Robust Optimization for the Loss-Averse Newsvendor Problem

Abstract

In economics and decision theory, loss aversion refers to people’s tendency to strongly prefer avoiding losses to acquiring gains. Many studies have revealed that losses are more powerful, psychologically, than gains. We initially introduce loss aversion into the decision framework of the robust newsvendor model, to provide the theoretical guidance and referential decision for loss-averse decision makers when only the mean and variance of the demand distribution are known. We obtain the explicit expression for the optimal order policy that maximizes the loss-averse newsvendor’s worst-case expected utility. We find that the robust optimal order policy for the loss-averse newsvendor is quite different from that for the risk-neutral newsvendor. Furthermore, the impacts of loss aversion level on the robust optimal order quantity and on the traditional optimal order quantity are roughly the same.

Appendices

Appendix 1

Proof of Theorem 3.1

Firstly, we prove the closed-form expression for U(q). Our proof is based on constructing a pair of primal-dual feasible solutions for (P) and (D), and make sure that they satisfy the complementary slackness condition. The optimality will then follow.

• Case 1: Three-point distribution. The remaining two bounds correspond to different three-point distribution.

  • (1a): Suppose that the smooth curve g(x) tangents the lines $l_{0}:y=\lambda [(p-s)x-(c-s)q]$, $l_{1}:y=(p-s)x-(c-s)q$ and $l_{2}:y=(p-c)q$. Let us denote these points as $x_{0}$, $x_{1}$ and $x_{2}$, where $0\le x_{0}<\frac{c-s}{p-s}q$, $\frac{c-s}{p-s}q\le x_{1}< q$ and $x_{2}\ge q$. Due to the tangency condition, they must satisfy the following system of equations:

    $$\begin{aligned}&g(x_{0})-u(\pi (q,x_{0}))=0, g'(x_{0})-u'(\pi (q,x_{0}))=0,\\&g(x_{1})-u(\pi (q,x_{1}))=0, g'(x_{1})-u'(\pi (q,x_{1}))=0,\\&g(x_{2})-u(\pi (q,x_{2}))=0, g'(x_{2})-u'(\pi (q,x_{2}))=0. \end{aligned}$$

    From the above system of equations, it is easy to get

    $$\begin{aligned} x_{0}= & {} \tfrac{\lambda (c-s)-(\lambda -1)(p-c)}{\lambda (p-s)} q, \\ x_{1}= & {} \tfrac{\lambda (c-s)+(\lambda -1)(p-c)}{\lambda (p-s)} q, \\ x_{2}= & {} \tfrac{\lambda (c-s)+(\lambda +1)(p-c)}{\lambda (p-s)} q. \end{aligned}$$

    It is easy to see that $x_{0}<\frac{c-s}{p-s}q\le x_{1}<q\le x_{2}$. And $x_{0}\ge 0$ is ensured if $\frac{c-s}{p-s}\ge \frac{\lambda -1}{2\lambda -1}$. Note that the three-point distribution has to satisfy the mean and variance constraints. This can be obtained using the probabilities of these three points constructed explicitly as:

    $$\begin{aligned} p_{x_{0}}= & {} \tfrac{\sigma ^{2}+(\mu -x_{1})(\mu -x_{2})}{(x_{0}-x_{1})(x_{0}-x_{2})},\\ p_{x_{2}}= & {} \tfrac{\sigma ^{2}+(\mu -x_{0})(\mu -x_{1})}{(x_{2}-x_{0})(x_{2}-x_{1})},\\ p_{x_{1}}= & {} 1-p_{x_{0}}-p_{x_{2}}. \end{aligned}$$

    For the solution $F^{*}(x)$ ($(x_{i},p_{x_{i}}), i=1,2,3$ construct the three-point distribution $F^{*}(x)$) to be primal feasible, we need to ensure that the values of $p_{x_{i}}$ are nonnegative. This is ensured if $\max \{(x_{1}-\mu )(\mu -x_{0}),(x_{2}-\mu )(\mu -x_{1})\}\le \sigma ^{2}\le (x_{2}-\mu )(\mu -x_{0})$. With the values of $x_{0}$, $x_{1}$ and $x_{2}$, it is easy to check that

    $$\begin{aligned} y_{0}^{*}= & {} (p-c)q+y_{2}^{*}x_{2}^{2}=\tfrac{[3(p-c)+\lambda (p-s)][(p-c)-\lambda (p-s)]q}{4(p-c)},\\ y_{1}^{*}= & {} -2y_{2}^{*}x_{2}=\tfrac{(p-s)[(p-c)+\lambda (p-s)]}{2(p-c)},\\ y_{2}^{*}= & {} \tfrac{p-s}{2(x_{1}-x_{2})}=\tfrac{-\lambda (p-s)^{2}}{4(p-c)q}, \end{aligned}$$

