Managerial flexibility strategies under supply and demand risks: quantity postponement vs. price postponement

Abstract

The stochastic demand and the uncertain supply/yield are common marketing operation risks. To solve these risks, firms usually apply quantity postponement strategy or price postponement strategy to improve supply chain performance. In this paper, we consider a firm’s two-stage decisions model where the firm needs to make the price decision and the quantity decision under both supply and demand risks. And the risks response depending on different postponement strategies is the key factor for the firm. When the firm adopts the quantity postponement strategy, the expected yield is equal to the market demand. Furthermore, the demand risk does not affect the firm’s pricing decision, and the supply risk forces the firm to push up the product’s sale price and to decrease order quantity. However, when the firm adopts the price postponement strategy, the expected yield is higher than market demand which indicates the firm may confront a dilemma of overproduction. Meanwhile, both demand risk and supply risk cause the decrease of order quantity. In terms of the ability of mitigating both supply and demand risks, the two postponement strategies show different performance. The quantity postponement strategy can mitigate the demand risk, but the firm can not mitigate the supply risk by adopting price postponement strategy.

Appendix: Uncertain theory

Lemma 1

(Liu [17]) Let $\xi _1$ and $\xi _2$ be independent uncertain variables with regular uncertainty distributions $\varPhi _1$ and $\varPhi _2$, respectively. If the function  $f(x_1, x_2)$ is strictly increasing with respect to $x_1$ and  strictly decreasing with respect to $x_2$, then $\xi =f(\xi _1, \xi _2)$ is also an uncertain variable with the inverse uncertainty distribution

$$\begin{aligned} \varPsi ^{-1}(\alpha )=f\left( \varPhi _1^{-1}(\alpha ), \varPhi _{2}^{-1}(1-\alpha )\right) . \end{aligned}$$

(37)

Liu and Ha [18] proved that the uncertain variable $\xi =f(\xi _1, \xi _2)$ has an expected value

$$\begin{aligned} {{\text{E}}}\left[ \xi \right] =\int ^{1}_{0}f\left( \varPhi _1^{-1}(\alpha ), \varPhi _{2}^{-1}(1-\alpha )\right) {\text{d}}\alpha . \end{aligned}$$

(38)

Lemma 2

(Liu and Ha [18]) Let $\xi$ be an uncertain variable on the uncertainty space $\left( \varTheta , \mathcal {L}, \mathcal {M}\right)$ with uncertain distribution $\varPhi (\cdot )$. If  $g: \mathcal {R}\rightarrow \mathcal {R}$ is a monotone function, the expect value of $g(\xi )$ is

$$\begin{aligned} {{\text{E}}}\left[ g\left( \xi \right) \right] =\int ^{+\infty }_{-\infty }g\left( x\right) {\text{d}}\varPhi (x). \end{aligned}$$

(39)

Lemma 3

Let $\xi$ be an uncertain variable with uncertainty distribution function $\varPhi (\cdot )$. Assume that the support of $\xi$ is $[r_{1}, r_{2}]$,  then we have

$$\begin{aligned} {{\text{E}}}\left[ \min \{\xi ,z\}\right] =z-\int ^{z}_{r_{1}}\varPhi (x){\text{d}}x, \end{aligned}$$

(40)

where $0\le r_{1}\le z\le r_{2}$.

Lemma 4

Let $\xi _{1}$ and $\xi _{2}$ be independent uncertain variables with uncertainty distribution functions $\varPhi _{1}(\cdot )$ and  $\varPhi _{2}(\cdot )$, respectively. Then the minimum

$$\begin{aligned} \xi =\text {min}\{\xi _{1},\xi _{2}\} \end{aligned}$$

(41)

has an expected value

$$\begin{aligned} \text {E}\left[ \xi \right] =\int _{-\infty }^{+\infty }x{\text{d}}\varPsi (x), \end{aligned}$$

(42)

where $\varPsi (x)=\text {max}\{\varPhi _{1}(x),\varPhi _{2}(x)\}$.

Based on the Lemma 4, we further get the following corollary.

