Customers' abandonment strategy in an M/G/1 queue
aa r X i v : . [ m a t h . P R ] A ug Customers’ abandonment strategy in an
M /G/
Eliran Sherzer and Yoav Kerner
Abstract
We consider an
M/G/
Understanding customers’ abandonments from a queue is of interest forservice providers and customers. There are many applications concerningcustomers’ abandonments since real-world customers are unwilling to waitfor excessive lengths of time. Applications are presented more broadly byMandelbaum and Shimkin [14]. Traditional queueing theory has dealt withthe analysis of queues under the assumption of a given patience distribu-tion, and many studies have addressed models which include abandonments,starting with Barrer [5], who studied the queue length distribution in the
M/M/s + D w case (where D indicates deterministic patience). Sufficientconditions for the existence of the steady-state virtual waiting time distri-bution in the G/G/ G w were later obtained by Baccelli [2, 3]. Boxmaet al. [7] showed how to determine the busy-period distribution for variouschoices of the patience time distribution. Brandt and Brandt [8] also studied1he distribution of the busy period. In particular, they gave an explicit rep-resentation of the Laplace-Stieltjes transform of the workload and the busyperiod, in the case of phase-type distributed impatience. Yet, these studiesassumed that the patience distribution was given. Here, we investigate howthese distributions are constructed by rational behavior of customers, whichare affected by factors such as individual costs and preferences. Essentially,the patience of each customer is based on an individual optimization, thatis, the perceived balance between the costs of waiting and the benefits ofservice. Since the behaviour of others has an influence on the individual(abandonments of others shorten the individual’s required waiting time), themutual interactions lead us to look at the system in the standard form of agame, and to seek the Nash equilibrium. Therefore, the patience distributionis no longer given, but instead it results from a cost/reward model, and fromthe strategic behaviour implied by it.The seminal study that viewed queues as economic systems, and studiedthe strategic behaviors within them, came from Naor [16]. Naor consid-ered an observable M/M/ R from service, and a constant waiting time rate C . He showed that underself-optimization, customers will join the queue if the number of customerspresent upon arrival is less than a threshold, n n = ⌊ RµC ⌋ . Moreover, once acustomer joins the queue, he stays until served. Hassin and Haviv [12] alsostudied an M/M/
R > T time units, otherwise it equals 0. Theyshowed that the pure equilibrium strategy is as follows: join the queue withsome probability p and otherwise balk, and if one joined then he waits T time units. The reason for this result is that the virtual waiting time followsan increasing failure rate (IFR) pattern. That is, the remaining waiting timestochastically decreases along with the time passed. This is true for every M/M/
M/M/m queue and heterogeneous customers whose waiting costs and re-wards may vary between the different customer types. Their results indicatethat, depending on the reward to cost ratio, a customer’s best response isto either abandon the queue upon arrival unless of course one of the servers2s available, in which case he enters service immediately or, given the IFRproperty, to never abandon and wait until receiving service. Moreover, theyproposed a case in which the solution is richer, in which customers are dis-charged without knowing it, (see, e.g. [17]). In their proposed model, eachcustomer will never be served with probability 1 − q . Thus, the longer acustomer has already waited, the higher the posterior probability that hehas been discharged. It turns out that the system has an eventually decreas-ing (and in fact unimodal) hazard rate function, which makes finite aban-donments rational. Thus, Mandelbaum and Shimkin showed that the bestresponse is to either abandon at arrival, unless, of course, one of the serversis available or to abandon after a finite time T , which is determined by theratio of the waiting cost and the reward. A follow up study by Mandelbaumand Shimkin [15] considered a nonlinear waiting cost in an unobservable M/M/m queue. They provided conditions for the existence and uniquenessof the equilibrium, and suggested procedures for its computation.Another important study, by Haviv and Ritov [13], explored an
M/M/ T time units in the queue ( T = 0 or T = ∞ are possible).Also, they showed that having a mixed Nash equilibrium may occur whenthe ratio of the cost and reward functions satisfies certain conditions. In thismodel, the waiting time has an IFR, and thus the remaining waiting timereduces with the elapsed waiting time. However, the waiting cost is convex,and therefore waiting becomes more expensive. The latter balances the im-provement in the waiting time and the remaining waiting costs.Afeche and Sarhangian [1] also study customers abandonments in an ob-servable M/M/ . In this study, we relax this assumption. Inparticular, we consider an observable M/G/ Except [6] and [9]. However, they focus on discrete-time model.
