soinc456

soinc456 - n = 0 , 1 , 2 , ... Solution 1. This is a...

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Solutions to In-Class Homework Problems 4, 5, 6. 4. Suppose we have one server and unlimited space for the waiting room. Thea r r iva lra tei s λ and the service rate is μ . Write down the Kolmogorov backward equations for P ij ( t ) . Solution: This is a birth-death process with λ n = λ and μ n = μ .Thu s ,th e equation is P 0 0 j ( t )= λ [ P 1 j ( t ) P 0 j ( t )] P 0 ij ( t )= λ P i +1 ,j ( t )+ μP i 1 ,j ( t ) ( λ + μ ) P ij ( t ) ,i> 0 . 5. Problem 13, Ch. 6. Solution: Let X ( t ) denote the number of customers in the shop at time t ,t h e n { X ( t ) ,t 0 } is a birth and death process with state space { 0 , 1 , 2 } and rates λ 0 = λ 1 =3 1 = μ 2 =4 . The limiting probabilities of the Markov chain satisfy 4 P 1 =3 P 0 4 P 2 =3 P 1 P 0 + P 1 + P 2 =1 , yielding P 0 = 16 37 ,P 1 = 12 37 ,P 2 = 9 37 . The average number of customers in the shop is P 1 +2 P 2 = 12 37 +2 × 9 37 = 30 37 . 6. People come to the server according to Poisson process at rate λ per
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Unformatted text preview: n = 0 , 1 , 2 , ... Solution 1. This is a birth-death process with λ n = λ and μ n = μ . Thus, equation in the notes gives P = 1 1 + P ∞ n =1 λ ··· λ n − 1 μ 1 ··· μ n = 1 1 + P ∞ n =1 λ n μ n = 1 1 + λ /μ 1 − λ μ = 1 − λ μ 1 − λ μ + λ /μ = 1 − λ μ , 1 and P n = λ λ 1 · · · λ n − 1 μ 1 · · · μ n P = λ n μ n P = μ λ μ ¶ n μ 1 − λ μ ¶ . Solution 2: This is a birth-death process with λ n = λ and μ n = μ . Thus, the balance equations are λ P = μP 1 λ P 1 = μP 2 · · · λ P n = μP n +1 · · · Thus, P 1 = λ μ P , P 2 = λ μ P 1 = μ λ μ ¶ 2 P , ..., P n = μ λ μ ¶ n P , ... Therefore, P = (1 + λ μ + · · · + μ λ μ ¶ n + · · · ) − 1 = Ã 1 1 − λ μ ! − 1 = 1 − λ μ , P n = μ λ μ ¶ n P = μ λ μ ¶ n μ 1 − λ μ ¶ . 2...
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This document was uploaded on 10/18/2011.

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soinc456 - n = 0 , 1 , 2 , ... Solution 1. This is a...

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