Heavy-traffic analysis of k-limited polling systems
aa r X i v : . [ m a t h . P R ] A ug Heavy-traffic analysis of k -limited polling systems M.A.A. Boon ∗ [email protected] E.M.M. Winands † [email protected], 2014 Abstract
In this paper we study a two-queue polling model with zero switch-over times and k -limited service (serve at most k i customers during one visit period to queue i , i = 1 , )in each queue. The arrival processes at the two queues are Poisson, and the service timesare exponentially distributed. By increasing the arrival intensities until one of the queuesbecomes critically loaded, we derive exact heavy-traffic limits for the joint queue-lengthdistribution using a singular-perturbation technique. It turns out that the number ofcustomers in the stable queue has the same distribution as the number of customers ina vacation system with Erlang- k distributed vacations. The queue-length distributionof the critically loaded queue, after applying an appropriate scaling, is exponentiallydistributed. Finally, we show that the two queue-length processes are independent inheavy traffic. Keywords: polling model, queue lengths, heavy traffic, perturbation
Mathematics Subject Classification:
This paper considers a two-queue k -limited polling model with exponentially distributed ser-vice times and zero switch-over times. Under the k -limited strategy the server continuesworking until either a predefined number of k i customers is served at queue i or until thequeue becomes empty, whichever occurs first. The interest for this model is fueled by a num-ber of different application areas. That is, the k -limited strategy did not only prove its meritin communication systems (see, e.g., [5, 8]), but also in the field of logistics (see, e.g. [28]),and vehicle actuated traffic signals [4]. In the present paper, we consider the heavy-trafficscenario, in which one of the queues becomes critically loaded with the other queue remainingstable.Although the number of papers on polling systems is impressive, hardly any exact results forpolling systems with the k -limited service policy have been obtained. This can be explained ∗ Eurandom and Department of Mathematics and Computer Science, Eindhoven University of Technology,P.O. Box 513, 5600MB Eindhoven, The Netherlands † University of Amsterdam, Korteweg-de Vries Institute for Mathematics, Science Park 904, 1098 XH Am-sterdam, The Netherlands
1y the fact that the k -limited strategy does not satisfy a well-known branching property forpolling systems, independently discovered by Fuhrmann [15] and Resing [26], which signifi-cantly increases the analytical complexity. For this reason, most of the papers in the pollingliterature focus on branching-type service disciplines, like exhaustive service (serve all cus-tomers in a queue until it is empty, before switching to the next queue) or gated service (serveall customers present at the server’s arrival to a queue, before switching to the next queue).Groenendijk [16] and Ibe [17] give an explicit Laplace-Stieltjes Transform for the waiting-time distribution in a two-queue -limited/exhaustive system. For two-queue systems whereboth queues are served according to the -limited discipline, the problem of finding the queuelength distribution can be shown to translate into a boundary value problem [6, 7, 11, 13].For general k , an exact evaluation for the queue-length distribution is known in two-queueexhaustive/ k -limited systems (see [21, 24, 25, 28]). The situation where both queues followthe k -limited discipline has not been solved yet.When studying the literature on heavy-traffic results for polling systems, a remarkable obser-vation is that distributional heavy-traffic limits have only been rigorously proven for systemswith branching-type service disciplines (there is only exception as described later). Even forthese systems, the proof is limited to exhaustive systems consisting of two queues [9, 10]or systems with Poisson arrivals [27]; all other results are based on conjectures. That is, aproof for the general setting is already a challenge for almost years since the publicationof the Coffman/Puhalskii/Reiman papers [9, 10]. Moreover, the only paper analysing theheavy-traffic behaviour for the k -limited discipline is the paper of Lee [21] for the two-queueexhaustive/ k -limited system without setup times. More specifically, he studies the limitingregime where the exhaustive queue remains stable, while the k -limited queue becomes crit-ically loaded in the limit. Lee [21] uses results for this system under stable conditions as abasis to derive heavy-traffic asymptotics by exploiting the fact that significant simplificationsresult as the load increases. However, corresponding steady-state results are not available forthe k -limited system under consideration in the current paper, which impels us to look for adifferent family of techniques.The main contribution of the present paper is that we derive heavy-traffic asymptotics for k -limited polling models via the singular-perturbation technique. That is, by increasing thearrival intensities until one of the queues becomes critically loaded, we derive heavy-trafficlimits for the joint queue-length distribution in a two-queue k -limited polling model. In thisway, we derive the lowest-order asymptotic to the joint queue-length distribution in terms ofa small positive parameter measuring the closeness of the system to instability.Furthermore, the results obtained in the present paper provide new insights into the heavy-traffic behaviour of k -limited polling systems. It is shown that the number of customersin the stable queue has the same distribution as the number of customers in a vacationsystem with Erlang- k distributed vacations, while the scaled queue-length distribution ofthe critically loaded queue is exponentially distributed. Finally, we prove that the two queue-length processes are independent in heavy traffic. The singular-perturbation technique can alsobe extended to an N -queue system ( N ≥ ) with one queue becoming critically loaded. In thislimiting regime the stable queues have the same joint distribution of a k -limited polling modelwith N − queues and an extended switch-over time. These results do not only generalisethose derived in [21] (where the special 2-queue case is studied in which only one queue isserved according to the k -limited service policy), but are also obtained via a fundamentally2ifferent singular-perturbation approach. We would like to note that the observed heavy-traffic behavior is different from the behaviour seen for branching policies in which the cycletimes normally converges to a Gamma distribution and the individual workloads follow froman averaging principle.Finally, to our opinion, the novel application of the singular perturbation technique to pollingsystems is interesting in itself. That is, the merits of this technique are in its intrinsic simplicityand its intuitively appealing derivation, although it requires some distributional assumptions.The singular perturbation framework is known as a powerful tool to determine asymptoticsin all kinds of queueing models, but it has never been applied to polling systems. We wouldlike to refer to Knessl and Tier [19] for an excellent survey of applications of the perturbationtechnique to queueing models. It is noteworthy that our paper is inspired by the manner inwhich Morrison and Borst [23] apply this technique to a model with interacting queues.The paper is structured as follows. In the next section we introduce the model and notation. InSection 3 we apply a perturbation technique to study the system under heavy-traffic conditionsand derive the limiting scaled joint queue-length distribution. In Section 4 we interpret theresults and give some suggestions on further research. The appendices contain some lengthyderivations required for the analysis in Section 3. We consider a polling model consisting of two queues, Q and Q , that are alternately visitedby a single server. Throughout this paper, the subscript i will always be used to refer to oneof the queues, meaning that it always takes on the values 1 or 2. When a server arrives at Q i ,it serves at most k i customers. When k i customers have been served or Q i becomes empty,whichever occurs first, the server switches to the other queue. We assume that switchingfrom one queue to the other requires no time. If the other queue turns out to be empty, theserver switches back and serves, again, at most k i customers. If both queues are empty, theserver waits until the first arrival and switches to the corresponding queue (say, Q j ) to startanother visit period of at most k j customers, j = 1 , . Customers arrive at Q i according toa Poisson process with intensity λ i . We assume that the service times of customers in Q i are independent and exponentially distributed with parameter µ i . We denote the load of thesystem by ρ = ρ + ρ , where ρ i = λ i /µ i . For a polling model without switch-over times, thestability condition is ρ < [12, 14]. Furthermore, we assume that λ k < λ k . (2.1)This assumption is discussed in more detail in the next section.The number of customers in Q i at time t , t ≥ , is denoted by N i ( t ) . In order to describethe queue length process as a Markov process, we use the approach of Blanc [3], introducinga supplementary variable H ( t ) which takes on values , , . . . , k + k . The variable H ( t ) isused to determine the server position ( Q or Q ) at time t and the number of customers thatcan be served before the server has to switch to the next queue. When ≤ H ( t ) ≤ k , thismeans that the server is serving Q at time t , and that the customer in service is the H ( t ) -thcustomer being served during the present visit period. If k + 1 ≤ H ( t ) ≤ k + k , the serveris serving the ( H ( t ) − k )-th customer in Q . Now, ( N ( t ) , N ( t ) , H ( t )) is a Markov process.3ssuming that ρ < , define the stationary probabilities p ( n , n , h ) := lim t →∞ P (cid:0) N ( t ) = n , N ( t ) = n , H ( t ) = h (cid:1) , and define the steady-state queue lengths N i . We study the heavy-traffic limit of the joint queue-length process ( N , N ) by increasing thearrival rate λ , while keeping λ fixed. When ρ tends to 1, Assumption (2.1) implies that Q will become critically loaded, whereas Q remains stable due to the fact that at most k customers are served during each visit period at Q . In case λ /k = λ /k in the limit, bothqueues would become critically loaded simultaneously and the system behaviour is differentfrom the limiting behaviour found in the present paper. In fact, the limiting queue-lengthbehaviour for that specific case remains an open problem. We discuss this topic briefly inSection 4.We use a single-perturbation method to find the queue-length distributions in heavy-traffic ( ρ ↑ . First, we write down the balance equations of our model and apply a perturbation to thearrival rate of Q to these equations, in the case that this queue is close to becoming criticallyloaded. By solving the system of balance equations for the lowest (zeroth) order terms, wefind the queue length distribution of the stable queue, Q . By analysing the equations for thefirst-order and second-order terms, we also obtain a differential equation for the scaled numberof customers in Q , which we can solve to show that this number converges to an exponentialdistribution. The balance equations for a polling model with exponentially distributed service times and k -limited service at each of the queues are given by Blanc [3]. For completeness, we present4hese equations for our two-queue model below. ( λ + λ + µ ) p (0 , n , k + 1) = λ p (0 , n − , k + 1) + µ p (0 , n + 1 , k + k )+ k X h =1 µ p (1 , n , h ) , (3.1a) ( λ + λ + µ ) p (0 , n , h ) = λ p (0 , n − , h ) + µ p (0 , n + 1 , h − , (3.1b) ( λ + λ + µ ) p (1 , n ,
1) = λ p (1 , n − ,
1) + µ p (1 , n + 1 , k + k ) , (3.1c) ( λ + λ + µ ) p (1 , n , h ) = λ p (1 , n − , h ) + µ p (2 , n , h − , (3.1d) ( λ + λ + µ ) p ( n + 1 , n ,
1) = λ p ( n , n ,
