The optimal order decisions of a risk-averse newsvendor under backlogging

Abstract

In this paper, we are concerned with the optimal order quantities for a risk-averse newsvendor with backlogging, where it is assumed that all or part of the unsatisfied demands can be fullfilled by backordering. The optimal order quantities are derived under the famous conditional value-at-risk (CVaR) criterion and mean-CVaR criterion on controlling the profit loss due to the uncertainty of market demands. With the optimal order quantities to the proposed models, several important monotone properties are obtained and their relationships with the existing results are discussed. Moreover, it is shown in this paper that low risk means low expected profit while high expected profit comes with high risk. Finally, we present some managerial insights for the risk-averse newsvendor model under backlogging.

Appendix

Proof of Theorem 4.1

Define

$$\begin{aligned} h(q,v) = v-\frac{1}{1-\alpha }\texttt {E\;}[v-P(q, \xi )]^+. \end{aligned}$$

By the definition of the profit function $P(q, \xi )$ in (1), it can be easily checked that

$$\begin{aligned} h(q,v)= & {} v -\frac{1}{1-\alpha }\int _0^{q}[v-(p-r)t +(c-r)q]^+dF(t)\nonumber \\&-\frac{1}{1-\alpha }\int _q^{+\infty }[v-(p-c+b -w(p_1)(p_1-c_s+b))q\nonumber \\&-(w(p_1)(p_1-c_s+b)-b )t]^+dF(t). ~~ \end{aligned}$$

(8)

It follows from Rockafellar and Uryasev (2002) that h(q, v) is jointly concave in (q, v).

We note that the CVaR model (4) is equivalent to

$$\begin{aligned} \max _{q\ge 0}[\max _{v \in R} h(q,v)].~~~~~~~~~~~~~~ \end{aligned}$$

(9)

To solve this problem, we distinguish between two cases: $b\le \frac{ w(p_1)(p_1-c_s) }{1-w(p_1)}$ and $b\ge \frac{ w(p_1)(p_1-c_s) }{1-w(p_1)}$ respectively.

(i) $b\le \frac{ w(p_1)(p_1-c_s) }{1-w(p_1)}$. The proof below aims to find the optimal solution of the inner problem of (9) when a particular q is given. We consider the following three cases.

Case 1 $v<(r-c)q$. For this case, it follows from (8) that

$$\begin{aligned} h(q, v) = v, \end{aligned}$$

and hence

$$\begin{aligned} \frac{\partial h(q,v)}{\partial v} = 1>0. \end{aligned}$$

It follows from the concavity of $h(q, \cdot )$ that the maximum of $h(q, \cdot )$ cannot be attained for this case.

Case 2 $(r-c)q \le v < (p-c)q$. For this case, it follows from (8) that

$$\begin{aligned} h(q,v)=v-\frac{1}{1-\alpha }\int _{0}^{\frac{v+(c-r)q}{p-r}}[v-(p-r)t+(c-r)q] dF(t) \end{aligned}$$

and

$$\begin{aligned} \frac{\partial h(q,v)}{\partial v}=1-\frac{1}{1-\alpha }F \left( \frac{v+(c-r)q}{p-r} \right) . \end{aligned}$$

Obviously, it satisfies $\frac{\partial h(q,v)}{\partial v}|_{v=(r-c)q}=1>0$. If $q\ge F^{-1}(1-\alpha )$, we must have

$$\begin{aligned} \frac{\partial h(q,v)}{\partial v}|_{v=(p-c)q}=1-\frac{1}{1-\alpha }F(q)\le 0. \end{aligned}$$

By the continuous differentiability of $h(q, \cdot )$, there exists $v^* \in [(r-c)q, (p-c)q)$ such that

$$\begin{aligned} \frac{\partial h(q,v)}{\partial v}|_{v= v^*} = 1-\frac{1}{1-\alpha }F\left( \frac{v^*+(c-r)q}{p-r}\right) =0, \end{aligned}$$

which implies

$$\begin{aligned} v^* = (p-r)F^{-1}(1-\alpha )-(c-r)q. \end{aligned}$$

It follows from the concavity of $h(q, \cdot )$ that the maximum of $h(q, \cdot )$ is attained at $v^*$ when $q\ge F^{-1}(1-\alpha )$.

