Stochastic models in product form: the (E)RCAT methodology

Transcription

1 Stochastic models in product form: the (E)RCAT methodology 1 Maria Grazia Vigliotti 2 1 Dipartimento di Informatica Università Ca Foscari di Venezia 2 Department of Computing Imperial College London

2 Second part: sketch 1 Introduction to the Extended Reversed Compound Agent Theorem (ERCAT) [Harrison 04a] 2 Applications for cooperations of pairs of automata which do not yield structural conditions of RCAT are shown 3 Special attention is devoted to queueing networks with finite and

3 Part II Extended Reversed Compound Agent Theorem (ERCAT)

4 1 by 2 3 Solution of the running 4 Open networks of exponential with finite and 5 6

5 A system in Boucherie s product-form λ1 Q1 λ2 Q2 µ1 µ2 Two exponential Q 1 and Q 2 with independent Poisson arrival streams with rate λ 1 and λ 2 Service rates are µ 1 and µ 2 If one of the enters in state 0 the other one is blocked (i.e. no arrivals or service completions occur) The model is known to be in Boucherie s product-form

6 (d,x d) Process representation (d,x d) (c,λ 1) (c,λ 1) (c,λ 1) Q (a,µ 1) (b,x b) (c,x c) (a,µ 1) (b,x b) (c,x c) (a,µ 1) (d,λ 2) (d,λ 2) (d,λ 2) Q (b,µ 2) (b,µ 2) (b,µ 2) (a,x a) (a,x a) Are (G)RCAT structural conditions satisfied? NO!

7 (d,x d) Process representation (d,x d) (c,λ 1) (c,λ 1) (c,λ 1) Q (a,µ 1) (b,x b) (c,x c) (a,µ 1) (b,x b) (c,x c) (a,µ 1) (d,λ 2) (d,λ 2) (d,λ 2) Q (b,µ 2) (b,µ 2) (b,µ 2) (a,x a) (a,x a) Are (G)RCAT structural conditions satisfied? NO!

8 A remark about reversed rates Consider two states n i and n j of cooperating automaton S Assume that there exists an active transition from n i to n j labelled by a with rate γ and that in n j this is the only incoming transition with this label (G)RCAT requires to compute the value of: K a = π(n i) π(n j ) γ K a may be interpreted as the rate of the transition from n j to n i in the reversed process of S We refer to K a as the reversed rate of transitions labelled by a

9 Joint state space ERCAT requires to check a rate equation for each state of the irreducible subset of the joint process Often, states can be opportunely clustered and hence the computation becomes feasible The computational complexity is higher than the standard (G)RCAT Let (s 1, s 2 ) be a state of the irreducible subset of the joint process

10 Fundamental definitions P (s 1,s 2 ) : outgoing passive labels from s 1 or s 2 P (s 1,s 2 ) : incoming passive labels into s 1 or s 2 A (s 1,s 2 ) : outgoing active labels from s 1 or s 2 A (s 1,s 2 ) : incoming active labels into s 1 or s 2 α (s 1,s 2 ) (a): rate of active transition labelled by a outgoing from (s 1, s 2 ) β (s 1,s 2 ) (a): reversed rate of the passive transition labelled by a incoming into (s 1, s 2 )

11 Theorem (ERCAT) ERCAT formulation Given two models Q 1 and Q 2 in which RCAT structural conditions are not satisfied but the reversed rates of the active transitions are constant, their cooperation is in product-form if the following rate equation is satisfied for each state (s 1, s 2 ) of the irreducible subset of states of the joint process: a P (s 1,s 2 ) x a = x a a A (s 1,s 2 ) (s β 1,s 2 ) a a P (s 1,s 2 ) A (s 1,s 2 ) α (s1,s2) a a A (s 1,s 2 ) P (s 1,s 2 )

12 1 by 2 3 Solution of the running 4 Open networks of exponential with finite and 5 6

13 Q 1 0 Q 2 (c,λ 1) (d,λ 2) (a,µ 1) P (0,0) = {} A (0,0) = {a, b} 0 (b,µ 2) State (0,0) P (0,0) A (0,0) = {} A (0,0) P (0,0) = {c, d} Note that: x a x b = α (0,0) c α (0,0) d Ok x a = λ 1, α (0,0) c = λ 1, x b = λ 2, α (0,0) d = λ 2

14 Q 1 0 Q 2 (c,λ 1) (d,λ 2) (a,µ 1) P (0,0) = {} A (0,0) = {a, b} 0 (b,µ 2) State (0,0) P (0,0) A (0,0) = {} A (0,0) P (0,0) = {c, d} Note that: x a x b = α (0,0) c α (0,0) d Ok x a = λ 1, α (0,0) c = λ 1, x b = λ 2, α (0,0) d = λ 2

