# Probabilstic Dynamic Programming

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A vendor sells sweatshirts at football games. They are equally likely (0.5,0.5) to sell 200 or 400 sweatshirts per game. Each order placed to the supplier costs $500 plus $5 per quantity ordered. The vendor sells each sweatshirt for $8. The is a holding cost of $2 (inventory costs) for each shirt leftover after each game. The maximum inventory that the vendor can store is 400. The number of shirts that can be ordered from the supplier must be a multiple of 100. Determine an ordering policy that maximizes expected profits earned during the first three games of the season. Assume that any leftover sweatshirts have a value of $6.

Here is where I started:

State 1 = Game 3

State 2 = Game 2

State 3 = Game 1

Decisions:

Buy 100

Buy 200

Buy 300

Buy 400

I- inventory at beginning of each game

Profits:=[(0.5*(200)*$8)-($500+($5*order))-($2*(order-200+I))]+ [(0.5*(400)*$8)-($500+($5*order))-($2*(order-400+I))]

I am not sure I have the above formula correct?

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##### Solution Summary

This solution discusses probabalistic dynamic programming.

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Please refer to the attachment.

Your formula looks plausible. However, I noticed some minor mistakes.

- It's impossible to order more than 400, because ...

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