7 of the customers enter service example 415 machine

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

This document was uploaded on 03/06/2014 for the course MATH 4740 at Cornell University (Engineering School).

Ask a homework question - tutors are online