Allocating Capacity with Demand Competition: Fixed Factor Allocation

Abstract

In practice, a supplier with limited capacity often puts capacity on allocation, i.e., rationing capacity through quantity competition of retailers rather than through a pricing mechanism. Capacity allocation is a common occurrence in industries in which capacity expansion is costly and time consuming and price is given exogenously (e.g., steel and paper). A supplier can use his prior beliefs on his own and the retailers’ needs to construct a capacity allocation mechanism for allocation of his capacity among retailers.

Appendix

Proof of Lemma 4.1

Note that $G_{12}(m_1,m_2)$ in Eq. (4.12) can be written as follows.

$$\begin{aligned} G_{12}^*= & {} \max _{m_1\in (K-m_2,K]}G_{12}(m_1, m_2)\nonumber \\= & {} \left\{ \begin{array}{ll} \pi _1^*=\max _{m_1\in (K-m_2,K]}\pi _1, ~~~~~~~~~~~~~~~~~~~~~~~~\\ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\hbox {if } m_2\le (1-\alpha )K;\\ \pi ^*_{23}=\max \Big \{\pi ^*_2=\max _{m_1\in (K-m_2,\alpha K]}\pi _2; \qquad \pi _3^*=\max _{m_1\in (\alpha K,K]}\pi _3 \Big \}, \\ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ ~~~~~~~~~~~~~~~~~~~~~~~~\hbox {if }m_2\in ((1-\alpha )K,K]. \end{array} \right. \end{aligned}$$

(A.1)

where

$$\begin{aligned} \pi _1= & {} (M-K-w)(K-m_2),\\ \pi _2= & {} (M-K-w)m_1,\\ \pi _3= & {} (M-K-w)\alpha K . \end{aligned}$$

Namely, we divide $G_{12}(m_1, m_2)$ into two sub-scenarios based on retailer 2’s order quantity: sub-scenario 1: $m_2\le (1-\alpha )K$, and sub-scenario 2: $m_2\in ((1-\alpha )K,K]$.

Immediately, we can gain the optimal value $\pi _1^*=(M- K-w)(K-m_2)$ and optimal solution $m_{12-1}^*\in (K-m_2,K]$ with the sub-scenario 1: $m_2\le (1-\alpha )K$.

Under the sub-scenario 2: $m_2\in ((1-\alpha )K,K]$, we have

$$\begin{aligned} m_{12-2}^*= & {} \arg \max _{m_1\in (K-m_2,\alpha K]}\pi _2=\left\{ \begin{array}{ll} (K-m_2)^+, &{} \hbox {if }M-w\le K; \nonumber \\ \alpha K, &{} \hbox {if }M-w>K. \end{array}\right. \\ \pi _2^*= & {} \left\{ \begin{array}{ll} (M-K-w)(K-m_2)^+, &{} \hbox {if }M-w\le K; \\ (M-K-w)\alpha K, &{} \hbox {if }M-w>K. \end{array}\right. \end{aligned}$$

(A.2)

and

$$\begin{aligned} m_{12-3}^*= & {} \arg \max _{m_1\in (K-m_2,\alpha K]}\pi _3\in [\alpha K,K],\nonumber \\ \pi ^*_3= & {} \max _{m_1\in (\alpha K, K]}\pi _3=(M-K-w)\alpha K. \end{aligned}$$

(A.3)

Comparing $\pi _2^*$ in Eq. (A.2) with $\pi _3^*$ in Eq. (A.3), we can directly gain the following result with condition $m_2\in ((1-\alpha )K,K]$ holding:

$$\begin{aligned} m_{23}^*= & {} \arg \max _{m_1}\{\pi ^*_2; \pi ^*_3\}=\left\{ \begin{array}{ll} (K-m_2)^+ &{} \hbox {if } M-w\le K; \\ \in [\alpha K,K] &{} \hbox {if }M-w>K. \end{array} \right. \\ \pi ^*_{23}= & {} \max \{\pi ^*_2; \pi ^*_3\} =\left\{ \begin{array}{ll} (M-K-w)(K-m_2)^+, &{} \hbox {if } M-w\le K; \\ (M-K-w)\alpha K, &{} \hbox {if }M-w>K. \end{array} \right. \end{aligned}$$

Consequently, by the analysis of sub-scenario 1 and sub-scenario 2, we prove the lemma.   $\square$

Proof of Theorem 4.1

To prove the theorem, we first introduce Lemma 4.5.

Lemma 4.5

Under condition $M-w\in [K,(1+\alpha )K]$, if $m_2\le \hat{\alpha }$ or $m_2\ge M-w+2\sqrt{(M-w-K)\alpha K}$, then $\frac{(M-w-m_2)^2}{4}\ge (M-w-K)\alpha K$. Furthermore, we have

(a) $M-w+2\sqrt{(M-w-K)\alpha K}>2K-(M-w)$;

(b) $\hat{\alpha }>(1-\alpha )K$;

(c) $\hat{\alpha }<2K-(M-w)$.