1) + λ p ( n + 1 , n − , µ p ( n + 1 , n + 1 , k + k ) , (3.1e) ( λ + λ + µ ) p ( n + 1 , n , h ) = λ p ( n , n , h ) + λ p ( n + 1 , n − , h )+ µ p ( n + 2 , n , h − , (3.1f) ( λ + λ + µ ) p ( n , n , k + 1) = λ p ( n − , n , k + 1) + λ p ( n , n − , k + 1)+ µ p ( n + 1 , n , k ) , (3.1g) ( λ + λ + µ ) p ( n , n , h ) = λ p ( n − , n , h ) + λ p ( n , n − , h )+ µ p ( n , n + 1 , h − , (3.1h)for n = 1 , , . . . ; n = 2 , , . . . ; h = 2 , , . . . , k , and h = k + 2 , . . . , k + k . Note that(3.1a)-(3.1h) are not all balance equations. We have omitted all equations for n = 0 and n = 1 , since it will turn out that these do not play a role after the perturbation. Theintuitive explanation is that N ( t ) will tend to infinity as Q becomes critically loaded andthe probabilities p ( n , n , h ) become negligible for low values of n . From the stability condition we have that the system becomes unstable as λ /µ ↑ − λ /µ ,which means that the arrival rate λ approaches µ (1 − λ /µ ) . Therefore we will assume that λ = µ (cid:18) − λ µ (cid:19) − δω, ω > , < δ ≪ . (3.2)At the end of this section we will take an appropriate choice for the constant ω , which willinfluence the limit of the scaled queue length in Q .Let ξ = δn , and p ( n , ξ/δ, h ) = δφ n ,h ( ξ, δ ) , < ξ = O (1) , h = 1 , , . . . , k + k . (3.3)Note that once ω is chosen, δ and the scaled variable ξ will be uniquely defined. The nextstep is to substitute (3.2) and (3.3) in the balance equations (3.1a)-(3.1h), and take the Taylorseries expansion with respect to δ . For reasons of compactness, we only show the intermediateresults for Equation (3.1h) as an illustration: ( λ + µ ) φ n ,h ( ξ, δ ) − λ φ n − ,h ( ξ, δ ) − µ φ n ,h − ( ξ + δ, δ ) = (cid:16) µ (cid:0) − λ µ (cid:1) − δω (cid:17)(cid:0) φ n ,h ( ξ − δ, δ ) − φ n ,h ( ξ, δ ) (cid:1) . ( λ + µ ) φ n ,h ( ξ, δ ) − λ φ n − ,h ( ξ, δ ) − µ (cid:18) φ n ,h − ( ξ, δ ) + δ ∂φ n ,h − ( ξ, δ ) ∂ξ + δ ∂ φ n ,h − ( ξ, δ ) ∂ξ (cid:19) = − (cid:16) µ (cid:0) − λ µ (cid:1) − δω (cid:17) (cid:18) δ ∂φ n ,h ( ξ, δ ) ∂ξ − δ ∂ φ n ,h ( ξ, δ ) ∂ξ (cid:19) + O ( δ ) . (3.4)Note that λ (or µ (cid:0) − λ µ (cid:1) − δω after the substitution) only plays a role in this equation for O ( δ ) terms and higher. It is readily verified that this is the case for all balance equations. Wenow expand in powers of δ , and let φ n ,h ( ξ, δ ) = φ (0) n ,h ( ξ ) + δφ (1) n ,h ( ξ ) + O ( δ ) . (3.5)One would expect that also δ φ (2) n ,h ( ξ ) should be introduced in order to analyse the second-order terms. However, Proposition 3.3 implies that the above power expansion suffices. Wealso define the corresponding generating functions e Q h ( z, ξ, δ ) := ∞ X n =0 φ n ,h ( ξ, δ ) z n , e Q ( j ) h ( z, ξ ) := ∞ X n =0 φ ( j ) n ,h ( ξ ) z n , j = 0 , , , . . . . In the next subsections we first equate the lowest order terms of the resulting equations tofind an expression for (the generating function of) φ (0) n,h ( ξ ) , and subsequently we equate thefirst-order and second-order terms to find the scaled queue-length distribution of Q . Equating the lowest-order terms of the balance equations, after substituting (3.2), (3.3), and(3.5), results in the following equations. ( λ + µ ) φ (0)0 ,k +1 ( ξ ) = µ φ (0)0 ,k + k ( ξ ) + k X h =1 µ φ (0)1 ,h ( ξ ) , (3.6a) ( λ + µ ) φ (0)0 ,h ( ξ ) = µ φ (0)0 ,h − ( ξ ) , (3.6b) ( λ + µ ) φ (0)1 , ( ξ ) = µ φ (0)1 ,k + k ( ξ ) , (3.6c) ( λ + µ ) φ (0)1 ,h ( ξ ) = µ φ (0)2 ,h − ( ξ ) , (3.6d) ( λ + µ ) φ (0) n +1 , ( ξ ) = λ φ (0) n , ( ξ ) + µ φ (0) n +1 ,k + k ( ξ ) , (3.6e) ( λ + µ ) φ (0) n +1 ,h ( ξ ) = λ φ (0) n ,h ( ξ ) + µ φ (0) n +2 ,h − ( ξ ) , (3.6f) ( λ + µ ) φ (0) n ,k +1 ( ξ ) = λ φ (0) n − ,k +1 ( ξ ) + µ φ (0) n +1 ,k ( ξ ) , (3.6g) ( λ + µ ) φ (0) n ,h ( ξ ) = λ φ (0) n − ,h ( ξ ) + µ φ (0) n ,h − ( ξ ) , (3.6h)for n = 1 , , . . . ; h = 2 , , . . . , k , and h = k + 2 , . . . , k + k .6ote that P ∞ n =0 P k + k h =1 φ (0) n ,h ( ξ ) = 1 . For this reason we introduce P ( ξ ) and π (0) n ,h , with φ (0) n ,h ( ξ ) = π (0) n ,h P ( ξ ) , and ∞ X n =0 k + k X h =1 π (0) n ,h = 1 , (3.7)for n = 0 , , , . . . and h = 1 , , . . . , k + k . Careful inspection of these balance equationsreveals that equations (3.6a)-(3.6h) describe the behaviour of a single-server vacation queuewith the following properties:P1. the arrival process is Poisson with intensity λ ,P2. the service times are exponentially distributed with mean /µ ,P3. the service discipline is k -limited service with service limit k ,P4. the vacations are Erlang- k distributed with parameter µ ,P5. whenever the server finds the system empty upon return from a vacation, it immediatelystarts another vacation.This system has been studied in the literature (cf. [22]) and, in general, no closed-formexpressions for the steady-state queue-length probabilities can be obtained. However, it ispossible to find the probability generating function (PGF) of the queue-length distribution.Define e L (0) ( z ) := k + k X h =1 e L (0) h ( z ) , where e L (0) h ( z ) := ∞ X n =0 π (0) n,h z n , h = 1 , . . . , k + k , and e G ( z ) := µ λ (1 − z ) + µ , e H ( z ) := µ λ (1 − z ) + µ . (3.8)It is easily seen that e G ( z ) and e H ( z ) are the PGFs of the number of arrivals during respectivelyone service, and during one stage of the vacation (which consists of k exponential stages). Itfollows from (3.6a)-(3.6h) that e L (0) k + k ( z ) = e H ( z ) k " π (0)0 ,k + k (cid:18) − (cid:16) e G ( z ) /z (cid:17) k (cid:19) + µ µ k − X h =1 π (0)1 ,h (cid:18) − (cid:16) e G ( z ) /z (cid:17) k − h (cid:19) − (cid:0) e G ( z ) /z (cid:1) k e H ( z ) k , (3.9) e L (0) h ( z ) = zµ µ (cid:0) e G ( z ) /z (cid:1) h he L (0) k + k ( z ) − π (0)0 ,k + k i − h − X h =1 π (0)1 ,h z (cid:16) e G ( z ) /z (cid:17) h − h , (3.10) e L (0) h ( z ) = e H ( z ) h − k " µ µ k − X h =1 π (0)1 ,h + π (0)0 ,k + k + µ zµ e L (0) k ( z ) , (3.11)for h = 1 , . . . , k and h = k + 1 , . . . , k + k − . A derivation of (3.9)-(3.11) can befound in Appendix A. These equations still contain k unknowns: π (0)1 , , π (0)1 , , . . . , π (0)1 ,k − , and π (0)0 ,k + k . See Appendix A for more details on how to eliminate them using Rouché’s Theorem.Foregoing the derivation of the limiting behaviour of Q we already would like to mentionthat these unknowns do not play a role therein.7e conclude from equating the lowest-order terms of the balance equations (3.1a)-(3.1h), aftersubstituting (3.2), (3.3), and (3.5), that ∞ X n =0 k + k X h =1 φ (0) n ,h ( ξ ) z n = e L (0) ( z ) P ( ξ ) , (3.12)where P ( ξ ) still has to be determined. Consequently, in heavy traffic, the queue length of thestable queue ( Q ) has the same distribution as the queue length in a vacation system withErlang( k ) distributed vacations with parameter µ , exponential service times with parameter µ , and k -limited service. The assumption that we have Poisson arrivals and first-come-first-served service implies that we can use the distributional form of Little’s law to obtain(the Laplace-Stieltjes transform of) the waiting-time distribution of customers in Q (see, forexample, Keilson and Servi [18]). Remark 3.1