    is a dual feasible solution and satisfies the complementarity slackness condition with the primal feasible solution $F^{*}(x)$ that we had identified before. Therefore (P) and (D) have the same optimal objective value, which is equal to

    $$\begin{aligned} U(q)= & {} \lambda [(p-s)x_{0}-(c-s)q]p_{x_{0}}+[(p-s)x_{1}-(c-s)q]p_{x_{1}}\\&+\,(p-c)qp_{x_{2}}\\= & {} y_{0}^{*} +\mu y_{1}^{*}+ (\mu ^{2}+\sigma ^{2}) y_{2}^{*}\\= & {} \tfrac{(p-s)[\lambda (p-s)+(p-c)]\mu }{2(p-c)}+\big [(p-c)-\tfrac{(\lambda (p-s)+(p-c))^{2}}{4\lambda (p-c)}\big ]q\\&-\tfrac{\lambda (p-s)^{2}\big (\mu ^{2}+\sigma ^{2}\big )}{4(p-c)q}. \end{aligned}$$

  • (1b): Suppose that g(x) intersects $l_{0}$ at the origin and tangents $l_{1}$ and $l_{2}$. The following proof is similar to that of (1a).

• Case 2: Two-point distribution. The remaining five bounds correspond to different two-point distributions.

  • (2a): Suppose that g(x) tangents the lines $l_{0}$ and $l_{1}$ only. Let these tangent points be $\bar{x}_{0}$ and $\bar{x}_{1}$, respectively. Due to the tangency condition, we can get

    $$\begin{aligned} \bar{x}_{0}+\bar{x}_{1}= & {} 2\tfrac{c-s}{p-s}q. \end{aligned}$$

    (2)

For the solution to be primal feasible, we need to construct the two-point distribution $F^{*}(x)$ as follows by using a positive variable $\tau$:

$$\begin{aligned} \bar{x}_{0}= & {} \mu -\sigma \tau , \hbox { with probability}\; p_{\bar{x}_{0}}=\frac{1}{\tau ^{2}+1};\nonumber \\ \bar{x}_{1}= & {} \mu +\sigma /\tau , \hbox { with probability}\; p_{\bar{x}_{1}}=\frac{\tau ^{2}}{\tau ^{2}+1}. \end{aligned}$$

(3)

Substituting (3) into the above Eq. (2), we deduce that

$$\begin{aligned} \tau =\tfrac{(\mu -\frac{c-s}{p-s}q)+\sqrt{(\mu -\frac{c-s}{p-s}q)^{2}+\sigma ^{2}}}{\sigma }. \end{aligned}$$

It is easy to see that $\bar{x}_{0}<\frac{c-s}{p-s}q<\bar{x}_{1}$ and prove the following results:

• 1. $\frac{c-s}{p-s}\ge \frac{\lambda -1}{2\lambda -1}$ and $\sigma ^{2}<(x_{1}-\mu )(\mu -x_{0})\Rightarrow \sigma ^{2}<(\omega -\mu )(\mu -\hat{x}_{0})$;

• 2. $\frac{c-s}{p-s}< \frac{\lambda -1}{2\lambda -1}$ and $\sigma ^{2}<(\omega -\mu )(\mu -\hat{x}_{0})\Rightarrow \sigma ^{2}<(x_{1}-\mu )(\mu -x_{0})$;

• 3. $\sigma ^{2}<(\omega -\mu )(\mu -\hat{x}_{0})\Leftrightarrow \bar{x}_{0}>0$;

• 4. $\sigma ^{2}<(x_{1}-\mu )(\mu -x_{0})\Leftrightarrow \sqrt{(\mu -\frac{c-s}{p-s}q)^{2}+\sigma ^{2}}<\frac{\lambda -1}{\lambda }\cdot \frac{p-c}{p-s}q\Rightarrow \bar{x}_{1}<q$.

