Case 2 (Sales Estimation): For the coming season, Specialty Toys, Inc., plans to

Question

Case 2 (Sales Estimation): For the coming season, Specialty Toys, Inc., plans to introduce a new product called Weather Teddy. Specialty’s managers claimed Weather Teddy gave predictions that were as good as many local TV weather forecasters! Now, Specialty faces the decision of how many Weather Teddy units to order from manufacturers for the coming holiday season. Members of the management team suggested four order quantities of 15,000, 18,000, 24,000, or 28,000 units. The team asks you for an analysis of the stock-out probabilities for various order quantities, an estimate of the profit potential, and to help make an order recommendation. Specialty expects to sell Weather teddy for $24 based on the unit cost of $20. If inventory remains after holiday season, Specialty will sell all surplus inventory for $5 per unit. After reviewing the sales history of similar products, Specialty’s senior sales forecaster predicted that the demand would be normally distributed with the mean as 20,000 units and that demand would be between 5,000 and 35,000 units with a 90% probability. The managerial report should answer:

1. Use the sales forecaster prediction to calculate the standard deviation for the demand distribution.

2. Compute the probability of quantity demanded less than 7,500 units.

3. Compute the probability of a stock-out (i.e. quantity demanded is greater than the order quantity) for each order quantity suggested by the management team.

4. One of Specialty’s managers felt strongly that the order quantity should have a 70% chance of meeting quantity demanded (i.e. only a 30% chance of stock-out). What quantity should be ordered under this policy?

5. Given the order quantity answered in Question 4, if the quantity demanded in the coming season will be only 20,000 units, then how much the profit will be earned?

Explanation / Answer

= 10000/1.645

= 6079

2) P(X<)=

3) Probability of stock out with an order of K units is P(X > K) = P(Z > (K-20000)/6079), where Z is distributed as standard normal

Order (K)

(K-20000)/6079

P(X > K)

15000

-0.822503

0.794605

18000

-0.329001

0.628923

24000

0.658002

0.255268

28000

1.316005

0.094086

4) The order quantity to meet 70% demand is found by solving

P(X < K) =0.70

P(Z < (K-20000)/6079 ) = 0.70

(K-20000)/6079 = 0.5244

K = 20000 + 6079 * 0.5244

    = 20000 + 3188

    = 23188.

5) The projected profit is

Order quantity from answer 4 is 23188

Sales is 20000

Profit = 4*20000-15*3188

= 32,180

Order (K)

(K-20000)/6079

P(X > K)

15000

-0.822503

0.794605

18000

-0.329001

0.628923

24000

0.658002

0.255268

28000

1.316005

0.094086

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