Discrete-Event Control of Stochastic Networks: by Eitan Altman

By Eitan Altman

Opening new instructions in learn in either discrete occasion dynamic platforms in addition to in stochastic regulate, this quantity specializes in a large category of keep an eye on and of optimization difficulties over sequences of integer numbers. this can be a counterpart of convex optimization within the atmosphere of discrete optimization. the idea built is utilized to the keep an eye on of stochastic discrete-event dynamic structures. a few purposes are admission, routing, carrier allocation and holiday regulate in queueing networks. natural and utilized mathematicians will get pleasure from studying the ebook because it brings jointly many disciplines in arithmetic: combinatorics, stochastic methods, stochastic keep an eye on and optimization, discrete occasion dynamic platforms, algebra.

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Additional info for Discrete-Event Control of Stochastic Networks: Multimodularity and Regularity

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This implies the existence of vk−1 ak vk−1 vk−1 ak vk−1 . Now, we finish the proof by noticing that the letter aK is surrounded by vK−1 and by noting that vK−1 vK−1 is necessarily surrounded by “aK ”. Proposition 16. The projection w of a sequence w over the alphabet A−{a} is w where all a’s have been removed. 5, then w is balanced implies that w is balanced. 52 2 Balanced Sequences Proof. Choose two words v1 , v2 of length n in w . Let v1 and v2 be any two words in w whose projections over the alphabet A − {a} are v1 and v2 , respectively.

Note that W has the following set of rates, (p1 /k, · · · , p1 /k, p2 , · · · , pK ). Next, we show that W is balanced. Since W is balanced, for an arbitrary integer m, the number of “a1 ”s in an interval of length m is n or n + 1, for some n. Now, for W , the number of “bi ”s in such an interval is either (n − 1)/k or (n + 1)/k . This proves that W is balanced. For the general case and distinct rates, it is natural to give the following conjecture (due to Fraenkel for bracket sequences): Conjecture 1.

This makes the marking M (t) semiMarkovian but the state space problem and the difficulty to study the trajectories are even more acute. 62 3 Stochastic Event Graphs P = P \ {π} ∪ {p1 , · · · pk }, Q = Q ∪ {q1 , · · · qk−1 }, k−1 E = E \ {(π, j), (i, π)} (qn , pn+1 ), (pn+1 , qn+1 ), n=0 if p ∈ P \ {π} Mp0 = Mp0 , Mp0 = 1, if p ∈ P \ P σq (n) = σq (n), σq (n) = 0, if q ∈ Q, if q ∈ Q \ Q, with the convention that q0 = i and qk = j. By repeating the expansion for all places with more than one token ini˜ Q, ˜ E, ˜ M˜ 0 , σ ˜ ) with a maximal marking tially, one gets an event graph G˜ = (P, m = 1 and such that the behavior of the initial graph is preserved.

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