Stochastic

# 1 1 3 1 1 2 2 4 2 2 4 4 3 4 3 3 3 0 3 solving

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Unformatted text preview: the day with 3 units, so the transition 28 CHAPTER 1. MARKOV CHAINS probability has constant rows 0 1 2 3 0 .1 .1 .1 .1 1 .2 .2 .2 .2 2 .4 .4 .4 .4 3 .3 .3 .3 .3 In this case it is clear that the stationary distribution is ⇡ (0) = .1, ⇡ (1) = .2, ⇡ (2) = .4, and ⇡ (3) = .3. If we end the day with k units then we sold 3 k and have to keep k over night. Thus our long run sales under this scheme are .1(36) + .2(24) + .4(12) = 3.6 + 4.8 + 4.8 = 13.2 dollars per day while the inventory holding costs are 2(.2) + 4(.4) + 6(.3) = .4 + 1.6 + 1.8 = 3.8 for a net proﬁt of 9.4 dollars per day. is Suppose we use a 1,3 inventory policy. In this case the transition probability 0 1 2 3 0 .1 .1 .3 .1 1 .2 .2 .4 .2 2 .4 .4 .3 .4 3 .3 .3 0 .3 Solving for the stationary distribution we get ⇡ (0) = 19/110 ⇡ (1) = 30/110 ⇡ (2) = 40/110 ⇡ (3) = 21/110 To compute the proﬁt we make from sales note that if we always had enough stock then by the calculation in the ﬁrst case, we would make 13.2 dollars per day. However, when Xn = 2 and the demand is 3, an event with probability (4/11) · 0.1 = 0.03636, we lo...
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## This document was uploaded on 03/06/2014 for the course MATH 4740 at Cornell.

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