Question

Flextrola, Inc. , an electronics systems integrator, is planning to sell its next generation product using components sourced from its supplier. 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 its supplier, Flextrola only has one opportunity to place an order prior to the beginning of its "selling season. " Assume that Flextrola’s demand during the season is normally distributed with a mean of 1000 and a standard deviation of 600. (A normal table is provided at the end of this exam for your convenience. )
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.


Required:

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 its forecast?

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

d. What are Flextrola’s expected sales?

e. How many units of inventory can Flextrola expect to sell in the secondary electronics market?

Answer 1

With a mean of 1000 and a standard deviation of 600, the probability that the demand is going to be withing 25 percent of its forecast is 0.3230.

a. Mean = 1000

sd = 600

p(1000x 1-25%) - p(1000x 1+25%)

using the z test

d-μ/σ

$p(z < \frac{1250-1000}{600} )-p(z < \frac{-1000}{600} )\\\\(z < 0.4167)-(z < -0.4167)$

find values using excel sheet formula

NORMSDIST(0.4167) - NORMSDIST(-0.4167)

=m0.6615 - 0.3385

= 0.3230

b. The probability that the forecast would be more than 40 percent

1000x 1+40%

= p(D>1400)

= 1- NormDist(0.667)

= 0.225

c. Cu = 121-72 = 49

Co = 72.50 = 22

The critical ratio calculation

49/22 +49 = 0.6901

Normsinv(0.6901) = 0.496

1000+0.496x600

= 1297

The units that Flextrola has to order is 1297.

d. The expected sales of Flextrola

1200-1000/600

= 0.3333

loss function from z = 0.333 is 0.254

600x0.254 = 152. 4

1000-152.4 = 847.6

the expected sales are 847.6

e 1200- 847.6

= 352. 4

The units of inventory that can be sold is 352.4

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