Supply commitment contract in capacity allocation games

Abstract

This paper considers a supply chain with one supplier and two retailers. The supplier has limited normal capacity. When the orders of two retailers exceed the capacity, the supplier can satisfy the retailers with the limited normal capacity, or he can start an emergent production to increase his capacity in a short time with a high unit production cost. When the cost of quickly expanding capacity is lower than the market price, it is profitable for the whole system to start the emergent production. But the supplier has no incentives to do additional production if his cost is higher than the wholesale price. To resolve this issue, we propose a supply commitment contract that the supplier will fully fill a retailer’s order if the retailer pays a premium to the supplier. We consider a practical setting that each retailer has private information on her average demand. Two contracting scenarios of the supplier are compared: contracting with the two retailers simultaneously (the simultaneous scenario) and sequentially (the sequential scenario). It is shown that the retailers will adopt the threshold strategy based on their private demand information to pay the premium. By numerically studies, we further illustrate the supplier’s optimal contract choices and the consequent system performance. First, the suppliers always prefer the retailers to move simultaneous although the supplier’s profit is only slightly higher than that of the sequential scenario; second, the supplier sets a lower wholesale price and a higher commitment fee in the sequential scenario than those of simultaneous scenario; finally, the sequential scenario is a better choice from the perspectives of both the whole supply chain system and the retailers. Our research recommends the sequential moving in the supply commitment contract that it benefits the whole supply chain system with only slightly less efficient for the supplier.

Appendix

1.1 Proof of Theorem 1

We prove this theorem by showing that, ordering $q_i=\mu _i+q^*$ is a dominate strategy of Ri.

If Ri has purchased the guaranteed supply, then her order will be fully satisfied, thus she will place an order that maximizes her expected profit $pE(q_i\wedge (\mu _i+\epsilon _i))-wq_i=\pi (q_i-\mu _i)+(p-w)\mu _i$. Note that

$$\begin{aligned} \begin{aligned}&\frac{d\pi }{dq}(q)=p(1-G(q))-w,\\&\frac{d^2\pi }{dq^2}(q)=-pg(q))\le 0. \end{aligned} \end{aligned}$$

We know that, $\pi$ is concave and achieves maximum at $q^*$, in which $G(q^*)=(p-w)/p$. Since $\epsilon _i$ is supported on $[\epsilon _*,\epsilon ^*]$, thus G is strictly increasing on $[\epsilon _*,\epsilon ^*]$, which means $q^*$ is unique. And Ri ordering $q_i=\mu _i+q^*$ is optimal.

If Ri hasn’t purchase the guaranteed supply, but Rj has purchased the guaranteed supply, then Rj will be satisfied first. Assume their order quantities are $q_i$ and $q_j$, then Ri will receive $q_i\wedge (K-q_j)^+$ products. The profit of Ri is then $\pi (q_i\wedge (K-q_j)^+-\mu _i)+(p-w)\mu _i$. Since $\pi$ is concave and achieves maximum at $q^*$, we know that $\pi (q_i\wedge (K-q_j)^+-\mu _i)\le \pi ((\mu _i+q^*)\wedge (K-q_j)^+-\mu _i)$, therefore, Ri ordering $q_i=\mu _i+q^*$ is optimal.

If both Ri and Rj haven’t purchase the guaranteed supply, then the capacity is allocated uniformly. Assume their order quantities are $q_i$ and $q_j$, then Ri will receive $q_i\wedge \left( K-q_j\wedge 0.5K\right)$ products. The profit of Ri is then $\pi (q_i\wedge \left( K-q_j\wedge 0.5K\right) \wedge (K-q_j)^+-\mu _i)+(p-w)\mu _i$ which is maximized at $q_i=\mu _i+q^*$.

