Single-period. This type of problem can be recognized when a probability distribution or empirical distribution for demand of a “perishable” item is given along with unit shortage and excess costs, or information that can be used to calculate them. Use the following steps to solve the problem. 1.

Question

Single-period. This type of problem can be recognized when a probability distribution or
empirical distribution for demand of a “perishable” item is given along with unit shortage and excess costs, or information that can be used to calculate them. Use the following steps to solve the problem.
1. Compute the optimal service level using Formula 12–17.
2. If the given distribution is uniform or normal, use that to obtain the exact stocking level.
3. If the distribution is Poisson or empirical, SL will fall between two cumulative frequencies.
Round up to the higher frequency to find the optimal stocking level. Note: If the distribution is empirical, first obtain the cumulative frequencies or probabilities.
A firm that installs cable TV systems uses a certain piece of equipment for which it carries two spare parts as optimal. The parts cost $500 each and have no salvage value or useful life after one period. Part failures can be modeled by a Poisson distribution with a mean of two failures during the useful life of the equipment. Holding and disposal costs are negligible. Estimate the apparent range of shortage cost.

Accepted Answer

[latex]C_s[/latex] is unknown [latex]C_e[/latex] = $500
The Poisson table (Appendix B, Table C) provides these values for a mean of 2.0:

Number of Failures	Cumulative Probability
0	.135
1	.406
2	.677
3	.857
4	.947
5	.983
:	:

For the optimal stocking level, the service level must usually be rounded up to a feasible stocking level. Hence, you know that the service level must have been between .406 and .677 in order to make two units the optimal level. By setting the service level equal first to .406 and then to .677, you can establish bounds on the possible range of shortage costs.

[latex]\frac{C_s}{C_s+\$5000}[/latex] = .406, so [latex]C_s[/latex]  = .406 ( $500 +  [latex]C_s[/latex])

Solving, you find [latex]C_s[/latex] = $341.75. Similarly,

[latex]\frac{C_s}{C_s+\$500}[/latex] = .677, so [latex]C_s[/latex]  = .677 ( $500 +  [latex]C_s[/latex] )

Solving, you find [latex]C_s[/latex]= $1,047.99. Hence, the apparent range of shortage cost is $341.75 to $1,047.99.

Question 12.p.8: Single-period. This type of problem can be recognized when a...

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