Question

2. The taco stand in the atrium of the new College of Business building carried two items, fish tacos and chicken tacos. The fish tacos sell for $15 and are made out of $5 of ingredients and the chicken tacos sell for $10 and are made out of $4 of ingredients. Some days the taco stand owner has only chicken at his disposal, so he makes nothing but chicken tacos, and some days the opposite is true and he makes only fish tacos. Thus, he is able to estimate demand for chicken tacos at 2500 per day with a standard deviation of 600 and the demand for fish tacos at 2000 per day with a standard deviation of 500. Any fish or chicken tacos that do not sell at the end of the day can be sold for $1 each as bait. On days when both proteins are available, the taco stand manager prefers to make a few of each kind. All tacos are made in his home kitchen and then transported to campus. Due to time constraints and the capacity of his pickup truck bed, he is limited to beginning each day with only 3000 tacos.

a. Suppose the taco stand manager could wake up a little earlier and borrow his buddy's full size pickup to transport tacos to campus. If time and capacity were not an issue, how many total tacos should he bring to campus each day?

b. Suppose the taco stand manager could wake up a little earlier and borrow his buddy's full size pickup to transport tacos to campus. If time and capacity were not an issue, what would the expected profit be per day?

c. The taco stand manager decides to make exactly the same quantity of chicken tacos as fish tacos. If time and capacity are not an issue, and none of the other parameters in the scenario are changed, what would the retail price of chicken tacos need to be to make the optimal order quantities identical?

Answer 1

2a)

For Fish Taco -

Cost of Shortage Cs = Selling Price - Manufacturing cost = 15-5 = 10

Cost of Excess Ce = Manufacturing Cost - Salvage Cost = 5-1 = 4

Critical Ratio = Cs/(Cs+Ce) = 10/(10+4) = 0.71429

z = NORM.S.INV(0.71429) = 0.56595

Optimal Quantity = Mean + z*Sd = 2000 + 0.56595*500 = 2282.98 = 2283 units

For Chicken Taco -

Cost of Shortage Cs = Selling Price - Manufacturing cost = 10-4 = 6

Cost of Excess Ce = Manufacturing Cost - Salvage Cost = 4-1 = 3

Critical Ratio = Cs/(Cs+Ce) = 6/(3+6) = 0.66667

z = NORM.S.INV(0.66667) = 0.43074

Optimal Quantity = Mean + z*Sd = 2500 + 0.43074*600 = 2758.44 = 2758 units

Total Taco = 2283+2758 = 5041 Tacos

2b)

Per unit profit from Chicken Taco = 10-4 = 6 per unit

Per unit profit from Fish Taco = 15-5 = 10 per unit

Total per Day Profit = 2283*10 + 2758*6 = $39378

2c)

For Fish Taco -

Cost of Shortage Cs = Selling Price - Manufacturing cost = 15-5 = 10

Cost of Excess Ce = Manufacturing Cost - Salvage Cost = 5-1 = 4

Critical Ratio = Cs/(Cs+Ce) = 10/(10+4) = 0.71429

z = NORM.S.INV(0.71429) = 0.56595

Optimal Quantity = Mean + z*Sd = 2000 + 0.56595*500 = 2282.98 = 2283 units

Unit of Chicken Taco will be 2283 units

Quantity = Mean +z*Sd

2283 = 2500+z*600

z = (2283-2500)/600

z = -0.36167

Critical Ratio = NORM.S.DIST(-0.36167,1) = 0.3588

Let assume p will be the selling price of Chicken Taco.

Cost of Shortage Cs = p-4

Cost of Excess, Ce = 4-1 = 3

Critical Ratio = (p-4)/(p-4+3)

0.3588 = (p-4)/(p-1)

0.3588*p - 0.3588*1 = p-4

(1-0.3588)*p = 4 - 0.3588*1

p = (4-0.3588)/(1-0.3588)

p = 5.67873

Retail price = 5.68 per unit