15 (c,x c) State (0,n), n>0 (c,λ 1) (d,λ 2) (d,λ 2) Q 1 0 n Q 2 (a,µ 1) (b,µ 2) (b,µ 2) (a,x a) P (0,n) = {a, c} A (0,n) = {a, b, d} P (0,n) A (0,n) = {c} A (0,n) P (0,n) = {b, d} x a + x c x a x b x d = β (0,n) c Note that: x b = λ 2, x c = µ 1, x d = µ 2, β (0,n) c α (0,n) b = µ 1, α (0,n) b α (0,n) d Ok! = µ 2, α (0,n) d = λ 2

16 (c,x c) State (0,n), n>0 (c,λ 1) (d,λ 2) (d,λ 2) Q 1 0 n Q 2 (a,µ 1) (b,µ 2) (b,µ 2) (a,x a) P (0,n) = {a, c} A (0,n) = {a, b, d} P (0,n) A (0,n) = {c} A (0,n) P (0,n) = {b, d} x a + x c x a x b x d = β (0,n) c Note that: x b = λ 2, x c = µ 1, x d = µ 2, β (0,n) c α (0,n) b = µ 1, α (0,n) b α (0,n) d Ok! = µ 2, α (0,n) d = λ 2

17 0=0 State (m,n), m,n>0 (d,x d) (c,x c) (c,λ 1) (c,λ 1) (d,λ 2) (d,λ 2) m Q 1 n Q 2 (a,µ 1) (a,µ 1) (b,µ 2) (b,µ 2) (b,x b) (a,x a) P (m,n) = {a, b, c, d} A (m,n) = {a, b, c, d} P (m,n) A (m,n) = {} A (m,n) P (m,n) = {} Note that states (m, 0) with m > 0 are similar to (0, n), n > 0.

18 of the running The model, as expected, is in product-form: π(m, n) ( λ1 µ 1 ) m ( λ2 Note that state (0, 0) is either the only ergodic state or does not belong to the irreducible subset Hence, the normalising constant distinguishes this solution from the case of independent Every Boucherie s product-form with full can be studied by ERCAT [Harrison 04a] µ 2 ) n

19 1 by 2 3 Solution of the running 4 Open networks of exponential with finite and 5 6

20 Queues with finite and Repetitive Service (RS) We consider a network of, Q 1,..., Q N with finite B i and service rate µ i At a job completion at Q i the customer goes to Q j with probability P ij. If Q j if saturated the customer service is restarted and a new target station is selected at job completion In open networks λ i is the arrival rate at Q i and leave the system with probability 1 j P ij. Arrivals at saturated are not allowed

21 Q 1 Q 2 λ 1 p 1 λ 2 µ 1 µ 2 p 1 q q Example Q1 0 λ1 (a, qµ2) (b, xb) λ1 (a, qµ2) 1 B1 (b, xb) (1 p)µ1 (1 p)µ1 λ2 λ2 (b, pµ1) (b, pµ1) Q2 0 1 B2 (a, xa) (a, xa) (1 q)µ2 (1 q)µ2

22 Notes Differently from ordinary queueing networks we use active transitions to model synchronised arrivals and passive to model synchronised departures Which states shall we consider? 1 (0, 0) 2 (0, K ) with 0 < K < B 2 (and symmetrically we obtain (K, 0) with 0 < K < B 1 ) 3 (0, B 2 ) 4 (K, B 2 ) with 0 < K < B 1 (and symmetrically we obtain (0, K ) with 0 < K < B 2 ) 5 (B 1, B 2 ) Note that α (, ) a = qµ 2, α (, ) b = pµ 1 and also β (, ) a = β a and β (, ) b = β b

23 State (0, B 2 ) λ 1 (a,qµ 2 ) λ2 (b,pµ 1 ) Q 1 0 Q 2 B 2 (b,x b ) (a,x a ) (1 p)µ 1 (1 q)µ 2 P (0,B2) = {a} A (0,B2) = {b} A (0,B2) P (0,B2) = {} P (0,B2) A (0,B2) = {} x a x b = 0 x a = x b (1)

