1. Probability

Each week Border's buys and sells Sporting Weekly. They buy each copy wholesale for 80¢ and sell it retail for $2.00. At the end of the week they can sell any unsold copies for a salvage value of 40¢. Suppose the probability, Pr{d}, that people want to buy d copies in any given week is given by Pr{d} = { (d+1)/110 for d=0,1,2,....9 (20-d)/110 for d=10,11,12,...,19 How many copies should Border's buy wholesale each week in order to maximize the expected profit? The formula 1 + 2 + … + n = (n+1)/2 might be useful to find the cumulative distribution function for the demand. 2. Originally Posted by Sally_Math Each week Border's buys and sells Sporting Weekly. They buy each copy wholesale for 80¢ and sell it retail for$2.00. At the end of the week they can sell any unsold copies for a salvage value of 40¢. Suppose the probability, Pr{d}, that people want to buy d copies in any given week is given by
Pr{d} = { (d+1)/110 for d=0,1,2,....9
(20-d)/110 for d=10,11,12,...,19
How many copies should Border's buy wholesale each week in order to maximize the expected profit?

The formula 1 + 2 + … + n = (n+1)/2 might be useful to find the cumulative distribution function for the demand.

Here is one possible approach:

Start by considering concrete cases. Let N be the number of copies bought wholesale by Borders. Let P be the profit

N = 0: P = 0.
Therefore E(P) = 0.

N = 1: P = -0.4, 1.2.
Therefore E(P) = (-0.4) Pr(d = 0) + (1.2) Pr(d > 0) = ....

N = 2: P = -0.8, 0.8, 2.4.
Therefore E(P) = (-0.8) Pr(d = 0) + (0.8) Pr(d = 1) + (2.4) Pr(d > 1) = ....

N = 3: P = -1.2, 0.4, 2.0, 3.6.
Therefore E(P) = (-1.2) Pr(d = 0) + (0.4) Pr(d = 1) + (2.0) Pr(d = 2) + (3.6) Pr(d > 2) = ....

Continue in this way until you get to a value of N where E(P) is less than for the previous value of N.

(If you're clever enough to see a pattern in the above calculations, you can come up with a general formula for E(P) and then use a spreadsheet to calculate the values of E(P) for all values of N).