv 2 Þ  2 þ lb hðlv 2 ð1  r22 Þ þ hÞðlv 2 ð1  r33 Þ þ hÞ lb h  1 ðh þ kq2  r33 lv 2 Þr 23 ð2  r 22  r 33 Þ ð1  r 22 Þðh þ r23 lv 2 Þ þ r 23 ðh þ kq2  r 33 lv 2 Þ þ : lb ð1  r22 Þð1  r33 Þ

ð4:30Þ

Hence, from (4.27), the expected net benefit of a customer who joins in the first-stage working vacation state, denoted by Uð1; q1 Þ, is Uð1; q1 Þ ¼ R  CE½W 1 , which does not depend on q0 and q2 . Then from (4.30), his expected net benefit in the second-stage working vacation state, denoted by Uð2; q1 ; q2 Þ, is Uð2; q1 ; q2 Þ ¼ R  CE½W 2 , which only depends on q1 and q2 but does not depend on q0 . On the other hand, from (4.24), his expected net benefit in busy state, denoted by Uð0; q0 ; q1 ; q2 Þ, is Uð0; q0 ; q1 ; q2 Þ ¼ R  CE½W 0 , which depends on all of q0 , q1 and q2 . Solving the equations Uð1; q1 Þ ¼ 0, Uð2; q1 ; q2 Þ ¼ 0 and Uð0; q0 ; q1 ; q2 Þ ¼ 0, we can get the root ðq ð0Þ; q ð1Þ; q ð2ÞÞ. As for the issue of uniqueness of ðq ð0Þ; q ð1Þ; q ð2ÞÞ, it is nearly impossible to take theoretical analysis because of complexity of the expressions. Therefore, we make some numerical experiments and observe that there indeed exists a unique feasible and positive root for a fixed k. This indicates that large customer arrivals just mainly exists negative externality. So the customers’ equilibrium strategy is ðqe ð0Þ; qe ð1Þ; qe ð2ÞÞ ¼ ðminfq ð0Þ; 1g; minfq ð1Þ; 1g; minfq ð2Þ; 1gÞ in partially observable case. Although the server’s service rates have the relations of lv 1 < lb and lv 2 < lb , Fig. 8 shows that qe ð2Þ is the lowest joining probability for a fixed k, and qe ð0Þ is not necessarily higher than qe ð1Þ. The reason is that lv 1 is not so slow compared with lb so that it can satisfy customers’ service requirement. Moreover, shorter queue length also attracts customers to join although the system is in vacation state. However, during the second-stage vacation, lv 2 is too slow and the queue length maybe contains some residual customers who do not complete service during the first-stage vacation. Of course, the cases of qe ð2Þ > qe ð0Þ or qe ð1Þ < qe ð2Þ also possibly occur, which depends on the values of lb ; lv 1 and lv 2 . Besides Fig. 8, Figs. 9–11 show the sensitivity of equilibrium mixed strategies with respect to h; lv 1 and lv 2 , respectively. Fig. 9 shows that both qe ð1Þ and qe ð2Þ increase with h but the increasing rate of qe ð1Þ is much faster than qe ð2Þð< 0:5Þ, which is resulted by lv 1 ¼ 2lv 2 and the cumulated customers in the first-stage vacation. Fig. 10 shows that qe ð1Þ increases with lv 1 whereas qe ð2Þ decreases with lv 1 . Moreover, qe ð1Þ may be greater than qe ð2Þ although lv 1 < lv 2 because of the shorter expected queue length in the first-stage vacation. Similar to Fig. 9, Fig. 11 also shows that both qe ð1Þ and qe ð2Þ increase with lv 2 . The reason is that higher value of lv 2 benefits customers arriving in the two-stage vacations. As for qe ð0Þ, Figs. 9–11 all

1

q (0) e q (1) e q (2)

0.9

e

0.8 0.7

qe

0.6 0.5 0.4 0.3 0.2 0.1

0

1

2

3

4

5

6

7

8

λ Fig. 8. Equilibrium mixed strategies for the partially observable case when R ¼ 10; C ¼ 1; lb ¼ 3; lv 1 ¼ 2,

lv 2 ¼ 1; h ¼ 0:1.

