Question

FlorU football programs are printed 1 week prior to each home game. Attendance averages 90,000 screaming and loyal Tators fans, of whom two-thirds usually buy the program, following a normal distribution with standard deviation of 5000 programs.  A program sells for $4 each. Unsold programs are sent to a recycling center that pays 10 cents per program. The cost to print each program is $1.

1. How many programs should be ordered per game to maximize expected profit?
2. What is the stockout risk for this order size?
3. How sensitive is the order quantity in (a) to the following estimates?
   1. The standard deviation of demand
   2. The selling price of a program
   3. The cost of recycling an unsold program

Answer by considering at least 4 values of each quantity and calculating the corresponding order quantity.  Draw a graph showing the effect of the above variables with order quantity on the Y axis.  Comment briefly (1-2 lines) on each graph.

Answer 1

The inventory management model to be used here is Single-Period Model.

Mean, = (2/3) x 90,000 = 60,000 programs.

Standard deviation, = 5,000 programs.

Cost per unit = $ 1.

Revenue per unit = $ 4.

Salvage value per unit = $ 0.10

Excess cost, C_e = Cost per unit – Salvage value per unit = 1 – 0.10 = $ 0.9

Shortage cost, C_s = Revenue per unit - Cost per unit = 4 – 1 = $ 3.

a.

Service Level, SL = C_s / (C_s + C_e­) = 3 / (3+ 0.9) = 3 / 3.9 = 0.7692

Corresponding z value from Standard Normal Distribution Table = 0.74

Optimal order quantity, S_O = + z. = 60,000 + (0.74 x 5,000) = 63,700 programs.

Number of programs to be ordered per game to maximize expected profit = 63,700.

b.

Stockout Risk for this order quantity = 1 – SL = 1 – 0.7692 = 0.2308 = 23.08%

c.

Sensitivity to Standard Deviation of Demand:

(i) s = 6,000 programs.

Optimal order quantity, S_O = + z. = 60,000 + (0.74 x 6,000) = 64,440 programs.

(ii) s = 7,000 programs.

Optimal order quantity, S_O = + z. = 60,000 + (0.74 x 7,000) = 65,180 programs.

(iii) s = 8,000 programs.

Optimal order quantity, S_O = + z. = 60,000 + (0.74 x 8,000) = 65,920 programs.

(iv) s = 9,000 programs.

Optimal order quantity, S_O = + z. = 60,000 + (0.74 x 9,000) = 66,660 programs.

Sensitivity to Selling Price of Program:

(i) Selling Price or Revenue per unit = $ 5.

Shortage cost, C_s = Revenue per unit - Cost per unit = 5 – 1 = $ 4.

Service Level, SL = C_s / (C_s + C_e­) = 4 / (4 + 0.9) = 4 / 4.9 = 0.8163

Corresponding z value from Standard Normal Distribution Table = 0.9

Optimal order quantity, S_O = + z. = 60,000 + (0.9 x 5,000) = 64,500 programs.

(ii) Selling Price or Revenue per unit = $ 6.

Shortage cost, C_s = Revenue per unit - Cost per unit = 6 – 1 = $ 5.

Service Level, SL = C_s / (C_s + C_e­) = 5 / (5 + 0.9) = 5 / 5.9 = 0.8475

Corresponding z value from Standard Normal Distribution Table = 1.03

Optimal order quantity, S_O = + z. = 60,000 + (1.03 x 5,000) = 65,150 programs.

(iii) Selling Price or Revenue per unit = $ 7.

Shortage cost, C_s = Revenue per unit - Cost per unit = 7 – 1 = $ 6.

Service Level, SL = C_s / (C_s + C_e­) = 6 / (6 + 0.9) = 6 / 6.9 = 0.8696

Corresponding z value from Standard Normal Distribution Table = 1.12

Optimal order quantity, S_O = + z. = 60,000 + (1.12 x 5,000) = 65,600 programs.

(iv) Selling Price or Revenue per unit = $ 8.

Shortage cost, C_s = Revenue per unit - Cost per unit = 8 – 1 = $ 7.

Service Level, SL = C_s / (C_s + C_e­) = 7 / (7 + 0.9) = 7 / 7.9 = 0.8861

Corresponding z value from Standard Normal Distribution Table = 1.21

Optimal order quantity, S_O = + z. = 60,000 + (1.21 x 5,000) = 66,050 programs.

Sensitivity to Cost of Recycling an Unsold Program:

(i) Salvage value per unit = $ 0.20

Excess cost, C_e = Cost per unit – Salvage value per unit = 1 – 0.20 = $ 0.8

Shortage cost, C_s = Revenue per unit - Cost per unit = 4 – 1 = $ 3.

Service Level, SL = C_s / (C_s + C_e­) = 3 / (3 + 0.8) = 3 / 3.8 = 0.7895

Corresponding z value from Standard Normal Distribution Table = 0.8

Optimal order quantity, S_O = + z. = 60,000 + (0.8 x 5,000) = 64,000 programs.

(ii) Salvage value per unit = $ 0.30

Excess cost, C_e = Cost per unit – Salvage value per unit = 1 – 0.30 = $ 0.7

Shortage cost, C_s = Revenue per unit - Cost per unit = 4 – 1 = $ 3.

Service Level, SL = C_s / (C_s + C_e­) = 3 / (3 + 0.7) = 3 / 3.7 = 0.8108

Corresponding z value from Standard Normal Distribution Table = 0.84

Optimal order quantity, S_O = + z. = 60,000 + (0.84 x 5,000) = 64,200 programs.

(iii) Salvage value per unit = $ 0.40

Excess cost, C_e = Cost per unit – Salvage value per unit = 1 – 0.40 = $ 0.6

Shortage cost, C_s = Revenue per unit - Cost per unit = 4 – 1 = $ 3.

Service Level, SL = C_s / (C_s + C_e­) = 3 / (3+ 0.6) = 3 / 3.6 = 0.8333

Corresponding z value from Standard Normal Distribution Table = 0.97

Optimal order quantity, S_O = + z. = 60,000 + (0.97 x 5,000) = 64,850 programs.

(iv) Salvage value per unit = $ 0.50

Excess cost, C_e = Cost per unit – Salvage value per unit = 1 – 0.50 = $ 0.50

Shortage cost, C_s = Revenue per unit - Cost per unit = 4 – 1 = $ 3.

Service Level, SL = C_s / (C_s + C_e­) = 3 / (3 + 0.5) = 3 / 3.5 = 0.8571

Corresponding z value from Standard Normal Distribution Table = 1.07

Optimal order quantity, S_O = + z. = 60,000 + (1.07 x 5,000) = 65,350 programs.

The graph of variation of Optimal Order Quantity with Standard Deviation of Demand is given below:

From the above graph, it can be seen that the variation of Optimal Order Quantity with Standard Deviation of Demand is exactly a straight line, since the Order Quantity changes by equal amount with change in Standard Deviation of Demand.