In this paper we assume that λ /k < λ /k , causing Q to become criticallyloaded when λ is being increased. We have implicitly used this assumption when solving thebalance equations (3.1a)-(3.1h). It is well-known that the vacation system described by theseequations is stable if and only if λ E [ C ] < k , (3.13)where E [ C ] is the mean cycle time, i.e., the mean length of one visit period plus one vacation.Denoting the length of a vacation by S , we have E [ C ] = E [ S ] / (1 − ρ ) = k / ( µ (1 − λ /µ )) .When substituting this in (3.13), we indeed obtain exactly the same inequality as (2.1) aftersubstituting (3.2) and letting δ ↓ . Remark 3.2
Another interesting observation is that one could consider more general ways ofvarying the arrival rates in order to let Q become critically loaded. To this end, we introduce λ ∗ and λ ∗ such that λ ∗ /µ + λ ∗ /µ = 1 . Additionally, we assume that λ ∗ k < k µ + k µ , or equivalently: λ ∗ k > k µ + k µ . (3.14)We now let λ → λ ∗ and λ → λ ∗ for δ ↓ , with λ µ + λ µ = 1 − δω ∗ , ω ∗ > , < δ ≪ . (3.15)Any arbitrary way in which we let λ and λ approach respectively λ ∗ and λ ∗ , for δ ↓ , willcause Q to become critically loaded (because of assumption (3.14)). All results obtained inthis paper will still be valid, by choosing ω ∗ = ω/µ . In this section we study, and solve, the system of equations that results from equating thefirst-order terms of the perturbed balance equations. For notational reasons, we define ψ (1) n ,h ( ξ ) := φ (1) n ,h ( ξ ) + φ ′ (0) n ,h ( ξ ) , where φ ′ (0) n ,h ( ξ ) := dφ (0) n ,h ( ξ ) dξ , n = 0 , , . . . and h = 1 , , . . . , k + k . The resulting set of equations for the probabilities φ (1) n ,h ( ξ ) is given below. ( λ + µ ) φ (1)0 ,k +1 ( ξ ) = µ ψ (1)0 ,k + k ( ξ ) + k X h =1 µ φ (1)1 ,h ( ξ ) − µ (cid:18) − λ µ (cid:19) φ ′ (0)0 ,k +1 ( ξ ) , (3.16a) ( λ + µ ) φ (1)0 ,h ( ξ ) = µ ψ (1)0 ,h − ( ξ ) − µ (cid:18) − λ µ (cid:19) φ ′ (0)0 ,h ( ξ ) , (3.16b) ( λ + µ ) φ (1)1 , ( ξ ) = µ ψ (1)1 ,k + k ( ξ ) − µ (cid:18) − λ µ (cid:19) φ ′ (0)1 , ( ξ ) , (3.16c) ( λ + µ ) φ (1)1 ,h ( ξ ) = µ φ (1)2 ,h − ( ξ ) − µ (cid:18) − λ µ (cid:19) φ ′ (0)1 ,h ( ξ ) , (3.16d) ( λ + µ ) φ (1) n +1 , ( ξ ) = λ φ (1) n , ( ξ ) + µ ψ (1) n +1 ,k + k ( ξ ) − µ (cid:18) − λ µ (cid:19) φ ′ (0) n +1 , ( ξ ) , (3.16e) ( λ + µ ) φ (1) n +1 ,h ( ξ ) = λ φ (1) n ,h ( ξ ) + µ φ (1) n +2 ,h − ( ξ ) − µ (cid:18) − λ µ (cid:19) φ ′ (0) n +1 ,h ( ξ ) , (3.16f) ( λ + µ ) φ (1) n ,k +1 ( ξ ) = λ φ (1) n − ,k +1 ( ξ ) + µ φ (1) n +1 ,k ( ξ ) − µ (cid:18) − λ µ (cid:19) φ ′ (0) n ,k +1 ( ξ ) , (3.16g) ( λ + µ ) φ (1) n ,h ( ξ ) = λ φ (1) n − ,h ( ξ ) + µ ψ (1) n ,h − ( ξ ) − µ (cid:18) − λ µ (cid:19) φ ′ (0) n ,h ( ξ ) , (3.16h)for n = 1 , , . . . ; h = 2 , , . . . , k , and h = k + 2 , . . . , k + k . The solution to this systemof equations, in terms of generating functions, can be found in Appendix B. This solution isused to derive the following relation λ µ ∞ X n =0 k + k X h =1 φ (1) n,h ( ξ ) − ∞ X n =0 k X h =1 φ (1) n,h ( ξ ) = λ µ µ P ′ ( ξ ) , (3.17)which turns out to play a key role in determining the HT limit of the joint queue-lengthdistribution (see the next section). In order to find an expression for P ( ξ ) and, consequently, solve (3.12), we consider the Taylorseries of all perturbed balance equations. In this section we show that, fortunately, we onlyneed to consider the sum of all these equations (such as Equation (3.4)) over all n = 0 , , , . . . and h = 1 , , . . . , k + k , and we consecutively consider the O (1) , O ( δ ) , and O ( δ ) terms.Using the results we have obtained so far, we prove that:1. all O (1) , i.e., all φ n ,h ( ξ, δ ) , terms cancel immediately,2. the O ( δ ) terms cancel after expanding φ n ,h ( ξ, δ ) in powers of δ (i.e., substituting (3.5)),3. the equation that results from equating the O ( δ ) terms, can be solved to find P ( ξ ) .The above three results are proven in Propositions 3.3, 3.4, and 3.5.9 roposition 3.3 After taking the summation over all n = 0 , , , . . . and h = 1 , , . . . , k + k of the Taylor series of all perturbed balance equations, the O (1) terms cancel. Proof:
We can follow the generating function approach, used in Section 3.3, but replace the proba-bilities π (0) n ,h , by φ n ,h ( ξ, δ ) . This results in the same set of equations as (3.9)-(3.11), but withterms e Q h ( z, ξ, δ ) and φ n ,h ( ξ, δ ) instead of e L (0) h ( z ) and π (0) n ,h . Substituting z = 1 yields: µ e Q h (1 , ξ, δ ) = µ e Q h − (1 , ξ, δ ) , (3.18) µ e Q k +1 (1 , ξ, δ ) = µ e Q k (1 , ξ, δ ) + µ k − X h =1 φ ,h ( ξ, δ ) + µ φ ,k + k ( ξ, δ ) , (3.19) µ e Q h (1 , ξ, δ ) = µ e Q h − (1 , ξ, δ ) − µ φ ,h − ( ξ, δ ) , (3.20) µ e Q (1 , ξ, δ ) = µ e Q k + k (1 , ξ, δ ) − µ φ ,k + k ( ξ, δ ) , (3.21)for h = 2 , . . . , k and h = k + 2 , . . . , k + k . The summation of (3.18)-(3.21) over all h = 1 , , . . . , k + k cancels all terms. (cid:3) Proposition 3.4
After taking the summation over all n = 0 , , , . . . and h = 1 , , . . . , k + k of the Taylor series of all perturbed balance equations, substituting (3.5) and using the resultsfrom Appendix A, the O ( δ ) terms cancel. Proof:
Define φ ′ n ,h ( ξ, δ ) := ∂φ n ,h ( ξ, δ ) ∂ξ , for n = 0 , , . . . and h = 1 , , . . . , k + k . Given the fact that the O (1) terms cancel, takingthe O ( δ ) terms leads to the following equations: µ φ ′ ,k + k ( ξ, δ ) − µ (cid:18) − λ µ (cid:19) φ ′ ,k +1 ( ξ, δ ) , µ φ ′ ,h − ( ξ, δ ) − µ (cid:18) − λ µ (cid:19) φ ′ ,h ( ξ, δ ) , µ φ ′ ,k + k ( ξ, δ ) − µ (cid:18) − λ µ (cid:19) φ ′ , ( ξ, δ ) , − µ (cid:18) − λ µ (cid:19) φ ′ ,h ( ξ, δ ) , µ φ ′ n +1 ,k + k ( ξ, δ ) − µ (cid:18) − λ µ (cid:19) φ ′ n +1 , ( ξ, δ ) , − µ (cid:18) − λ µ (cid:19) φ ′ n +1 ,h ( ξ, δ ) , − µ (cid:18) − λ µ (cid:19) φ ′ n ,k +1 ( ξ, δ ) , µ φ ′ n ,h − ( ξ, δ ) − µ (cid:18) − λ µ (cid:19) φ ′ n ,h ( ξ, δ ) , h = 2 , . . . , k and h = k + 2 , . . . , k + k . Taking the generating functions of theseequations and substituting z = 1 results in the following set of equations: µ φ ′ ,k + k ( ξ, δ ) − µ (cid:18) − λ µ (cid:19) e Q ′ k +1 (1 , ξ, δ ) , (3.22) µ e Q ′ h − (1 , ξ, δ ) − µ (cid:18) − λ µ (cid:19) e Q ′ h (1 , ξ, δ ) , (3.23) µ (cid:16) e Q ′ k + k (1 , ξ, δ ) − φ ′ ,k + k ( ξ, δ ) (cid:17) − µ (cid:18) − λ µ (cid:19) e Q ′ (1 , ξ, δ ) , (3.24) − µ (cid:18) − λ µ (cid:19) e Q ′ h (1 , ξ, δ ) , (3.25)for h = 2 , . . . , k and h = k + 2 , . . . , k + k . The PGF e Q ′ h ( z, ξ, δ ) is the derivative of e Q h ( z, ξ, δ ) with respect to ξ .The summation of (3.22)-(3.25) over all h = 1 , , . . . , k + k yields: µ k X h =1 e Q ′ h (1 , ξ, δ ) = µ λ µ k + k X h =1 e Q ′ h (1 , ξ, δ ) . (3.26)Apparently, the O ( δ ) terms do not cancel (yet). However, after substituting (3.5), taking the O ( δ ) terms, and using (3.7), we obtain: µ k X h =1 e L (0) h (1) P ′ ( ξ ) = µ λ µ k + k X h =1 e L (0) h (1) P ′ ( ξ ) . (3.27)Since the (at this moment still unknown) terms P ′ ( ξ ) cancel out, and since P k + k h =1 e L (0) h (1) = 1 ,Equation (3.27) reduces to k X h =1 e L (0) h (1) = ρ , which is indeed true (see (A.5)). (cid:3) Proposition 3.5
Taking the summation over all n = 0 , , , . . . and h = 1 , , . . . , k + k ofthe Taylor series of all perturbed balance equations and equating the O ( δ ) terms, yields thefollowing differential equation for P ( ξ ) : ωP ′ ( ξ ) = − (cid:18) µ + λ µ ( µ − µ ) µ (cid:19) P ′′ ( ξ ) . (3.28) Proof:
Define φ ′′ n ,h ( ξ, δ ) := ∂ φ (0) n ,h ( ξ, δ ) ∂ξ , for n = 0 , , . . . and h = 1 , , . . . , k + k . As before, the O (1) terms cancel. From the proofof the Proposition 3.4 we have learned to include O ( δ ) terms as well, because multiplied by11 φ ′ (1) n,h ( ξ ) these terms are O ( δ ) as well. This leads to the following equations: µ (cid:20) φ ′ (1)0 ,k + k ( ξ ) + 12 φ ′′ (0)0 ,k + k ( ξ ) (cid:21) − µ (cid:18) − λ µ (cid:19) (cid:20) φ ′ (1)0 ,k +1 ( ξ ) − φ ′′ (0)0 ,k +1 ( ξ ) (cid:21) + ωφ ′ (0)0 ,k +1 ( ξ, δ ) , µ (cid:20) φ ′ (1)0 ,h − ( ξ ) + 12 φ ′′ (0)0 ,h − ( ξ ) (cid:21) − µ (cid:18) − λ µ (cid:19) (cid:20) φ ′ (1)0 ,h ( ξ ) − φ ′′ (0)0 ,h ( ξ ) (cid:21) + ωφ ′ (0)0 ,h ( ξ ) , µ (cid:20) φ ′ (1)1 ,k + k ( ξ ) + 12 φ ′′ (0)1 ,k + k ( ξ ) (cid:21) − µ (cid:18) − λ µ (cid:19) (cid:20) φ ′ (1)1 , ( ξ ) − φ ′′ (0)1 , ( ξ ) (cid:21) + ωφ ′ (0)1 , ( ξ ) , − µ (cid:18) − λ µ (cid:19) (cid:20) φ ′ (1)1 ,h ( ξ ) − φ ′′ (0)1 ,h ( ξ ) (cid:21) + ωφ ′ (0)1 ,h ( ξ ) , µ (cid:20) φ ′ (1) n +1 ,k + k ( ξ ) + 12 φ ′′ (0) n +1 ,k + k ( ξ ) (cid:21) − µ (cid:18) − λ µ (cid:19) (cid:20) φ ′ (1) n +1 , ( ξ ) − φ ′′ (0) n +1 , ( ξ ) (cid:21) + ωφ ′ (0) n +1 , ( ξ ) , − µ (cid:18) − λ µ (cid:19) (cid:20) φ ′ (1) n +1 ,h ( ξ ) − φ ′′ (0) n +1 ,h ( ξ ) (cid:21) + ωφ ′ (0) n +1 ,h ( ξ ) , − µ (cid:18) − λ µ (cid:19) (cid:20) φ ′ (1) n ,k +1 ( ξ ) − φ ′′ (0) n ,k +1 ( ξ ) (cid:21) + ωφ ′ (0) n ,k +1 ( ξ ) , µ (cid:20) φ ′ (1) n ,h − ( ξ ) + 12 φ ′′ (0) n ,h − ( ξ ) (cid:21) − µ (cid:18) − λ µ (cid:19) (cid:20) φ ′ (1) n ,h ( ξ ) − φ ′′ (0) n ,h ( ξ ) (cid:21) + ωφ ′ (0) n ,h ( ξ ) , for h = 2 , . . . , k and h = k + 2 , . . . , k + k . As we have done already a couple oftimes before, we can use the generating functions to easily sum all of these equations. Eachequation contains the following three types of terms: φ ′ (0) n ,h ( ξ ) , φ ′ (1) n ,h ( ξ ) , and φ ′′ (0) n ,h ( ξ ) . Wedenote the corresponding generating functions with e Q ′ (0) h ( z, ξ ) , e Q ′ (1) h ( z, ξ ) , and e Q ′′ (0) h ( z, ξ ) .After summing all equations, we obtain the following equation: µ λ µ k + k X h =1 e Q ′ (1) h (1 , ξ ) − µ k X h =1 e Q ′ (1) h (1 , ξ ) (3.29a) + 12 µ k + k X h = k +1 e Q ′′ (0) h (1 , ξ ) + 12 µ (cid:18) − λ µ (cid:19) k + k X h =1 e Q ′′ (0) h (1 , ξ ) (3.29b) + ω k + k X h =1 e Q ′ (0) h (1 , ξ ) (3.29c) = 0 . Note that the derivation of (3.29a) and (3.29b) follows closely the manner in which (3.26) hasbeen derived. The last term (3.29c) follows directly from collecting the ω -terms.12sing the results obtained in appendices the equation can be rewritten to λ µ µ P ′′ ( ξ ) + µ (cid:18) − λ µ (cid:19) P ′′ ( ξ ) + ωP ′ ( ξ ) = 0 . (3.30)The proof that (3.29a) is the same as the first term in (3.30) can be found in Appendix B.The second and third term follow from (A.5) and the fact that e Q (0) h (1 , ξ ) = e L (0) h (1) P ( ξ ) .Equation (3.30) can be rewritten to (3.28), which concludes this proof. (cid:3) Now we can finally present the density of the scaled number of customers in Q , denoted by P ( ξ ) . It is obtained by solving the differential equation (3.28): P ( ξ ) = η e − ηξ , (3.31)with η = ω (cid:20) µ + λ µ ( µ − µ ) µ (cid:21) − . We have used that R ∞ P ( ξ ) d ξ = 1 and that P ∞ n =0 P ∞ n =0 P k + k h =1 p ( n , n , h ) = 1 . Withoutloss of generality, we may take ω = µ , which means that η = 1 − λ µ + µ λ µ . (3.32)We motivate this choice for ω by noting that we consider the scaled queue length δN for δ ↓ . By choosing ω = µ , and using (3.2), our scaling becomes equivalent to considering thescaled queue length (1 − ρ ) N , which is commonly used. Finally, by applying the multiclassdistributional law of Bertsimas and Mourtzinou [2] it directly follows that the scaled waitingtime at Q also follows an exponential distribution with parameter λ η . The analysis of the present section has the following immediate consequence for the joint(scaled) queue-length distribution in heavy traffic, which is the main result of this paper.
Main result:
For λ /k < λ /k and λ = µ ((1 − ρ ) − δ ) , we have: lim δ ↓ P [ N ≤ n , δN ≤ ξ ] = L ( n ) (cid:16) − e − ηξ (cid:17) , (3.33)where L ( · ) is the cumulative probability distribution of the queue length of a queueing systemwith multiple vacations satisfying Properties P1–P5, and η is given by (3.32).13 Final remarks and suggestions for further research
Interpretation.