Corollary 4

• (1) When there exist even points  $0\le r_{1}=x_{0}<x_{1}<\cdots<x_{2n}<x_{2n+1}=r_{2}<+\infty$,  if  $x\in \left[ x_{2i},x_{2i+1}\right) , i=0,1,\ldots ,n$,  then  $\varPhi _{1}(x)\ge \varPhi _{2}(x)$.  Otherwise , $\varPhi _{1}(x)\le \varPhi _{2}(x)$,  when  $x\in \left[ x_{2i-1},x_{2i}\right) , i=1,2,\ldots ,n$.  Based on these assumptions, we have

  $$\begin{aligned} {{\text{E}}}\left[ \min \{\xi _{1},\xi _{2}\}\right] =r_{2}-\sum _{i=0}^{n}\int _{x_{2i}}^{x_{2i+1}}\varPhi _{1}(x){\text{d}}x -\sum _{i=1}^{n}\int _{x_{2i-1}}^{x_{2i}}\varPhi _{2}(x){\text{d}}x. \end{aligned}$$

  (43)

• (2) When there exist uneven points  $0\le r_{1}=x_{0}<x_{1}<\cdots<{x_{2n-1}}<x_{2n}=r_{2}<+\infty$, if  $x\in \left[ x_{2i-2},x_{2i-1}\right) , {i=1,2,\ldots ,n}$,  then  $\varPhi _{1}(x)\ge \varPhi _{2}(x)$.  Otherwise,  $\varPhi _{1}(x)\le \varPhi _{2}(x)$,  when  $x\in \left[ x_{2i-1},x_{2i}\right] , i=1,2,\ldots ,n$.  Based on these assumptions, we have

  $$\begin{aligned} {{\text{E}}}\left[ \min \{\xi _{1},\xi _{2}\}\right] =r_{2}-\sum _{i=1}^{n}\int _{x_{2i-2}}^{x_{2i-1}}\varPhi _{1}(x){\text{d}}x -\sum _{i=1}^{n}\int _{x_{2i-1}}^{x_{2i}}\varPhi _{2}(x){\text{d}}x. \end{aligned}$$

  (44)

Proof

Let $\eta =\min \{\xi _{1},\xi _{2}\}$. According to operational law for calculating the uncertainty distributions, we obtain

$$\begin{aligned} \mathcal {M}\{\eta \le x\}&=\mathcal {M}\{\min \{\xi _{1},\xi _{2}\}\le x\}\nonumber \\&=\mathcal {M}\{\{\xi _{1}\le x\}\cap \{\xi _{2}\le x\}\}\nonumber \\&=\mathcal {M}\{\xi _{1}\le x\}\wedge \mathcal {M}\{\xi _{2}\le x\}. \end{aligned}$$

(45)

Furthermore, when there exist even points $0\le r_{1}=x_{0}<x_{1}<\cdots<x_{2n}<x_{2n+1}=r_{2}<+\infty$, if $x\in \left[ x_{2i},x_{2i+1}\right] , i=0,1,\ldots ,n$, then $\varPhi _{1}(x)\ge \varPhi _{2}(x)$. Otherwise, $\varPhi _{1}(x)\le \varPhi _{2}(x)$, when $x\in \left[ x_{2i-1},x_{2i}\right] , i=1,2,\ldots ,n$. Then we have

$$\begin{aligned} \varPsi (x)=\left\{ \begin{array}{lcl} 0,&{}&{}{{\text {if }}}x<x_{0}\\ \varPhi _{1}(x),&{}&{}{{\text {if }}}x_{1}>x\ge x_{0}\\ \varPhi _{2}(x),&{}&{}{{\text {if }}}x_{2}>x\ge x_{1}\\ \vdots &{}&{}\vdots \\ \varPhi _{1}(x),&{}&{}{{\text {if }}}x_{2n+1}>x\ge x_{2n}\\ 1,&{}&{}{\text {if }}x\ge x_{2n+1}.\\ \end{array} \right. \end{aligned}$$

(46)