Case 3. $v \ge (p-c)q$. For this case, it follows from (8) that

$$\begin{aligned} h(q,v)= & {} v-\frac{1}{1-\alpha }\int _0^{q}[v-(p-r)t+(c-r)q] dF(t) -\frac{1}{1-\alpha }\int _q^{\frac{v-(p-c+b-w(p_1)(p_1-c_s+b))q }{w(p_1)(p_1-c_s+b)-b}}[v- \\&(p-c+b-w(p_1)(p_1-c_s+b))q-(w(p_1)(p_1-c_s+b)-b)t]dF(t),~~~~~~~~~~~~~~~~ \end{aligned}$$

and

$$\begin{aligned} \frac{\partial h(q,v)}{\partial v}=1-\frac{1}{1-\alpha } F \left( \frac{v-(p-c+b-w(p_1)(p_1-c_s+b))q }{w(p_1)(p_1-c_s+b)-b} \right) . \end{aligned}$$

Note that $\alpha >0$, we have

$$\begin{aligned} \lim _{v \rightarrow +\infty } \frac{\partial h(q,v)}{\partial v} < 0. \end{aligned}$$

If $q\le F^{-1}(1-\alpha )$, we have

$$\begin{aligned} \frac{\partial h(q,v)}{\partial v}|_{v=(p-c)q} = 1-\frac{1}{1-\alpha }F(q) \ge 0. \end{aligned}$$

Due to the continuous differentiability of $h(q, \cdot )$, there must exist $v^*$ such that

$$\begin{aligned} \frac{\partial h(q,v)}{\partial v}|_{v= v^*} =1-\frac{1}{1-\alpha } F \left( \frac{v^*-(p-c+b-w(p_1)(p_1-c_s+b))q }{w(p_1)(p_1-c_s+b)-b} \right) =0, \end{aligned}$$

which implies

$$\begin{aligned} v^*=(w(p_1)(p_1-c_s+b)-b)F^{-1}(1-\alpha )+(p-c+b-w(p_1)(p_1-c_s+b))q. \end{aligned}$$

It follows from the concavity of $h(q, \cdot )$ that the maximum of $h(q, \cdot )$ is attained at $v^*$ when $q\le F^{-1}(1-\alpha )$.

Putting together the three cases, we have proved that the optimal solution $v^*$ to the inner problem of (9) is given by

$$\begin{aligned} v^{*} = \left\{ \begin{array}{llll} (p-r)F^{-1}(1-\alpha )-(c-r)q&{}~~q\ge F^{-1}(1-\alpha ) ,\\ (w(p_1)(p_1-c_s+b)-b)F^{-1}(1-\alpha )\\ +(p-c+b-w(p_1)(p_1-c_s+b))q &{}~~ q \le F^{-1}(1-\alpha ). \\ \end{array}\right. \end{aligned}$$

(10)

Therefore, the problem (9) is reduced to

$$\begin{aligned} \max _{q \ge 0} h(q, v^*).~~ \end{aligned}$$

(11)

Since $v^*$ is a linear function of q, $h(q, v^*)$ is concave in q due to the fact that h(q, v) is jointly concave in (q, v). The rest of the proof is to show that the maximum of $h(q, v^*)$ is attained.

We consider two situations. One is $q\ge F^{-1}(1-\alpha )$. It follows from (8) and (10) that

$$\begin{aligned}&h(q,v^*) = (p-r)F^{-1}(1-\alpha )- (c-r)q\\&-\frac{1}{1 - \alpha }\int _{0 }^{ F^{-1}(1-\alpha )} [(p-r)(F^{-1}(1-\alpha )-t)]dF(t) \end{aligned}$$

and

$$\begin{aligned} \frac{\partial h(q,v^*)}{\partial q}=-(c-r)<0. \end{aligned}$$

By the concavity of $h(q, v^*)$ in q, the maximum of $h(q, v^*)$ cannot be achieved for this situation.

The remaining situation is $q\le F^{-1}(1-\alpha )$. It follows from (8) and (10) that

$$\begin{aligned} h(q,v^*)= & {} (w(p_1)(p_1-c_s+b)-b)F^{-1}(1-\alpha )+(p-c+b-w(p_1)(p_1-c_s+b))q~~~~~~~~~~~~~~~~ \\&- \frac{1}{1-\alpha }\int _{0 }^{q} [(w(p_1)(p_1-c_s+b)-b)F^{-1}(1-\alpha )\\&-(p-r)t+(p-r+b-w(p_1)(p_1-c_s+b))q]dF(t) \\&- \frac{1}{1-\alpha }\int _{q }^{F^{-1}(1-\alpha )}[ (w(p_1)(p_1-c_s+b)-b)(F^{-1}(1-\alpha )-t)]dF(t) \end{aligned}$$

and

$$\begin{aligned} \frac{\partial h(q,v^*)}{\partial q}=p-c+b-w(p_1)(p_1-c_s+b)-\frac{p-r+b-w(p_1)(p_1-c_s+b)}{1-\alpha }F(q). \end{aligned}$$

