Question

Wholemark is an internet order business that sells one popular New Year greeting card once a year. The cost of the paper on the which the card is printed is $0.05 per card, and the cost of printing is $0.15 per card. The company receives $2.15 per card sold. Since the cards have the current year printed on them, unsold cards have no salvage value. Their customers are from the four areas: Los Angeles, Santa Monica, Hollywood, and Pasadena. Based on past data, the number of customers from the each of the four regions is normally distributed with mean 2,000 and standard deviation of 500. (Assume these four are independent.)
What is the optimal production quality for the card? (Use Excel's NORMSINV{} function to find the Z-score. Round intermediate calculations to four decimal places. Submit your answer to the nearest whole number.)

Answer 1

Answer:

The optimal production quantity is 9,322 cards.

Step-by-step explanation:

The information provided is:

Cost of the paper = $0.05 per card

Cost of printing = $0.15 per card

Selling price = $2.15 per card

Number of region (n) = 4

Mean demand = 2000

Standard deviation = 500

Compute the total cost per card as follows:

Total cost per card = Cost of the paper + Cost of printing

                                = $0.05 + $0.15

                                = $0.20

Compute the total demand as follows:

Total demand = Mean × n

                       = 2000 × 4

                       = 8000

Compute the standard deviation of total demand as follows:

$SD_{\text{total demand}}=\sqrt{500^{2}\times 4}=1000$

Compute the profit earned per card as follows:

Profit = Selling Price - Total Cost Price

         = $2.15 - $0.20

         = $1.95

The loss incurred per card is:

Loss = Total Cost Price = $0.20

Compute the optimal probability as follows:

$\text{Optimal probability}=\frac{\text{Profit}}{\text{Profit+Loss}}$

                               $=\frac{1.95}{1.95+0.20}\\\\=\frac{1.95}{2.15}\\\\=0.9069767\\\\\approx 0.907$

Use Excel's NORMSINV{0.907} function to find the Z-score.

z = 1.322

Compute the optimal production quantity for the card as follows:

$\text{Optimal Production Quantity}=\text{Total Demand}+(z\times SD_{\text{total demand}})$

                                               $=8000+(1.322\times 1000)\\=8000+1322\\=9322$

Thus, the optimal production quantity is 9,322 cards.