4106-08-h5-solu

# 4106-08-h5-solu - HMWK 5 Problems from the Text Ch 6 Page...

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HMWK 5 Problems from the Text Ch. 6, Page 409: 12, 13, 22 1. Chapter 6, Problem 12. λ n = + θ, n < N, λ n = nλ, n N, μ n = nμ, n 0 . 3. (b) Fix N = 3. Let P i be the limiting probability that the system is in state i. We are looking for i =3 P i . We have λ k P k = μ k +1 P k +1 . This yields P 1 = θ μ P 0 P 2 = λ + θ 2 μ P 1 = θ ( λ + θ ) 2 μ 2 P 0 P 3 = 2 λ + θ 2 μ P 2 = θ ( λ + θ )(2 λ + θ ) 6 μ 3 P 0 For k 4, we get P k = ( k - 1) λ P k - 1 , which implies P k = ( k - 1)( k - 2) ··· 3 k ( k - 1) ··· 4 ( λ μ ) k - 3 P 3 = 3 k ( λ μ ) k - 3 P 3 ; Therefore k =3 P k = 3( μ λ ) 3 P 3 k =3 1 k ( λ μ ) k . But k =1 1 k ( λ μ ) k = log ( 1 1 - λ μ ) if λ μ < 1 . So k =3 P k = 3( μ λ ) 3 P 3 [ log ( μ μ - λ ) - λ μ - 1 2 ( λ μ ) 2 ] = 3( μ λ ) 3 [ log ( μ μ - λ ) - λ μ - 1 2 ( λ μ ) 2 ] θ ( λ + θ )(2 λ + θ ) 6 μ 3 P 0 . Now i =0 ( P i ) = 1 implies P 0 = [1 + θ μ + θ ( λ + θ ) 2 μ 2 + θ ( λ + θ )(2 λ + θ ) 2 λ 3 [ log ( μ μ - λ ) - λ μ - 1 2 ( λ μ ) 2 ]] - 1 . And ﬁnally, k =3 P k = [[ 1 2 λ 3 ][ log ( μ μ - λ ) - λ μ - 1 2 ( λ μ ) 2 ] θ ( λ + θ )(2 λ + θ ) / [1+ θ μ + θ ( λ + θ ) 2 μ 2 + θ ( λ + θ )(2 λ + θ ) 2 λ 3 [ log ( μ μ - λ ) - λ μ -

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4106-08-h5-solu - HMWK 5 Problems from the Text Ch 6 Page...

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