The corresponding dual solution which satisfies the complementarity slackness condition with the primal feasible solution $F^{*}(x)$ is

$$\begin{aligned} y_{0}^{*}= & {} -(c-s)q+y_{2}^{*}\bar{x}_{1}^{2}=-(c-s)q-\tfrac{(\lambda -1)(p-s)\big [\frac{c-s}{p-s}q+\sqrt{(\mu -\frac{c-s}{p-s}q)^{2}+\sigma ^{2}}\big ]^{2}}{4\sqrt{(\mu -\frac{c-s}{p-s}q)^{2}+\sigma ^{2}}},\\ y_{1}^{*}= & {} (p-s)-2y_{2}^{*}\bar{x}_{1}=\tfrac{(\lambda +1)(p-s)}{2}+\tfrac{(\lambda -1)(c-s)q}{2\sqrt{(\mu -\frac{c-s}{p-s}q)^{2}+\sigma ^{2}}},\\ y_{2}^{*}= & {} \tfrac{(\lambda -1)(p-s)}{2(\bar{x}_{0}-\bar{x}_{1})}=\tfrac{-(\lambda -1)(p-s)}{4\sqrt{(\mu -\frac{c-s}{p-s}q)^{2}+\sigma ^{2}}}. \end{aligned}$$

In this case, we still need to guarantee that the solution $(y_{0}^{*},y_{1}^{*},y_{2}^{*})$ also satisfies the dual feasibility condition by checking $y_{0}^{*}+y_{1}^{*}x+y_{2}^{*}x^{2}<(p-c)q$ for all $x\ge q$. Let $\varDelta$ be the discriminant of the quadratic function $y_{2}^{*}x^{2}+y_{1}^{*}x+[y_{0}^{*}-(p-c)q]$. If $\frac{c-s}{p-s}\ge \frac{\lambda -1}{2\lambda -1}$ and $\sigma ^{2}<(x_{1}-\mu )(\mu -x_{0})$, or else if $\frac{c-s}{p-s}<\frac{\lambda -1}{2\lambda -1}$ and $\sigma ^{2}<(\omega -\mu )(\mu -\hat{x}_{0})$, then we have

$$\begin{aligned} \varDelta= & {} (y_{1}^{*})^{2}-4y_{2}^{*}[y_{0}^{*}-(p-c)q]= \lambda (p-s)^{2}-\tfrac{(\lambda -1)(p-s)(p-c)q}{\sqrt{(\mu -\frac{c-s}{p-s}q)^{2}+\sigma ^{2}}}\\< & {} \lambda (p-s)^{2}-\tfrac{(\lambda -1)(p-s)(p-c)q}{\frac{\lambda -1}{\lambda }\cdot \frac{p-c}{p-s}q}\;\;(\hbox {from the results 2, 4})= 0. \end{aligned}$$

Since $\varDelta <0$ and $y_{2}^{*}<0$, the dual feasibility condition is satisfied. Thus the optimality holds by this pair of solutions.

• (2b): Suppose that g(x) tangents the lines $l_{1}$ and $l_{2}$. The following proof is similar to that of (2a).

• (2c): Suppose that g(x) tangents $l_{0}$ and $l_{2}$. The following proof is similar to that of (2a).

• (2d): Suppose that g(x) intersects $l_{0}$ at the origin and tangents $l_{1}$. The following proof is similar to that of (2a).

• (2e): Suppose that g(x) intersects $l_{0}$ at the origin and g(x) tangents $l_{2}$. The following proof is similar to that of (2a).

Secondly, we prove the differentiability of U(q) on $[0,+\infty ]$. We denote by $U_{1a}(q)$, $U_{1b}(q)$, $U_{2a}(q)$, $U_{2b}(q)$, $U_{2c}(q)$, $U_{2d}(q)$, $U_{2e}(q)$, respectively, the tight lower bound U(q) of cases (1a), (1b), (2a), (2b), (2c), (2d), (2e) in Theorem 3.1.

Since U(q) is a piecewise function made up of seven differentiable cases, we just need to show the differentiability of adjoining points. We indicate the proof for the differentiability of the adjoining point between case (1b) and case (2e) only, and the other proof is similar. From $\sigma ^{2}=(\hat{x}_{2}(q)-\mu )(\mu -\hat{x}_{0}(q))$, we can get that the adjoining point between case (1b) and case (2e) is

$$\begin{aligned} \hat{q}=\tfrac{(p-s)(\mu ^{2}+\sigma ^{2})}{2\mu \big [(p-c)+(\lambda -1)(c-s)-\sqrt{(\lambda -1)(c-s)((p-c)+\lambda (c-s))}\big ]}. \end{aligned}$$

It is easy to verify that

$$\begin{aligned} U_{1b}(\hat{q})=\lim _{q\rightarrow \hat{q}_{-}}U_{2e}(q)=\tfrac{(p-c)\mu ^{2}-\lambda (c-s)\sigma ^{2}}{\mu ^{2}+\sigma ^{2}}\hat{q}, \end{aligned}$$

and

$$\begin{aligned} {U}_{1b}^{'}(\hat{q})=\lim _{q\rightarrow \hat{q}_{-}}U_{2e}^{'}(q)=\tfrac{(p-c)\mu ^{2}-\lambda (c-s)\sigma ^{2}}{\mu ^{2}+\sigma ^{2}}. \end{aligned}$$

Finally, we prove the concavity of U(q) on $[0,+\infty ]$.

