Question

Joe’s Slop House is a little hole in the wall restaurant that serves slow-cooked barbecue beef sandwiches. The beef for the sandwiches cooks overnight, so each evening Joe decides how much beef to cook for the next day. The beef costs $2.0 a pound and Joe sells each sandwich (filled with 1 pound of beef) for $4.5. At the end of the day, any leftover beef is given to the neighborhood church for their soup kitchen. The tax deduction for this is worth $0.5 a pound. Daily demand for the beef can be approximated by a normal distribution with a mean of 270 pounds and a standard deviation of 40 pounds.

- Cost of Shortage (Cs):

- Cost of Excess (Ce):

- What is the optimal service level?

- What is the corresponding z value?

- How many pounds of beef should Joe cook for the next day?

Answer 1

Solution:

(a) Shortage cost (Cs) i.e. cost of underbuying is calculated as,

Cs = Selling price - Cost price

Cs  = $4.5 - $2.0

Cs = $2.5

(b) Excess cost (Ce) i.e. cost of overbuying is calculated as,

Ce = Cost price - Salvage value

Ce = $2.0 - $0.5

Ce = $1.5

(c) Optimal service level is calculated as,

Service level = Cs / (Cs + Ce)

Service level = $2.5 / ($2.5 + $1.5)

Service level = 0.625 or 62.5%

(d) Z-value:

Z-value can be determined using NORMSINV function in Excel.

Z-value = NORMSINV (Service level)

Z-value = NORMSINV (0.625)

Z-value = 0.3186

(e) Amount (Q) of beef to cook for next day is calculated as,

Q = Average Demand + (Z-value x Standard deviation)

Q = 270 + (0.3186 x 40)

Q = 282.74 or 283 (Rounding off to the nearest whole number)

Quantity for next day = 283 pounds