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Robust Optimization for the Loss-Averse Newsvendor Problem

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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.

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References

  1. Kahneman, D., Tversky, A.: Prospect theory: an analysis of decisions under risk. Econometrica 47(2), 263–291 (1979)

    Article  MATH  Google Scholar 

  2. Fiegenbaum, A., Thomas, H.: Attitudes toward risk and the risk-return paradox: prospect theory explanations. Acad. Manag. J. 31, 85–106 (1988)

    Article  Google Scholar 

  3. Putler, D.: Incorporating reference price effects into a theory of household choice. Mark. Sci. 11, 287–309 (1992)

    Article  Google Scholar 

  4. Benartzi, S., Thaler, R.H.: Myopic loss aversion and the equity premium puzzle. Q. J. Econ. 110, 73–92 (1995)

    Article  MATH  Google Scholar 

  5. Camerer, C., Babcock, L., Loewenstein, G., Thaler, R.H.: Labor supply of New York City cabdrivers: one day at a time. Q. J. Econ. 112(2), 407–442 (1997)

    Article  Google Scholar 

  6. Bowman, D., Minehart, D., Rabin, M.: Loss aversion in a consumption–savings model. J. Econ. Behav. Org. 38, 155–178 (1999)

    Article  Google Scholar 

  7. Genesove, D., Mayer, C.: Loss aversion and seller behavior: evidence from the housing market. Q. J. Econ. 116, 1233–1260 (2001)

    Article  MATH  Google Scholar 

  8. MacCrimmon, K.R., Wehrung, D.A.: Taking Risks: The Management of Uncertainty. Free Press, New York (1986)

    Google Scholar 

  9. Duxbury, D., Summers, B.: Financial risk perception: Are individuals variance averse or loss averse? Econ. Lett. 84, 21–28 (2004)

    Article  Google Scholar 

  10. Brooks, P., Zank, H.: Loss averse behavior. J. Risk Uncertain. 31, 301–325 (2005)

    Article  MATH  Google Scholar 

  11. Schweitzer, M.E., Cachon, G.P.: Decision bias in the newsvendor problem with a known demand distribution: experimental evidence. Manag. Sci. 46, 404–420 (2000)

    Article  MATH  Google Scholar 

  12. Wang, C.X., Webster, S.: The loss-averse newsvendor problem. Omega 37, 93–105 (2009)

    Article  Google Scholar 

  13. Scarf, H.: A min-max solution of an inventory problem. In: Arrow, K., Karlin, S., Scarf, H. (eds.) Studies in The Mathematical Theory of Inventory and Production, pp. 201–209. Stanford University Press, Stanford, CA (1958)

    Google Scholar 

  14. Gallego, G., Moon, I.: The distribution free newsboy problem: review and extensions. J. Oper. Res. Soc. 44(8), 825–834 (1993)

    Article  MATH  Google Scholar 

  15. Moon, I., Choi, S.: Distribution free newsboy problem with balking. J. Oper. Res. Soc. 46(4), 537–542 (1995)

    Article  MATH  Google Scholar 

  16. Moon, I., Choi, S.: Distribution free procedures for make-to-order (MTO), make-in-advance (MIA), and composite policies. Int. J. Prod. Econ. 48(1), 21–28 (1997)

    Article  Google Scholar 

  17. Alfares, H.K., Elmorra, H.H.: The distribution-free newsboy problem: extension to the shortage penalty case. Int. J. Prod. Econ. 93–94, 465–477 (2005)

    Article  Google Scholar 

  18. Mostard, J., Koster, R., Teunter, R.: The distribution-free newsboy problem with resalable returns. Int. J. Prod. Econ. 97(3), 329–342 (2005)

    Article  Google Scholar 

  19. Yue, J., Chen, B., Wang, M.: Expected value of distribution information for the newsvendor problem. Oper. Res. 54(6), 1128–1136 (2006)

    Article  MathSciNet  MATH  Google Scholar 

  20. Liao, Y., Banerjee, A., Yan, C.: A distribution-free newsvendor model with balking and lost sales penalty. Int. J. Prod. Econ. 133, 224–227 (2011)

    Article  Google Scholar 

  21. Lee, C., Hsu, S.: The effect of advertising on the distribution-free newsboy problem. Int. J. Prod. Econ. 129, 217–224 (2011)

    Article  Google Scholar 

  22. Pal, B., Sana, S.S., Chaudhuri, K.: A distribution-free newsvendor problem with nonlinear holding cost. Int. J. Syst. Sci. 46(7), 1269C1277 (2013)

    MathSciNet  MATH  Google Scholar 

  23. Zhu, Z., Zhang, J., Ye, Y.: Newsvendor optimization with limited distribution information. Opt. Methods Softw. 28(3), 640–667 (2013)

    Article  MathSciNet  MATH  Google Scholar 

  24. Kysang, K., Taesu, C.: A minimax distribution-free procedure for a newsvendor problem with free shipping. Eur. J. Oper. Res. 232, 234–240 (2014)

    Article  MathSciNet  MATH  Google Scholar 

  25. Tversky, A., Kahneman, D.: Loss aversion in riskless choice: a reference-dependent model. Q. J. Econ. 106, 1039–1061 (1991)

    Article  Google Scholar 

  26. Barberis, N., Huang, M.: Mental accounting, loss aversion, and individual stock returns. J. Finance 56, 1247–1292 (2001)

    Article  Google Scholar 

  27. Natarajan, K., Zhou, L.: A mean-variance bound for a three-piece linear function. Prob. Eng. Inf. Sci. 21, 611–621 (2007)

    Article  MathSciNet  MATH  Google Scholar 

  28. Silver, E.A., Peterson, R.: Decision Systems for Inventory Management and Production Planning, 2nd edn. Wiley, New York (1985)

    Google Scholar 

Download references

Acknowledgments

This research was partially supported by the Fundamental Research Funds for the Central Universities of China (Grant No. CDJSK100211) and by the Scientific Research Foundation and the Project Spark of the Chongqing University of Technology.

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Correspondence to Guang-Ya Chen.

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.

  1. Case 1:

    Three-point distribution. The remaining two bounds correspond to different three-point distribution.

    1. (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

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

  2. Case 2:

    Two-point distribution. The remaining five bounds correspond to different two-point distributions.

    1. (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. 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. 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. 3.

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

  4. 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.

  1. (2b):

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

  2. (2c):

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

  3. (2d):

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

  4. (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 \([0,+\infty ]\).

  1. (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 \([0,+\infty ]\).

  2. (1b),

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

  3. (2d):

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

  4. (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

  1. (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 \(q^{*}=q_{1a}\).

  2. (1b):

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

  3. (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.

  4. (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 \(q^{*}=q_{2b}\).

  5. (2c):

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

  6. (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.

  7. (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:

  1. (1a):

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

  2. (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 \(q^{*}=q_{2b}\).

  3. (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. 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. 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. 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

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

\(\square \)

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Yu, H., Zhai, J. & Chen, GY. Robust Optimization for the Loss-Averse Newsvendor Problem. J Optim Theory Appl 171, 1008–1032 (2016). https://doi.org/10.1007/s10957-016-0870-9

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  • DOI: https://doi.org/10.1007/s10957-016-0870-9

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