Question
Can you tell me how the redbox is derived? Thank you so much!!
Suppose that customers arrive at a train depot in accordance with a renewal process having a mean interarrival time mu . Whenever there are N customers waiting in the depot, a train leaves. If the depot incurs a cost at the rate of nc dollars per unit time whenever there are n customers waiting, what is the average cost incurred by the depot? If we say that a cycle is completed whenever a train leaves, then the preceding is a renewal reward process. The expected length of a cycle is the expected time required for N customers to arrive and, since the mean interarrival time is mu , this equals E[length of cycle] = N mu If we let Tn denote the time between the nth and (n+ 1)st arrival in a cycle, then the expected cost of a cycle may be expressed as E[cost of a cycle] = E[c T1 + 2c T2 + . . . + (N - 1)cTN -1 ] which, since E[Tn]= mu , equals Hence, the average cost incurred by the depot is c mu N(N - 1)/2N mu = c(N - 1)/2
Explanation / Answer
E[cost of cycle] = E[cT1 + 2cT2 + ... + (N-1)cTN-1]
Expectations are linear, so we can reduce it as follows:
Factor out c:
E[cost of cycle] = c*E[T1 + 2T2 + ... + (N-1)TN-1]
Break up terms:
= c*{E[T1] + E[2T2] + ... + E[(N-1)TN-1]}
Factor out individual multipliers:
= c*{E[T1] + 2E[T2] + ... + (N-1)E[TN-1]}
Since E[Tn] = , we can substitute that in for all E[Tn], where n = 1, 2, ..., N-1
= c*{ + 2 + ... + (N-1)}
Factor out :
= c*[1 + 2 + ... + (N-1)]
The summation in brackets is a basic summation for a triangular number. It is equivalent to (N-1)N/2:
= c*N*(N-1)/2
And that's the solution!