Thus, we can obtain

$$\begin{aligned} {{\text{E}}}\left[ \min \{\xi _{1},\xi _{2}\}\right]&=\int _{0}^{+\infty }\mathcal {M}\{\eta \ge x\}{\text{d}}x-\int _{-\infty }^{0}\mathcal {M}\{\eta \le x\}{\text{d}}x\nonumber \\&=\int _{0}^{+\infty }(1-\varPsi (x)){\text{d}}x-\int _{-\infty }^{0}\varPsi (x){\text{d}}x\nonumber \\&=\int _{0}^{x_{0}}(1-0){\text{d}}x+\sum _{i=0}^{n}\int _{x_{2i}}^{x_{2i+1}}(1-\varPhi _{1}(x)){\text{d}}x \nonumber \\&\quad +\sum _{i=1}^{n}\int _{x_{2i-1}}^{x_{2i}}(1-\varPhi _{2}(x)){\text{d}}x+\int _{x_{2n+1}}^{+\infty }(1-1){\text{d}}x\nonumber \\&=x_{2n+1}-\sum _{i=0}^{n}\int _{x_{2i}}^{x_{2i+1}}\varPhi _{1}(x){\text{d}}x -\sum _{i=1}^{n}\int _{x_{2i-1}}^{x_{2i}}\varPhi _{2}(x){\text{d}}x\nonumber \\&=r_{2}-\sum _{i=0}^{n}\int _{x_{2i}}^{x_{2i+1}}\varPhi _{1}(x){\text{d}}x -\sum _{i=1}^{n}\int _{x_{2i-1}}^{x_{2i}}\varPhi _{2}(x){\text{d}}x. \end{aligned}$$

(47)

When $\varPhi _{1}(x)$ and $\varPhi _{2}(x)$ are regular, and their inverse uncertainty distributions are $\varPhi _{1}^{-1}(\alpha )$ and $\varPhi _{2}^{-1}(\alpha )$. Then we have

$$\begin{aligned} \text {E}\left[ \text {min}\{\xi _{1},\xi _{2}\}\right]&=\int _{0}^{1}\text {min} \{\varPhi _{1}^{-1}(\alpha ),\varPhi _{2}^{-1}(\alpha )\}{\text{d}}\alpha \nonumber \\&=\sum _{i=0}^{n}\int _{\alpha _{2i}}^{\alpha _{2i+1}}\varPhi _{2}^{-1}(\alpha ){\text{d}}\alpha +\sum _{i=1}^{n}\int _{\alpha _{2i- 1}}^{\alpha _{2i}}\varPhi _{1}^{-1}(\alpha ){\text{d}}\alpha , \end{aligned}$$

(48)

where $\alpha _{i}=\varPhi _{1}(x_{i})=\varPhi _{2}(x_{i}),\ i=0,1,\ldots ,n$.

The other case is similar and omitted. $\square$

According to the Lemma 1, the Corollary 4 can also rewritten as follows.

Corollary 5

If  $\varPhi _{1}(x)$  and $\varPhi _{2}(x)$ are regular, and their inverse uncertainty distributions are  $\varPhi _{1}^{-1}(\alpha )$  and  $\varPhi _{2}^{-1}(\alpha )$. Then we can be written the expectation

$$\begin{aligned} {{\text{E}}}\left[ \min \{\xi _{1},\xi _{2}\}\right] =\sum _{i=0}^{n}\int _{\alpha _{2i}}^{\alpha _{2i+1}}\varPhi _{2}^{-1}(\alpha ){\text{d}}\alpha +\sum _{i=1}^{n}\int _{\alpha _{2i-1}}^{\alpha _{2i}}\varPhi _{1}^{-1}(\alpha ){\text{d}}\alpha , \end{aligned}$$

(49)

where  $\alpha _{i}=\varPhi _{1}(x_{i})=\varPhi _{2}(x_{i}),\ i=0,1,\ldots ,2n+1$. Or the expectation can be also rewritten as

$$\begin{aligned} {{\text{E}}}\left[ \min \{\xi _{1},\xi _{2}\}\right] =\sum _{i=1}^{n}\int _{\alpha _{2i-2}}^{\alpha _{2i-1}}\varPhi _{2}^{-1}(\alpha ){\text{d}}\alpha +\sum _{i=1}^{n}\int _{\alpha _{2i-1}}^{\alpha _{2i}}\varPhi _{1}^{-1}(\alpha ){\text{d}}\alpha , \end{aligned}$$

(50)

where  $\alpha _{i}=\varPhi _{1}(x_{i})=\varPhi _{2}(x_{i}),\ i=0,1,\ldots ,2n$.

