University of Florida football programs are printed 1 week prior to each home game. Attendance averages 75 comma 000 screaming and loyal Gators fans, of whom two-thirds usually buy the program, following a normal distribution, for $5 each. Unsold programs are sent to a recycling center that pays only 10 cents per program. The standard deviation is 10 comma 000 programs, and the cost to print each program is $2. Refer to the standard normal tableLOADING... for z-values.
(a) What is the cost of underestimating demand for each program?
(b) What is the overage cost per program?
(c) How many programs should be ordered per game?
(d) What is the stockout risk for this order size?

Question

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Respuesta :

andromache andromache · Answer 1 · 2020-05-03T14:30:15+02:00

Answer :

a) Cost of underestimating demand = $3

b) Average cost per program =$1.90

c) number of program ordered 51,503

d) Stock out risk = 0.3878

Explaination :

As per the data given in the question,

Total purchased program = (2 ÷ 3) × 75,000 = 50,000

Unsold program = 10% × 50,000 = 5,000

a) Cost of underestimating demand = cost of each program - cost to print each program

= $5 - $2

= $3

b)Average cost per program = cost to print each program - amount got for sending it for recycling

= $2 - $0.10

= $1.90

c) Service level = Cost of underestimating demand ÷ (Cost of underestimating demand + Average cost per program)

= $3 ÷ ($3 + $1.90)

= 0.6122

So, Z is 0.3005

Therefore number of program ordered = 50,000 + 0.3005 × 5,000

= 51,502.5

= 51,503

d) Stock out risk = 1 - Service level

= 1 - 0.6122

= 0.3878

We simply applied the above formulas