In this case the average number of people in the

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Unformatted text preview: nd r2 = 2 + p1 r1 (4.28) 146 CHAPTER 4. CONTINUOUS TIME MARKOV CHAINS Plugging in the values for this example and solving gives ✓ ◆ 1 1 1 1 r1 = 1 + r2 and r2 = 2 + r1 = 2 + 1 + r2 4 2 2 4 So (7/8)r2 = 5/2 or r2 = 20/7, and r1 = 1 + 20/28 = 11/7. Since r1 = 11/7 < 4 = µ1 and r2 = 20/7 < 3.5 = µ2 this analysis suggests that there will be a stationary distribution. To prove that there is one, we return to the general situation and suppose that the ri we find from solving (4.28) satisfy ri < µi . Thinking of two independent M/M/1 queues with arrival rates ri , we let ↵i = ri /µi and guess: mn Theorem 4.13. If ⇡ (m, n) = c↵1 ↵2 where c = (1 stationary distribution. ↵1 )(1 ↵2 ) then ⇡ is a Proof. The first step in checking ⇡ Q = 0 is to compute the rate matrix Q. To do this it is useful to draw a picture. Here, we have assumed that m and n are both positive. To make the picture slightly less cluttered, we have only labeled half of the arrows and have used qi = 1 pi . (m 1, n + 1) (a) (m (m, n + 1) @@ I 6 µ2 q2 @@ @ µ2 p2 @ @@ R@ ? @ 1, n) @ - (m, n) 1 @ (c) @ @ @ µ1 q1 - (m + 1, n) 6 @@...
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