Three newsvendor functions, three optimal solutions—which is correct?

Question

Here is a typical setup for the newsvendor problem:

The newsvendor buys newspapers for $c$ each, sells them for $r$ each, and salvages unsold newspapers for $v$ each. The demand distribution has pdf $f(x)$ and cdf $F(x)$. What is the optimal order quantity, $Q^*$?

Each unsold unit (newspapers that the newsvendor buys but is unable to sell) incurs a holding cost of $h = c - v$. Each unmet demand (newspapers that customers demanded but the newsvendor did not have enough inventory to supply) incurs a stockout cost of $p = r - c$.

(Holding and stockout costs are sometimes called overage and underage costs and denoted $c_o$ and $c_u$, respectively.)

Here are three ways to approach formulating the objective function (minimizing the newsvendor's expected cost or maximizing expected profit). Each leads to a different "optimal" solution.

CAUTION: Two of these approaches are wrong! Keep reading.

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Approach 1: Minimize the expected holding and stockout costs: $$g(Q) = \mathbb{E}\left[h(Q-D)^+ + p(D-Q)^+\right],$$ where $D$ is the random demand and $a^+ = \max\{0,a\}$; or, $$g(Q) = h\int_0^Q (Q-x)f(x)dx + \int_Q^\infty (x-Q)f(x)dx.$$ Taking the derivative, setting it equal to 0, and solving for $Q$, we get: $$Q^* = F^{-1}\left(\frac{p}{p+h}\right) = F^{-1}\left(\frac{r-c}{r-v}\right).$$

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Approach 2: We purchase $Q$ units at a cost of $c$ each. Then we incur a cost of $h$ per unsold item and $p$ per unmet demand, so: $$\begin{align} g(Q) & = cQ + \mathbb{E}\left[h(Q-D)^+ + p(D-Q)^+\right] \\ & = cQ + h\int_0^Q (Q-x)f(x)dx + \int_Q^\infty (x-Q)f(x)dx. \end{align}$$ Then $$Q^* = F^{-1}\left(\frac{p-c}{p+h}\right) = F^{-1}\left(\frac{r - 2c}{r-v}\right).$$

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Approach 3: We earn a revenue of $r$ per unit sold, and then we incur holding and stockout costs as above. Therefore, the expected profit is as follows: $$\begin{align} \pi(Q) & = r\mathbb{E}[\text{units sold}] - \mathbb{E}\left[h(Q-D)^+ + p(D-Q)^+\right] \\ & = r\left(\int_0^Q xf(x)dx + (1-F(Q))Q \right) - h\int_0^Q (Q-x)f(x)dx - \int_Q^\infty (x-Q)f(x)dx. \end{align}$$ This gives: $$Q^* = F^{-1}\left(\frac{r+p}{r+h+p}\right) = F^{-1}\left(\frac{2r-c}{2r-v}\right).$$

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All three approaches seem logical. Which one is right, and why?

Answer 1

This question provides a good example about a common problem we have when teaching newsvendor concepts. In some of the most useful problem settings, it can be tricky for students to properly specify holding (overage) and stockout (underage) costs. It just isn't always intuitive and it can be a disservice to students to force them to think this way in every case.

Here is another way to introduce the particular variant in this question; my claim is not that it is easier to understand, but rather that it provides a slightly different perspective that does not require determining overage and underage for a standard form newsvendor. If demand were known, profit is maximized by selecting $Q$ to minimize: $$f(Q) = cQ - r \min(Q,d) - v \max(Q-d,0) \; .$$

Note now that $\min(Q,d) = d - \max(d-Q,0)$ and thus: $$f(Q) = -rd + cQ + r \max(d-Q,0) - v \max(Q-d,0) \; .$$

To minimize expected costs, we seek to minimize: $$g(Q) = k + c Q + r E[\max(D-Q,0)] - v E[\max(Q-D,0)] \; $$ where $k$ in this case is a constant equal to expected revenue earned if we could satisfy all demand. In my view, this form is also useful as an alternative standard form for newsvendor. We pay $c$ per unit item to procure each item, we forego revenue of $r$ for each unit of demand that exceeds supply, and we receive a salvage value of $v$ for each unit of supply that exceeds demand. When $D$ is a continuous random variable, it is not difficult to derive the minimizer of $g(Q)$ directly and to show that at the optimal quantity $Q^*$: $$F_D(Q^*) = \frac{r - c}{c - v}$$.

Answer 2

Only approach 1 is correct! The other two approaches double-count one of the costs.

In particular, approach 2 double-counts the cost of purchasing units. The cost of purchasing units is already "baked into" the holding and stockout costs, so it is incorrect to include it explicitly in the objective function. In other words, in approach 2 we are charging $c$ twice for each unit purchased, which is why the "optimal" solution for this approach is the same as the optimal solution for approach 1 but with $c$ replaced with $2c$.

Approach 3, meanwhile, double-counts the revenue. The revenue is already baked into the stockout cost, so it is incorrect to include it explicitly in the objective function. We are pretending we earn $p$ twice for each unit sold, which is why the "optimal" solution for this approach is the same as the optimal solution for approach 1 but with $r$ replaced with $2r$.