Proof. Under condition $M-w\in [K,(1+\alpha )K]$, inequality $\frac{(M-w-m_2)^2}{4}\ge (M-w-K)\alpha K$ is equivalent to

$$\begin{aligned} m_2^2-2(M-w)m_2+(M-w)^2-4(M-w-K)\alpha K\ge 0, \end{aligned}$$

and is also equivalent to

$$\begin{aligned} m_2\ge M-w+2\sqrt{(M-w-K)\alpha K} \quad \text{ or } \quad m_2\le M-w-2\sqrt{(M-w-K)\alpha K}. \end{aligned}$$

Note that $M-w+2\sqrt{(M-w-K)\alpha K}>2K-(M-w)$ is equivalent to $2(M-w-K)+2\sqrt{(M-w-K)\alpha K}>0$, and that $M-w-2\sqrt{(M-w-K)\alpha K}>(1-\alpha )K$ is equivalent to $M-w-(1-\alpha )K>2\sqrt{(M-w-K)\alpha K}$. Further, we have that $M-w-2\sqrt{(M-w-K)\alpha K}<2K-(M-w)$. Thus, the lemma holds.   $\square$

Now we prove the theorem. By Eq. (4.10), we compare $G_{11}^*$ and $G_{12}^*$ according the range of w, which can be divided into two cases:

Case 1: $w\ge M-K$. In the case, we have

$$\begin{aligned} G_{11}^*= & {} \left\{ \begin{array}{ll} 0, \quad m_{11}^*=0 &{} \hbox {if }m_2\in (M-w,K]; \\ \frac{(M-w-m_2)^2}{4}, \quad m_{11}^*=\frac{M-w-m_2}{2} &{} \hbox {if }m_2\in [0,M-w]. \end{array} \right. \\ G_{12}^*= & {} \left\{ \begin{array}{ll} (M-K-w)(K-m_2), \quad m_{12}^*\in [K-m_2, K] &{} \hbox {if }m_2\in [0, (1-\alpha )K]; \\ (M-K-w)(K-m_2)^+,\quad m_{12}^*=(K-m_2)^+ &{} \hbox {if }m_2\in ((1-\alpha )K, K]. \end{array} \right. \end{aligned}$$

Case 1 can be divided into the following two subcases to analyze.

Case 1.a. $\boldsymbol{w\in (M-(1-\alpha )K,M]}$.

In this subcase, $0\le M-w<(1-\alpha )K<K$. Thus, we have:

if $0\le m_2<M-w$, then $G_{11}^*=\frac{(M-w-m_2)^2}{4}>G_{12}^*=(M-K-w)(K-m_2)$. Hence, $\Pi _1(w,m_2)=\frac{(M-w-m_2)^2}{4}$ and $m_1(m_2)=m_{11}^*=\frac{M-w-m_2}{2}$;

if $m_2\in [M-w,K]$, then $G_{11}^*=0$ and $G_{12}^*$ is negative. Hence, $\Pi _1(w,m_2)=0$ and $m_1(m_2)=m_{11}^*=0$.

Case 1.a. is illustrated as follows.

Case 1.b. $\boldsymbol{w\in [M-K,M-(1-\alpha )K]}$.

In this subcase, we have $0<(1-\alpha )K<M-w<K$. Thus, we have:

if $m_2\in (0, (1-\alpha )K]$, then $G_{11}^*=\frac{(M-w-m_2)^2}{4}$ and $G_{12}^*=(M-K-w)(K-m_2)$. We have that $G_{11}^*> G_{12}^*$. Hence, $\Pi _1(w,m_2)=\frac{(M-w-m_2)^2}{4}$ and $m_1(m_2)=m_{11}^*=\frac{M-w-m_2}{2}$;

if $m_2\in ((1-\alpha )K,M-w]$, then $G_{11}^*=\frac{(M-w-m_2)^2}{4}$, and $G_{12}^*=(M-K-w)(K-m_2)^+$ is negative. Thus, we have that $G_{11}^*> G_{12}^*$. Hence, $\Pi _1(w,m_2)=\frac{(M-w-m_2)^2}{4}$ and $m_1(m_2)=m_{11}^*=\frac{M-w-m_2}{2}$;

if $m_2\in (M-w,K]$, then $G_{11}^*=0$ and $G_{12}^*=(M-K-w)(K-m_2)^+$ is negative. Thus, we have that $G_{11}^*> G_{12}^*$. Hence, $\Pi _1(w,m_2)=0$ and $m_1(m_2)=m_{11}^*=0$.

Case 1.b. is illustrated as follows.

Combining Cases 1.a. and 1.b., we have (i).

Case 2: $w<M-K$. In this subcase, we have

$$\begin{aligned} G_{11}^*= & {} \left\{ \begin{array}{ll} \frac{(M-w-m_2)^2}{4}, \quad m_{11}^*=\frac{M-w-m_2}{2} &{} \hbox {if }m_2\le 2K-(M-w),\\ (M-K-w)(K-m_2), \quad m_{12}^*=K-m_2 &{} \hbox {if }m_2\in (2K-(M-w), K]. \end{array} \right. \\ G_{12}^*= & {} \left\{ \begin{array}{ll} (M-K-w)(K-m_2), \quad m_1^*\in [K-m_2, K] &{} \hbox {if }m_2\in [0, (1-\alpha )K]; \\ (M- K-w)\alpha K , \quad m_1^*\in [\alpha K,K] &{} \hbox {if } m_2\in ((1-\alpha )K ,K]. \end{array} \right. \end{aligned}$$

The case can be divided into the following three subcases to analyze.

Case 2.a. $\boldsymbol{w\in (M-(1+\alpha )K,M-K)}$.

