Please provide the math/work: Flextrola, Inc., an electronics systems integrator

Question

Please provide the math/work:

Flextrola, Inc., an electronics systems integrator, is planning to design a key component for its next-generation product with Solectrics. Flextrola will integrate the component with some software and then sell it to consumers. Given the short life cycles of such products and the long lead times quoted by Solectrics, Flextrola only has one opportunity to place an order with Solectrics prior to the beginning of its selling season. Flextrola’s demand during the season is normally distributed with a mean of 1000 and a standard deviation of 600.

Solectrics’ production cost for the component is $52 per unit, and it plans to sell the component for $72 per unit to Flextrola. Flextrola incurs essentially no cost associated with the software integration and handling of each unit. Flextrola sells these units to consumers for $121 each. Flextrola can sell unsold inventory at the end of the season in a secondary electronics market for $50 each. The existing contract specifies that once Flextrola places the order, no changes are allowed to it. Also, Solectrics does not accept any returns of unsold inventory, so Flextrola must dispose of excess inventory in the secondary market.

a.

What is the probability that Flextrola’s demand will be within 25 percent of its forecast?

b.

What is the probability that Flextrola’s demand will be more than 40 percent greater than Flextrola’s forecast?

c.

Under this contract, how many units should Flextrola order to maximize its expected profit?

a.

What is the probability that Flextrola’s demand will be within 25 percent of its forecast?

b.

What is the probability that Flextrola’s demand will be more than 40 percent greater than Flextrola’s forecast?

c.

Under this contract, how many units should Flextrola order to maximize its expected profit?

Explanation / Answer

a) upper bound of 25% of mean forecast = 1000*(1+25%) = 1250

z-stat = (1250-1000)/600 = 0.4167

Corresponding F(z) = NORMSDIST(0.4167) = 0.6615

lower bound of 25% of mean forecast = 1000*(1-25%) = 750

z-stat = (750-1000)/600 = -0.4167

Corresponding F(z) = NORMSDIST(-0.4167) = 0.3385

Probability that demand for this book will be within 20 percent of the mean forecast = 0.6615 - 0.3385 = 0.323 or 32.3 %

b) 40% greater than mean forecast = 1000*(1+40%) = 1400

z-stat = (1400-1000)/600 = 0.6667

F(z) = NORMSDIST(z) = NORMSDIST(0.6667) = 0.7475

Therefore, probability that demand will be more than 40% greater than forecast = 1 - 0.7475 = 25.25 %

c) Underage cost, Cu = selling price - purchase price = 121 - 72 = 49

Overage cost, Co = purchase price - salvage price = 72 - 50 = 22

Optimal service level, F(z) = Cu/(Cu+Co) = 49/(49+22) = 0.69

z-stat = NORMSINV(0.69) = 0.4962

Optimal order quantity that maximizes expected profit = mean + z * Std dev

= 1000 + 0.4962 * 600

= 1298

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