The above discussion indicates that, ordering $q_i=\mu _i+q^*$ is a dominate strategy for Ri. Hence, R1 and R2 with private information $\mu _1$ and $\mu _2$ submitting order quantities $\mu _1+q^*$ and $\mu _2+q^*$ forms a Bayesian Nash equilibrium. $\square$

1.2 Proof of Lemma 1

Assume Rj adopts strategy $s_j$. If Ri with private information $\mu _i$ does not purchase the guaranteed supply, then her expected profit is

$$\begin{aligned} \begin{aligned} \int _{{\mathbb {R}}}s_j(\mu _j)\pi _{0,1}(\mu _i,\mu _j;K)f(\mu _j)d\mu _j+\int _{{\mathbb {R}}}(1-s_j(\mu _j))\pi _{0,0}(\mu _i,\mu _j;K)f(\mu _j)d\mu _j. \end{aligned} \end{aligned}$$

If Ri with private information $\mu _i$ purchases the guaranteed supply, then her expected profit is $\pi _1(\mu _i)-T$. Hence, Ri will purchase the guaranteed supply if and only if

$$\begin{aligned} \begin{aligned} T&\le \int _{{\mathbb {R}}}s_j(\mu _j)[\pi _1(\mu _i)-\pi _{0,1}(\mu _i,\mu _j;K)]f(\mu _j)d\mu _j\\&\quad +\int _{{\mathbb {R}}}(1-s_j(\mu _j))[\pi _1(\mu _i)-\pi _{0,0}(\mu _i,\mu _j;K)]f(\mu _j)d\mu _j\\&\doteq \varGamma (\mu _i,s_j;K). \end{aligned} \end{aligned}$$

Note that,

$$\begin{aligned} \begin{aligned} \pi _1(\mu _i)-\pi _{0,1}(\mu _i,\mu _j;K)&=\pi (q^*)-\pi \left( (\mu _i+q^*)\wedge (K-\mu _j-q^*)^+-\mu _i\right) \\&=\pi (q^*)-\pi \left( q^*\wedge [(K-\mu _j-q^*)^+-\mu _i]\right) ,\\ \pi _1(\mu _i)-\pi _{0,0}(\mu _i,\mu _j;K)&=\pi (q^*)-\pi \left( (\mu _i+q^*)\wedge \left( K-(\mu _j+q^*)\wedge 0.5K\right) -\mu _i\right) \\&=\pi (q^*)-\pi \left( q^*\wedge [\left( K-(\mu _j+q^*)\wedge 0.5K\right) -\mu _i]\right) \end{aligned} \end{aligned}$$

are both increasing in $\mu _i$, and both tend to 0 (or $\pi (q^*)-\pi (-\infty )=\infty$) as $\mu _i\rightarrow -\infty$ (or $\mu _i\rightarrow \infty$). Hence $\varGamma (\mu _i,s_j;K)$ is increasing in $\mu _i$, $\varGamma (-\infty ,s_j;K)=0$ and $\varGamma (\infty ,s_j;K)=\infty$. Therefore, for any $T>0$, there exists a $\mu _0(s_j)$ such that $\varGamma (\mu _i,s_j;K)<T$ for all $\mu _i<\mu _0(s_j)$ and $\varGamma (\mu _i,s_j;K)\ge T$ for all $\mu _i\ge \mu _0(s_j)$. This result indicates that, Ri’s best response to $s_j$ is a step function. $\square$

1.3 Proof of Theorem 2

Before proving the theorem, we first give a lemma.

Lemma 4

$\varOmega (x)$ is a function such that $d \varOmega /d x\ge 0$ and for any $\varOmega (x_0)=c$ (in which c is a fixed constant) it holds $d \varOmega /d x(x_0)>0$, then $x_0$ is unique (assume $x_0$ exists).