In this subcase, we know that $0<(1-\alpha )K<\hat{\alpha }<2K-(M-w)<K$. By Lemma 4.5, we consider the following conditions:

if $m_2\in [0,(1-\alpha )K)$, then $G^*_{11}=\frac{(M-w-m_2)^2}{4}$ and $G_{12}^*= (M-K-w)(K-m_2)$, which implies $G^*_{11}>G_{12}^*$. Therefore, we have $\Pi _1(w,m_2)=\frac{(M-w-m_2)^2}{4}$ and $m_1(m_2)=\frac{M-w-m_2}{2}$;

if $m_2\in [(1-\alpha )K,\hat{\alpha }]$, then $G^*_{11}= \frac{(M-w-m_2)^2}{4}$ and $G_{12}^*= (M-K-w)\alpha K$, which implies $G^*_{11}>G_{12}^*$ by Lemma 4.5. Therefore, we have $\Pi _1(w,m_2)=\frac{(M-w-m_2)^2}{4}$ and $m_1(m_2)=\frac{M-w-m_2}{2}$;

if $m_2\in (\hat{\alpha },2K-(M-w))$, then $G^*_{11}=\frac{(M-w-m_2)^2}{4}$ and $G_{12}^*= (M-K-w)\alpha K$, which implies $G^*_{11}\le G_{12}^*$. Therefore, we have $\Pi _1(w,m_2)=(M-K-w)\alpha K$ and $m_1(m_2)\in [\alpha K,K]$;

if $m_2\in [2K-(M-w),K]$, then $G^*_{11}= (M-K-w)(K-m_2)$ and $G_{12}^*= (M-K-w)\alpha K$, which implies $G^*_{11}<G_{12}^*$. Therefore, we have $\Pi _1(w,m_2)=(M-K-w)\alpha K$ and $m_1(m_2)\in [\alpha K,K]$.

Case 2.a. is illustrated as follows.

From the analysis of Case 2.a., we have (ii).

Case 2.b. $\boldsymbol{w\in (M-2K,M-(1+\alpha )K]}$.

In this subcase, $0<2K-(M-w)\le (1-\alpha )K<K$. Thus, we have

if $m_2\in [0,2K-(M-w))$, then $G_{11}^*=\frac{(M-w-m_2)^2}{4}$ and $G^*_{12}=(M-K-w)(K-m_2)$, which implies $G_{11}^*>G^*_{12}$. Therefore, we have $\Pi _1(w,m_2)=\frac{(M-w-m_2)^2}{4}$ and $m_1(m_2)=\frac{M-w-m_2}{2}$;

if $m_2\in [2K-(M-w),(1-\alpha )K)$, then $G_{11}^*=(M-K-w)(K-m_2)$ and $G^*_{12}=(M-K-w)(K-m_2)$, which implies $G_{11}^*=G^*_{12}$. Therefore, we have $\Pi _1(w,m_2)=(M-K-w)(K-m_2)$ and $m_1(m_2)\in [K-m_2,K]$;

if $m_2\in [(1-\alpha )K,K]$, then $G_{11}^*=(M-K-w)(K-m_2)$ and $G^*_{12}=(M- K-w)\alpha K$, which implies $G_{11}^*<G^*_{12}$. Therefore, we have $\Pi _1(w,m_2)=(M-K-w)\alpha K$ and $m_1(m_2)\in [\alpha K,K]$.

Case 2.b. is illustrated as follows.

From the analysis of Case 2.b., we have (iii).

Case 2.c. $\boldsymbol{w\le M-2K}$.

In this subcase, $0<(1-\alpha )K<K$, then we have

if $m_2\in [0,(1-\alpha )K)$, then we can obtain that $G_{11}^*=(M-K-w)(K-m_2)$ and $G_{12}^*=(M-K-w)(K-m_2)$, which implies that $G_{11}^*=G_{12}^*$. Therefore, we have $\Pi _1(w,m_2)=(M-K-w)(K-m_2)$ and $m_1(m_2)\in [K-m_2,K]$;

if $m_2\in [(1-\alpha )K,K]$, then we can obtain that $G_{11}^*=(M-K-w)(K-m_2)$ and $G_{12}^*=(M- K-w )\alpha K$, which implies that $G_{11}^*\le G_{12}^*$. Therefore, we have $\Pi _1(w,m_2)=(M-K-w)\alpha K$ and $m_1(m_2)\in [\alpha K,K]$.

Case 2.c. is illustrated as follows.

From the analysis if Case 2.c., we have (iv).   $\square$

Proof of Lemma 4.2

Note that $G_{22}(m_1, m_2)$ can be written as follows.

$$\begin{aligned} G_{22}^*= & {} \max _{m_2\in (K-m_1,K]} G_{22}(m_1, m_2)\nonumber \\= & {} \left\{ \begin{array}{ll} \hat{\pi }_1^*=\max _{m_2\in (K-m_1,K]}\hat{\pi }_1, \qquad \\ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ \hbox {if } m_1\le \alpha K;\\ \hat{\pi }^*_{23}=\max \Big \{\hat{\pi }^*_2=\max _{m_2\in (K-m_1,(1-\alpha ) K]}\hat{\pi }_2;\qquad \hat{\pi }_3^*=\max _{m_2\in ((1-\alpha ) K,K]}\hat{\pi }_3 \Big \}, \\ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ \hbox {if }m_1\in ( \alpha K,K]. \end{array} \right. \qquad \end{aligned}$$

(A.4)

where

$$\begin{aligned} \hat{\pi }_1= & {} (M-K-w) (K-m_1),\\ \hat{\pi }_2= & {} (M-K-w)m_2,\\ \hat{\pi }_3= & {} (M-K-w)(1-\alpha ) K . \end{aligned}$$

It is easy to obtain that

$$\begin{aligned} m_{21}^*= & {} \arg \max _{m_2\in (K-m_1,K]}\hat{\pi }_1\in (K-m_1,K],\nonumber \\ \hat{\pi }_1^*= & {} (M-K-w) (K-m_1). \end{aligned}$$

(A.5)

