Queueing Networks and Insensitivity

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Lukáš Adam 29. 10. 2012 1 / 40

Table of contents 1 Jackson networks 2 Insensitivity in Erlang s Loss System 3 Quasi-Reversibility and Single-Node Symmetric Queues 4 Quasi-Reversibility in Networks 5 The Arrival Theorem 2 / 40

Reminder Birth death process M/M/1 One type of customer One node 3 / 40

Jackson networks Series of K nodes, not necessarily linear. Arrivals from external sources are independent Poisson processes with intensities α 1,..., α K. A customer having completed service at node k goes to node l with probability γ kl and leaves the system with probability γ k0. A single exponential server at each node with corresponding service rates δ 1,..., δ K. 4 / 40

Closed and open networks Closed network No external inflow of outflow of customers. The number of customers in the network is constant. α k = 0, γ k0 = 0. Open network Opposite to closed network. At least one external inflow α k is nonzero. At least one probability of external outflow γ k0 is nonzero. 5 / 40

Examples 6 / 40

Proposition Some propositions Any ergodic birth-death process is time reversible. Definition Consider now a doubly infinite stationary version {X t } <t< and define X t = X t = lim s t X s. Then a departure for {X t } at time s corresponds to an arrival for X t at time s. Proposition If the queue is either ergodic birth-death queue with Poisson arrivals or stationary M/G/ queue, then the departure is a Poisson process. 7 / 40

Arrivals are Poisson by assumption, thus the departures are Poisson as well. Departures from a node form arrivals to another node. Denote X (k) t number of customers at node k at time t, X t = (X (1) t,..., X (K) t ) and various states n = n 1... n K. Possible evolution of n n n kl n ( ) k n (+) k with intensities on the right hand side. δ k γ kl δ k γ k0 α k 8 / 40

Traffic equations Throughput rate β k, the common rate of the input and output processes. The input rate is the sum of rate from external arrivals and arrivals from other nodes. The external rate is α k. Rate from node l to k is equal to β l γ lk. Traffic equations β k = α k + K β l γ lk. l=1 9 / 40

Open networks Assumptions Every node k may receive external input, thus either α k > 0 or α k1 γ k1k 2... γ kn 1k n γ knk > 0. Every node k may produce external ouput, thus either γ k0 > 0 or γ kk1... γ kn 1k n γ kn0 > 0 These conditions imply irreducibility. Proposition Traffic equations have unique nonnegative solution, which is moreover positive. 10 / 40

Theorem Assume that ρ k = β k δ k < 1 for all k. Then {X t } is ergodic with stationary distribution K π n = π n1...n K = (1 ρ k )ρ n k k=1 11 / 40

Closed networks Constant state space E = {n; k n k = N}. Traffic equations simplify to β = βγ, where Γ = (γ kl ) kl 0. Theorem If Γ is irreducible, then {X t } is ergodic with stationary distribution π N = K k=1 ρ n k k. 12 / 40

Limiting state for closed networks Assume ρ 1 = 1, ρ 2 < 1,..., ρ K < 1. Define η (N) n = P e (X (1) t = n) = n 2+ +n K =N n π nn2...n K θ n (N) 2...n K = P e (X (2) t = n 2,..., X (K) t = n K ) = π n1n 2...n K Theorem Taking the limit N we obtain η (N) n 0 K (1 ρ k )ρ n k θ (N) n 2...n K k=2 k. 13 / 40

Example For high enough N to maximize θ n (N) 2...n K hence to minimize n k. means to maximize ρ n k k, In the limiting case the longest queue should be at the node with highest ρ k. Example with K = 3, ρ 1 = 1, ρ 2 = ρ 3 = 1 2. k=0 1 2 k+1 (k + 1) = 1. 1 0 4 1 1 8 1 2 16 1 3 32 P e (X (1) t N 3) 13 16. 0 1 2 3 1 8 1 16 1 32 1 16 1 32 1 32 14 / 40

Erlang s Loss System K lines with intensity of arrivals β. Duration of a call follows a phasetype distribution B with p phases, initial vector α and phase generator T. Then the exit rate vector is t = T 1. State space E = {i = n 1... n p ; n k {0,..., K}, i = n 1 + + n p K}, where n k gives the number of lines where the call is currently being handled in phase k. Possible evolution of i i i rs i r ( ) i (+) r n r t rs n r t r βα r 15 / 40

Loss probability Theorem Let µ B = αt 1 1 denote the mean of B and let η = βµ B. Then E K = n 1 + +n p=k π n1...n p = η K K!. 1 + + ηk K! 16 / 40

Quasi-Reversibility and Single-Node Symmetric Queues Simple queue (K = 1) with multiple customer classes c C, which is of finite or countable number. Assumptions The time evolution of the queue can be modelled by an ergodic Markov process {X t } with a finite or countable state space E. There are A c, D c E E such that i j when ij A c D c and that A c D c =. Interpretation X t = i, X t = j corresponds to an arrival of a customer of class c when ij A c and to a departure when ij D c. Usually ij A c if and only if ji D c. 17 / 40

