Single Period Inventory

4. Single Period Inventory

We buy some item wholesale and sell it retail. The demand for the item varies, so we treat it a random variable. We want to know how many to buy wholesale so as to maximize the expected profit.

Example 1: Each day a newsstand buys and sells The Wall Street Journal. They buy each copy wholesale for 60¢ and sell it retail for $1.00. At the end of the day they can sell any unsold copies for a salvage value of 40¢. Suppose the probability, Pr{D = k}, that people want to buy k copies in any given day is given by Pr{D = 0} = 0.21, Pr{D = 1} = 0.26, Pr{D = 2} = 0.32, Pr{D = 3} = 0.16, Pr{D = 4} = 0.05. How many copies should the newsstand buy wholesale each day in order to maximize the expected profit?

Theory: Let w = wholesale cost we pay for each item = $0.60 in the example r = retail price we receive for each item we sell to the public = $1.00 in the example s = salvage amount that we get for each item that we buy wholesale that we are unable to sell at the retail price = $0.40 in the example. We assume s < w. D = demand (the amount the public wants to buy) (a random variable)

fk = Pr{ D = k } for k = 0, 1, 2, … = probability mass function of D

(f0 = 0.21, f1= 0.26, f2 = 0.32, f3 = 0.16, f4 = 0.05 in the example)

y = number bought wholesale (a variable, but not a random variable)

Ny = number sold retail assuming one buys y wholesale (a random variable) =

Ry = rNy = revenue from retail sales (a random variable)

Sy = s(y - Ny) = revenue from the salvage of items not sold retail (a random variable)

cy = wy = cost (amount one pays wholesale for the y items) (not a random variable)

Py = Ry + Sy - cy = profit (a random variable that is a function of y)

We want to choose y to maximize p(y) = the expected profit

= E(Py) = E(Ry + Sy - cy) = E(rNy + s(y - Ny) - wy) = E( (r – s)Ny - (w – s)y )

= (r – s)E(Ny) - (w – s)y One way to think of this last formula is as follows. For each item you buy wholesale you can potentially lose w – s. However, for each that you sell retail, you reduce this potential loss by r – s.

In order to find where p(y) is maximized we consider p(y) = marginal profit = change in profit by buying one more item wholesale = p(y+1) – p(y) (the discrete analogue of ) = [(r – s)E(Ny+1) - (w – s)(y+1)] – [(r – s)E(Ny) - (w – s)y]

= (r – s)[ E(Ny+1) - E(Ny) ] - (w – s)

One has

4 - 1 E(Ny) = = + y = + y

If one uses the first formula for E(Ny+1) and the second for E(Ny) one obtains E(Ny+1) - E(Ny) = + (y + 1) - - y = = 1 – Fy where

Fk = cumulative distribution function of D = Pr{ D  k } =

The formula E(Ny+1) - E(Ny) = can be seem intuitively as follows. If one stocks one additional paper then the expected increase in sales is one more for each day the demand is greater than y. This is just the probability that the demand is greater than y which is . Thus

p(y) = (r – s)(1 – Fy) - (w – s)

= (r-w) - (r-s) Fy

Since Fy is increasing with y, it follows that p(y) is decreasing with y. The maximum of p(y) occurs at the first value of y such that p(y)  0. This is the first value of y such that

(r-w) - (r-s) Fy  0 or

Fy 

So profit is maximized if one orders y items wholesale where y is the first value of k such that the cumulative distribution function Fk is greater than or equal to .

In our example (r-w)/(r-s) = (1.00 – 0.60)/(1.00 – 0.40) = 2/3 and F0 = 0.21, F1 = 0.47, F2 = 0.79, F3 = 0.95,

F4 = 1. So the first value of k such that Fk is greater than or equal to 2/3 is k = 2. So the maximum expected profit occurs when we order y = 2 newspapers wholesale each day.

Problem 1. A news vendor buys newspapers at the start of every day to sell on the street corner. Each newspaper costs 15 cents and is sold for 25 cents. Any unsold newspapers will be bought back by the supplier for 10 cents. After observing demand for a few months the news vendor has determined that the demand follows a uniform distribution between 21 and 40 newspapers, i.e. the probability of selling k newspapers is 1/20 for k = 21, 22, …, 40 and the probability is 0 for any other value of k. Suppose the news vendor buys q newspapers at the start of the day. a. Suppose the news vendor would be able to sell k newspapers where k is greater than or equal to q. What is his profit? b. Suppose the news vendor sells k newspapers where k is less than q. What is his profit? c. What is the expected profit. For this part you can express your answer using summation notation. d. Express the sum in part c in closed form. You can use the fact that 1 + 2 + 3 + … + n = n(n+1)/2. e. Find q to maximize the result in part d.