With condition $m_1\in ( \alpha K,K]$, we have

$$\begin{aligned} \hat{\pi }^*_2= & {} \max _{m_2\in (K-m_1,(1-\alpha ) K]}\hat{\pi }_2\\= & {} \left\{ \begin{array}{ll} (M-K-w)(K-m_1)^+, &{} \hbox {if }M-w\le K; \\ (M-K-w)(1-\alpha )K, &{} \hbox {if }M-w>K. \end{array}\right. \end{aligned}$$

$$\begin{aligned} \hat{\pi }^*_3= & {} \max _{m_2\in ((1-\alpha ) K, K]}\hat{\pi }_3=(M-K-w)(1-\alpha ) K , \quad m_{2}^*\in ((1-\alpha ) K,K]. \end{aligned}$$

Comparing $\pi _2^*$ with $\pi _3^*$, we can directly gain the following result with condition $m_1\in (\alpha K,K]$ holding:

$$\begin{aligned} m_{23}^*= & {} \arg \max _{m_2}\{\hat{\pi }^*_2; \hat{\pi }^*_3\}\\= & {} \left\{ \begin{array}{ll} (K-m_1)^+ &{} \hbox {if } M-w\le K; \\ \in [(1-\alpha )K,K] &{} \hbox {if }M-w>K. \end{array} \right. \\ \hat{\pi }^*_{23}= & {} \max \{\hat{\pi }^*_2; \hat{\pi }^*_3\}\\= & {} \left\{ \begin{array}{ll} (M-K-w)(K-m_1)^+, &{} \hbox {if } M-w\le K; \\ (M-K-w)(1-\alpha )K, &{} \hbox {if }M-w>K. \end{array} \right. \end{aligned}$$

Hence, the lemma holds.   $\square$

Proof of Theorem 4.2 First, we would like to introduce the following result: inequality $\frac{(M-w-m_1)^2}{4}>(M-K-w)(1-\alpha )K$ is equivalent to $m_1<\beta$ or $m_1>M-w+2\sqrt{(M-K-w)(1-\alpha )K}$. Note that inequality $\frac{(M-w-m_1)^2}{4}>(M-K-w)(1-\alpha )K$ is equivalent to $M-w-m_1>2\sqrt{(M-K-w)(1-\alpha )K}$ or $M-w-m_1<-2\sqrt{(M-K-w)(1-\alpha )K}$, and thus the result holds. This result will be implicitly used in the following proof.

With Eq. (4.13) and Lemma 4.2, we can divide the problem into the following two cases to compare $G_{21}^*$ and $G_{22}^*$ according the range of w

Case 1: $w\ge M-K$. In the case, we have

$$\begin{aligned} G_{21}^*= & {} \left\{ \begin{array}{ll} 0, \quad m_{21}^*=0 &{} \hbox {if }m_1\in (M-w,K]; \\ \frac{(M-w-m_1)^2}{4}, \quad m_{21}^*=\frac{M-w-m_1}{2} &{} \hbox {if }m_1\in [0,M-w]. \end{array} \right. \\ G_{22}^*= & {} \left\{ \begin{array}{ll} (M-K-w)(K-m_1), \quad m_{22}^*\in [K-m_1, K] &{} \hbox {if }m_1\in [0, \alpha K]; \\ (M-K-w)(K-m_1)^+,\quad m_{22}^*=(K-m_1)^+ &{} \hbox {if }m_1\in (\alpha K, K]. \end{array} \right. \end{aligned}$$

Case 1 can be divided into the following two subcases to analyze.

Case 1.a. $\boldsymbol{w\in (M-\alpha K,M]}$.

In this subcase, $0\le M-w<\alpha K<K$. Thus, we have:

if $0\le m_1<M-w$, then $G_{21}^*=\frac{(M-w-m_1)^2}{4}>G_{22}^*=(M-K-w)(K-m_1)$. Hence, $\Pi _2(w,m_1)=\frac{(M-w-m_1)^2}{4}$ and $m_2(m_1)=m_{21}^*=\frac{M-w-m_1}{2}$;

if $m_1\in [M-w,K]$, then $G_{21}^*=0$ and $G_{22}^*$ is negative. Hence, $\Pi _1(w,m_1)=0$ and $m_2(m_1)=m_{21}^*=0$.

Case 1.a. is illustrated as follows.

Case 1.b. $\boldsymbol{w\in [M-K,M-\alpha K]}$.

In this subcase, we have $0<\alpha K<M-w<K$. Thus, we have:

if $m_1\in (0, \alpha K]$, then $G_{21}^*=\frac{(M-w-m_1)^2}{4}$ and $G_{22}^*=(M-K-w)(K-m_1)$. We have that $G_{21}^*> G_{22}^*$. Hence, $\Pi _2(w,m_1)=\frac{(M-w-m_1)^2}{4}$ and $m_2(m_1)=m_{21}^*=\frac{M-w-m_1}{2}$;

if $m_1\in (\alpha K,M-w]$, then $G_{21}^*=\frac{(M-w-m_1)^2}{4}$, and $G_{22}^*=(M-K-w)(K-m_1)^+$ is negative. Thus, we have that $G_{21}^*> G_{22}^*$. Hence, $\Pi _2(w,m_1)=\frac{(M-w-m_1)^2}{4}$ and $m_2(m_1)=m_{21}^*=\frac{M-w-m_1}{2}$;

if $m_1\in (M-w,K]$, then $G_{21}^*=0$ and $G_{22}^*=(M-K-w)(K-m_1)^+$ is negative. Thus, we have that $G_{21}^*> G_{22}^*$. Hence, $\Pi _2(w,m_1)=0$ and $m_2(m_1)=m_{21}^*=0$.

