to its size. However, a renewal interval of average size is not size-biased. {| class="toccolours collapsible collapsed" width="80%" style="text-align:left" !Proof |- |Observe that the last jump-time before t is ; and that the renewal interval containing t is . Then since both and are greater than or equal to for all values of s. |} Superposition Unless the renewal process is a Poisson process, the superposition (sum) of two independent renewal processes is not a renewal process. However, such processes can be described within a larger class of processes called the Markov-renewal processes. However, the cumulative distribution function of the first inter-event time in the superposition process is given by where Rk(t) and αk > 0 are the CDF of the inter-event times and the arrival rate of process k. Example application Eric the entrepreneur has n machines, each having an operational lifetime uniformly distributed between zero and two years. Eric may let each machine run until it fails with replacement cost €2600; alternatively he may replace a machine at any time while it is still functional at a cost of €200. What is his optimal replacement policy? {| class="toccolours collapsible collapsed" width="80%" style="text-align:left" !Solution |- |The lifetime of the n machines can be modeled as n independent concurrent renewal-reward processes, so it is sufficient to consider the case n=1. Denote this process by . The successive lifetimes S of the replacement machines are independent and identically distributed, so the optimal policy is the same for all replacement machines in the process. If Eric decides at the start of a machine's life to replace it at time 0 < t < 2 but the machine happens to fail before that time then the lifetime S of the machine is uniformly distributed on [0, t] and thus has expectation 0.5t. So the overall expected lifetime of the machine is: and the expected cost W per machine is: So by the strong law of large numbers, his long-term average cost per unit time is: then differentiating with respect to t: this implies that the turning points satisfy: and thus We take the only solution t in [0, 2]: t = 2/3. This is indeed a minimum (and not a maximum) since the cost per unit time tends to infinity as t tends to zero, meaning that the cost is decreasing as t increases, until the point 2/3 where it starts to increase.