We consider a single server queueing model in which the arrivals are ac-cording to a Poisson process with rate λ and service times are iid generallydistributed. We denote the service time by X , and E [ X ] by ¯ x . We also usethe notation f ( · ) for the pdf and F ( · ) for the cdf, with ¯ F ( · ) = 1 − F ( · ), andthe hazard function is denoted by h ( · ). The service discipline is FCFS. Eachcustomer can see his position in the queue at any moment, and he is ableto observe service completions and abandonments made by others. How-ever, customers are not aware of events that occurred prior to their arrival.Of course, they do not anticipate future service times. Customers can keeptrack of time, and are allowed to abandon at any moment. All customersare homonomous in their reward from service, which is denoted by V , and alinear waiting cost, which is denoted by C . In order to complete the modeldescription we first present the following definition. For a non-negative ran-dom variable X with cdf F ( · ), the MRL function is defined as follows: m X ( x ) = E [ X − x | X > x ] = Z ∞ F ( x + t ) F ( x ) dt, x ≥ m X ( ∞ ) = lim x →∞ m X ( x ). We distinguish between service distributionsin which the MRL is bounded, and those in which it is unbounded. Weassume that the service time has an increasing mean residual life (IMRL) [4].5 .2 Utility function For each individual the expected utility function balances the reward fromservice with the expected waiting cost. Customers always take into accounttheir future costs, while the time they already waited is considered a sunkcost. Each customer wishes to maximize his own expected utility; as a result,his best response is to stay as long as the expected utility is positive. Wefirst consider a case where abandonments are not allowed. This will assist uncontract the utility function with abandonments. Let G n ( t ) be the expectedutility function value from staying until being served, for an individual thathas n customers in front of him in the system when the current service ageis t . Clearly, G n ( t ) = ( V n = 0 V − C ( E [ X − t | X > t ] + ( n − x ) n ≥ G n ( t ) for n ≥ Proposition 1.
For any value of n ≥ , G n ( t ) solves the sequence of differ-ential equations for n ≥ , G ′ n ( t ) = C − h ( t )( G n − (0) − G n ( t )) . (1) Proof.
First, one can see that G n − (0) − G n ( t ) = C E [ X − t | X > t ] and hencethe RHS of Equation (1) equals C − Ch ( t ) E [ X − t | X > t ]. We also note that E [ X − t | X > t ] = R ∞ s = t ¯ F ( s ) dsF ( t ) . Hence, the derivative of G n ( t ) is dG n ( t ) dt = − C − F ( t ) F ( t ) + f ( t ) R ∞ s = t ¯ F ( s ) dsF ( t ) = C − Ch ( t ) E [ X − t | X > t ]which matched the derivative presented in the proposition, and hence theproof is concluded.
Remark 2.1.
The intuition behind the proposition is as follows: Since thereis a linear waiting cost, then clearly, C should be in the derivative. Moreover,if service completion occurs, then the expected utility changes from G n ( t ) to G n − (0) , which happens with rate h ( t ) . G n ( t ). The minor differencebetween the two functions shows how the possibility of abandoning laterreflects in the expected utility function. We distinguish between customerswho already observed a service completion (called type i customers) andthose who haven’t observed a service completion (type ii customers) . Dueto the fact that keeping track of time is possible, type i customers know theexact service age. However, since the current service age is unknown uponarrival, type ii customers can only estimate it. Let U n ( t ) be the expectedutility of a type i customer taking the optimal action in the next moment,given that there are n others in front of him in the system and t time unitselapsed since the last service completion. Proposition 2. U n ( t ) = ( G n ( t )) + , n ≥ Proof. n = 0 implies that an individual is already in service, and hence hasno cost. Clearly, the optimal action in the next moment is to stay. For n ≥
1, we first prove the proposition for n = 1 and then prove for n ≥ n = 1, we have U ( t ) = ( h ( t ) dtU (0) + (1 − h ( t ) dt ) U ( t + dt ) − Cdt + o ( dt )) + , t ≥ . (2)If the RHS of (2) is negative, then the best response is to abandon, andhence the expected utility is zero. If the RHS is positive, then we have thefollowing differential equation: U ′ ( t ) = C − h ( t )( U (0) − U ( t ))where clearly, U (0) = V . This differential equation coincides with Proposi-tion 1, and hence U ( t ) = ( V − C ( E [ X − t | X > t ] + ¯ x )) + . We first assumethat U n − ( t ) = ( G n − ( t )) + . Based on our assumption, clearly, U n − (0) =( V − C ( n − x ) + . Combining with U n ( t ) = h ( t ) dtU n − (0) + (1 − h ( t ) dt ) U n ( t + dt ) − Cdt + o ( dt ) Upon arrival, everyone are type ii customers, which potentially may switch to type i due to being present in the queue while service completion occurred. U ′ n ( t ) = C − h ( t )( U n − (0) − U n ( t ))and solving the differential equation complete the proof. Remark 2.2.
One can see that the only way the possibility of abandon-ing later reflects in the expected utility function is by abandoning once theexpected utility becomes negative. Consequently, only myopic decisions areunder consideration.
Let ˆ U n ( t ) be the expected utility of a type ii customer taking the optimalaction in the next moment, given that n other in front of him in the systemand t time units elapsed since his arrival. Formally, let N be the numberof customers in the system and A ( t ) be the service age, both upon arrival.Finally, let X n ( t ) follow the distribution of the service residual, given thatupon the arrival of a tagged customer there were n customers in the systemand t time units elapsed since then. That is, X n ( t ) d = { X − A ( t ) | N = n } . Proposition 3. ˆ U n ( t ) = ( V − C ( E [ X n ( t ) − t | X n ( t ) > t ] + ( n − x )) + n ≥ Proof.