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W. Sun et al. / Applied Mathematics and Computation 248 (2014) 195–214 1

q (0) e q (1) e q (2)

0.9

e

0.8 0.7

q

e

0.6 0.5 0.4 0.3 0.2 0.1 0

0.1

0.2

0.3

0.4

0.5

0.6

θ

0.7

0.8

0.9

1

Fig. 9. Sensitivity of joining probabilities for the unobservable case with respect to h when R ¼ 5; C ¼ 1,

lb ¼ 3; k ¼ 5; lv 1 ¼ 1, lv 2 ¼ 0:5.

1 0.9 0.8 qe(0) qe(1) qe(2)

0.7

qe

0.6 0.5 0.4 0.3 0.2 0.1

0

0.5

1

1.5

μv

2

2.5

3

1

Fig. 10. Sensitivity of joining probabilities for the partially observable case with respect to

lv 1 when R ¼ 10, C ¼ 1; lb ¼ 3; k ¼ 3; lv 2 ¼ 1, h ¼ 0:1.

show that qe ð0Þ nearly has little change compared with qe ð1Þ and qe ð2Þ, and qe ð0Þ is not necessarily higher than qe ð1Þ or qe ð2Þ in case the vacation environment is better enough. Then, we consider the equilibrium social welfare. Based on (4.17), the unconditional expected queue length, denoted by E½Lpo  is

E½Lpo  ¼

1 X kðpk0 þ pk1 þ pk2 Þ k¼0

ðh þ kq2  r33 lv 2 Þ kq p00 ðr23 r31 r22 ð2  r11  r 22 Þð2  r 11  r22  r 33 Þ þ r21 ð1  r11 r22 ÞÞ ¼ 0 h ðh þ r 23 lv 2 Þð1  r 11 Þ2 ð1  r 22 Þ2 þ

hr11 kq0 ð1  r 11 Þ2

þ

r 31 ð1  r 11 r 33 Þ ð1  r 11 Þ2 ð1  r 33 Þ2

þ

r 22 ðh þ kq2  r 33 lv 2 Þ ðh þ r 23 lv 2 Þð1  r22 Þ2

þ

r33 ð1  r33 Þ2

þ

r 23 ðh þ kq2  r33 lv 2 Þð1  r 22 r 33 Þ ðh þ r23 lv 2 Þð1  r 22 Þ2 ð1  r33 Þ2

! :

ð4:31Þ Hence, the social welfare per time unit for ðq0 ; q1 ; q2 Þ, denoted by U s ðq0 ; q1 ; q2 Þ, is U s ðq0 ; q1 ; q2 Þ ¼ kðp0 q0 þ p1 q1 þp2 q2 ÞR  CE½Lpo , where p0 ; p1 ; p2 are given by (4.18), (4.19), (4.20). When all the customers follow the equilibrium mixed strategy ðqe ð0Þ; qe ð1Þ; qe ð2ÞÞ, the equilibrium social welfare per time unit can be expressed as U s ðqe ð0Þ; qe ð1Þ; qe ð2ÞÞ. Fig. 12 shows the tendency of U s ðqe ð0Þ; qe ð1Þ; qe ð2ÞÞ with k for the partially observable case.4 Similar to U s ðne ð0Þ; ne ð1Þ; ne ð2ÞÞ; U s ðqe ð0Þ; qe ð1Þ; qe ð2ÞÞ also first increases then decreases. 4

The fitting curve in Fig. 9 is not so smooth because of the calculating error.

