$20$ points$)$ Determining reject allowances, decisions over time, and probabilistic Dynamic Program$-$

ming.

A firm receives an order to supply one hard$$to$$manufacture item. The firm may need to produce more

than one item in order to obtain a single item that is acceptable to the customer. The number of extra

items produced is called the reject allowance. Each item produced will be acceptable with probability

$1/2$ or defective with probability $1/2.$ Defective items cannot be reworked; they must be trashed. The

number of acceptable items in a lot of size L is distributed Binomial with number of trials equal to

L and probability of success on each trial equal to $1/2.$ Consequently, the probability of having no

acceptable items in a lot of size L is $(1/2)$L$.$

The marginal $($per$$unit$)$ production cost is $$100.$ There is a set$$up cost of $$300$ for each production

run. The firm has enough time to make up to three production runs before the order is due. If no

acceptable item is produced by the end of the third run then the firm incurs a penalty cost of $$1600.$

Determine the lot size policy $($i$.$e$.$ how many items should be produced in each of the three runs$)$ that

minimizes total expected cost to the firm. We can solve this problem using Dynamic Programming.

Here are some thoughts to get started: Let each stage be represented by the number of production

runs left to go$,$ i$.$e$.$ n $=1,2,3.$ The decision to be made is xn$,$ the lot size for stage n$.$ The state of

the system is Sn$,$ the number of items still needed at the beginning of stage n$.$ Note that Sn in $\{0,1\}.$

With n $=3$ stages to go$,$ S$3=1.$ If at least one acceptable item is produced then Sn $=0$ and no

additional costs are incurred