For n = 0 there is no cost, similar to U n ( t ). For n ≥
1, we haveˆ U n ( t ) = U n − (0) h n ( t ) dt + (1 − h n ( t ) dt ) ˆ U n ( t + dt ) − Cdt + o ( dt )where h n ( t ) is the corresponding hazard function of X n ( t ). The differentialequation is ˆ U ′ n ( t ) = C − h n ( t )( ˆ U n − (0) − ˆ U n ( t ))For both U n ( t ) and ˆ U n ( t ) there are differential equations with the same struc-ture. Hence, their solutions also have the same structure, where in this case, X is replaced by X n ( t ). Remark 2.3.
The fact that service time has an IMRL implies that both U n ( t ) and ˆ U n ( t ) are decreasing with t . Remark 2.4. the two expected utility are similar and they differ only by thecustomers type. This is due to the fact that in both cases the expected waitingtime possesses the IMRL property. .3 Strategy profile As mentioned, customers’ best response differs depending on the number ofcustomers in front of them and which of the two different customer typesthey belong to. Therefore, each pair of queue length and customer typeneeds to be considered separately. Also, customers may balk, and clearlywhen the queue is long enough customers will not join. This happens whenthe expected utility is negative from the moment one arrives. Hence, underthe assumption of rationality of customers, there is a maximum number ofcustomers in the system, and it is denoted by n max . We show how to obtainit in section 3.2.The customers’ strategy profile is as follows. All customers join if thesystem length is less than n max . Of course, joining customers will adapt theirstrategy according to the expected utility function, which is defined by thecustomer’s type and the number of customers in front of them in the queue.Both U n ( t ) and ˆ U n ( t ) are monotonically decreasing with t , and therefore thebest response is unique and hence pure. Let T n be the time a type i customerwho has n customers in front of him in the system is willing to wait fromthe moment of service completion until the next service completion occurs,or abandonment is made by the customer in front of him. Let S n be thetime that a type ii customer who has n customers in front of him in thesystem is willing to wait from his arrival point until a service completionoccurs, or abandonment is made by the customer in front of him. For anytype of customer if, while waiting, service completion occurs, he updates hisexpected utility function and consequently his best response. However if,while waiting, the customer in front of him abandons, his best response is toabandon as well. This is because the customer in front, who abandoned first,gathered more information and his abandonment puts the customer behindin the exact same position in the queue, to which the best response was toabandon.In conclusion, we have two sequences that define customers’ strategies.The first one is { T , T , ..., T n max − . } and the second is { S , S , ..., S n max − } .The largest index value of the second sequence is A n max − , because it is thelargest value observed upon arrival for which one would be willing to join.That is, if a customer joined and observed n max customers in the system, hewould no longer join. In the first sequence, the largest index is obtained whena customer in the n max th position in the queue observes service completion.In this scenario, he has n max − T n and S n . Using Propositions 2 and 3 we show how to obtain customers’ best responses,for both type i and type ii customers, given N = n . Specifically, we showhow to obtain the sequences { T , T , ..., T n max − . } and { S , S , ..., S n max − } .We begin with type i customers. Before presenting and proving the followinglemmas, we note that if the MRL is bounded it is possible that finite thresh-olds are not possible. That is, even after waiting a long time, one still benefitfrom staying. Thus, there is condition on lim t →∞ m X ( t ) in order to compute thethreshold values. Lemma 1. If lim t →∞ m X ( t ) > VC − ( n − x , and based on the expected utilityfunction from Proposition 2, T n is the value of t that solves G n ( t ) = V − C ( m X ( t ) + ( n − x ) = 0 1 ≤ n ≤ n max − Otherwise, T n = ∞ .Proof. Since n < n max −
2, the expected utility for t = 0 is positive andof course, customers will be willing to wait as long their expected utility ispositive. Recall that the service time follows IMRL and hence, the expectedutility is monotone decreasing. Moreover, the condition lim t →∞ m X ( t ) < VC − ( n − x , means that for some value of t the expected utility will be negative.Therefore, by combining the last two arguments, Equation (3) has a uniqueand finite solution. Otherwise, the expected utility will be positive for every t > T n forany possible n can be easily obtained. Lemma 2. If lim t →∞ m X n ( t ) ( t ) > VC − ( n − x , and based on the expected utilityfunction from Proposition 3, S n is the value of t that solves V − C ( m X n ( t ) + ( n − x ) = 0 , ≤ n ≤ n max − Otherwise, S n = ∞ . roof. We follow the same line of argument as Lemma 1.Solving (4) is not straightforward, mainly because obtaining the distributionof X n ( t ) is challenging. As mentioned, X n ( t ) d = { X − A ( t ) | N = n } . Thatis, in order to obtain the distribution of X n ( t ), one must first obtain thedistribution of A ( t ) | N = n . Let R ( t, a ) follow the distribution of the residualservice time, given that the service age upon arrival is a and t time units haveelapsed since the customer’s arrival. That is, R ( t, a ) d = X − ( t + a ) | X > ( a + t ).Therefore, by using the law of total probability, the following equation isequivalent to (4): V − C Z a ( E [ R ( t, a ) | N = n ] + ( n − x ) f A ( t ) | N = n ( a ) da = 0 , n ≤ n max − f A ( t ) | N = n ( a ) is unknown, and will be derived in the following sections. Proposition 4. n max = sup ( n ∈ N : n ≤ VC − R a E [ R (0 , a ) | N = n ] f A (0) | N = n ( a ) da + ¯ x ¯ x ) + 1(6) Proof.