211

W. Sun et al. / Applied Mathematics and Computation 248 (2014) 195–214 1 0.9 0.8 0.7

qe

0.6 0.5 0.4 0.3 0.2

qe(0) qe(1) qe(2)

0.1 0

0

0.5

1

1.5

μv

2

2.5

3

2

Fig. 11. Sensitivity of joining probabilities for the partially observable case with respect to

lv 2 when R ¼ 10, C ¼ 1; lb ¼ 3; k ¼ 3; lv 1 ¼ 1, h ¼ 0:1.

12

Us(qe(0),qe(1),qe(2))

10

8

6

4

2

0

0

0.5

1

1.5

2

2.5

3

3.5

λ Fig. 12. Equilibrium social welfare for the partially observable case when R ¼ 10; C ¼ 1; lb ¼ 3; lv 1 ¼ 2; lv 2 ¼ 1, h ¼ 0:1.

q

0,1 v1

q

0,2 b

2,1

...

n,1

v1

v1

q

q

1,2 v2

2,2

...

n,2

v2

q

0,0

q

q

1,1

n+1, 2

...

n+1, 0

...

q

2,0 b

...

v2

q

1,0

n+1, 1

...

n,0 b

Fig. 13. Transition rate diagram for unobservable queues.

W. Sun et al. / Applied Mathematics and Computation 248 (2014) 195–214 1

0.9

q

e

0.8

0.7

0.6

0.5

0.4

0

1

2

3

4

5

6

λ Fig. 14. Equilibrium mixed strategy for the unobservable case when R ¼ 10; C ¼ 1; lb ¼ 3; lv 1 ¼ 2; lv 2 ¼ 1, h ¼ 0:1.

1 0.9 0.8

qe

0.7 0.6 0.5 0.4 0.3 0.2

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

θ Fig. 15. Sensitivity of joining probability for the unobservable case with respect to h when R ¼ 5; C ¼ 1,

lb ¼ 3; k ¼ 3; lv 1 ¼ 1, lv 2 ¼ 0:5.

1 0.9 0.8 0.7

qe

212

0.6 0.5 0.4 0.3

0

0.5

1

1.5

2

2.5

3

μv

1

Fig. 16. Sensitivity of joining probability for the unobservable case with respect to

lv 1 when R ¼ 10; C ¼ 1, lb ¼ 3; k ¼ 3; lv 2 ¼ 1; h ¼ 0:1.

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W. Sun et al. / Applied Mathematics and Computation 248 (2014) 195–214 0.95 0.9 0.85 0.8

qe

0.75 0.7 0.65 0.6 0.55 0.5 0.45

0

0.5

1

1.5

2

2.5

3

μ

v

2

Fig. 17. Sensitivity of joining probability for the unobservable case with respect to

lv 2 when R ¼ 10; C ¼ 1, lb ¼ 3; k ¼ 3; lv 1 ¼ 1; h ¼ 0:1.

10 9 8 7

Us(qe)

6 5 4 3 2 1 0

0

0.4

0.8

1.2

1.6

2.0

2.4 2.6

λ Fig. 18. Equilibrium social welfare for the unobservable case when R ¼ 10; C ¼ 1; lb ¼ 3; lv 1 ¼ 2; lv 2 ¼ 1; h ¼ 0:1.

5. The unobservable queues Finally, we discuss the customers’ equilibrium behavior in the unobservable case, i.e., an arriving customer can observe neither the server’s current state IðtÞ nor the system occupancy NðtÞ at time t. So the customers’ decision is equivalent to select a joining probability q ð0 6 q  1Þ, and their equilibrium mixed strategy is denoted by qe . The transition diagram is depicted in Fig. 13. Replacing all qi ði ¼ 0; 1; 2Þ with q in (4.31), including those in p00 and fr ij 2 Rji; j ¼ 1; 2; 3g, we can directly get the expression of the expected queue length, denoted by E½Luo , in unobservable queues. Therefore, the expected sojourn time of a joining customer, denoted by E½W, is

