By L. Breuer, Dieter Baum

ISBN-10: 1402036302

ISBN-13: 9781402036309

ISBN-10: 1402036310

ISBN-13: 9781402036316

I have not learn the ebook. although, from its content material i do know it's a stable publication. apparently every person loves "Fundamentals of Queueing Theory". even though, it has too many pages. 464 pages. i am unable to think somebody can end it more often than not until he/she makes use of that e-book for textbook and has a weekly lecture approximately this booklet. hence, in case you have sturdy mathematical ability and plan to self-study queueing concept, this is often the ebook for you.

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**Additional info for An introduction to queueing theory and matrix-analytic methods**

**Example text**

It is immediately evident that these are solvable only by πi = 0 for all i ∈ E, which means that there is no stationary distribution π. 11 Let the underlying Markov chain X in the definition of the Markov process Y be irreducible and positive recurrent. Further assume that ˇ := inf{λi : i ∈ E} > 0. Then there is a unique stationary distribution for λ Y. 18, the transition matrix P of X admits a unique stationary distribution ν with νP = ν. The generator G is defined by G = Λ(P − IE ), with Λ = diag(λi : i ∈ E).

2. Clearly, the structure of the matrix G shows that the process Q is irreducible and hence there is at most one stationary distribution π for Q. 3) n=0 of equations, where the latter is simply the normalization of the distribution π. The first two equations are the global balance equations and can be illustrated by the following scheme: λ λ 0 λ 1 µ λ ... 2. λ µ m µ ... µ Transition rates for the M/M/1 queue This gives the rates of jumps between the states of the system. If we encircle any one state, then the sum of the rates belonging to the arcs reaching into this state must equal the sum of the rates which belong to the arcs that go out of this state.

Hence πP = π implies first π0 · (1 − p) = π0 ⇒ π0 = 0 since 0 < p < 1. Assume that πn = 0 for any n ∈ N0 . This and the condition πP = π further imply for πn+1 πn · p + πn+1 · (1 − p) = πn+1 ⇒ πn+1 = 0 22 AN INTRODUCTION TO QUEUEING THEORY which completes an induction argument proving πn = 0 for all n ∈ N0 . Hence the Bernoulli process does not have a stationary distribution. 21 The solution of πP = π and P = j∈E πj = 1 is unique for 1−p p p 1−p with 0 < p < 1. Thus there are transition matrices which have exactly one stationary distribution.

### An introduction to queueing theory and matrix-analytic methods by L. Breuer, Dieter Baum

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