Example 1

Let $\xi _{1}$ and $\xi _{2}$ be independent uncertain variables with uncertainty distribution functions $\varPhi _{1}(\cdot )$ and $\varPhi _{2}(\cdot )$, respectively. If there exists a point $x_{1}$ which satisfies $\varPhi _{1}(x_{1})=\varPhi _{2}(x_{1})$ and $0\le r_{1}=x_{0}<x_{1}<x_{2}=r_{2}<+\infty$, assume that $\varPhi _{1}(x)\ge \varPhi _{2}(x)$ when $x\in \left[ x_{0},x_{1}\right]$, otherwise, $\varPhi _{1}(x)\le \varPhi _{2}(x)$ when $x\in \left[ x_{1},x_{2}\right]$. Then the minimum

$$\begin{aligned} \xi =\text {min}\{\xi _{1},\xi _{2}\} \end{aligned}$$

(51)

has an expected value

$$\begin{aligned} \text {E}\left[ \text {min}\{\xi _{1},\xi _{2}\}\right] =r_{2}-\int _{r_{1}}^{x_{1}}\varPhi _{1}(x){\text{d}}x-\int _{x_{1}}^{r_{2}}\varPhi _{2}(x){\text{d}}x. \end{aligned}$$

(52)

Lemma 5

Let $\xi$ be an uncertain variable with uncertainty distribution function $\varPhi (\cdot )$. Assume that the support of $\xi$ is $[r_{1}, r_{2}]$, we have

$$\begin{aligned} {{\text{E}}}\left[ \max \{\xi ,z\}\right] =r_{2}-\int ^{r_{2}}_{z}\varPhi (x){\text{d}}x, \end{aligned}$$

(53)

where $0\le r_{1}\le z\le r_{2}$.

Proof

Let $\eta =\max \{\xi ,z\}$. According to operational law for calculating the uncertainty distributions, we obtain

$$\begin{aligned} \mathcal {M}\{\eta \le x\}&=\mathcal {M}\{\max \{\xi ,z\}\le x\}\nonumber \\&=\mathcal {M}\{\{\xi \le x\}\cap \{z\le x\}\}\nonumber \\&=\mathcal {M}\{\xi \le x\}\wedge \mathcal {M}\{z\le x\}. \end{aligned}$$

(54)

Furthermore, it is clear that

$$\begin{aligned} \mathcal {M}\{z\le x\}=\left\{ \begin{array}{lcl} 0,&{}&{}{\text {if }}x<z\\ 1,&{}&{}{\text {if }}x\ge z.\\ \end{array} \right. \end{aligned}$$

(55)

Substituting Eq.(55) into Eq.(45) yields

$$\begin{aligned} \mathcal {M}\{\eta \le x\}=\left\{ \begin{array}{lcl} 0,&{}&{}{\text {if }}x<z\\ \mathcal {M}\{\xi \le x\},&{}&{}{\text {if }}x\ge z.\\ \end{array} \right. \end{aligned}$$

(56)

Thus, for arbitrarily small positive number $\varepsilon$, we can obtain

$$\begin{aligned} {{\text{E}}}\left[ \max \{\xi ,z\}\right]&=\int _{-\infty }^{+\infty }x{\text{d}}\mathcal {M}\{\eta \le x\}=\int _{r_{1}}^{r_{2}}x{\text{d}}\mathcal {M}\{\eta \le x\}\nonumber \\&=\lim _{\varepsilon \rightarrow 0}\int _{r_{1}}^{z-\varepsilon }x{\text{d}}\mathcal {M}\{\eta \ge x\}+ \lim _{\varepsilon \rightarrow 0}\int _{z-\varepsilon }^{z}x{\text{d}}\mathcal {M}\{\eta \le x\}\nonumber \\&\quad +\int _{z}^{r_{2}}x{\text{d}}\mathcal {M}\{\eta \le x\}\nonumber \\&=\lim _{\varepsilon \rightarrow 0}\int _{z-\varepsilon }^{z}x{\text{d}}\mathcal {M}\{\eta \le x\}+ \int _{z}^{r_{2}}x{\text{d}}\mathcal {M}\{\xi \le x\}\nonumber \\&=z\mathcal {M}\{\eta \le x\}-\lim _{\varepsilon \rightarrow 0}(z-\varepsilon )\mathcal {M}\{\eta \le z-\varepsilon \}+r_{2}-z\mathcal {M}\{\xi \le z\}\nonumber \\&\quad -\int _{z}^{r_{2}}\mathcal {M}\{\xi \le x\}{\text{d}}x\nonumber \\&=z\mathcal {M}\{\xi \le x\}+r_{2}-z\mathcal {M}\{\xi \le z\}-\int _{z}^{r_{2}}\mathcal {M}\{\xi \le x\}{\text{d}}x\nonumber \\&=r_{2}-\int _{z}^{r_{2}}\varPhi (x){\text{d}}x. \end{aligned}$$