The main result (3.33) derived in the preceding section has the followingintuitively appealing interpretation:1. The number of customers in the stable queue has the same distribution as the numberof customers in a k -limited vacation system with Erlang- k distributed vacations.2. The scaled number of customers in the critically loaded queue is exponentially dis-tributed with parameter η .3. The number of customers in the stable queue and the (scaled) number of customers inthe critically loaded queue are independent.Below we explain these properties heuristically.Property 1 can be explained by the fact that if Q is in heavy-traffic, then exactly k customersare served at this queue during each cycle. If we place an outside observer at Q , then, fromhis perspective, this queue behaves like a k -limited vacation model in heavy-traffic, where thevacation distribution is given by the convolutions of k exponentially distributed service timesdistributions in Q .For Property 2, we note that the total workload in the system equals the amount of work inan M/G/ queue in which the two customer classes are combined into one customer class witharrival rate λ + λ and hyperexponentially distributed service times, i.e., the service time iswith probability λ i / ( λ + λ ) exponentially distributed with parameter µ i , i = 1 , . Based onstandard heavy-traffic results for the M/G/ queue, this implies that the distribution of thescaled total workload converges to an exponential distribution with mean ρ E [ R ] , where R isa residual service time. For a hyperexponential distribution, we have E [ R ] = λ /µ + λ /µ ρ , which implies that the total (scaled) asymptotic workload is exponentially distributed withparameter λ /µ + λ /µ . In heavy traffic, almost all customers are located in Q so the totalnumber of customers at this queue is exponentially distributed with mean µ ( λ /µ + λ /µ ) .Since λ ↑ µ (1 − λ /µ ) , the scaled number of customers in Q is exponentially distributedwith parameter η . Using the multiclass distributional law of Bertsimas and Mourtzinou [2],it can be shown that the scaled asymptotic waiting time of customers in Q is exponentiallydistributed with parameter λ η .Finally, Property 3 follows from the time-scale separation in heavy-traffic which implies thatthe dynamics of the stable queue evolve at a much faster time scale than the dynamics ofthe critically loaded queue. Since the amount of “memory” of the stable queue asymptoticallyvanishes compared to that of the critically loaded queue, the number of customers in Q andthe scaled queue length of Q become independent in the limit. Two critically loaded queues.
In the current paper we have analysed the heavy-trafficbehaviour in case only Q becomes critically loaded, i.e., when Assumption (2.1) is satisfied.The limiting regime in which both queues become saturated simultaneously ( λ /k = λ /k ) ,14hows fundamentally different system behaviour. That is, for general ρ the waiting time at Q is an (unknown) function of the visit time at Q and / (1 − ρ ) . This implies that O (1 − ρ ) variations in the visit time at Q are relevant for the heavy-traffic behaviour at Q . Morecolloquially, it is not sufficient anymore to use a scaling that implies that exactly k customersare served at Q during each cycle, i.e., the probability that there are served less than k customers cannot be neglected, when analysing the asymptotic behaviour of Q . Further research.
The analysis in this paper allows different kind of extensions. Firstly,one could consider phase-type interarrival-time or service-time distributions. The approach in-troduced in the present paper may be extended, without fundamentally changing the analysis,to such a system. Another extension could be the introduction of switch-over times wheneverthe server switches between queues. Such an extension requires more severe adaptations tothe approach and the analysis, and is the topic of a forthcoming paper. Finally, we want tomention that the singular-perturbation technique can also be applied to derive the HT analysisof a system consisting of more than two (say N ) queues, with one queue becoming criticallyloaded. Following the lines of the current paper, one can show that in this limiting regime thestable queues have the same joint queue-length distribution as in a polling model with N − queues and an extended switch-over time, whereas the scaled queue-length distribution of thecritically loaded queue is again exponentially distributed. As such, the results of the presentpaper provide a theoretical basis for the transformation of large polling systems into smallersystems for approximation purposes, cf. LaPadula and Levy [20]. Acknowledgement
The authors would like to thank Sem Borst and Onno Boxma for interesting discussions andvaluable comments on earlier drafts of this paper.
AppendixA A vacation model with k -limited service In this appendix we study a queueing model with multiple vacations and k -limited service.The main goal is to find the PGF of the queue-length distribution, as to prove (3.9)-(3.11).At the end of this appendix some additional properties of this queue-length distribution aregiven, which will be used in Section 3.5 and in Appendix B.The service times in this vacation model are exponentially distributed with parameter µ , andthe vacation length is Erlang( k ) distributed with parameter µ . The service discipline is k -limited service, with at most k customers being served during one visit period. Although thequeue-length distribution for the case with generally distributed service and vacation timeshas been studied by Lee [22], we provide the proof here to keep the paper self-contained, butalso because our state space is slightly different and we do not look at embedded epochs,yielding slightly different expressions than in [22].The starting point is to obtain generating functions from the balance equations (3.6a)-(3.6h).15ultiplying Equation (3.6h) with z n , summing over all n = 1 , , . . . , and adding Equation(3.6b), yields the following equation: e L (0) h ( z ) = e H ( z ) e L (0) h − ( z ) , (A.1)for h = k + 2 , . . . , k + k , where e H ( z ) is defined in (3.8). The interpretation of (A.1)is that the number of customers in the system during a certain vacation stage is simply thenumber of customers present at the previous stage of the vacation, plus the arrivals during one(exponentially distributed) stage. Obviously, no customers leave the system during a vacation.Multiplying Equation (3.6g) with z n , summing over all n = 1 , , . . . , and adding Equation(3.6a), yields the following equation: e L (0) k +1 ( z ) = e H ( z ) µ µ k − X h =1 π (0)1 ,h + π (0)0 ,k + k + µ µ z e L (0) k ( z ) ! . (A.2)Multiplying Equation (3.6f) with z n +1 , summing over all n = 1 , , . . . , and adding Equation(3.6d) multiplied by z , yields the following equation: e L (0) h ( z ) = e G ( z ) z (cid:16)e L (0) h − ( z ) − π (0)1 ,h − z (cid:17) , (A.3)for h = 2 , , . . . , k , where e G ( z ) is defined in (3.8).Multiplying Equation (3.6e) with z n +1 , summing over all n = 1 , , . . . , and adding Equation(3.6c) multiplied by z , yields the following equation: e L (0)1 ( z ) = e G ( z ) µ µ (cid:16) e L (0) k + k ( z ) − π (0)0 ,k + k (cid:17) . (A.4)We now have k + k equations, each of which expresses e L (0) h ( z ) in terms of e L (0) h − ( z ) (and e L (0)1 ( z ) in terms of e L (0) k + k ( z ) ). Finally, we can solve these equations to find the expressionsfor e L (0) h ( z ) , for h = 1 , , . . . , k + k . The results are given in (3.9)-(3.11).Note that there are still k unknowns: π (0)1 ,h , for h = 1 , . . . , k − , and π (0)0 ,k + k . These canbe found using the roots of the denominator of (3.9). Rouché’s Theorem implies that thedenominator has k roots on and inside the unit circle. The requirement that (3.9) shouldbe regular inside the unit circle, implies that the numerator of (3.9) should have these sameroots [1]. Hence, we have a set of equations involving the roots and the numerator of (3.9) toeliminate these k unknowns. Some additional properties.