Case 1.b. is illustrated as follows.

Combining Cases 1.a. and 1.b., we have (i).

Case 2: $w<M-K$. In this subcase, we have

$$\begin{aligned} G_{21}^*= & {} \left\{ \begin{array}{ll} \frac{(M-w-m_1)^2}{4}, \quad m_2^*=\frac{M-w-m_1}{2} &{} \hbox {if }m_1\le 2K-(M-w),\\ (M-K-w)(K-m_1), \quad m_2^*=K-m_1 &{} \hbox {if }m_1\in (2K-(M-w), K]. \end{array} \right. \\ G_{22}^*= & {} \left\{ \begin{array}{ll} (M-K-w)(K-m_1), \quad m_2^*\in (K-m_1, K] &{} \hbox {if }m_1\in [0, \alpha K]; \\ (M- K-w)(1-\alpha ) K , \quad m_2^*\in [(1-\alpha ) K,K] &{} \hbox {if } m_1\in (\alpha K ,K]. \end{array} \right. \end{aligned}$$

The case can be divided into the following three subcases to analyze.

Case 2.a. $\boldsymbol{M-w\in (M-(2-\alpha )K,M-K)}$.

In this subcase, we know that $0<\alpha K<\beta<2K-(M-w)<K$. By Lemma 4.5, we consider the following conditions:

if $m_1\in [0,\alpha K)$, then $G^*_{21}=\frac{(M-w-m_1)^2}{4}$ and $G_{22}^*= (M-K-w)(K-m_1)$, which implies $G^*_{21}>G_{22}^*$. Therefore, we have $\Pi _2(w,m_1)=\frac{(M-w-m_1)^2}{4}$ and $m_2(m_1)=\frac{M-w-m_1}{2}$;

if $m_1\in [\alpha K,\beta ]$, then $G^*_{21}= \frac{(M-w-m_1)^2}{4}$ and $G_{22}^*= (M-K-w)(1-\alpha ) K$, which implies $G^*_{21}\ge G_{22}^*$ by Lemma 4.5. Therefore, we have $\Pi _2(w,m_1)=\frac{(M-w-m_1)^2}{4}$ and $m_2(m_1)=\frac{M-w-m_1}{2}$;

if $m_1\in (\beta ,2K-(M-w))$, then $G^*_{21}=\frac{(M-w-m_1)^2}{4}$ and $G_{22}^*= (M-K-w)(1-\alpha ) K$, which implies $G^*_{21}< G_{22}^*$. Therefore, we have $\Pi _2(w,m_1){=}(M-K-w)(1-\alpha )K$ and $m_2(m_1)\in [(1-\alpha ) K,K]$;

if $m_1\in [2K-(M-w),K]$, then $G^*_{21}= (M-K-w)(K-m_1)$ and $G_{22}^*= (M-K-w)(1-\alpha ) K$, which implies $G^*_{21}<G_{22}^*$. Therefore, we have $\Pi _2(w,m_1)=(M-K-w)(1-\alpha )K$ and $m_2(m_1)\in [(1-\alpha ) K,K]$.

Case 2.a. is illustrated as follows.

From the analysis of Case 2.a., we have (ii).

Case 2.b. $\boldsymbol{w\in (M-2K,M-(2-\alpha )K]}$.

In this subcase, $0<2K-(M-w)\le (2-\alpha )K<K$. Thus, we have

if $m_1\in [0,2K-(M-w))$, then $G_{21}^*=\frac{(M-w-m_1)^2}{4}$ and $G^*_{22}=(M-K-w)(K-m_1)$, which implies $G_{21}^*>G^*_{22}$. Therefore, we have $\Pi _2(w,m_1)=\frac{(M-w-m_1)^2}{4}$ and $m_2(m_1)=\frac{M-w-m_1}{2}$;

if $m_1\in [2K-(M-w),\alpha K)$, then $G_{21}^*=(M-K-w)(K-m_1)$ and $G^*_{22}=(M-K-w)(K-m_1)$, which implies $G_{21}^*=G^*_{22}$. Therefore, we have $\Pi _2(w,m_1)=(M-K-w)(K-m_1)$ and $m_2(m_1)\in [K-m_1,K]$;

if $m_1\in [\alpha K,K]$, then $G_{21}^*=(M-K-w)(K-m_1)$ and $G^*_{22}=(M- K-w)(1-\alpha ) K$, which implies $G_{21}^*<G^*_{22}$. Therefore, we have $\Pi _2(w,m_1)=(M-K-w)(1-\alpha )K$ and $m_2(m_1)\in [(1-\alpha ) K,K]$.

Case 2.b. is illustrated as follows.

From the analysis of Case 2.b., we have (iii).

Case 2.c. $\boldsymbol{w\le M-2K}$.

In this subcase, $0<\alpha K<K$, then we have

if $m_1\in [0,\alpha K)$, then we can obtain that $G_{21}^*=(M-K-w)(K-m_1)$ and $G_{22}^*=(M- K-w)(K-m_1)$, which implies that $G_{21}^*=G_{22}^*$. Therefore, we have $\Pi _2(w,m_1)=(M-K-w)(K-m_1)$ and $m_2(m_1)\in [K-m_1,K]$;

if $m_1\in [\alpha K,K]$, then we can obtain that $G_{21}^*=(M-K-w)(K-m_1)$ and $G_{22}^*=(M- K-w )(1-\alpha ) K$, which implies that $G_{21}^*\le G_{22}^*$. Therefore, we have $\Pi _2(w,m_1)=(M-K-w)(1-\alpha ) K$ and $m_2(m_1)\in [(1-\alpha ) K,K]$.

