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Hw7_sol - Homework 7 Solutions Chapter 4 8.38 In this case...

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Homework 7: Solutions Chapter 4 8.38 In this case the equations are r 1 = λ + (1 - p ) r 2 r 2 = r 1 which solves to give that each r i = λ/p . For stability we must have that λ/p < min { μ 1 , μ 2 } . 8.42 In this case the equations are r 1 = 10 + 2 r 2 / 5 r 2 = 5 + r 1 / 4 r 3 = 3 r 1 / 4 + 3 r 2 / 5 Inserting the second equation in the first gives r 1 = 12 + r 1 / 10 so r 1 = 40 / 3. Using the second and third equations now we have r 2 = 5 + 10 / 3 = 25 / 3 r 3 = 10 + 5 The problem told us that μ 1 = 25, μ 2 = 30, and μ 3 = 20. Thus r i < μ i for all i and there is a stationary distribution consisting of independent shifted geometrics with failure probabilities r i i . 8.44 This is a special case of Example 7.1 in which the service rates are μ 1 = 1 / 4, μ 2 = 1 / 8, and the routing matrix is p ( i, j ) = 0 1 1 / 2 1 / 2 The stationary distribution has π 1 = 1 / 3 and π 2 = 2 / 3. These queues are single servers so ψ 1 ( n ) = 1 / 4 n ψ 2 ( n ) = 1 / 8 n Plugging into the formula in (7.3) we have π ( n, 4 - n ) = c (4 / 3) n (16 / 3) 4 - n = c 0 (1 / 4) n where c 0 is chosen to make the sum equal to 1.
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