We seek the largest integer value of n that obeys ˆ U n (0) >
0. Thus,after extracting n from equation (5) and applying the supremum we get theexpression in (6).We observe that once f A (0) | N = n ( a ) is derived, n max can be computed. Our motivation for using a Markov chain is mainly to obtain the pdf of theservice age for a tagged customer who observes n others in the system uponarrival, with t time units having elapsed since then. This will eventuallyallow us to find a strategy that holds for the Nash equilibrium.11 .3.1 Markov chain state space First, we give some general notation for the steady states of the Markovchain: B = { ( k, a, w k +1 , w k +2 , ..., w n − ) } where • k is the number of waiting customers that observed service completion; • a is the age of the current service; • w i is the waiting time of the i th customer in the system; and • n is the number of customers in the system.A general steady-state density is denoted by p ( k, a, w k +1 , w k +2 , ..., w n − ), andthe probability density of having n customers in the system and a service ageof a is denoted by π ( n, a ). It can be derived from the steady states of theMarkov chain that π ( n, a ) = n − X k =0 Z w p ( k, a, w k +1 , ..., w n − ) dw, n ≥ , a ∈ R + , We denote the marginal probability of having n customers in the system by π n , which is derived as π n = R a π ( n, a ) da . We first indicate some general andrather trivial relationships. From the arrival order, we get w n − < w n − .... 2. This is because, if a > T k , then the k th customer will have alreadyabandoned by now. There are no constraints on a for k = 0. That is, ifno-one observed service completion, the first in the queue could arrive at anyvalue of a . By the same argument, w i < S i , for k + 1 ≤ i ≤ n − . For further analysis we present the following definition: let the state structure be the combination set of ( n, k ), where n is the number of customers inthe system and k is the number of waiting customers that observed servicecompletion. Since the queue length is limited, the Markov process has alimited number of different state structures.12 roposition 5. The total amount of Markov chain state structures is |B ( k,n ) | = n max ( n max + 1)2 Proof. We first claim that if there are n customers in system, then there are n different state structures for 1 ≤ n ≤ n max − 1. What determines the numberof state structures for a given n , is the number of different possible values of k , where 0 ≤ k ≤ n − 1. That is, the values of k can be from zero to n − n = n max there are n max − ≤ k ≤ n max − 2, while the state { n max − , a } is not possible.Finally, n = 0 means an empty system, with just one state structure. Thetotal number can be computed as an arithmetic progression. It is equivalentto the sum P n max i =1 i , where each value of i represents the amount of statestructures for a given n , except i = n max , which in this case includes casesfor both n = 0 and n = n max . From here the result is straightforward. Due to the complexity of the process we begin with a simple example. Let usconsider state (0 , a ), which refers to an active server with a current serviceage a and an empty queue. For a < S , p (0 , a ) = p (0 , e − λa ¯ F ( a ) (7)Equation (7) justifies the following. State (0 , a ) will always follow state (0 , , 0) occurred, and during the intervening a time unitsthere were no service completions and no arrivals. Thus, the probabilitydensity of p (0 , a ) is as for p (0 , a > S , we allow arrivals tooccur from the beginning of service, assuming that they will have abandonedby the time the service reaches age a . For mS ≤ a < ( m + 1) S , where m ∈ { , , , ... } , it is possible that a customer will arrive and abandon after S time units. If, during the stay of the new arrival, more customers arrive,then they will not be in the queue once he abandons. This is because, if theystayed until then, they would abandon as well, since the customer aheadof them abandoned. This process can happen no more than m times, for13 < ( m + 1) S . For example, if m = 1, which means S < a < S , then twocases are possible: no arrivals at all and no service completion, or one arrivalwho abandoned before the state reaches (0 , a ) and no service completion.From basic probability we obtain p (0 , a ) = p (0 , e − λa + λe − λa ( a − S )) ¯ F ( a )Thus, p (0 , a ) is equal to p (0 , 0) times the probability that no service comple-tion occurs and there were from 0 to m arrival events followed by abandon-ments. We next give a general expression. Let g ( k, n, m, a ) be g ( k, n, m, a ) = (cid:16)P mj =0 λ j e − λ ( a − jSn ) ( a − jS n ) j j ! (cid:17) F ( a ) 1 ≤ k +1 = n ≤ n max − (cid:16)P mj =0 λ j e − λ ( wn − − jSn ) ( w n − − jS n ) j j ! (cid:17) F ( a ) F ( a − w n − ) ≤ k +1 < n ≤ n max − F ( a ) F ( a − w n − ) n = n max and m ∈ ( N k = 0 { , , , ... j T k − S n S n k } ≤ k ≤ n max − a is the current service age, w n − represents the waiting time of thelast joining type ii customer who observed n − n is the number of customers in the system, and m relates tothe possible