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Unformatted text preview: Homework 7: Solutions Chapter 4 8.38 In this case the equations are r 1 = + (1 p ) r 2 r 2 = r 1 which solves to give that each r i = /p . For stability we must have that /p < min { 1 , 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 1 = 25, 2 = 30, and 3 = 20. Thus r i < 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 1 = 1 / 4, 2 = 1 / 8, and the routing matrix is p ( i,j ) = 1 1 / 2 1 / 2 The stationary distribution has 1 = 1 / 3 and 2 = 2 / 3. These queues are single servers so 1 ( n ) = 1 / 4 n 2 ( n ) = 1 / 8 n Plugging into the formula in (7.3) we havePlugging into the formula in (7....
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 Spring '10
 DURRETT
 Equations

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