• (1a): For any $q\ge 0$, we can calculate that

  $$\begin{aligned} {U}_{1a}^{''}(q)=-\tfrac{\lambda (p-s)^{2}(\mu ^{2}+\sigma ^{2})}{2(p-c)q^{3}}<0. \end{aligned}$$

  So, $U_{1a}(q)$ is concave on $[0,+\infty ]$.

• (1b), (2a), (2b), (2c): These proofs are similar to that of (1a).

• (2d): It is obvious that $U_{2d}(q)$ is a linear function, so $U_{2d}(q)$ is a concave function.

• (2e): This proof is similar to that of (2d). Since the differentiable function U(q) is a piecewise function made up of seven concave cases, U(q) is concave on $[0,+\infty ]$. $\square$

Appendix 2

Proof of Theorem 3.2

• (1a): Setting

  $$\begin{aligned} {U}_{1a}^{'}(q)=(p-c)-\tfrac{[\lambda (c-s)+(\lambda +1)(p-c)]^{2}}{4\lambda (p-c)}+\tfrac{\lambda (p-s)^{2}(\mu ^{2}+\sigma ^{2})}{4(p-c)q^{2}}=0, \end{aligned}$$

  we can get a positive stationary point

  $$\begin{aligned} q_{1a}=\lambda (p-s)\sqrt{\tfrac{\mu ^{2}+\sigma ^{2}}{(\lambda (c-s)+(\lambda +1)(p-c))^{2}-4\lambda (p-c)^{2}}} \end{aligned}$$

  Since $U_{1a}(q)$ is concave, $\hbox {arg}\max \limits _{q}U_{1a}(q)=q_{1a}$. Thus, if $q_{1a}$ satisfies conditions (1a), then $q^{*}=q_{1a}$.

• (1b): This proof is similar to that of (1a).

• (2a): For any $q\ge 0$,

  $$\begin{aligned} {U}_{2a}^{'}(q)=-\tfrac{(\lambda +1)(c-s)}{2}\Big [1-\tfrac{(\lambda -1)(\mu -\frac{c-s}{p-s}q)}{(\lambda +1)\sqrt{(\mu -\frac{c-s}{p-s}q)^{2}+\sigma ^{2}}}\Big ]<0. \end{aligned}$$

  Obviously, $U_{2a}(q)$ is monotone decreasing in q on $[0,+\infty ]$, and $\hbox {arg}\max \limits _{q}U_{2a}(q)=0$. But $q=0$ does not satisfy condition (2a), so $q^{*}$ can not be attained under this case.

• (2b): Setting

  $$\begin{aligned} {U}_{2b}^{'}(q)=\tfrac{(p-c)-(c-s)}{2}+\tfrac{(p-s)(\mu -q)}{2\sqrt{(\mu -q)^{2}+\sigma ^{2}}}=0, \end{aligned}$$

  we can get a unique stationary point

  $$\begin{aligned} q_{2b}=\mu +\tfrac{\sigma }{2}\Big (\sqrt{\tfrac{p-c}{c-s}}-\sqrt{\tfrac{c-s}{p-c}}\Big ). \end{aligned}$$

  If $\tfrac{\mu }{\sigma }<\tfrac{(c-s)-(p-c)}{2(p-c)(c-s)}$, then $q_{2b}<0$. Since $U_{2b}(q)$ is concave, $\hbox {arg}\max \limits _{q}U_{2b}(q)=0$. But $q=0$ does not satisfy condition (2b). If $\tfrac{\mu }{\sigma }\ge \tfrac{(c-s)-(p-c)}{2(p-c)(c-s)}$, then $q_{2b}\ge 0$. Since $U_{2b}(q)$ is concave, $\hbox {arg}\max \limits _{q}U_{2b}(q)=q_{2b}$. To sum up, if $q_{2b}$ satisfies conditions (2b), then $q^{*}=q_{2b}$.

• (2c): This proof is similar to that of (2b).