Proof

Assume there is another $x_1$ such that $\varOmega (x_1)=c$. Without loss of generality, assume $x_0<x_1$. Since $\varOmega$ is increasing, then we have $\varOmega (x)=c$ for all $x\in [x_0,x_1]$. Therefore, $d \varOmega /d x(x)=0$ for all $x\in [x_0,x_1]$, which contradicts with $d \varOmega /d x(x)>0$ whenever $\varOmega (x)=c$! Hence, $x_0$ is unique. $\square$

Assume thresholds $\mu _1^*$ and $\mu _2^*$ form a Bayesian Nash equilibrium, then R1 with private information $\mu _1^*$ will be indifference in purchasing the guaranteed supply or not purchasing it. Therefore, we have the following equation

$$\begin{aligned} \begin{aligned} \pi _1(\mu _1^*)-T=\int _{t\ge \mu _2^*}\pi _{0,1}(\mu _1^*,t;K)f(t)dt+\int _{t<\mu _2^*}\pi _{0,0}(\mu _1^*,t;K)f(t)dt, \end{aligned} \end{aligned}$$

which is equivalent to $T=H(\mu _1^*,\mu _2^*;K)$. Similar, $T=H(\mu _2^*,\mu _1^*;K)$ also must hold. Note that $H(-\infty ,y;K)=0$ and $H(\infty ,y;K)=\infty$, there exist L(y; K) such that $H(L(y;K),y;K)=T$ for any $T>0$. Since $\pi _1(x)-\pi _{0,1}(x,t)$ and $\pi _1(x)-\pi _{0,0}(x,t)$ are both increasing in x (see the proof of Theorem 1), we know that H(x, y) is increasing in x. In the following, we show $(\partial H/\partial x) (L(y;K),y;K)>0$, which means L(y; K) is unique according to Lemma 4. Note that

$$\begin{aligned} \begin{aligned}&\frac{\partial H}{\partial x}(L(y),y;K)\\&\quad =\int _{-\infty }^y \frac{d\pi }{dq}(q^*\wedge [\left( K-(t+q^*)\wedge 0.5K\right) -L(y;K)])f(t)dt\\&\qquad +\int _y^{\infty }\frac{d\pi }{dq}(q^*\wedge [(K-t-q^*)^+-L(y;K)])f(t)dt\ge 0. \end{aligned} \end{aligned}$$

If $(\partial H/\partial x)(L(y;K),y;K)=0$, then $q^*\wedge [\left( K-(t+q^*)\wedge 0.5K\right) -L(y;K)]=q^*$ if $t\in [\mu _*,\mu ^*]\cap (-\infty ,y]$, and $q^*\wedge [(K-t-q^*)^+-L(y;K)]=q^*$ if $t\in [\mu _*,\mu ^*]\cap [y,\infty )$. Hence

$$\begin{aligned} \begin{aligned}&H(L(y;K),y;K)\\&\quad =\int _{-\infty }^y [\pi _1(L(y;K))-\pi _{0,0}(L(y;K),t;K)]f(t)dt\\&\qquad -\int _y^{\infty } [\pi _1(L(y;K))-\pi _{0,1}(L(y;K),t;K)]f(t)dt\\&\quad =\int _{-\infty }^y [\pi (q^*)-\pi \left( q^*\wedge [\left( K-(t+q^*)\wedge 0.5K\right) -L(y;K)]\right) ]f(t)dt\\&\qquad -\int _y^{\infty } [\pi (q^*)-\pi \left( q^*\wedge [(K-t-q^*)^+-L(y;K)]\right) ]f(t)dt\\&\quad =\int _{-\infty }^y [\pi (q^*)-\pi (q^*)]f(t)dt-\int _y^{\infty } [\pi (q^*)-\pi (q^*)]f(t)dt=0, \end{aligned} \end{aligned}$$

which contradicts with $H(L(y;K),y;K)=T>0$! Therefore, $(\partial H/\partial x)(L(y;K),y;K)>0$ and L(y; K) is unique.

Further we have

$$\begin{aligned} \begin{aligned} \frac{\partial H}{\partial y}(x,y;K)=&\,\pi _{0,1}(x,y;K)f(y)-\pi _{0,0}(x,y;K)f(y)\\ =&\,f(y)[\pi \left( q^*\wedge [(K-y-q^*)^+-x]\right) \\&-\pi \left( q^*\wedge [\left( K-(y+q^*)\wedge 0.5K\right) -x]\right) ]\\ \le&\,0. \end{aligned} \end{aligned}$$

Thus H(x, y) is decreasing in y and L(y; K) is increasing in y.

