Question

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 900 and a standard deviation of 700.

Solectrics’ production cost for the component is $52 per unit, and it plans to sell the component for $70 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 $124 each. Flextrola can sell unsold inventory at the end of the season in a secondary electronics market for $46 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% of its forecast?

B) What is the probability that Flextrola's demand will be more than 40% greater than Flextrola's forecast?

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

D) If Flextrola orders 1200 units, how many units of inventory can Flextrola expect to sell in the secondary electronics market?

E) If Flextrola orders 1200 units, what are expected sales?

F) If Flextrola orders 1200 units, what is expected profit?

Answer 1

a) The Upper and lower limit for 25% of forecast are 900*125% and 900*75% = 1125 and 675

z value for upper limit = (1125-900)/700 = 0.3214, F(z) =NORMSDIST(0.3214) =0.6261

z value for lower limit = (675-1400)/600 = -0.3214, F(z) =NORMSDIST(-0.3214) =0.3739

Therefore, probability of demand being between 675 and 1125 = 0.6261 - 0.3739 = 0.2522

b) 40% greater than forecast = 900*140% = 1260

z value = (1260-900)/700 =0.5143

F(z) = NORMSDIST(0.5143) =0.6965

Probability of demand being greater than 1260 = 1-0.6965 = 0.3035

c) For Flextrola,

Shortage of Underage cost, Cu = 124-70 = 54 (also known as marginal profit)

Excess or Overage cost, Co = 70-46 = 24 (also known as marginal loss)

Critical ratio = Cu/(Cu+Co) = 54/(54+24) = 0.6923   this is F(z)

Lookup above value of F(z) in the table, for corresponding z value = 0.5024 (using interpolation

Units Flextrola should order = μ +zσ = 900+0.5024*700 = 1252

d) z value = (1200-900)/700 = 0.4286

Corresponding I(z) value taken from the table = 0.6493

Expected inventory, V = σ*I(z) = 700*0.6493 = 455

Flextrola can expect to sell 455 units in the secondary market

e) Expected sales, S = Q - σ*I(z) = 1200 - 455 = 745

f) Expected profit = S*Cu - V*Co = 745*54 - 455*24 = $ 29,310