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

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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...
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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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