Since $H(\mu _i^*,\mu _j^*;K)=T$, we know that $\mu _i^*=L(\mu _j^*;K)$. If $\mu _i^*<\mu _j^*$, then we have

$$\begin{aligned} \begin{aligned} T=H(\mu _i^*,\mu _j^*;K)\le H(\mu _j^*,\mu _j^*;K)\le H(\mu _j^*,\mu _i^*;K)=T. \end{aligned} \end{aligned}$$

Therefore, it must hold $H(\mu _j^*,\mu _j^*)=T$, which contradicts with $\mu _i^*=L(\mu _j^*;K)$ is unique! Hence $\mu _i^*$ must equal to $\mu _j^*$. This result indicates that, the $2\times 2$ Bayesian Nash game has only symmetric equilibria. Note that $H(-\infty ,-\infty ;K)=0$ and $H(\infty ,\infty ;K)=\infty$, hence $T=H(\mu _0^*,\mu _0^*;K)$ has at least ont solution. Further, it is easy to check that, for any $\mu _0^*$ satisfies $T=H(\mu _0^*,\mu _0^*;K)$, $(\mu _0^*,\mu _0^*)$ forms a Bayesian Nash equilibrium. Finally, we prove the three properties. (The expected profit of Ri can be computed easily.)

• (I) If $T\le H(\mu _*,\mu _*;K)$, then $T=H(\mu _0^*,\mu _0^*;K)$ has a solution with $\mu _0^*\le \mu _*$, which means both retailers always purchasing the guaranteed supply forms a Bayesian Nash equilibrium.

• (II) If $T\ge H(\mu ^*,\mu ^*;K)$, then $T=H(\mu _0^*,\mu _0^*;K)$ has a solution with $\mu _0^*\ge \mu ^*$, which means both retailers never purchasing the guaranteed supply forms a Bayesian Nash equilibrium.

• (III) If $T\in (H(\mu _*,\mu _*;K),H(\mu ^*,\mu ^*;K))$, then $T=H(\mu _0^*,\mu _0^*;K)$ has a solution with $\mu _0^*\in (\mu _*,\mu ^*)$, which means there exists a Bayesian Nash equilibrium such that both retailer will chose nontrivial thresholds. $\square$

1.4 Proof of Lemma 2

Case 1 R2 observes that R1 has purchased the guaranteed supply.

R2’s belief of R1’s private information is $m_b$. If R2 (with private information $\mu _2$) purchases the guaranteed supply, then her expected profit is $\pi _1(\mu _2)-T$. If R2 does not purchase the guaranteed supply, then her expected profit is

$$\begin{aligned} \begin{aligned} \int _{{\mathbb {R}}}\pi _{0,1}(\mu _2,\mu _1;K)m_b(d\mu _1). \end{aligned} \end{aligned}$$

Hence, R2’s threshold $\mu _{2b}^*$ satisfies

$$\begin{aligned} \begin{aligned} T=\pi _1(\mu _{2b}^*)-\int _{{\mathbb {R}}}\pi _{0,1}(\mu _{2b}^*,t;K)m_b(dt). \end{aligned} \end{aligned}$$

Further, we show that $\mu _{2b}^*$ is unique. Defining $H_b(x;m_b,K)=\pi _1(x)-\int _{{\mathbb {R}}}\pi _{0,1}(x,t;K)m_b(dt)$, then $H_b(\mu _{2b}^*;m_b,K)=T$. Note that

$$\begin{aligned} \begin{aligned} \left. \frac{dH_b}{dx}(x;m_b,K)\right| _{x=\mu _{2b}^*}=\int _{{\mathbb {R}}} \frac{d\pi }{dq}(q^*\wedge \left[ (K-t-q^*)^+-\mu _{2b}^*\right] )m_b(dt)\ge 0. \end{aligned} \end{aligned}$$