Example State space E = {i = c 1... c n ; n N, c k C}. X t = c 1... c n means that there are n customers in the system. Customer of the type c 1 is being served, the first one in the queue is of type c 2 and so on. ij A c if and only if i = c 1... c n and j = c 1... c n c. Similar for ij D c. 18 / 40

Quasi reversibility Definition Define random sets N (+) c (t) = {s t; X s X s A c } the arrival process of class c customers after time t and N ( ) c (t) = {0 s t; X s X s D c } the departure prior to time t. Then the the queue is called quasi-reversible if X t and all N c (+) (t) and N c ( ) (t) are independent in the steady state for any t 0. 19 / 40

Proposition For a stationary quasi reversible queue, the arrival processes N c (+) (0), c C are independent Poisson processes, and so are the departure processes N c ( ) ( ), c C. Proposition The timereverse { X (t)} of a quasi reversible queue is itself quasi reversible corresponding to à c = {ij; ji D c } and D c = {ij; ji A c }. 20 / 40

Conditions for quasi reversibility Proposition For a stationary quasi reversible queue, the rates µ (+) c N c (+) (0) and N c ( ) ( ) are given by µ (+) c = j;ij A c λ(i, j) µ ( ) c = 1 π j i;ij D c π i λ(i, j). and µ ( ) c of In particular, the first right hand side does not depend on i and the second one on j. Proposition If the nondepandance of the right hand sides is fulfilled, then the queue is quasi reversible. 21 / 40

Example Consider a single classed queue modelled by a birth death process. A = {01, 12,... }, B = {10, 21,... }. In order for the queue to be quasi reversible it is necessary and sufficent that β N = β. j;ij A c λ(i, j) = β i 1 π j i;ij D c π i λ(i, j) = 1 π j π j+1 δ j+1 and use local balance equation π j β j = π j+1 δ j+1 to obtain only β i. 22 / 40

Example: multiclass M/M/1 queue Let β = c C β c denote the overall arrival rate, assume tha β < and δ c = δ and denote by p c = βc β the probability that an arriving customer is of class c and the traffic intensity is ρ = β δ. The goal is to corfirm that π c1...c n = (1 ρ)ρ n p c1... p cn is a stationary distribution and based on this to prove the quasi reversibility of the queue. Lemma (Kelly) Let Λ and Λ be nonexplosive intensity matrices and π a distribution such that π i λ(i, j) = π j λ (j, i). Then Λ and Λ are ergodic with stationary distribution π for both, and further Λ coincides with the intensity matrix of the time-reversed process { X t }. 23 / 40

Λ has only two positive members on every row λ(c 1... c n c, c 1... c n ) = δ λ(c 1... c n, cc 1... c n ) = β c We need to verify that π i λ(i, j) = π j λ(j, i). It is necessary to check two cases. We will show only the first one The stationarity is proven. π i λ(i, j) π j λ(j, i) = π iβ c pi j δ = 1 ρp c β c δ = β ρδ = 1. 24 / 40

For the quasi reversibility consider again 1 π j π i λ(i, j) = 1 π j+c δ = ρ p c δ = ρδ = β π j i;ij D c c C c C And the second condition j;ij A c λ(i, j) = β. Hence the system is quasi reversible by one of the previous propositions. 25 / 40

More general model Assume that different customer classes have independent Poisson arrivals with rate β c and let β = c C β c denote the overall arrival rate and assume β <. Assume further that class c customers have a phase type service time distribution B c with initial vector α c = (α cr ) r=1,...,pc and phase generator T c = (t crs ) Then the exit rate vector is t c = T c 1 = (t cr ), the mean service time is m c = α c Tc 1 1 and the overall traffic intensity is ρ = c C ρ c, where ρ c = β c m c. Define ν c as the equilibrium distribution of the phase of service of a class c customer and p cr as the probability that a randomly selected customer in service is of class c and in phase r of service ν c = (ν cr ) = α ctc 1 m c, p cr = β cm c ρ ν cr 26 / 40

A state of the system has the form i = c 1 r 1... c n r n, which means, where are n = n(i) customers. Customer k is of class c k and in phase r k. Server works ar rate φ(n) depending on the number of customers in the system, devoting ω(n, k) of its capacity to customer k. Thus ω(n, k) = 1. k Customers arriving to a system with n customers take position k {1,..., n + 1} with probability ω(n + 1, k) independently of their class. Upon their arrival, their current phase is chosen based on α c. Thus, ω has two functions, both position allocation and service allocation. This coined the term symmetric to this type of queue. 27 / 40

Main theorem Theorem Let Φ(n) = n φ(k), δ = k=1 n=0 ρ n Φ(n). If δ <, then the symmetric multiclass M/PH/ queue is ergodic and the stationary distribution is given by ρn π i = π c1 r 1...c nr n = δ 1 Φ(n) n p ck r k. k=1 Further the queue is quasi-reversible with µ (+) c = µ ( ) c. 28 / 40