• (2d): It is easy to verify that if $U_{2d}(q)$ is monotone decreasing in q, $\hbox {arg}\max \limits _{q}U_{2d}(q)=0$. But $q=0$ does not satisfy condition (2d), so $q^{*}$ can not be attained under this case.

• (2e): It is obvious that

  $$\begin{aligned}&U_{2e}(q)\hbox { is monotone increasing in }q,\quad \hbox {if}\; \big (\tfrac{\mu }{\sigma }\big )^{2}>\frac{\lambda (c-s)}{p-c},\\&U_{2e}(q)\hbox { is monotone decreasing in }q, \quad \hbox {if}\; \big (\tfrac{\mu }{\sigma }\big )^{2}\le \frac{\lambda (c-s)}{p-c}. \end{aligned}$$

If $\big (\tfrac{\mu }{\sigma }\big )^{2}>\frac{\lambda (c-s)}{p-c}$, $q^{*}$ can not be attained under this case.

If $\big (\tfrac{\mu }{\sigma }\big )^{2}\le \frac{\lambda (c-s)}{p-c}$, then $\hbox {arg}\max \limits _{q\in [0,+\infty ]}U_{2e}(q)=q_{2e}=0$. It is obvious that $q_{2e}$ satisfies condition (2e), so $q^{*}=q_{2e}$. $\Box$

Appendix 3

Proof of Corollary 3.1

Let $\lambda =1$, then $\frac{c-s}{p-s}\ge \frac{\lambda -1}{2\lambda -1}=0$, $x_{0}(q)=x_{1}(q)=\frac{c-s}{p-s}q$, $x_{2}(q)=\frac{(c-s)+2(p-c)}{p-s}q$ and $q_{2b}=q_{2c}=\mu +\tfrac{\sigma }{2}\Big (\sqrt{\tfrac{p-c}{c-s}}-\sqrt{\tfrac{c-s}{p-c}}\Big )$. Furthermore, the five cases of the robust optimal order quantity $q^{*}$ in Theorem 3.2 can reduce to the following three cases:

• (1a): If $\sigma ^{2}=(x_{2}(q_{1a})-\mu )(\mu -x_{0}(q_{1b}))$, then $q^{*}=q_{1a}$.

• (2b), (2c): If $\sigma ^{2}<(x_{2}(q_{2b})-\mu )(\mu -x_{1}(q_{2b}))$, or else if $(x_{2}(q_{2b})-\mu )(\mu -x_{0}(q_{2b}))<\sigma ^{2}\le (\nu (q_{2b})-\mu )(\mu -\hat{x}_{0}(q_{2b}))$, then $q^{*}=q_{2b}$.

• (2e): If $\big (\tfrac{\mu }{\sigma }\big )^{2}\le \frac{c-s}{p-c}$, then $q^{*}=q_{2e}=0$.

Moreover, it is easy to verify that

• 1. $\big (\tfrac{\mu }{\sigma }\big )^{2}>\frac{c-s}{p-c}\Rightarrow \sigma ^{2}\le (\nu (q_{2b})-\mu )(\mu -\hat{x}_{0}(q_{2b}))$.

• 2. From the result 2 in the proof of Theorem 3.1 (2b), we can get

  $$\begin{aligned} \sigma ^{2}=(x_{2}(q_{2b})-\mu )(\mu -x_{1}(q_{2b}))\Leftrightarrow \mu =\tfrac{[3(p-c)+(c-s)]\sigma }{2(p-c)}\sqrt{\tfrac{c-s}{p-c}}. \end{aligned}$$

• 3. Substitute $\mu =\tfrac{[3(p-c)+(c-s)]\sigma }{2(p-c)}\sqrt{\tfrac{c-s}{p-c}}$ into $q_{1a}$ and $q_{2b}$, we can obtain that

  $$\begin{aligned} q_{1a}=q_{2b}=\tfrac{(p-s)^{2}\sigma }{2(p-c)\sqrt{(p-c)(c-s)}}. \end{aligned}$$

To sum up, from the results 1, 2, 3, we can obtain that

$$\begin{aligned} {q^{*}=}\left\{ \begin{array}{ll} \mu +\tfrac{\sigma }{2}\Big (\sqrt{\tfrac{p-c}{c-s}}-\sqrt{\tfrac{c-s}{p-c}}\Big ), &{} \hbox {if }\big (\frac{\mu }{\sigma }\big )^{2}>\frac{c-s}{p-c};\\ 0, &{} \hbox {if } \big (\frac{\mu }{\sigma }\big )^{2}\le \frac{c-s}{p-c}. \end{array}\right. \end{aligned}$$

$\square$