If $(dH_b/dx) (\mu _{2b}^*;m_b,K)=0$, then $q^*\wedge \left[ (K-t-q^*)^+-\mu _{2b}^*\right] =q^*$ for all $t\in supp(m_b)$ (the support of $m_b$). Therefore,

$$\begin{aligned} \begin{aligned} H_b(\mu _{2b}^*;m_b,K)=&\int _{{\mathbb {R}}}[\pi _1(\mu _{2b}^*)-\pi _{0,1}(\mu _{2b}^*,t;K)]m_b(dt)\\ =&\int _{{\mathbb {R}}}[\pi (q^*)-\pi (q^*\wedge \left[ (K-t-q^*)^+-\mu _{2b}^*\right] )]m_b(dt)=0, \end{aligned} \end{aligned}$$

which contradicts with $H_b(\mu _{2b}^*,m_b)=T>0$! Thus, $(dH_b/dx) (\mu _{2b}^*;m_b)>0$, according to Lemma 4, $\mu _{2b}^*$ is unique.

Case 2 R2 observes that R1 hasn’t purchased the guaranteed supply.

The prove is very similar to that in the above case, hence we provide a sketch of the prove.

The threshold $\mu _{2n}^*$ should make R2 be indifferent in purchasing or not purchasing the guaranteed supply, i.e.,

$$\begin{aligned} \begin{aligned} T=\pi _1(\mu _{2n}^*)-\int _{{\mathbb {R}}}\pi _{0,0}(\mu _{2n}^*,t;K)m_n(dt). \end{aligned} \end{aligned}$$

Define $H_n(x;m_n,K)=\pi _1(x)-\int _{{\mathbb {R}}}\pi _{0,0}(x,t)m_n(dt)$, then $H_n(\mu _{2n}^*;m_n,K)=T$. Similar to the above case, we can check that $dH_n/dx (\mu _{2n}^*;m_n,K)\ge 0$ and $dH_n/dx (\mu _{2n}^*;m_n,K)=0$ will lead to a contradiction. Hence, $\mu _{2n}^*$ is unique. $\square$

1.5 Proof of Lemma 3

Be aware of R2’s belief $(m_b,m_n)$, R1 can predict R2’s decision $(\mu _{2b}^*,\mu _{2n}^*)$ correctly. If R1 (with private information $\mu _1$) purchases the guaranteed supply, then her expected profit is $\pi _1(\mu _1)-T$. If R1 does not purchase the guaranteed supply, then her expected profit is

$$\begin{aligned} \begin{aligned}&\int _{-\infty }^{\mu _{2n}^*} \pi _{0,0}(\mu _1,\mu _2;K)f(\mu _2)d\mu _2+\int _{\mu _{2n}^*}^{\infty } \pi _{0,1}(\mu _1,\mu _2;K)f(\mu _2)d\mu _2\\&\quad =\pi _1(\mu _1)-H(\mu _1,\mu _{2n}^*;K). \end{aligned} \end{aligned}$$

Hence, R1 will chose threshold $\mu _{1c}^*$ that satisfies $H(\mu _{1c}^*,\mu _{2n}^*;K)=T$, i.e., $\mu _{1c}^*=L(\mu _{2n}^*;K)$. $\square$

1.6 Proof of Theorem 3

We prove the three properties fist, then the existence of the PBE follows directly from the three properties. (The expected profit of Ri can be computed easily.)

(I) It is known that $H(\mu _*,\eta ;K)$ is decreasing in $\eta$ and $\pi _1(\eta )-\pi _{0,0}(\eta ,\mu ^*;K)$ is increasing in $\eta$ (see the proof of Theorem 1). Further, we have

$$\begin{aligned} \begin{aligned} H(\mu _*,\mu _*;K)=&\,\pi _1(\mu _*)-\int _{{\mathbb {R}}}\pi _{0,1}(\mu _*,t;K)f(t)dt\\ \ge&\,\pi _1(\mu _*)-\int _{{\mathbb {R}}}\pi _{0,0}(\mu _*,t;K)f(t)dt\\ =&\,\pi _1(\mu _*)-\pi _{0,0}(\mu _*,\mu ^*;K), \end{aligned} \end{aligned}$$

