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In: Accounting

(All answers were generated using 1,000 trials and native Excel functionality.) In preparing for the upcoming...

(All answers were generated using 1,000 trials and native Excel functionality.)

In preparing for the upcoming holiday season, Fresh Toy Company (FTC) designed a new doll called The Dougie that teaches children how to dance. The fixed cost to produce the doll is $100,000. The variable cost, which includes material, labor, and shipping costs, is $34 per doll. During the holiday selling season, FTC will sell the dolls for $42 each. If FTC overproduces the dolls, the excess dolls will be sold in January through a distributor who has agreed to pay FTC $10 per doll. Demand for new toys during the holiday selling season is uncertain. The normal probability distribution with an average of 60,000 dolls and a standard deviation of 15,000 is assumed to be a good description of the demand. FTC has tentatively decided to produce 60,000 units (the same as average demand), but it wants to conduct an analysis regarding this production quantity before finalizing the decision.

(a) Create a what-if spreadsheet model using formulas that relate the values of production quantity, demand, sales, revenue from sales, amount of surplus, revenue from sales of surplus, total cost, and net profit. What is the profit when demand is equal to its average (60,000 units)?
$
(b) Modeling demand as a normal random variable with a mean of 60,000 and a standard deviation of 15,000, simulate the sales of The Dougie doll using a production quantity of 60,000 units. What is the estimate of the average profit associated with the production quantity of 60,000 dolls? Round your answer in whole dollar.
$
How does this compare to the profit corresponding to the average demand (as computed in part a)?
The input in the box below will not be graded, but may be reviewed and considered by your instructor.
(c) Before making a final decision on the production quantity, management wants an analysis of a more aggressive 70,000-unit production quantity and a more conservative 50,000-unit production quantity. Run your simulation with these two production quantities. What is the average profit associated with each? Round your answer in whole dollar.
When ordering 50,000 units, the average profit is approximately $
When ordering 70,000 units, the average profit is approximately $
(d) Besides average profit, what other factors should FTC consider in determining a production quantity? Compare the four production quantities (40,000; 50,000; 60,000; and 70,000) using all these factors. What trade-offs occur?
If required, round Probability of a Loss to three decimal places and Probability of a Shortage to two decimal places. Round your answer in whole dollar.
Production
Quantity
Average
Net Profit
Profit Standard
Deviation
Maximum
Net Profit
Probability of
a Loss
Probability of
a Shortage
40,000 $ $ $
50,000 $ $ $
60,000 $ $ $
70,000 $ $ $
What is your recommendation?
The input in the box below will not be graded, but may be reviewed and considered by your instructor.

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Solutions

Expert Solution

FC = Fixed Cost $100,000
SP1 = Sales Price (during holiday season) $42
VC = Variable Cost $ 34
SP2 = Sales Price (in January – off season ) $10 per doll
Average demand 60000 toy
Standard Deviation 15000

Calculated Production forecast = ?

Average Profit:

Profit standard Deviation:

Maximum Profit:

Profit = Sales – (Variable Cost + Fixed Cost)

During the holiday season,

For 40,000 dolls, sales = 40,000 * 42 = $1,680,000

profit = $1,680,000 – (VC+FC)

VC = 34*40,000 = 1,360,000

FC = Fixed Cost = $100,000

TC = Total Cost = VC + FC = 1,360,000+100,000 = $1,460,000

profit = 1,680,000 – 1,460,000 = $220,000

Maximum Profit = $220,000

Average Profit:

Off season sale price * Demand = $10*40,000 = $400,000

average sales = ($400,000 + 1,680,000 ) / 2 = $1,040,000

Average profit = 1,040,000 – 73,000 = $967,000

Probability of a loss:

Probability = 1 - F(Z)

where F(Z) = (Qty – Miu / SD)

F(Z) = 60,000 - 40,000 / 15,000 = 1.33

Absolute value of (1 – F(Z)) = 0.33

Hence probability of loss = 0.33 = 1/3

Possibility of a Shortage = 1 – Probability of loss = 1 – 1/3 = 2/3


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