values of a : specifically, mS n < a < ( m + 1) S n . Lastly, k is thenumber of customers who observed service completion. Lemma 3. The function g ( k, n, m, a ) is represented differently in three cases.Case 1, with ≤ k +1 = n ≤ n max − , represents the probability that the cur-rent state is ( k, a ) , given that a time units ago the state was ( k, .Case 2, with ≤ k + 1 < n ≤ n max − , represents the probability that thecurrent state is now ( k, a, w k +1 ...w n − , w n − ) , given that w n − time units agothe state was ( k, a − w n − , w k +1 − w n − , ..., w n − − w n − , .Case 3 represents the probability that the current Markov state is ( k, a, w k +1 ..., w n max − , w n max − ) ,given that w n max − time units ago the Markov state was: ( k, a − w n max − , w k +1 − w n max − , ..., w n max − − w n max − , .Proof. Case 1: in order that the state ( k, 0) will be replaced by the state( k, a ) after a time units, we need to ensure that there will not be a service14ompletion during those a time units. Also, we need to ensure that there willnot be any new arrivals, or if there are, then they will have abandoned bythe time the Markov chain state reaches ( k, a ). The probability of no servicecompletion is ¯ F ( a ). The probability of not having new arrivals once the statereaches ( k, a ) is m X j =0 λ j e − λ ( a − jS n ) ( a − jS n ) j j !Of course, scenarios which include more than one abandonment made bythe ( n + 1) th customer in the system are under consideration, where j is thenumber of times it occurs. We also note that j ≤ m .Case 2: in order that the state ( k, a − w n − , w k +1 − w n − , ..., w n − − w n − , 0) willbe replaced by the state ( k, a, w k +1 ...w n − , w n − ) after w n − time units, weneed to ensure that there will not be service completion and no new arrivalswhich stayed during that time (similar to Case 1). The probability that therewill not be service completion is P ( X > a | X > a − w n − ), which is equiva-lent to ¯ F ( a )¯ F ( a − w n − ) . In Case 3, new arrivals are not a possibility anyway, andtherefore, in order that the state will be transposed from ( k, a − w n max − , w k +1 − w n max − , ..., w n max − − w n max − , 0) to ( k, a, w k +1 ..., w n max − , w n max − ), we only needto ensure that there will be no service completion, which is ¯ F ( a )¯ F ( a − w n ) as in Case2. Hence, from Lemma 3 we have, for mS n ≤ a ≤ ( m + 1) S n , p ( k, a, w k +1 , ..., w n − ) = p ( k, a − w n − , ..., w n − − w n − , g ( k, n, m, a ) (8)From (8) we see that it is possible to separate the expression of the steady-state densities of the Markov chain into two parts. The first one is also asteady-state density of the Markov chain, for which the last argument is set tobe 0. The second is the function g ( k, n, m, a ), which is computable given themodel parameters. Therefore, in order to obtain the steady-state densitiesof the Markov process we need to find those in which the last argument isset to be 0. From the balance equations, λπ = Z a p (0 , a ) h ( a ) da (9) λp ( k, a, w k +1 , ..., w n − ) = p ( k, a, w k +1 , ..., w n − , 0) (10)15 a n − X i =0 p ( i, a, w i +1 , ..., w n +1 ) h ( a ) da = p ( n, 0) (11)Recall from Proposition 5 that the number of state structures is n max ( n max +1)2 .Excluding the state 0 for each state structure, there is single state for whichthe argument is 0. Therefore, from equations (9) to (11) we have n max ( n max +1)2 equations. Where in fact from (9) consist one Equation, (10) consist n max ( n max − +2 and (11) consist n max − P n max n =0 π n = 1, all the steady states of the Markov process is derived. Remark 3.1. For numerical computations, we guess a value for π . Usingequations (8) to (11), we compute P n max n =0 π n . If the total sum is smaller than1, we guess a larger number for π and vice versa. We next give a special case where n max = 3. There are 6 different state struc-tures, and they are represented as { (0) , (0 , a ) , (0 , a, w ) , (1 , a ) , (1 , a, w ) , (0 , a, w , w ) } .From balance equations we state that λπ = Z ∞ a =0 p (0 , a ) h ( a ) da (12)Also, λp (0 , a ) = p (0 , a, 0) (13) λp (0 , a, w ) = p (0 , a, w , 0) (14) Z ∞ a =0 Z A ∧ aw =0 p (0 , a, w ) h ( a ) da = p (0 , 0) (15) Z ∞ a =0 Z S ∧ aw =0 Z S ∧ w w =0 p (0 , a, w , w ) h ( a ) da = p (1 , 0) (16) λp (1 , a ) = p (1 , a, 0) (17)Using equations (8) and (12) to (17), and applying the numerical procedurefrom Remark 3.1, all steady states can be computed.16 .4 The age distribution given the queue length We next show how to obtain f A ( t ) | N = n ( a ), while using the steady-state prob-ability densities of the Markov chain. Before doing so, we would like toemphasise it’s complexity. Suppose a tagged customer arrived and observedone customer in the queue and one in service. This could result in manycases, for example, the system was empty, then one arrived and enter serviceand then another another arrival occurred. A different example, could be justlike the previous one, only now, another customer arrived prior to the taggedcustomer’s arrival and then abandoned. Of