Case 2.c. is illustrated as follows.

From the analysis if Case 2.c., we have (iv).   $\square$

Proof of Theorem 4.3

Case 1: $w\in [0,M-2K)$ In this case, we have:

$$\begin{aligned}&m_1(m_2)\in \, {\left\{ \begin{array}{ll} {[}K-m_2,K], &{} \text {if } m_2\in [0,(1-\alpha )K); \\ {[}\alpha K,K], &{} \text {if } m_2\in {[}(1-\alpha )K,K]. \end{array}\right. }&\\ \\&m_2(m_1)\in \, {\left\{ \begin{array}{ll} {[}K-m_1,K], &{} \text {if } m_1\in [0, \alpha K); \\ {[}(1-\alpha ) K, K], &{} \text {if } m_1\in [ \alpha K ,K]. \end{array}\right. }&\end{aligned}$$

In this case, we can obtain the Nash equilibria $m_1^*\times m_2^*\in [\alpha K,K]\times [(1-\alpha )K,K]$.

Case 2: $w\in [M-2K, M-(1+\alpha )K)$. In this case, we have:

$$\begin{aligned} m_1(m_2){\left\{ \begin{array}{ll} =\frac{M-w-m_2}{2}, &{} \hbox {if } m_2<2K-(M-w);\\ \in [K-m_2,K], &{} \hbox {if } m_2\in [2K-(M-w),(1-\alpha )K);\\ \in [\alpha K,K], &{} \hbox {if } m_2\in [(1-\alpha )K,K]. \end{array}\right. } \end{aligned}$$

$$\begin{aligned} m_2(m_1){\left\{ \begin{array}{ll} =\frac{M-w-m_1}{2}, &{} \hbox {if } m_1<2K-(M-w);\\ \in [K-m_1,K], &{} \hbox {if } m_1\in [2K-(M-w), \alpha K);\\ \in [(1-\alpha ) K, K], &{} \hbox {if } m_1\in [ \alpha K ,K]. \end{array}\right. } \end{aligned}$$

In this case, we can prove that $2K-(M-w)<\frac{M-w}{3}$ and obtain the Nash equilibria $m_1^*\times m_2^*\in [\alpha K,K]\times [(1-\alpha ) K,K]$.

Case 3: $w\in [M-(1+\alpha )K,M-(2-\alpha )K)$. In this case, we have:

$$\begin{aligned} m_1(m_2){\left\{ \begin{array}{ll} =\frac{M-w-m_2}{2}, &{} \hbox {if } m_2\le \hat{\alpha };\\ \in [\alpha K,K], &{} \hbox {if } m_2\in (\hat{\alpha },K]. \end{array}\right. } \end{aligned}$$

$$\begin{aligned} m_2(m_1){\left\{ \begin{array}{ll} =\frac{M-w-m_1}{2}, &{} \hbox {if } m_1<2K-(M-w);\\ \in [K-m_1,K], &{} \hbox {if } m_1\in [2K-(M-w), \alpha K);\\ \in [(1-\alpha ) K, K], &{} \hbox {if } m_1\in [ \alpha K ,K]. \end{array}\right. } \end{aligned}$$

Note that if $w\ge M-\frac{9\alpha K-3\sqrt{(9\alpha -4)\alpha }K}{2}$, then there exist multiple Nash equilibria $(m_1^*,m_2^*)=(\frac{M-w}{3},\frac{M-w}{3})$ and $m_1^*\times m_2^*\in [\alpha K,K]\times [\hat{\alpha },K]$, where in the later set of equilibria we have that $g_1(m_1^*,m_2^*)=\alpha K$ and $g_2(m_1^*,m_2^*)=(1-\alpha )K$. Furthermore, the former equilibrium $(\frac{M-w}{3},\frac{M-w}{3})$ dominates any equilibrium belonging to $[\alpha K,K]\times [\hat{\alpha }K,K]$ in the sense of it gains larger profit for both retailers. If $w<M-\frac{9\alpha K-3\sqrt{(9\alpha -4)\alpha }K}{2}$, then we have that in equilibrium the two retailers’ total order size is no less than K, and the equilibria are $m_1^*\times m_2^*\in [\alpha K,K]\times [\hat{\alpha },K]$.

Case 4: $w\in [M-(2-\alpha )K, M-K)$. In this case, we have:

$$\begin{aligned} m_1(m_2){\left\{ \begin{array}{ll} =\frac{M-w-m_2}{2}, &{} \hbox {if } m_2\le \hat{\alpha };\\ \in [\alpha K,K], &{} \hbox {if } m_2\in (\hat{\alpha },K]. \end{array}\right. } \end{aligned}$$

$$\begin{aligned} m_2(m_1){\left\{ \begin{array}{ll} =\frac{M-w-m_1}{2}, &{} \hbox {if } m_1\le \beta ;\\ \in [(1-\alpha )K,K], &{} \hbox {if } m_1\in (\beta , K]; \end{array}\right. } \end{aligned}$$

Note that if $w\ge M-\frac{9\alpha K-3\sqrt{(9\alpha -4)\alpha }K}{2}$, then there exist multiple Nash equilibria $(m_1^*,m_2^*)=(\frac{M-w}{3},\frac{M-w}{3})$ and $m_1^*\times m_2^*\in [\beta ,K]\times [\hat{\alpha },K]$, where in the later set of equilibria we have that $g_1(m_1^*,m_2^*)=\alpha K$ and $g_2(m_1^*,m_2^*)=(1-\alpha )K$. Furthermore, the former equilibrium $(\frac{M-w}{3},\frac{M-w}{3})$ dominates any equilibrium belonging to $[\beta ,K]\times [\hat{\alpha },K]$ in the sense of it gains larger profit for both retailers. If $w<M-\frac{9\alpha K-3\sqrt{(9\alpha -4)\alpha }K}{2}$, then we have that in equilibrium the two retailers’ total order size is no less than K, and the equilibria are $m_1^*\times m_2^*\in [\beta ,K]\times [\hat{\alpha },K]$.

