5.4 The News Vendor Problem
The news vendor model is a classic inventory model that allows for the modeling of how much to order for a decision maker facing uncertain demand for an upcoming period of time. It takes on the news vendor moniker because of the context of selling newspapers at a newsstand. The vendor must anticipate how many papers to have on hand at the beginning of a day so as to not run short or have papers left over. This idea can be generalized to other more interesting situations (e.g. air plane seats). Discussion of the news vendor model is often presented in elementary textbooks on inventory theory. The reader is referred to (Muckstadt and Sapras 2010) for a discussion of this model.
The basic model is considered a single period model; however, it can be extended to consider an analysis covering multiple (or infinite) future periods of demand. The representation of the costs within the modeling offers a number of variations.
Let \(D\) be a random variable representing the demand for the period. Let \(F(d)= P[D \leq d]\) be the cumulative distribution function for the demand. Define \(G(q,D)\) as the profit at the end of the period when \(q\) units are ordered at the start of the period with \(D\) units of demand. The parameter \(q\) is the decision variable associated with this model. Depending on the value of \(q\) chosen by the news vendor, a profit or loss will occur.
There are two cases to consider \(D \geq q\) or \(D < q\). That is, when demand is greater than or equal to the amount ordered at the start of the period and when demand is less than the amount ordered at the start of the period. If \(D \geq q\) the news vendor runs out of items, and if \(D < q\) the news vendor will have items left over.
In order to develop a profit model for this situation, we need to define some model parameters. In addition, we need to determine the amount that we might be short and the amount we might have left over in order to determine the revenue or loss associated with a choice of \(q\). Let \(c\) be the purchase cost. This is the cost that the news vendor pays the supplier for the item. Let \(s\) be the selling price. This is the amount that the news vendor charges customers. Let \(u\) be the salvage value (i.e. the amount per unit that can be received for the item after the planning period). For example, the salvage value can be the amount that the news vendor gets when selling a left over paper to a recycling center. We assume that selling price \(>\) purchase cost \(>\) salvage value.
Consider the context of a news vendor planning for the day. What is the possible amount sold? If \(D\) is the demand for the day and we start with \(q\) items, then the most that can be sold is \(\min(D, q)\). That is, if \(D\) is bigger than \(q\), you can only sell \(q\). If \(D\) is smaller than \(q\), you can only sell \(D\).
What is the possible amount left over (available for salvage)? If \(D\) is the demand and we start with \(q\), then there are two cases: 1) demand is less than the amount ordered (\(D < q\)) or 2) demand is greater than or equal to the amount ordered (\(D \geq q\)). In case 1, we will have \(q-D\) items left over. In case 2, we will have \(0\) left over. Thus, the amount left over at the end of the day is \(\max(0, q-D)\). Since the news vendor buys \(q\) items for \(c\), the cost of the items for the day is \(c \times q\). Thus, the profit has the following form:
\[\begin{equation} G(D,q) = s \times \min(D, q) + u \times \max(0, q-D) - c \times q \tag{5.15} \end{equation}\]
In words, the profit is equal to the sales revenue plus the salvage revenue minus the ordering cost. The sales revenue is the selling price times the amount sold. The salvage revenue is the salvage value times the amount left over. The ordering cost is the cost of the items times the amount ordered. To find the optimal value of \(q\), we should try to maximize the expected profit. Since \(D\) is a random variable, \(G(D,q)\) is also a random variable. Taking the expected value of both sides of Equation (5.15), yields:
\[g(q) = E[G(D,q)] = sE[\min(D, q)] + uE[\max(0, q-D)] - cq\]
Whether or not this equation can be optimized depends on the form of the distribution of \(D\); however, simulation can be used to estimate this expected value for any given \(q\). Let’s look at an example.
Example 5.3 Sly’s convenience store sells BBQ wings for 25 cents a piece. They cost 15 cents a piece to make. The BBQ wings that are not sold on a given day are purchased by a local food pantry for 2 cents each. Assuming that Sly decides to make 30 wings a day, what is the expected revenue for the wings provided that the demand distribution is as show in Table 5.3.
| \(d_{i}\) | 5 | 10 | 40 | 45 | 50 | 55 | 60 |
|---|---|---|---|---|---|---|---|
| \(f(d_{i})\) | 0.1 | 0.2 | 0.3 | 0.2 | 0.1 | 0.05 | 0.05 |
| \(F(d_{i})\) | 0.1 | 0.3 | 0.6 | 0.8 | 0.9 | 0.95 | 1.0 |
The code for the news vendor problem will require the generation of the demand from BBQ Wing demand distribution. In order to generate from this distribution, the DEmpirical class should be used. Since the cumulative distribution has already been computed, it is straight forward to write the inputs for the DEmpirical class.
double q = 30; // order qty
double s = 0.25; //sales price
double c = 0.15; // unit cost
double u = 0.02; //salvage value
double[] values = {5, 10, 40, 45, 50, 55, 60};
double[] cdf = {0.1, 0.3, 0.6, 0.8, 0.9, 0.95, 1.0};
DEmpiricalRV dCDF = new DEmpiricalRV(values, cdf);
Statistic stat = new Statistic("Profit");
double n = 100; // sample size
for (int i = 1; i <= n; i++) {
double d = dCDF.getValue();
double amtSold = Math.min(d, q);
double amtLeft = Math.max(0, q - d);
double g = s * amtSold + u * amtLeft - c * q;
stat.collect(g);
}
System.out.printf("%s \t %f %n", "Count = ", stat.getCount());
System.out.printf("%s \t %f %n", "Average = ", stat.getAverage());
System.out.printf("%s \t %f %n", "Std. Dev. = ", stat.getStandardDeviation());
System.out.printf("%s \t %f %n", "Half-width = ", stat.getHalfWidth());
System.out.println(stat.getConfidenceLevel() * 100 + "% CI = " + stat.getConfidenceInterval());Running the model for 100 days results in the following output.
Count = 100.000000
Average = 1.677500
Std. Dev. = 2.201324
Half-width = 0.436790
95.0% CI = [1.2407095787617952, 2.1142904212382048]