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per unit time is
10 + 0.3T
T 0.05T 2
To maximize we take the derivative
0.3(T 0.05T 2 ) (10 + 0.3T )(1 0.1T )
(T 0.1T 2 )2
0.3T 0.015T 2 10 0.3T + T + 0.03T 2
(T 0.1T 2 )2
The numerator is 0.015T 2 + T 10 which is 0 when
1 ± 1 + 4(0.015)(10)
1 ± 1.6
We want the + root which is T = 8.83.
Using the idea of renewal reward processes, we can easily treat the following
extension of renewal processes.
Example 2.5. Alternating renewal processes. Let s1 , s2 , . . . be independent with a distribution F that has mean µF , and let u1 , u2 , . . . be independent
with distribution G that has mean µG . For a concrete example consider the
machine in Example 1.1 that works for an amount of time si before needing a
repair that takes ui units of time. However, to talk about things in general we
will say that the alternating renewal process spends an amount of time si in
state 1, an amount of time ui in state 2, and then repeats the cycle again.
Theorem 3.4. In an alternating renewal process, the limitin...
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This document was uploaded on 03/06/2014 for the course MATH 4740 at Cornell University (Engineering School).
- Spring '10
- The Land