($\pi _{0,0}(\mu _*,t;K)\equiv \pi ((\mu _*+q^*)\wedge 0.5K)+(p-w)\mu _*$ for $t\in [\mu _*,\mu ^*]$ is adopted in the above formula) and

$$\begin{aligned} \begin{aligned} H(\mu _*,\mu ^*;K)=&\,\pi _1(\mu _*)-\int _{{\mathbb {R}}}\pi _{0,0}(\mu _*,t;K)f(t)dt\\ =&\,\pi _1(\mu _*)-\pi _{0,0}(\mu _*,\mu ^*;K)\\ \le&\,\pi _1(\mu ^*)-\pi _{0,0}(\mu ^*,\mu ^*;K). \end{aligned} \end{aligned}$$

Thus there exists a $\eta _0\in [\mu _*,\mu ^*]$ such that $H(\mu _*,\eta _0;K)=\pi _1(\eta _0)-\pi _{0,0}(\eta _0,\mu ^*;K)$. In the following, we show that there is a PBE under which both retailers will always purchase the guaranteed supply when $T\le H(\mu _*,\eta _0;K)$.

Consider an equation $\pi _1(x)-\pi _{0,0}(x,\mu ^*;K)=T$. Since

$$\begin{aligned} \begin{aligned} \pi _1(\eta _0)-\pi _{0,0}(\eta _0,\mu ^*;K)&=H(\mu _*,\eta _0;K)\ge T,\\ \pi _1(-\infty )-\pi _{0,0}(-\infty ,\mu ^*;K)&=0<T, \end{aligned} \end{aligned}$$

there exists a $\eta _b\le \eta _0$ such that $\pi _1(\eta _b)-\pi _{0,0}(\eta _b,\mu ^*;K)=T$. In the following, we check that $(\mu _{1c}^*,\mu _{2b}^*,\mu _{2n}^*;m_b,m_n)=(L(\eta _b;K),L(\mu _*;K),\eta _b;(F,f),1_{\mu ^*})$ ($1_{\mu ^*}$ is the probability measure with 100% on $\mu ^*$) forms a PBE.

By the definition of L, we have $H(L(\mu _*;K),\mu _*;K)=T$, i.e., (3) holds for $\mu _{2b}^*=L(\mu _*;K)$ and $m_b=(F,f)$.

By the definition of $\eta _b$, we have $\pi _1(\eta _b)-\pi _{0,0}(\eta _b,\mu ^*;K)=T$, i.e., (4) holds for $\mu _{2n}^*=\eta _b$ and $m_n=1_{\mu ^*}$.

By the definition of L, (5) holds for $\mu _{1c}^*=L(\eta _b;K)$ and $\mu _{2n}^*=\eta _b$.

Finally, $H(L(\eta _b;K),\eta _b;K)=T\le H(\mu _*,\eta _0;K)\le H(\mu _*,\eta _b;K)$, which means $\mu _{1c}^*=L(\eta _b;K)\le \mu _*$. Therefore (6) holds for $\mu _{1c}^*=L(\eta _b;K)$, $m_b=(F,f)$ and $m_n=1_{\mu ^*}$.

Further, $H(L(\mu _*;K),\mu _*;K)=T\le H(\mu _*,\eta _0;K)\le H(\mu _*,\mu _*;K)$, which means $\mu _{2b}^*=L(\mu _*;K)\le \mu _*$. Together with $\mu _{1c}^*=L(\eta _b;K)\le \mu _*$, we know that, R1 will always purchase the guaranteed supply, R2 will always purchase the guaranteed supply after seeing R1 has purchased the guaranteed supply. Hence, both retailers always purchase the guaranteed supply in the PBE $(\mu _{1c}^*,\mu _{2b}^*,\mu _{2n}^*;m_b,m_n)=(L(\eta _b;K),L(\mu _*;K),\eta _b;(F,f),1_{\mu ^*})$.

(II) We check that

$$\begin{aligned} \begin{aligned} (\mu _{1c}^*,\mu _{2b}^*,\mu _{2n}^*;m_b,m_n)= (L(L(\mu ^*;K);K),L(\mu _*;K),L(\mu ^*;K);(F,f),(F,f)) \end{aligned} \end{aligned}$$

forms a PBE when $T\ge H(\mu ^*,\mu ^*;K)$.

