Question

A scientist has a machine for measuring ozone in the atmosphere that is located in the mountains just north of LA. At times of a Poisson process with rate 1, storms or animals disturb the equipment so that it can no longer collect data. The scientist comes every L units of time to check the equipment. If the equipment has been disturbed, she can usually x it quickly so we will assume the repairs take 0 time. (a) What is the limiting fraction of time the machine is working? (b) Suppose that the data that is being collected is worth a dollars per unit time, while each inspection costs c < a dollars. Find the best value of the inspection time L.

Answer 1

a) This is I'm aware of two possibilities for what "cycle rene can be here: one is to define the renewal event "scientist visits the set-up", and the other is to define the renewal event "scientist fixes the machine. The first makes cycle time be easy, the second makes working time per cycle be easy Let's first work through the solution approach where the renewal event is "scientist visits the set-up", and the reward per cycle is "time the set-up is working in a cycle". The expected length of a cycle is L; the expected reward per cycle takes a little more work, bccause the reward is min((1), L): if the scientist comes and the equiptment is stl working, then the reward in a cycle is L, not the (larger) value of E(1) in this case. Thus, lemin(r, L) dr density of ε(1)reward, given E(1)- We see that the expected working time per cycle is a little s we're truncating off the large values of the exponential. maller than 1, since Consequently, the proportion of working time is (1 - el)/L Now let the renewal event be the equipment is fixed". Then the cycle time may be L or 2L or 3L orin general, the number of visits a scientist makes to the machine in one cycle has a geometric distribution. On the other hand, the working ycle is simply (1). Here, omputing the mean cycle length will be a little harder Elcycle length] L (1-e2L e (1-el) +3L (e L)(e+. - L(1 -e). (1+2e1 +3(e1)+...) How do we find the infinite sum? If you know the mcan of the gcomctric distribution, then you just write down the answer: mean cycle length is L(1-e-ヶ, If you don't, here's how to go about t: let f(p) 2p 3p +4p3+... then Plugging in p e we get E[cycle length-La I,)2-1-e-L, -e so the fraction of working time is E(cycle length L/1 e LL

B)

(b) Now in a renewal cycle of length L, we get an expected profit of a(1 -e-L) units, and a cost of c units (once per inspection, so once per cycle). Thus, the monetary expected reward per cycle is a(1 - eL) - c, and the rate of accumulating profits is We want to maximize this over L, which requires taking the derivative and setting it to 0: 12 This can't actually be solved analytically (because of the Le* term), so we'l leave it at this expression. (Some computer algebra systems give an answer in terms of the Lambert W function, which solves LeL-r for L as a function of r; but since this isn't one of the standard functions, people rarely have an intuition for it.) In the other version of what a cycle is, notice that the cost of scientist visits per cycle isn't c, t's c times the expected number of visits! So there the benefit (or cost) per unit time would be computed as mcan bencfit per cycle a c/(1 -e-L) mcan cycle length Of course, the answer is still the same