Case 5: $w\in [M-K,M]$ In this case, we have:

$$\begin{aligned} m_1(m_2)={\left\{ \begin{array}{ll}\frac{M-w-m_2}{2}, &{} \hbox {if } m_2<M-w;\\ 0, &{} \hbox {if } m_2\in [M-w,K].\end{array}\right. } \end{aligned}$$

$$\begin{aligned} m_2(m_1)={\left\{ \begin{array}{ll}\frac{M-w-m_1}{2}, &{} \hbox {if } m_1<M-w;\\ 0, &{} \hbox {if } m_1\in [M-w,K]. \end{array}\right. } \end{aligned}$$

From the equations, we can gain the Nash equilibrium $(m_1^*,m_2^*)=(\frac{M-w}{3},\frac{M-w}{3})$. Combining Cases 1–5, the theorem holds.   $\square$

Proof of Lemma 4.3

(i) To prove $3K-3\sqrt{\sqrt{(9\alpha -4)\alpha }+1-3\alpha }$ is decreasing in $\alpha \in [\frac{1}{2},1]$ is equivalent to prove that $\sqrt{\sqrt{(9\alpha -4)\alpha }+1-3\alpha }$ is increasing in $\alpha \in [\frac{1}{2},1]$, which is further equivalent to

$$\begin{aligned} \sqrt{[9(\alpha +\Delta )-4](\alpha +\Delta )}+1-3(\alpha +\Delta ) >\sqrt{(9\alpha -4)\alpha }+1-3\alpha , \end{aligned}$$

where $\Delta >0$. After straightforward algebra operations, we have prove (i).

Part (ii) can be proven similarly.   $\square$

Proof of Theorem 4.4

In view of Eq. (4.14), the supplier will choose his optimal wholesale price to maximize her profit:

$$\begin{aligned} \max \left\{ \max _{w\in (0, M-\frac{9\alpha K-3\sqrt{(9\alpha -4)\alpha }K}{2})}wK; \max _{w\in [M-\frac{9\alpha K-3\sqrt{(9\alpha -4)\alpha }K}{2},M)}\frac{2(M-w)w}{3}\right\} . \end{aligned}$$

(A.6)

Note that

$$\begin{aligned} K( M-\frac{9\alpha K-3\sqrt{(9\alpha -4)\alpha }K}{2})> & {} (M-\frac{9\alpha K-3\sqrt{(9\alpha -4)\alpha }K}{2})(3\alpha -\sqrt{(9\alpha -4)\alpha })K,\quad \quad \end{aligned}$$

(A.7)

$$\begin{aligned} \arg \max _w \{\frac{2(M-w)w}{3}\}= & {} \frac{M}{2},\end{aligned}$$

(A.8)

$$\begin{aligned} \max _w \{\frac{2(M-w)w}{3}\}= & {} \frac{M^2}{6}. \end{aligned}$$

(A.9)

Therefore, if $\frac{M}{2}\le M-\frac{9\alpha K-3\sqrt{(9\alpha -4)\alpha }K}{2}$, i.e., $M\ge 9\alpha K-3\sqrt{(9\alpha -4)\alpha }K$, then by (A.7) and (A.8), we have $w^*(\alpha )=( M-\frac{9\alpha K-3\sqrt{(9\alpha -4)\alpha }K}{2})^-$ and ${\Pi }_{s}^*(\alpha )=K[( M-\frac{9\alpha K-3\sqrt{(9\alpha -4)\alpha }K}{2})^-]$.

When $\frac{M}{2}> M-\frac{9\alpha K-3\sqrt{(9\alpha -4)\alpha }K}{2}$, i.e., $M< 9\alpha K-3\sqrt{(9\alpha -4)\alpha }K$, we can show that

$$\begin{aligned} K[(M-\frac{9\alpha K-3\sqrt{(9\alpha -4)\alpha }K}{2})-] \le \frac{M^2}{6} ~~~~~~~~ ~~~~~~~~~~~~~~~~~~~~~\quad \nonumber \\ ~~~~~~~~ ~~~~~\text{ for } M\in (0, 3K-3\sqrt{\sqrt{(9\alpha -4)\alpha }+1-3\alpha }K];\quad \quad \end{aligned}$$

(A.10)

$$\begin{aligned} \text{ and } K[(M-\frac{9\alpha K-3\sqrt{(9\alpha -4)\alpha }K}{2})-] > \frac{M^2}{6} ~~~~~~~~ ~~~~~~~~~~~~~~~~~~~~~\quad \nonumber \\ ~~~~~~ ~~\text{ for } M\in ( 3K-3\sqrt{\sqrt{(9\alpha -4)\alpha }+1-3\alpha }K,9\alpha K-3\sqrt{(9\alpha -4)\alpha }K ). \quad \quad \end{aligned}$$

(A.11)

Hence, we have that

if $M\in (0, 3K-3\sqrt{\sqrt{(9\alpha -4)\alpha }+1-3\alpha }K]$, then by (A.8), (A.9) and (A.10), we have that $w^*(\alpha )= \frac{M}{2}$ and ${\Pi }_{s}^*(\alpha )= \frac{M^2}{6}$;

if $M\in (3K{-}3\sqrt{\sqrt{(9\alpha -4)\alpha }{+}1-3\alpha }K, 9\alpha K-3\sqrt{(9\alpha -4)\alpha }K)$, then by (A.8), (A.9) and (A.11), we have $w^*=M-\frac{9\alpha K-3\sqrt{(9\alpha -4)\alpha }K}{2}$ and ${\Pi }_{s}^*(\alpha )= K[(M-\frac{9\alpha K-3\sqrt{(9\alpha -4)\alpha }K}{2})^-]$.