Proof of the theorem Similar as in the multiclass M/M/1 queue. First it is necessary to compute transition matrix, the rightmost column giving the intensities i(k, s) φ(n)ω(n, k)t ck r k s i = c 1 r 1... c n r n i(k) φ(n)ω(n, k)t ck r k i(k, c, r) β c ω(n + 1, k)α cr. First case: customer k changes phase from r k from s. Second case: customer k leaves the queue. Third case: a new customer of class c in phase r arrives at position k. For arrivals and departures we get A c = {ij; j = i(k, c, r) for some k, c, r} D c = {ij; j = i(k) for some k}. 29 / 40

Lemma Sufficient criteria for a process to be nonexplosive are either of 1 sup i E λ(i) < 2 E is finite 3 {Y n } is recurrent. By the previous lemma we obtain nonexplosivity and may use Kelly s lemma. Hence, we need to prove that π i λ(i, j) = π j λ(j, i). 30 / 40

The reversed parameters are α cr = m c t cr ν cr t crs = ν cst csr t cr = ν cr α cr m c ν cr, which was shown in the previous presentation (III.5.7). Again, we need to check three cases, j = i(k, s), j = i(k) and j = i(k, c, r). First case Second case π i λ(i, j) π j λ(j, i) = ν c k r k t ck r k s = 1. ν ck s t ck sr k π i λ(i, j) π j λ(j, i) = ρ c k ν ck r k φ(n)ω(n, k)t ck r k = ρ c k = 1. φ(n) β ck ω(n, k) α ck r k β ck m k 31 / 40

Third case follows from the second one by interchanging the given system and the tilded one. We verified that π i is a stationary distribution. To verify quasi reversibility one could compute λ(i, j) = λ(i, i(k, c, s)) = = β j;ij A c k,c,s By symmetry the same relationship holds for tilded system, which is the same as the second condtition, which needs to be verified. 32 / 40

Examples Multiclass preemptive LIFO M/PH/1 queues. This corresponds to φ(n) = 1, ω(n, 1) = 1, all other ω(n, k) = 0. Multiclass PS (processor sharing) M/PH/1 queues. This corresponds to φ(n) = 1 and all ω(n, k) = 1 n. Multiclass M/PH/ queues. This corresponds to φ(n) = n and all ω(n, k) = 1 n. 33 / 40

Quasi-Reversibility in Networks Similar to Jackson networks with the addition of customer classes. K nodes, E (k) state space, A (k) c and D c (k) disjoint classes of transitions that we think of as arrivals and departures. Arrivals at node k are independent Poisson with intensities α kc. Probability of departures to different nodes are γ kc,ld, hence the customer class may change, and the probability of leaving the system is γ kc,0 = 1 l,d γ kc,ld. 34 / 40

Example Example, which shows the superiority over Jackson networks. Customer wants to select one of the possible routes R 1, R 2,... with probability q(r r ). After choosing, the class is the set of nodes c = l 1... l t to visit. Upon leaving the first node, the customer class changes to reflect this and becomes only l 2... l t, thus otherwise γ kc,ld = 0. γ l1(l 1...l t),l 2(l 2...l t) = 1, 35 / 40

Traffic equations Traffic equations are similar to the Jackson network case β kc = α kc + l,d β ld γ ld,kc. Assume existence, uniqueness is not required. Assume further that the ergodic intensity matrix (λ (k) (i, j)) ij with stationary distribution say (π (k) i ) i E (k), such that the corresponding Markov process on E (k) is quasi-reversible. The state space E = K k=1 E (k). 36 / 40

Stationary distribution The derivation method is the same as in previous cases. There are four possible cases, which may happen New customer arrives to the system. A customer leaves the system. A customer changes the node, but keeps his class. A customer changes the node and his class. Theorem The network of quasi-reversible nodes is ergodic with stationary distribution of product form π i = π i1...i K = π (1) i 1... π (K) i K. 37 / 40

PASTA property In a singlenode queue with Poisson arrivals, the PASTA property to be formalized in VII.6 implies that the steady-state distribution of the state of the queue seen by a customer just before his arrival is the same as the steady-state distribution of the state of the queue at an arbitrary point of time. We will show here that results of rather similar form hold for networks. In the network we have considered, a customer going to node l after being served at node k does so instantaneously. However, we may imagine that he observes the state of the rest of the network during the infinitesimal interval where the transition takes place. Typically, these transition instants do not form a Poisson process so PASTA does not apply. Nevertheless, we will show here that results of rather similar form hold. 38 / 40

Theorems Theorem In the model of previous section, the steadystate distribution η kc,ld of the state seen by a customer in transition from node k as a class c customer to node l as a class d customer coincides with the time-stationary distribution π. Proposition Let H = {kc} and K = {ld} be arbitrary subsets of {0,..., K} C. Then the steady-state distribution of a H customer in transition to become a K customer is π. 39 / 40

Theorem For a closed Jackson network with N customers, the steady-state distribution η N;kl of the state seen by a customer in transition from node k to node l coincides with the time-stationary distribution π N 1 of the network with one customer removed. That is, η N;kl n = C N 1 K m=1 ρ nm m when n 1 + + n K = N 1. 40 / 40