course, there can be many casesto consider and they get more complicated as the queue gets longer. Foreach scenario the age distribution will be computed differently. Although,it seems very complicated, we next show how via a few simple probabilityoperations it can be done.Let Y follow the distribution of the total amount of time a tagged customerwaited from arrival until either he abandons or there was service comple-tion, sampled by an outside inspector at an arbitrary moment. In fact, { A ( y ) | N = n } d = { A | N = n,Y = y } . From Bayes’ law, f A | N = n,Y = y ( a ) = f Y | N = n,A = a ( y ) f A | N = n ( a ) R a f Y | N = n,A = a ( y ) f A | N = n ( a ) da . (18)Since the presentation in a general case is implicit, we demonstrate using aspecial case of n max = 3. However, we can proceed similarly for any valueof n max . We show separately how to obtain f A | N = n ( a ) and f Y | N = n,A = a ( y )for both N = 1 and N = 2. f A | N = n ( a ) can be derived directly from thesteady-state densities of the Markov process, specifically for N = 1, f A | N =1 ( a ) = p (0 , a ) π and for N = 2, f A | N =2 ( a ) = p (1 , a ) + R S V aw =0 p (0 , a, w ) dw π We next derive f Y | N =1 ,A = a ( y ). Let Q be a random variable that representsthe inter-arrival times. Of course, Q ∼ Exp( λ ). Due to the PASTA property,an outside inspector sampling times is equivalent to customer arrival times.Therefore, Q | Q ≤ R (0 , a ) ∧ S is equivalent to Y | N =1 , A = a .17 emma 4. The conditional density of Y given A = a , N =1 is λe − λy ¯ F ( a + y ) F ( a ) P ( Q ≤ S ∧ R (0 , a )) and P ( Q ≤ S ∧ R (0 , a )) = Z S r =0 (1 − e − λr ) dr + Z ∞ r = S (1 − e − λS ) dr Proof. P ( Q ≤ y | Q ≤ R (0 , a ) ∧ S ) = P ( Q ≤ y, Q ≤ R (0 , a ) ∧ S ) P ( Q ≤ R (0 , a ) ∧ S )We give explicit expressions for both numerator and denominator.The numerator is P ( Q ≤ y, Q ≤ R (0 , a ) ∧ S ) = Z yr =0 P ( Q ≤ r ) f ( a + r ) F ( a ) dr + Z ∞ r = y P ( Q ≤ y ) f ( a + r ) F ( a ) dr and the denominator is P ( Q ≤ R (0 , a ) ∧ S ) = Z S r =0 P ( Q ≤ r ) f ( a + r ) F ( a ) dr + Z ∞ r = S P ( Q ≤ S ) f ( a + r ) F ( a ) dr After taking the derivative, we get the pdf of Q | Q ≤ R (0 , a ) ∧ S , and henceof Y | N =1 , A = a as well.We next derive f Y | N =2 ,A = a ( y ). Let W be a random variable taking value of w given that the state is (0 , a, w ), sampled at an arbitrary moment. The pdfof W is denoted by f W ( w ), and it is equivalent to p (0 ,a,w ) R S V au =0 p (0 ,a,u ) du . Whena customer arrives to a system given that N = 2, there are two possiblestate structures: i . p (1 , a ) and ii . p (0 , a, w ). Let I be an indicator thatreceives the value of 1 when the state structure is p (1 , a ), and receives 0when it is p (0 , a, w ). From simple probability considerations, P ( I = 1) = p (1 ,a ) p (1 ,a )+ R S V aw p (0 ,a,w ) dw . We indicate that Q | Q ≤ R (0 , a ) ∧ S ∧ ( I ( T − a ) +(1 − I )( S − W )) is equivalent to Y | N = 2 ,A = a .18 emma 5. The conditional density of Y given A = a , N = 2 and Y < ( I ( T − a ) + (1 − I )( S − W )) ∧ S ∧ R (0 , a ) is P ( I = 1) λe − λy ¯ F ( a + y ) F ( a ) P ( Q ≤ S ∧ R (0 , a ) ∧ ( T − a )) + P ( I = 0) Z S w =0 λe − λy ¯ F ( a + y ) F ( a ) P ( Q ≤ S ∧ R (0 , a ) ∧ ( S − w )) p (0 , a, w ) dw and P ( Q ≤ S ∧ R (0 , a ) ∧ ( T − a )) = Z ( T − a ) ∧ A r =0 P ( Q ≤ r ) f ( a + r )1 − F ( a ) dr + Z ∞ r = S ∧ ( T − a ) P ( Q ≤ ( T − a ) ∧ S ) f ( a + r ) F ( a ) dr and P ( Q ≤ S ∧ R (0 , a ) ∧ ( S − w )) = Z ( S − w ) ∧ S r =0 P ( Q ≤ r ) f ( a + r ) F ( a ) dr + Z ∞ r =( S − w ) ∧ S P ( Q ≤ ( S − w ) ∧ S ) f ( a + r ) F ( a ) dr The proof is given in Appendix A. In this section we refer to some trivial and nontrivial results concerning thevalues of the sequences { S , S , ..., S n max − } and { T , T , ..., T n max − } . Specif-ically, we describe the dependencies and boundaries between them. Intu-itively, the more customers are in front of you, the less you would be willingto wait. Lemma 6. If T i < ∞ , T n > T n +1 for ≤ n ≤ n max − .Proof. Recall that t in equation (3) refers to the time one waited in the queuesince service completion. Due to the fact that E [ X − t | X > t ] is increasingwith t , it follows straightforwardly that the larger the value of n the lower thevalue of t that solves the equation, and hence the lower the value of T n . Lemma 7. If S n < ∞ , S n > S n +1 for ≤ n ≤ n max − . roof. Suppose an individual observed n + 1 customers upon arrival. Thelast event prior to his arrival could be either an arrival, an abandonmentor a service completion. If it were an arrival, then clearly the one whoobserved n customers was in a better situation when he arrived. If it was anabandonment, then this individual (if he knew) should not join at all, andhence he is in a worse situation than the one who observed n upon arrival.The last case is that where the arriving customer is the first to arrive duringthe current service period. In this case, the age of the service time is his inter-arrival time. Yet, we claim that the greater the queue length, the smallerthe probability of such an event. This is because, after service completionwhen there are either n or n + 1 customers in the system everything is thesame except one thing. In the n + 1 system, the last customer can alsoreach his abandonment time (which didn’t exist in the n customers system).Thus, the probability that an arrival will occur prior to service completionor abandonment is indeed lower in the n + 1 customers system.We also observe that all { T , T , ..., T n max − } are obtained independently ofeverything else. This is a direct result from equation (3), where the value of T n is determined by the values of the model parameter and n . Moreover, weclaim that S i , for ∀ i ∈ { , , ..., n max − } , depends only on { S , S , ..., S i − } and { T , T , ..., T i } , and for i = n max − 1, is dependent only on { S , ..., S i − } and { T , T , ..., T n max − } . This means, that one who observed i customersupon arrival, is not effected in any way, by actions taken by customers whowere in the i + 2 th position or worse in the system (that is, have i + 1 ormore customers in from of them). This result is rather surprising since atfirst thought it seems that the entire history of a busy period would effect thevalue of S i . But the values of S i are not affected by { S i +1 , ..., S n max − } and { T i +1 , ...T n max − } . The intuition behind this is as follows. First, informationregarding actions that took place in previous service period are irrelevant forthe an arriving customer, because they have no impact (given the informationhe observes) on his waiting time. Now, let C i and C i +1 be two customers whofound i and i + 1 customers upon arrival, respectively, before an arrival anew customer, and all three arrived at the same service period. Also, assumethat the new customer found i customers in the system. Thus, C i and C i +1 abandoned before the arrival of the new customer. The abandonment of C i +1 could be triggered by either his loss of patience or by the loss of patience of C i .recall that if one abandons those after him abandon as well. The informationthat that the new customer faces is the same in both scenarios, because given20hat fact the C i abandoned, the threshold that C i +1 has no influence on thestate that the new customer faces. Theorem 1. The Nash equilibrium profile is defined by two finite sequencesof thresholds, { T , T , ...T n max − } and { S , S , ...S n max − } , each sequence foreach customer type. Within each sequence, from an individual point of view,each threshold (e.g., the time he waits in line before reneging) is determinedby the number of customers in front of him and of course the model input.Proof. Customers’ strategies are determined by their expected utility func-tions. Based on Propositions 2 and 3, we claim that there are two sequencesbecause the different customer types have different expected utility functions,and therefore their strategies are different as well. The sequence lengths area direct result of Proposition 4 and the definition of the steady states ofthe Markov chain. Finally, the fact that the Nash equilibrium profile withineach sequence is defined by thresholds was already proved in Lemma 3 andLemma 4. Remark 3.2. After obtaining the steady state density probabilities (see re-mark 3.1), we now use a different scheme to compute the threshold values.First, the values of T , T , ..., T n max − are computed independently directlyfrom Equation (3). Then, the values of S , S , ..., S n max − are computed re-cursively starting from S . Thus, we guess a value of S and while usingEquation (18) and employing the scheme in Lemma 4 while using the steadystate probabilities that were obtained as described in Remark 3.1. If the ex-pected utility which is computed according to Equation (4) is negative weguess a lower value of S and vice versa. After obtaining S this goes on un-til we obtain all values of S n for n ∈ { , , ..., n max − } . The computationsare very complex, which is due to two main reasons. The first is that wedon’t have closed form equations. As a result, there are numerous iterationsfor each value that is being computed. The second is that the density prob-abilities functions are very complicated and are calculated differently alongthe support as a result of the function g ( k, n, m, a ) . Finally we note that theboth the procedure that was described in Remark 3.1 and in current one werecomputed by using Wolfram-Mathematica software and we used a toleranceparameter ǫ = 10 − . We note that although we provided in this paper the density probability of f A | N = n for n = 1 and n = 2 only, by following the same line of thought it can obtained for every n ≤ n max − n = 3 (that is n max = 4) itgets extremely difficult. .6 Numerical result We present an example where the service distribution is hyperexponential,and the model parameters are V = 4 . c = 1, µ = 1, µ = 0 . p = 0 . λ = 3. The (symmetric) Nash equilibrium is n max = 3, T = 7 . S = 7 . 202 and S = 3 . 