03 0 0 0 0 02 01 0 1 0 0 0 04 0 01 0 1 0 where after

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

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: re working. When two are broken, they are replaced and the machine is back to working order the next day. Declaring the state space to be the parts that are broken {0, 1, 2, 3, 12, 13, 23}, we arrived at the following transition matrix: 0 1 2 3 12 13 23 0 .93 0 0 0 1 1 1 1 .01 .94 0 0 0 0 0 2 .02 0 .95 0 0 0 0 3 .04 0 0 .97 0 0 0 12 13 0 0 .02 .04 .01 0 0 .01 0 0 0 0 0 0 23 0 0 .04 .02 0 0 0 and we asked: If we are going to operate the machine for 1800 days (about 5 years) then how many parts of types 1, 2, and 3 will we use? To ﬁnd the stationary distribution we look at the last row of 0 B B B B B B B B @ .07 0 0 0 1 1 1 .01 .06 0 0 0 0 0 .02 0 .05 0 0 0 0 .04 0 0 .03 0 0 0 0 .02 .01 0 1 0 0 0 .04 0 .01 0 1 0 where after converting the results to fractions we have: 1 1 1C C 1C C 1C C 1C C 1A 1 1 ⇡ (0) = 3000/8910 ⇡ (1) = 500/8910 ⇡ (2) = 1200/8910 ⇡ (3) = 4000/8910 ⇡ (12) = 22/8910 ⇡ (13) = 60/8910 ⇡ (23) = 128/8910 We use up one part of type 1 on each visit to 12 or to 13, so on the average we use 82/8910 of a part per day. Over 1800 days we will use an averag...
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