24 λ 1 λ2 (a,qµ 2 ) (b,pµ 1 ) Q 1 0 Q 2 K (b,x b ) (a,x a ) (1 p)µ 1 (1 q)µ 2 State (0, K ) (b,pµ 1 ) (b,pµ 1 ) (a,x a ) (1 q)µ 2 P (0,K ) = {a} A (0,K ) P (0,K ) = {b} A (0,K ) = {b} P (0,K ) A (0,K ) = {a} i.e.: x a x b = β (0,K ) b α (0,K ) a (1) β b = α a (2) By symmetry, state (K, 0) gives β a = α b

25 State (0, 0) Q 1 0 Q 2 λ 1 (a,qµ 2) (b,x b) (1 p)µ 1 P (0,0) = {} A (0,0) = {} 0 λ2 (b,pµ 1) (a,x a) (1 q)µ 2 A (0,0) P (0,0) = {a, b} P (0,0) A (0,0) = {a, b} which is a consequence of (2) β (0,0) a + β (0,0) b = α (0,0) a + α (0,0) b

26 States (K, B 2 ) and (B 1, K ), (B 1, B 2 ) λ 1 λ 1 λ 2 (a,qµ 2 ) (a,qµ 2 ) (b,pµ 1 ) Q 1 K Q 2 B 2 (b,x b ) (1 p)µ 1 (b,x b ) (1 p)µ 1 For these states we have: P (, ) = {a, b} A (, ) = {a, b} (a,x a ) (1 q)µ 2 Since all the synchronising labels are present in both these sets, the rate equation for these states is an identity.

27 Conditions derived from the ERCAT rate equations The process analysis gives: x a = x b β b = α a = qµ 2 β a = α b = pµ 1 β b = x b(λ 1 + qµ 2 ) x b + (1 p)µ 1 β a = x a(λ 2 + pµ 1 ) x a + (1 q)µ 2 From which we straightforwardly derive: x a = (1 q)pµ 1µ 2 λ 2 x b = (1 p)qµ 1µ 2 λ 1 (3)

28 Product-form rate condition Since x a = x b by (1) we have the product-form rate condition: (1 p)qλ 2 = (1 q)pλ 1 Under this assumption expressions (3) for x a, x b satisfies: x a = (x b + (1 p)µ 1 )qµ 2 λ 1 + qµ 2 x b = (x a + (1 q)µ 2 )pµ 1 λ 2 + pµ 1 λ1 (a,qµ2) λ1 (a,qµ2) Q1 0 1 B1 (b,xb) (1 p)µ1 λ2 (b,pµ1) (b,xb) (1 p)µ1 λ2 (b,pµ1) Q2 0 1 B2 (a,xa) (1 q)µ2 (a,xa) (1 q)µ2

29 Generalisation ERCAT may be applied to a set of agent with pairwise cooperations (this is also known a MARCAT) In case of QN with RS and general topology in [Balsamo et al. 10] is proved that: Theorem A QN (open or closed) with finite stations and RS policy with reversible routing matrix always satisfies ERCAT rate equations. Product-form for reversible routing has been proved in [Akyildiz 87]

30 Closed QN with RS Consider a closed QN with RS policy Note that the ERCAT rate equation is an identity for state n when none of the stations is empty in n We immediately have the following result: Theorem (QN with strict non-empty condition) A closed QN with finite stations and RS is in product-form if the number of is such that none of the station can be empty (strict non-empty condition) In [Balsamo et al. 10] we prove that the same result for QN in which at most one station can be empty (non-empty condition)

31 1 by 2 3 Solution of the running 4 Open networks of exponential with finite and 5 6

32 Q 1 Q 2 λ p 10 1 λ 2 p 20 - µ 1 µ 2 p + 12 Exp. service times with parameters µ 1 and µ 2 Independent Poisson arrival processes A customer leaving Q 2 may: p + 21 Model description p 21 p 10 + p + 12 = 1 p p 21 +p 20 = 1 leave the system with pr. p 20 enter Q 1 as a standard customer with pr. p + 21 delete a customer in Q 1 with pr. p 21 While Q 2 is saturated, from Q 1 are deleted with rate ɛ

33 Underlying processes (a 21,x 21 ) λ 1 (a+ 21,µ 2p + 21 ) λ 1 (a + 21,µ 2p + 21 ) 0 1 B 1 Q 1 µ 1 p 10 (a+ 12,x+ 12 ) (a 21,x 21 ) µ 1 p 10 (a + 12,x+ 12 ) (a 21,x 21 ) λ 2 (a+ 12,µ 1p + 12 ) λ 2 (a + 12,µ 1p + 12 ) (a 21,ǫ) 0 1 B 2 Q 2 µ 2 p 20 (a+ 21,x+ 21 ) (a 21,µ 2p 21 ) µ 2 p 20 (a + 21,x+ 21 ) (a 21,µ 2p 21 )

