# Stochastic

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Unformatted text preview: Dn+1 > Xn we end the day with Xn+1 = 0. If Xn s then we begin the day with S units, and the reasoning is the same as in the previous case. Suppose now that an electronics store sells a video game system and uses an inventory policy with s = 1, S = 5. That is, if at the end of the day, the number of units they have on hand is 1 or 0, they order enough new units so their total on hand at the beginning of the next day is 5. If we assume that for k = P (Dn+1 = k ) 0 .3 1 .4 2 .2 3 .1 then we have the following transition matrix: 0 1 2 3 4 5 0 0 0 .3 .1 0 0 1 0 0 .4 .2 .1 0 2 .1 .1 .3 .4 .2 .1 3 .2 .2 0 .3 .4 .2 4 .4 .4 0 0 .3 .4 5 .3 .3 0 0 0 .3 To explain the entries, we note that when Xn 3 then Xn Dn+1 0. When Xn+1 = 2 this is almost true but p(2, 0) = P (Dn+1 = 2 or 3). When Xn = 1 or 0 we start the day with 5 units so the end result is the same as when Xn = 5. In this context we might be interested in: Q. Suppose we make \$12 proﬁt on each unit sold but it costs \$2 a day to store items....
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## This document was uploaded on 03/06/2014 for the course MATH 4740 at Cornell University (Engineering School).

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