13. In this example, the value of S is significantlylower than S . This is due to two reasons. The first is trivial, where oneneeds to wait for an extra customer. The second is that the age distributionis stochastically larger, which is a direct result from Lemma 7. In this study we show how to obtain the Nash equilibrium in an observable M/G/ { T , T , ...T n max − } represents theabandonment thresholds for customers that observed service completion, andthey are obtained by solving a linear equation. The values of the sequence { S , S , ..., S n max − } , represents the abandonment thresholds for customersthat didn’t observe service completion, and obtaining them is much moredifficult. They can be computed recursively based on the definition of theMarkov process. The reason we are able to obtain them recursively is be-cause customers’ decisions are not effected by future arrivals, (i.e. they aretransparent to them). In other words, from the point of view of a customerwho is in the n th position in the queue, the maximum length of the queue is n . Also, a numerical example is given in which both sequences are computed.Finally, we discuss three limitations of our model:1. We assumed that customers are homogenous with respect to their ser-vice value and waiting cost. Of course, considering heterogenous cus-tomers is more realistic. Perhaps, a future study may extend this re-sults, but, it is vital that a simpler solution will be obtained first tobuild the foundation of such study.2. We assumed that the time one already waited is considered to be a sunkcost and only future waiting time is considered. In real life, this maynot always be the case. Relaxing this assumption, may have a huge22mpact on our solution due to the following reason: If one abandonedin front of me, I no longer necessarily abandon as well. This is because,I will not be in the exact situation as he was. This may alter the entirestrategy profile.3. We assumed that the service distribution possess the IMRL property.The case where the service is with IFR was already argued for. Wenext discuss the case in which the hazard function is neither with IFRor DFR. If so, we believe that a general solution cannot be obtaineddue to the dependency on the service distribution. References [1] P. Afeche and M. Pavlin. Rational Abandonment from Priority Queues:Equilibrium Strategy and Pricing Implications. Management science forthcoming, 2015.[2] F. Baccelli, F. Boyer, and G. Hebuterne. Single server queues withimpatient customers. Appl. Probability , 16:887–905, 1984.[3] F. Baccelli and G. Hebuterne. On queues with impatient customers. Performance , pp. 159–179, 1981.[4] R. E Balow and F. Proschan. Statistical theory reliability and life test-ing: Probability models. Hold, Rinehart and Winston , New York, 1975.[5] D.Y. Barrer. Queuing with impatient customers and ordered service. Oper. Res , 5:650–656, 1957.[6] M. Cripps and C. Thomas. Strategic Experimentation in Queues. Menuscript , 2014.[7] O. Boxma, D. Perry, W. Stadje, and S. Zacks. The busy period of anM/G/1 queue with customer impatience. J. Appl. Probab , 45(1):130–145, 2010.[8] A. Brandt and M. Brandt. Workload and busy period for the M/GI/1with a general impatience mechanism. Queueing Systems , 75:189–209,2013. 239] L. Debo, R. Hassin, and U. S. Veeraraghavan. Learning quality throughservice outcomes. Menuscript , 2012.[10] R. Hassin and M. Haviv (2003). To Queue or Not to Queue: EquilibriumBehavior in Queueing Systems. Kluwer Academic Publishers, Boston.[11] R. Hassin (2016). Rational Queueing. Chapman & Hall.[12] R. Hassin and M. Haviv. Equilibrium strategies for queues with impa-tient customers. Oper. Res. Lett , 17:41–45, 1995.[13] M. Haviv and Y. Ritov. Homogeneous customers renege from invisiblequeues at random times under deteriorating waiting conditions. Queue-ing Systems , 38:495–508, 2001.[14] A. Mandelbaum and N. Shimkin. A model for rational abandonmentsfrom invisible queues. Queueing Systems , 36:141–173, 2000.[15] A. Mandelbaum and N. Shimkin. Rational abandonment from tele-queues: Nonlinear waiting costs with heterogeneous preferences. Queue-ing Systems , 47:117–146, 2004.[16] P. Naor. The regulation of queue size by levying tolls. Econometrica ,37(1):15–24, 1969.[17] C. Palm. Methods of judging the annoyance caused by congestion. Tele ,2:1–20, 1953. A Appendix A Proof. P ( Q ≤ y | Q ≤ R (0 , a ) ∧ A ∧ ( I ( T − a ) + (1 − I )( A − W )))= P ( Q ≤ y, Q ≤ R (0 , a ) ∧ A ∧ ( I ( T − a ) + (1 − I )( A − W ))) P ( Q ≤ R (0 , a ) ∧ A ∧ ( I ( T − a ) + (1 − I )( A − W )))24e give explicit expressions for both the numerator and the denominator.The numerator is P ( Q ≤ y, Q ≤ R (0 , a ) ∧ A ∧ ( I ( T − a ) + (1 − I )( A − W )))= Z A w =0 P ( I = 1) P ( Q ≤ y, Q ≤ R (0 , a ) ∧ A ∧ ( T − a ))+ P ( I = 0) Z A ∧ ( A − w ) a =0 P ( Q ≤ y, Q ≤ R (0 , a ) ∧ A ∧ ( A − w )) p (0 , a, w ) dw and the denominator is P ( Q ≤ R (0 , a ) ∧ A ∧ ( I ( T − a ) + (1 − I )( A − W )))= Z A w =0 P ( I = 1) P ( Q ≤ R (0 , a ) ∧ A ∧ ( T − a ))+ P ( I = 0) P ( Q ≤ R (0 , a ) ∧ A ∧ ( A − w )) p (0 , a, w ) dw After taking the derivative, we obtain the p.d.f. of Q | Q ≤ R (0 , a ) ∧ A ∧ ( I ( T − a ) + (1 − I )( A − W )), and hence of Y | N =2 , A = aa