In this paragraph we derive some results that are usedthroughout this paper, particularly in Section 3.5 and in Appendix B. From a balancing ar-gument, we know that the fraction of time that the system is in a vacation is − ρ (where, inthis system, ρ = λ /µ ). Conversely, the fraction of time that the system is serving customersis ρ . Hence, k X h =1 e L (0) h (1) = λ µ , k + k X h = k +1 e L (0) h (1) = k e L (0) k + k (1) = 1 − λ µ . (A.5)16oreover, from (A.1) we know that e L (0) h (1) = e L (0) h − (1) for h = k + 2 , . . . , k + k . It followsthat e L (0) h (1) = 1 k (cid:18) − λ µ (cid:19) , h = k + 1 , . . . , k + k . From (A.3) and (A.4) we now have e L (0) h (1) = µ µ (cid:18) k (cid:18) − λ µ (cid:19) − π (0)0 ,k + k (cid:19) − h − X i =1 π (0)1 ,i , h = 1 , . . . , k . (A.6)The following relation for the unknowns π (0)0 ,k + k and π (0)1 ,h ( h = 1 , . . . , k − can be derivedby combining all of these results: k π (0)0 ,k + k + µ µ k − X h =1 ( k − h ) π (0)1 ,h = 1 k (cid:20) k (cid:18) − λ µ (cid:19) − k λ µ (cid:21) . (A.7)This relation turns out to be crucial to derive many results in this paper, without having toknow the exact expressions for all of the individual probabilities. Remark A.1
A balancing argument has been the starting point to derive all of the aboveproperties. A rigorous way to derive these results, is by using L’Hôpital’s rule on (3.9) todetermine e L (0) k + k (1) , and subsequently deriving an expression for P k + k h =1 e L (0) h (1) , which weknow is equal to one. B The second-order perturbed balance equations
The main goal of this appendix is to prove that (3.29a) can be written as the first term in(3.30). If we rearrange the summations slightly, we can write the equation that we need toprove as follows: µ λ µ k + k X h = k +1 e Q ′ (1) h (1 , ξ ) − µ (cid:18) − λ µ (cid:19) k X h =1 e Q ′ (1) h (1 , ξ ) = λ µ µ P ′′ ( ξ ) . (B.1)This equation should follow from Equations (3.16a)-(3.16h). In order to prove it, we take thefollowing steps:1. First, we take the generating functions of Equations (3.16a)-(3.16h) to develop relationsfor e Q ′ (1) h ( z, ξ ) ( h = 1 , . . . , k + k ) .2. The next step involves solving these equations to find an expression for e Q ′ (1) k + k ( z, ξ ) .3. Step 3 is to reformulate (B.1) in terms of e Q ′ (1) k + k (1 , ξ ) . It turns out that in this stage allterms containing probabilities φ ′ (1) n,h ( ξ ) (or their generating functions) are eliminated.4. The last step involves some more algebraic manipulations which eliminate all termscontaining probabilities φ ′′ (0) n,h ( ξ ) and, eventually, prove (B.1).17 tep 1: Find relations for the generating functions. Multiplying Equation (3.16h)with z n , summing over all n = 1 , , . . . , and adding Equation (3.16b), yields the followingequation: e Q (1) h ( z, ξ ) = e H ( z ) (cid:18) e Q (1) h − ( z, ξ ) + e Q ′ (0) h − ( z, ξ ) − (cid:18) − λ µ (cid:19) e Q ′ (0) h ( z, ξ ) (cid:19) , (B.2)for h = k + 2 , . . . , k + k , where e H ( z ) is defined in (3.8).Multiplying Equation (3.16g) with z n , summing over all n = 1 , , . . . , and adding Equation(3.16a), yields the following equation: e Q (1) k +1 ( z, ξ ) = e H ( z ) µ µ k − X h =1 φ (1)1 ,h ( ξ ) + φ (1)0 ,k + k ( ξ ) + µ zµ e Q (1) k ( z, ξ ) + φ ′ (0)0 ,k + k ( ξ ) − (cid:18) − λ µ (cid:19) e Q ′ (0) k +1 ( z, ξ ) (cid:19) . (B.3)Multiplying Equation (3.16f) with z n +1 , summing over all n = 1 , , . . . , and adding Equation(3.16d) multiplied by z , yields the following equation: e Q (1) h ( z, ξ ) = e G ( z ) z (cid:18) e Q (1) h − ( z, ξ ) − φ (1)1 ,h − ( ξ ) z − zµ µ (cid:18) − λ µ (cid:19) e Q ′ (0) h ( z, ξ ) (cid:19) , (B.4)for h = 2 , , . . . , k , where e G ( z ) is defined in (3.8).Multiplying Equation (3.16e) with z n +1 , summing over all n = 1 , , . . . , and adding Equation(3.16c) multiplied by z , yields the following equation: e Q (1)1 ( z, ξ ) = µ µ e G ( z ) (cid:18) e Q (1) k + k ( z, ξ ) − φ (1)0 ,k + k ( ξ ) + e Q ′ (0) k + k ( z, ξ ) − φ ′ (0)0 ,k + k ( ξ ) − (cid:18) − λ µ (cid:19) e Q ′ (0)1 ( z, ξ ) (cid:19) . (B.5) Step 2: Solve these relations and determine e Q ′ (1) k + k ( z, ξ ) . Solving Equations (B.2)-(B.5) requires a lot of straightforward, but tedious, computations. For reasons of compactnesswe will present some relevant intermediate results in this appendix, but leave the exact deriva-tions to the reader. Firstly, one can use (B.2), combined with (A.1), to express e Q (1) k + k ( z, ξ ) in terms of e Q (1) k +1 ( z, ξ ) and e Q ′ (0) k + k ( z, ξ ) : e Q (1) k + k ( z, ξ ) = e H ( z ) k − e Q (1) k +1 ( z, ξ ) + ( k − (cid:18) − (cid:18) − λ µ (cid:19) e H ( z ) (cid:19) e Q ′ (0) k + k ( z, ξ ) . (B.6)Second, we use (B.3) to express e Q (1) k +1 ( z, ξ ) in terms of e Q (1) k ( z, ξ ) and e Q ′ (0) k +1 ( z, ξ ) . Subse-quently, we use (B.4) and (A.3) to express e Q (1) k ( z, ξ ) in terms of e Q (1)1 ( z, ξ ) and e Q ′ (0) h ( z, ξ ) : e Q (1) k ( z, ξ ) = e G ( z ) z ! k − e Q (1)1 ( z, ξ ) − k − X h =1 φ (1)1 ,h ( ξ ) z e G ( z ) z ! k − h − zµ µ (cid:18) − λ µ (cid:19) k X h =2 e G ( z ) z ! k − h +1 e Q ′ (0) h ( z, ξ ) . (B.7)18inally, we use (B.5) to express e Q (1)1 ( z, ξ ) in terms of e Q (1) k + k ( z, ξ ) again. After some rear-rangement of the terms, this leads to the following expression for e Q (1) k + k ( z, ξ ) : e Q (1) k + k ( z, ξ ) = A (1) ( z, ξ ) + A ′ (0) ( z, ξ ) D ( z ) , (B.8)where A (1) ( z, ξ ) = e H ( z ) k " φ (1)0 ,k + k ( ξ ) (cid:18) − (cid:16) e G ( z ) /z (cid:17) k (cid:19) + µ µ k − X h =1 φ (1)1 ,h ( ξ ) (cid:18) − (cid:16) e G ( z ) /z (cid:17) k − h (cid:19) , (B.9) A ′ (0) ( z, ξ ) = k e Q ′ (0) k + k ( z, ξ ) − e H ( z ) k µ µ k − X h =1 φ ′ (0)1 ,h ( ξ ) (cid:18) − (cid:16) e G ( z ) /z (cid:17) k − h (cid:19) − (cid:18) − λ µ (cid:19) " e H ( z ) k k X h =1 e Q ′ (0) h ( z, ξ ) (cid:16) e G ( z ) /z (cid:17) k − h +1 + k e H ( z ) e Q ′ (0) k + k ( z, ξ ) , (B.10) D ( z ) = 1 − (cid:0) e G ( z ) /z (cid:1) k e H ( z ) k . (B.11)Note that A (1) ( z, ξ ) only contains the probabilities φ (1) n,h ( ξ ) . All probabilities φ ′ (0)0 ,k + k ( ξ ) , andtheir generating functions, are contained in A ′ (0) ( z, ξ ) . Also note that A (1) ( z,ξ ) D ( z ) is exactly thesame expression as (3.9), but with constants π (0) n,h replaced by φ (1) n,h . The reason is that, if onewould ignore the probabilities φ ′ (0) n,h ( ξ ) in the balance equations (3.16a)-(3.16h), the system iscompletely equivalent to the system (3.6a)-(3.6h), which corresponds to the vacation systemstudied in Appendix A. Step 3: reformulate the original problem in terms of e Q ′ (1) k + k (1 , ξ ) . In order to solve(B.1), we need to determine P k h =1 e Q ′ (1) h (1 , ξ ) and P k + k h = k +1 e Q ′ (1) h (1 , ξ ) . After substituting z = 1 in Equations (B.2)-(B.5), we can express these sums in terms of e Q ′ (1) k + k (1 , ξ ) and e Q ′′ (0) h (1 , ξ ) : k X h =1 e Q ′ (1) h (1 , ξ ) = k µ µ (cid:16) e Q ′ (1) k + k (1 , ξ ) − φ ′ (1)0 ,k + k (cid:17) − k − X h =1 ( k − h ) φ ′ (1)1 ,h + k e Q ′′ (0)1 (1 , ξ ) − µ µ (cid:18) − λ µ (cid:19) k X h =1 ( k − h + 1) e Q ′′ (0) h (1 , ξ ) , (B.12) k + k X h = k +1 e Q ′ (1) h (1 , ξ ) = k e Q ′ (1) k + k (1 , ξ ) − λ µ (cid:18) k − (cid:19) (cid:18) − λ µ (cid:19) P ′′ ( ξ ) . (B.13)Since e Q ′′ (0) h (1 , ξ ) for h = 1 , . . . , k can be determined directly using (A.6), it only remains todetermine e Q ′ (1) k + k (1 , ξ ) . Fortunately, according to the following lemma we can focus on thepart lim z → A ′ (0) ( z,ξ ) D ( z ) only. 19 emma B.1 The probabilities φ ′ (1)0 ,k + k and φ ′ (1)0 ,h ( h = 1 , . . . , k − in the left hand side ofEquation (B.1) cancel out. Using (B.12) and (B.13), we can express this statement in a moreformal presentation: lim z → µ λ µ k A (1) ( z, ξ ) D ( z ) ! − µ (cid:18) − λ µ (cid:19) k µ µ A (1) ( z, ξ ) D ( z ) − φ (1)0 ,k + k ! − k − X h =1 ( k − h ) φ (1)1 ,h ! = 0 . (B.14) Proof:
Using L’Hôpital’s rule it can be shown that lim z → A (1) ( z, ξ ) D ( z ) = (cid:16) − λ µ (cid:17) C (1) k (cid:16) − λ µ (cid:17) − k λ µ , (B.15)where C (1) = k φ (1)0 ,k + k + µ µ k − X h =1 ( k − h ) φ (1)1 ,h . After substituting (B.15) in (B.14), it is easily shown that the left hand side of (B.14) indeedequals zero. (cid:3)
Step 4: solve the original problem.
We are almost ready to solve Equation (B.1), butwe compute two helpful intermediate results first.
Lemma B.2
Define Y := lim z → ∂∂z e Q ′ (0) k + k ( z, ξ ) . This can be written as Y = (cid:20) µ µ (cid:18) k (cid:18) − λ µ (cid:19) − k λ µ (cid:19)(cid:21) − × "(cid:18) − λ µ (cid:19) k − X h =1 h ( k − h ) φ ′ (0)1 ,h + λ µ (cid:18) − λ µ (cid:19) ( k + 1) − λ µ µ (cid:18) − λ µ (cid:19) ( k −
1) + 2 (cid:18) λ µ (cid:19) ! P ′ ( ξ ) . (B.16) Proof:
Equation (B.16) follows after differentiating (3.9) with respect to z , and subsequently applyingL’Hôpital’s rule twice. Obviously, some basic algebraic manipulations are required to obtainthe presentation in (B.16). (cid:3) emma B.3 Define X := lim z → A ′ (0) ( z,ξ ) D ( z ) . This can be written as X = − µ µ Y + (cid:20) k (cid:18) − λ µ (cid:19) − k λ µ (cid:21) − × (cid:26) φ ′ (0)0 ,k + k (cid:18) k (cid:18) − λ µ (cid:19) (cid:18) k λ µ + (cid:18) λ µ − k (cid:18) − λ µ (cid:19)(cid:19) µ µ (cid:19)(cid:19) + (cid:18) − λ µ (cid:19) (cid:18) k λ µ + λ µ − µ µ − k (cid:18) − λ µ (cid:19)(cid:19) k − X h =1 ( k − h ) φ ′ (0)1 ,h + (cid:18) − λ µ (cid:19) k − X h =1 h ( k − h ) φ ′ (0)1 ,h − (cid:18) − λ µ (cid:19) (cid:18) k k (cid:18) λ µ − k (cid:18) − λ µ (cid:19)(cid:19) µ µ + λ µ (cid:18) µ µ + k (cid:19)(cid:19) P ′ ( ξ ) ) . (B.17) Proof:
This equation follows from applying L’Hôpital’s rule to A ′ (0) ( z,ξ ) D ( z ) and, hence, differentiating(B.10) and (B.11) with respect to z . Substitution of z = 1 gives the desired result after, again,many algebraic manipulations. (cid:3) Finally, we are ready to present the main result of this appendix, which is the proof of Equation(B.1). Using Lemma B.1, and Equations (B.12) and (B.13), we can write the left hand sideof Equation (B.1) as µ λ µ (cid:18) k X ′ − λ µ (cid:18) k − (cid:19) (cid:18) − λ µ (cid:19) P ′′ ( ξ ) (cid:19) − µ (cid:18) − λ µ (cid:19) " k µ µ X ′ + k e Q ′′ (0)1 (1 , ξ ) − µ µ (cid:18) − λ µ (cid:19) k X h =1 ( k − h + 1) e Q ′′ (0) h (1 , ξ ) , where X ′ is the derivative of X with respect to ξ . Using Lemma B.3 and Lemma B.2 (andEquation (A.6) to determine e Q ′′ (0) h (1 , ξ ) for h = 1 , . . . , k ) we can show that the above expres-sion reduces to λ µ µ P ′′ ( ξ ) . References [1] I. J. B. F. Adan, J. S. H. van Leeuwaarden, and E. M. M. Winands. On the application ofRouché’s theorem in queueing theory.
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