E½W ¼

ðh þ kq  r 33 lv 2 Þ E½Luo  p00 ¼ ðr23 r 31 r 22 ð2  r 11  r 22 Þð2  r 11  r 22  r 33 Þ þ r 21 ð1  r11 r 22 ÞÞ kq h ðh þ r 23 lv Þð1  r 11 Þ2 ð1  r 22 Þ2 2 þ

hr 11 kqð1  r 11 Þ2

þ

r 31 ð1  r 11 r 33 Þ ð1  r 11 Þ2 ð1  r 33 Þ2

þ

r 22 ðh þ kq  r 33 lv 2 Þ ðh þ r 23 lv 2 Þð1  r22 Þ2

þ

r33 ð1  r33 Þ2

þ

r 23 ðh þ kq  r 33 lv 2 Þð1  r 22 r 33 Þ ðh þ r 23 lv 2 Þð1  r 22 Þ2 ð1  r 33 Þ2

! : ð5:1Þ

Hence, the customer’s expected net benefit, denoted by UðqÞ, is UðqÞ ¼ R  CE½W. Solving UðqÞ ¼ 0, we get the root q , and the customers’ equilibrium mixed strategy is qe ¼ minfq ; 1g. Similar to the partially observable case, we also numerically observe that UðqÞ ¼ 0 has a unique positive root for a fixed k on condition that q ¼ kq=lb < 1. Fig. 14 shows the decreasing

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trend of qe with k for the unobservable case. Then Figs. 15–17 show that qe increases with respect to h; lv 1 and lv 2 , respectively, which are consistent with intuition. Then according to (5.1), we can get the social welfare per time unit for q, denoted by U s ðqÞ, as U s ðqÞ ¼ kqðR  CE½WÞ. When all customers follow the above equilibrium mixed strategy qe , the equilibrium social welfare per time unit can be denoted by U s ðqe Þ. Similar to U s ðne ð0Þ; ne ð1Þ; ne ð2ÞÞ and U s ðqe ð0Þ; qe ð1Þ; qe ð2ÞÞ, Fig. 18 also shows the first-increase-thendecrease trend of U s ðqe Þ with k for the unobservable case. 6. Conclusions Weighing the cost of state conversion and the actual requirement of assistant work, usually in practice, the server has to take a fix number of vacations, and the lower service rates in successive vacation times can be adjusted but not always invariable. Therefore, we presented a new two-stage working vacation policy in this paper, which is an original case. Based on different precision levels of system information, we studied observable, partially observable and unobservable Markovian queues with two-stage working vacations, respectively, and derived customers’ equilibrium balking strategies and equilibrium social welfare. No matter which type of system information, we observed that customers’ positive equilibrium strategy is unique. By comparisons, customers’ equilibrium joining probabilities in vacation states are not necessarily smaller than that in busy state for the partially observable queues. Moreover, in observable/partially observable queues, the order of balking thresholds/joining probabilities in the two-stage working vacations depends on the relation of service rates. Acknowledgements The authors would like to thank the anonymous reviewers for the useful comments on this work, and the support from the National Natural Science Foundation of China (Nos. 71101124, 71301139, 71271187), the China Postdoctoral Science Foundation (No. 2014T70231), the Research Fund for the Doctoral Program of Higher Education (No. 20131333120001), the Humanity and Social Science Foundation of Ministry of Education of China(No. 12YJC790101), and the Natural Science Foundation of Hebei Province (Nos. G2012203068, F2013203136). References [1] A. Burnetas, A. Economou, Equilibrium customer strategies in a single server Markovian queue with setup times, Queueing Syst. 56 (3–4) (2007) 213– 228. [2] A. Economou, S. Kanta, Equilibrium balking strategies in the observable single-server queue with breakdowns and repairs, Oper. Res. Lett. 36 (6) (2008) 696–699. [3] A. Economou, A. Gómez-Corral, S. 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