# 7 of the customers enter service example 415 machine

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Unformatted text preview: minute to take the ball down the court and then they are on o↵ense. To ﬁnd the stationary distribution we 0 32 B0 5 ⇡0 ⇡1 ⇡2 ⇡3 B @1 0 6 0 want to solve 1 1 1 5 1C C= 0 2.5 1A 0 1 0 0 1 The answer can be found by reading the fourth row of the inverse of the matrix: 10 29 4 29 12 29 3 29 Thus in the long run the chain spends a fraction 4/29 in state 1, and 6/29 in state 3. To translate this into a more useful statistic, we note that the average time in state 1 is 1/5 and the average time in state 3 is 1/6, so the number of baskets per minute for the two teams are 4/29 20 = = 0.6896 1/5 29 3/29 18 = = 0.6206 1/6 29 Multiplying by 2 points per basket and 40 minutes per game yields 55.17 and 49.65 points per game respectively. Detailed balance condition. Generalizing from discrete time we can formulate this condition as: ⇡ (k )q (k, j ) = ⇡ (j )q (j, k ) for all j 6= k (4.16) The reason for interest in this concept is Theorem 4.5. If (4.16) holds, then ⇡ is a stationary distribution. Why is this tru...
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## This document was uploaded on 03/06/2014 for the course MATH 4740 at Cornell University (Engineering School).

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