(57)

The proof is complete. $\square$

Lemma 6

Let $\xi _{1}$ and $\xi _{2}$ be independent uncertain variables with uncertainty distribution function $\varPhi _{1}(\cdot )$ and  $\varPhi _{2}(\cdot )$, respectively. Then the maximum

$$\begin{aligned} \xi =\text {max}\{\xi _{1},\xi _{2}\} \end{aligned}$$

(58)

has an expected value

$$\begin{aligned} \text {E}\left[ \xi \right] =\int _{-\infty }^{+\infty }x{\text{d}}\varPsi (x), \end{aligned}$$

(59)

where $\varPsi (x)=\text {min}\{\varPhi _{1}(x),\varPhi _{2}(x)\}$.

Based on the Lemma 6, we further get the following corollary.

Corollary 6

• (1) When there exist even points  $0\le r_{1}=x_{0}<x_{1}<\cdots<x_{2n}<x_{2n+1}=r_{2}<+\infty$,  if  $x\in \left[ x_{2i},x_{2i+1}\right) , i=0,1,\ldots ,n$,  then  $\varPhi _{1}(x)\ge \varPhi _{2}(x)$.  Otherwise,  $\varPhi _{1}(x)\le \varPhi _{2}(x)$,  when  $x\in \left[ x_{2i-1},x_{2i}\right) , i=1,2,\ldots ,n$.  Based on these assumptions, we have

  $$\begin{aligned} {{\text{E}}}\left[ \max \{\xi _{1},\xi _{2}\}\right] =r_{2}-\sum _{i=0}^{n}\int _{x_{2i}}^{x_{2i+1}}\varPhi _{2}(x){\text{d}}x -\sum _{i=1}^{n}\int _{x_{2i-1}}^{x_{2i}}\varPhi _{1}(x){\text{d}}x. \end{aligned}$$

  (60)

• (2) When there exist uneven points  $0\le r_{1}=x_{0}<x_{1}<\cdots<{x_{2n-1}}<x_{2n}=r_{2}<+\infty$,  if  $x\in \left[ x_{2i-2},x_{2i-1}\right) , {i=1,2,\ldots ,n}$, then  $\varPhi _{1}(x)\ge \varPhi _{2}(x)$.  Otherwise,  $\varPhi _{1}(x)\le \varPhi _{2}(x)$,  when  $x\in \left[ x_{2i-1},x_{2i}\right) , i=1,2,\ldots ,n$.  Based on these assumptions, we have

  $$\begin{aligned} {{\text{E}}}\left[ \max \{\xi _{1},\xi _{2}\}\right] =r_{2}-\sum _{i=1}^{n}\int _{x_{2i-2}}^{x_{2i-1}}\varPhi _{2}(x){\text{d}}x -\sum _{i=1}^{n}\int _{x_{2i-1}}^{x_{2i}}\varPhi _{1}(x){\text{d}}x. \end{aligned}$$

  (61)

Proof

The proof is similar to Corollary 4 and omitted. $\square$

Example 2

Let $\xi _{1}$ and $\xi _{2}$ be independent uncertain variables with uncertainty distribution functions $\varPhi _{1}(\cdot )$ and $\varPhi _{2}(\cdot )$, respectively. If there exists a point $x_{1}$ which satisfies $\varPhi _{1}(x_{1})=\varPhi _{2}(x_{1})$ and $0\le r_{1}=x_{0}<x_{1}<x_{2}=r_{2}<+\infty$, assume that $\varPhi _{1}(x)\ge \varPhi _{2}(x)$ when $x\in \left[ x_{0},x_{1}\right]$. Otherwise, $\varPhi _{1}(x)\le \varPhi _{2}(x)$ when $x\in \left[ x_{1},x_{2}\right]$. Then the maximum

$$\begin{aligned} \xi =\text {max}\{\xi _{1},\xi _{2}\} \end{aligned}$$

(62)

has an expected value

$$\begin{aligned} \text {E}\left[ \text {max}\{\xi _{1},\xi _{2}\}\right] =r_{2}-\int _{r_{1}}^{x_{1}}\varPhi _{2}(x){\text{d}}x-\int _{x_{1}}^{r_{2}}\varPhi _{1}(x){\text{d}}x. \end{aligned}$$

(63)