By Jerzy A. Filar

ISBN-10: 1601980884

ISBN-13: 9781601980885

Managed Markov Chains, Graphs & Hamiltonicity summarizes a line of study that maps definite classical difficulties of discrete arithmetic - comparable to the Hamiltonian cycle and the touring Salesman difficulties - into convex domain names the place continuum research might be conducted.

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**Additional resources for Controlled Markov Chains, Graphs and Hamiltonicity **

**Example text**

In particular, the following was shown in [16] that g11 result holds. 6. Let f ∈ DS be any short cycle deterministic policy; and let k < N be the length of the cycle containing the home node 1. The following properties hold. ε (f ) depends only on ε and k and equals (a) The value g11 ε g11 (f ) = 1 + N + 1 N 1 N 2ε N −k 1 − (1 − N ε)k kN ε − 1 + (1 − N ε)k . 1 − (1 − N ε)k 32 Analysis in the Policy Space ε (f ) has a pole of order 1 at ε = 0, and the (b) The functional g11 initial terms of its Laurent expansion are: ε g11 (f ) = N (k − 1) (1 + k) N − k −1 2 k + kN − N ε + + ε 2 N k 2kN 12k N 2 (k − 1) (1 + k) 2 + ε + O ε3 .

0 ε 0 0 ··· 0 1−ε | 0 ··· 0 Pε (f ) = 0 ··· 0 0 | 0 ··· 0 1 0 , − − − − − − − − − − | P21 (f ) | P22 (f ) | where the dimension of the top left-hand block is m × m and, typically, the P21 (f ) block must contain one or more nonzero entries. It follows that the states m + 1, m + 2, . . , N are all transient. Now, if we let π(f ) = (π1 (f ), . . , πN (f )) denote the unique stationary distribution vector of Pε (f ), then πk (f ) = 0 for k ≥ m + 1.

F ) depends only on ε and k and equals (a) The value g11 ε g11 (f ) = 1 + N + 1 N 1 N 2ε N −k 1 − (1 − N ε)k kN ε − 1 + (1 − N ε)k . 1 − (1 − N ε)k 32 Analysis in the Policy Space ε (f ) has a pole of order 1 at ε = 0, and the (b) The functional g11 initial terms of its Laurent expansion are: ε g11 (f ) = N (k − 1) (1 + k) N − k −1 2 k + kN − N ε + + ε 2 N k 2kN 12k N 2 (k − 1) (1 + k) 2 + ε + O ε3 . 3. 6) and has the form: ε 1 − 3ε ε ε ε ε ε 1 − 3ε Pε (f ) = . ε ε ε 1 − 3ε ε ε 1 − 3ε ε The corresponding matrix 5/4 − ε −3/4 + 3 ε 1 I − Pε (f ) + J = −ε + 1/4 4 −ε + 1/4 −3/4 + 3 ε −ε + 1/4 −ε + 1/4 5/4 − ε −ε + 1/4 −ε + 1/4 −ε + 1/4 5/4 − ε −ε + 1/4 −3/4 + 3 ε .

### Controlled Markov Chains, Graphs and Hamiltonicity by Jerzy A. Filar

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