Consequently, the theorem holds.   $\square$

Proof of Theorem 4.5

The theorem follows Theorem 4.4 with straightforward algebra operations, and hence we omit the proof for preciseness.   $\square$

Proof of Lemma 4.4

The lemma can be proven using straightforward algebra operations, and hence we omit the proof for preciseness.   $\square$

Proof of Theorem 4.6

Part (i) follows Theorem 4.4 and Proposition 4.1.

If $M<3(2-\sqrt{2})K$, then retailers’ profits are $\Pi _1^*=M^2/36$, $\Pi _2^*=M^2/36$ and $\Pi _1^*+\Pi _2^*=M^2/18$ under proportional allocation, the same as under fixed factor allocation with $\alpha =\frac{\sqrt{2}}{2}$. If $M\ge 3(2-\sqrt{2})K$, then retailers’ profits are $\Pi _1^*=\frac{3\sqrt{2}-4}{2}K^2$, $\Pi _2^*=\frac{3\sqrt{2}-4}{2}K^2$ and $\Pi _1^*+\Pi _2^*=(3\sqrt{2}-4)K^2$ under proportional allocation. Under fixed factor allocation with $\alpha =\frac{\sqrt{2}}{2}$ , retailers’ profits are $\Pi _1^*(\alpha )=(3-2\sqrt{2})K^2>\Pi _1^*$, $\Pi _2^*(\alpha )=(5\sqrt{2}-7)K^2<\Pi _2^*$, and $\Pi _1^*(\alpha )+\Pi _2^*(\alpha )=(3\sqrt{2}-4)K^2 = \Pi _1^*+\Pi _2^*$. Hence, part (ii) of the theorem holds.   $\square$

Proof of Theorem 4.7

Following Eq. (4.15), the proof is divided into two cases.

Case 1. If $K\in [0,\ \frac{M}{3-3\sqrt{\sqrt{(9\alpha -4)\alpha }+1-3\alpha }})$, then the first order condition is

$$\begin{aligned} \frac{\partial \hat{\Pi }^*_s(K)}{\partial K}=(M-c)-\big (9\alpha -3\sqrt{(9\alpha -4)\alpha }\big )K=0. \end{aligned}$$

which follows from $0<\frac{(M-c)}{9\alpha -3\sqrt{(9\alpha -4)\alpha }}<\frac{M}{3-3\sqrt{\sqrt{(9\alpha -4)\alpha }+1-3\alpha } }$. Hence,

$$\begin{aligned}{} & {} \arg \max _{K\in [0, \ \frac{M}{3-3\sqrt{\sqrt{(9\alpha -4)\alpha }+1-3\alpha } })} \{(M-\frac{9\alpha -3\sqrt{(9\alpha -4)\alpha }}{2}K)K-cK \} =\frac{ M-c }{9\alpha -3\sqrt{(9\alpha -4)\alpha }},\quad \quad \\{} & {} \max _{K\in [0, \frac{M}{3-3\sqrt{\sqrt{(9\alpha -4)\alpha }+1-3\alpha } })} \{(M-\frac{9\alpha -3\sqrt{(9\alpha -4)\alpha }}{2}K)K-cK \} =\frac{(M-c)^2}{2(9\alpha -3\sqrt{(9\alpha -4)\alpha })}. \quad \end{aligned}$$

Case 2. It is clear that

$$\begin{aligned}{} & {} \arg \max _{K\in [\frac{M}{3-3\sqrt{\sqrt{(9\alpha -4)\alpha }+1-3\alpha }}, \ \infty )} \{\frac{M^2}{6}-cK \} =\frac{M}{3-3\sqrt{\sqrt{(9\alpha -4)\alpha }+1-3\alpha } }, \\{} & {} \max _{K\in [\frac{M}{3-3\sqrt{\sqrt{(9\alpha -4)\alpha }+1-3\alpha } }, \ \infty )} \{\frac{M^2}{6}-cK \}=\frac{M^2}{6}-\frac{cM}{3-3\sqrt{\sqrt{(9\alpha -4)\alpha }+1-3\alpha } }.\quad \end{aligned}$$

Note that

$$\begin{aligned} \frac{(M-c)^2}{2(9\alpha -3\sqrt{(9\alpha -4)\alpha })}-\Big (\frac{M^2}{6}-\frac{cM}{3-3\sqrt{\sqrt{(9\alpha -4)\alpha }+1-3\alpha }}\Big )\\ =\big [M\big (\sqrt{\sqrt{(9\alpha -4)\alpha }-3\alpha +1}\big )+c\big ]^2>0. \end{aligned}$$

So, the optimal capacity is that $K^*(\alpha )=\frac{ M-c }{9\alpha -3\sqrt{(9\alpha -4)\alpha }}$. Using Theorem 4.4 and Eq. (4.15), the optimal wholesale price and profit are $(\frac{M+c}{2})^-$ and $\frac{(M-c)^2}{2(9\alpha -3\sqrt{(9\alpha -4)\alpha })}$, respectively.

Therefore, the theorem holds.   $\square$