Since it satisfies

$$\begin{aligned} \frac{\partial h(q,v^*)}{\partial q}|_{q=0}= & {} p-c+b-w(p_1)(p_1-c_s+b)\ge 0,\\&\frac{\partial h(q,v^*)}{\partial q}|_{q=F^{-1}(1-\alpha )}= -(c-r)\le 0 . \end{aligned}$$

There must exist $q_1^*\in (0, F^{-1}(1-\alpha ))$ such that

$$\begin{aligned}&\frac{\partial h(q_1^*,v^*)}{\partial q}=p-c+b-w(p_1)(p_1-c_s+b)\\&\qquad -\frac{p-r+b-w(p_1)(p_1-c_s+b)}{1-\alpha }F(q_1^*)=0, \end{aligned}$$

which gives

$$\begin{aligned} q_1^*=F^{-1} \left( \frac{(1-\alpha )(p-c+b-w(p_1)(p_1-c_s+b)}{p-r+b-w(p_1)(p_1-c_s+b)} \right) . \end{aligned}$$

It follows from the concavity of $h(q, v^*)$ that $q^*_1$ is the optimal solution of (11), and hence is also the optimal solution of (9) for the case of i).

(ii) $b\ge \frac{w(p_1)(p_1-c_s) }{1-w(p_1)}$. The proof is similar to the case of i), which is omitted herein. This completes the proof. $\square$

Proof of Corollory. 4.2

Since $F(\cdot )$ is monotonically increasing, it follows from the formula of $q_1^*$ that $q_1^*$, as a function of c, $p_1$, b, r and $c_s$, has the corresponding increasing or decreasing properties as stated in the corollary. This completes the proof. $\square$

Proof of Corollory. 4.3

The proof is similar to that of Corollary. 4.2, which is omitted herein. $\square$

Proof of Corollory 4.4

It follows from (1) that

$$\begin{aligned} \texttt {E\;}[P(q, \xi )]= & {} (p-c+b-w(p_1)(p_1-c_s+b))q+(w(p_1)(p_1-c_s+b)-b)E(\xi )\\&-(p-r+b-w(p_1)(p_1-c_s+b))\int _0^q(q-t)dF(t).~~ \end{aligned}$$

We calculate

$$\begin{aligned} \frac{\partial \texttt {E\;}[P(q^*_1,\xi )]}{\partial \alpha }= & {} [p-c+b-w(p_1)(p_1-c_s+b)\nonumber \\&-(p-r+b-w(p_1)(p_1-c_s+b))F(q^*_1)]\frac{\partial q^*_1}{\partial \alpha }. ~ \end{aligned}$$

(12)

If $b \le \frac{w(p_1)(p_1-c_s) }{1-w(p_1)}$, it follows with Theorem 4.1 that $q^*_1=F^{-1} \Big ( \frac{(1-\alpha )(p-c+b-w(p_1)(p_1-c_s+b))}{p-r+b-w(p_1)(p_1-c_s+b)}\Big )$ and $\frac{\partial q^*_1 }{\partial \alpha }\le 0$. We recall that

$$\begin{aligned} q^*_0=F^{-1} \left( \frac{p-c+b-w(p_1)(p_1-c_s+b)}{p-r+b-w(p_1)(p_1-c_s+b)} \right) , \end{aligned}$$

which gives

$$\begin{aligned} p-c+b -w(p_1)(p_1-c_s+b)-(p-r+b-w(p_1)(p_1-c_s+b))F(q^*_0)=0. \end{aligned}$$

We also recall $q_1^* \le q_0^*$. The monotonicity of $F(\cdot )$ implies

$$\begin{aligned}&p-c+b -w(p_1)(p_1-c_s+b)-(p-r+b-w(p_1)(p_1-c_s+b))F(q^*_1)\ge ~~\\&p-c +b-w(p_1)(p_1-c_s+b)-(p-r+b-w(p_1)(p_1-c_s+b))F(q^*_0)=0. \end{aligned}$$

It follows with (12) and $\frac{\partial q^*_1 }{\partial \alpha }\le 0$ that

$$\begin{aligned} \frac{\partial \texttt {E\;}[P(q^*_1, \xi )]}{\partial \alpha } \le 0, \end{aligned}$$

which proves that $\texttt {E\;}[P(q^*_1, \xi )]$ is decreasing in the confidence level $\alpha$ when $b \le \frac{w(p_1)(p_1-c_s) }{1-w(p_1)}$. The proof for the case of $b\ge \frac{w(p_1)(p_1-c_s) }{1-w(p_1)}$ is similar to that for the case of $b \le \frac{w(p_1)(p_1-c_s) }{1-w(p_1)}$, which is omitted herein. This completes the proof. $\square$

Proof of Theorem 4.5

The proof is similar to that for Theorem 4.1, which is omitted herein. $\square$