This is a sequel to my last post about the Umbrella Problem. As a Midwesterner, one of the problems with using an umbrella is that they get blown inside out, since it’s really windy in the Midwest. As a result, umbrellas have a short lifespan. **Here, I extend the Umbrella problem to consider an umbrella being destroyed with probability q.** (Only when the umbrella is used, of course. It still rains with probability p.)

In the Markov chain, let our state reflect **the total number of umbrellas** **(this part is new) and the number of available umbrellas at the professors current location (either home or her office)**. Then the Markov chain has Ux(U+1) states, given by (U,A), where A=number of available umbrellas. The transition probability matrix for U = 2 is:

(0,0) (1,0) (1,1) (2,0) (2,1) (2,2) P= (0,0) [1 0 0 0 0 0 <-- when there are no umbrellas available at all (1,0) 0 0 1 0 0 0 <--in this state and in (2,0), we don't currently have any umbrellas to move (1,1) qp 1-p (1-q)p 0 0 0 <-- in this state and in (2,1) and (2,2), we can move or lose am umbrella (2,0) 0 0 0 0 0 1 (2,1) 0 qp 0 0 1-p (1-q)p (2,2) 0 qp 0 1-p (1-q)p 0]

This Markov chain has an absorbing state of (0,0), meaning that in the long-run, we will have zero umbrellas with certainty. Therefore, we cannot identify the steady state proportion of time the professor gets wet as we did before.

**However, Midwesterners periodically replenish their umbrella supplies.** This would lead to the design of a (u, U) inventory model, where the number of umbrellas must be replenished to have U umbrellas when their number dwindles down to a mere u umbrellas.Let’s consider a (0,2) inventory model, which means that the professor immediately buys two umbrellas when her last umbrella is destroyed. The Markov chain is then ergodic (all states are in a single, recurrent class, aperiodic). The changes are in boldface:

(0,0) (1,0) (1,1) (2,0) (2,1) (2,2) P= (0,0) [00 0 0 01(1,0) 0 0 1 0 0 0 (1,1) qp 1-p (1-q)p 0 0 0 (2,0) 0 0 0 0 0 1 (2,1) 0 qp 0 0 1-p (1-q)p (2,2) 0 qp 0 1-p (1-q)p 0]

Now we can analyze this Markov chain by solving for the limiting distribution π in the usual way. The probability that the professor gets wet is now p(π[0,0]+π[1,0]+π[2,0]). This assumes that when the umbrella brakes, the professor is able to stay dry. That may be a dubious assumption, but Midwesterners are crafty like that.

A sensitivity analysis on p with q = 0.1 yields the following results:

A sensitivity analysis on q with p = 0.25 yields the following results:

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