34 Observe that label a 21 does not change ERCAT rate condition: a 21 exits from every state of Q 1 (passive): a 21 P(s 1,s 2 ) 0 s 1 B 1, 0 s 2 B 2 a 21 enters into every state of Q 2 (active): a 21 A(s 1,s 2 ) 0 s 1 B 1, 0 s 2 B 2 ERCAT rate condition: x a a P (s 1,s 2 ) = x a a A (s 1,s 2 ) (s β 1,s 2 ) a a P (s 1,s 2 ) A (s 1,s 2 ) Key-idea α (s1,s2) a a A (s 1,s 2 ) P (s 1,s 2 )

35 We still have x + 12 = x + 21, where: x + 12 = x + 21 = x 21 = Product-form conditions µ 1 p + 12 λ 2 + µ 1 p + (µ 2 p 20 + x µ 2p + 21 ) 12 µ 2 p + 21 λ 1 + µ 2 p + (µ 1 p 10 + x x 21 ) 21 µ 2 p 21 µ 2 p 21 + x µ 2p 20 (λ 2 + µ 1 p + 12 ) After some algebra we derive the condition: ( λ 1 p + 12 (1 p 21 ) = λ 2p + 21 p 10 + λ 2 µ 1 p 21 1 p + 21 )

36 Explicit expressions of x + 12, x + 21, x 21, ɛ We derive the expressions for x + 21 = x + 12 and x 21 : x + 21 = µ 2p + 21 λ 1 ( x 21 = λ 2p 21 1 p + 21 µ 1 p 10 + λ 2p 21 1 p + 21 ) Constant reverse rates of the active transitions: ɛ = x 21

37 Product-form expression Product-form expression: ( λ1 + µ 2 p + 21 π(n 1, n 2 ) µ 1 p 10 + x x 21 ( ) n1 λ 2 + µ 1 p + 12 µ 2 p 20 + x µ 2p 21 ) n2 for 0 n 1 B 1 and 0 n 2 B 2

38 1 by 2 3 Solution of the running 4 Open networks of exponential with finite and 5 6

39 Final remarks In the second part of the tutorial we have shown how to overcome some limitations of original RCAT and GRCAT formulation In case structural conditions of (G)RCAT are not satisfy we may apply ERCAT Application of (G)RCAT or ERCAT may be done algorithmically, however the computational cost of ERCAT is higher than that of (G)RCAT

40 Applications Other models than those presented here may be studied by RCAT and its extensions (e.g. product-form Stochastic Petri Nets) New product-form may be derived The solution of the traffic equations may be efficiently computed by means of the algorithm presented in [Marin et al. 09] Numerical and iterative algorithm Product-form of models expressed in terms of different formalisms may be derived.

41 Appendix: Reversible routing matrix Consider a queueing network with N stations and fixed routing probability matrix P = [p ij ], 1 i, j N p i0 is the probability of leaving the network after a job completion at station i e i is the (relative) visit ratio to station i λ i is the arrival rate at station i Definition (Reversible routing matrix) The routing matrix P is said reversible if: { e i p ij = e j p ji for 1 i, j N λ i = e i p i0 for 1 i N

42 For Further Reading I B. Pittel:Closed exponential networks of with saturation: The Jackson-type stationary distribution and its asymptotic analysis, Math. of Op. Res., vol. 4, n. 4, pp , 1979 I.F. Akyildiz: Exact product form solution for queueing networks with, IEEE Trans. on Computers, vol. C-36-1, pp , 1987 P.G. Harrison: Turning back time in Markovian process algebra, Theoretical Computer Science, vol. 290, n. 3, pp , 2003

43 For Further Reading II P.G. Harrison: Reversed processes, product forms and a non-product form, Linear Algebra and its App., vol. 386, pp , P.G. Harrison: Compositional reversed Markov processes, with applications to G-networks, Perf. Eval., vol. 57, n. 3, pp , 2004 A. Marin and S. R. Bulò: A general algorithm to compute the steady-state solution of product-form cooperating Markov chains, in Proc. of MASCOTS 2009, pp , 2009

44 For Further Reading III A. Marin, M.G. Vigliotti: A general result for deriving product-form solutions of Markovian models, Proc. of WOSP/SIPEW Int. Conf. on Perf. Eval. S. Balsamo, P. G. Harrison, A. Marin: A unifying approach to product-forms in networks with finite constraints, Proc. of the 2010 ACM SIGMETRICS, pp , 2010