Rearranging gives 0 2 the root we

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: are lost and hence 57/65’s or 87.7% of the customers enter service. Example 4.15. Machine repair model. A factory has three machines in use and one repairman. Suppose each machine works for an exponential amount of time with mean 60 days between breakdowns, but each breakdown requires an exponential repair time with mean 4 days. What is the long-run fraction of time all three machines are working? Solution. Let Xt be the number of working machines. Since there is one repairman we have q (i, i +1) = 1/4 for i = 0, 1, 2. On the other hand, the failure rate is proportional to the number of machines working, so q (i, i 1) = i/60 for i = 1, 2, 3. Setting ⇡ (0) = c and plugging into the recursion (4.17) gives 1/4 1/60 1/4 1 · ⇡ (1) = ⇡ (2) = µ2 2/60 1/4 2 ⇡ (3) = · ⇡ (2) = µ3 3/60 ⇡ (1) = 0 µ1 · ⇡ (0) = · c = 15c 225c 2 225c 1125c · = 2 2 · 15c = Adding up the ⇡ ’s gives (1125 + 225 + 30 + 2)c/2 = 1382c/2 so c = 2/1480 and we have ⇡ (3) = 1125 1382 ⇡ (2) = 225 1382 ⇡ (1) = 30 1382 ⇡ (0) = 2 1382 Thus in the long run all three machines are working 1125/1382 = 0.8140 of the time. Example 4.16. M/M/1 queue. In this case q (n,...
View Full Document

Ask a homework question - tutors are online