According to the definition of L, we know that

$$\begin{aligned} \begin{aligned} H(L(\mu _*;K),\mu _*;K)&=H(L(\mu ^*;K),\mu ^*;K) \\&=H(L(L(\mu ^*;K);K),L(\mu ^*;K);K)=T, \end{aligned} \end{aligned}$$

that is, (3), (4) and (5) hold.

Further, $H(L(\mu ^*;K),\mu ^*;K)=T\ge H(\mu ^*,\mu ^*;K)$, which means $L(\mu ^*;K)\ge \mu ^*$. Note that L is increasing, hence $L(L(\mu ^*;K);K)\ge L(\mu ^*;K)\ge \mu ^*$. Therefore (6) holds for $\mu _{1c}^*=L(L(\mu ^*;K);K)$ and $m_b=m_n=(F,f)$.

Finally, we know $L(L(\mu ^*;K);K)=\mu _{1c}^*\ge L(\mu ^*;K)=\mu _{2n}^*\ge \mu ^*$, i.e., R1 will never purchase the guaranteed supply, and R2 will never purchase the guaranteed supply after seeing R1 hasn’t purchased the guaranteed supply. Hence, both retailers never purchase the guaranteed supply in the PBE

$$\begin{aligned} \begin{aligned} (\mu _{1c}^*,\mu _{2b}^*,\mu _{2n}^*;m_b,m_n)= (L(L(\mu ^*;K);K),L(\mu _*;K),L(\mu ^*;K);(F,f),(F,f)). \end{aligned} \end{aligned}$$

(III) We first show the existence of $\eta _1$. Since

$$\begin{aligned} \begin{aligned} H(\mu _*,\eta _0;K)=\pi _1(\eta _0)-\pi _{0,0}(\eta _0,\mu ^*;K)\ge \pi _1(\eta _0)-\pi _{0,0}(\eta _0,\mu _*;K), \end{aligned} \end{aligned}$$

and

$$\begin{aligned} \begin{aligned} H(\mu _*,\mu ^*;K)=&\pi _1(\mu _*)-\int _{{\mathbb {R}}}\pi _{0,0}(\mu _*,t;K)f(t)dt\\ =&\pi _1(\mu _*)-\pi _{0,0}(\mu _*,\mu _*;K)\\ \le&\pi _1(\mu ^*)-\pi _{0,0}(\mu ^*,\mu _*;K). \end{aligned} \end{aligned}$$

Thus there exist an $\eta _1\in [\eta _0,\mu ^*]$ such that $H(\mu _*,\eta _1;K)=\pi _1(\eta _1)-\pi _{0,0}(\eta _1,\mu _*;K)$.

Next, we show the existence of a nontrivial PBE. To do so, we aim to reformulate (3), (4), (5) and (6) to a single equation of $\mu _{2n}^*$. Then we show the existence of such $\mu _{2n}^*$, and prove that the corresponding solution $(\mu _{1c}^*,\mu _{2b}^*,\mu _{2n}^*;m_b,m_n)$ forms a nontrivial PBE.

Since $T\in (H(\mu _*,\eta _1;K),H(\mu ^*,\mu ^*;K))$, we have $H(L(\eta _1;K),\eta _1;K)>H(\mu _*,\eta _1;K)$ and $H(L(\mu ^*;K),\mu ^*;K)<H(\mu ^*,\mu ^*;K)$. Hence $L(\eta _1;K)>\mu _*$ and $L(\mu ^*;K)<\mu ^*$. Note that L is an increasing function, we have $L(x;K)\in (\mu _*,\mu ^*)$ for all $x\in [\eta _1,\mu ^*]$. Note that $H(\mu _{1c}^*,\mu _{2n}^*;K)=T$ must hold, hence $\mu _{1c}^*=L(\mu _{2n}^*;K)\in (\mu _*,\mu ^*)$ for all $\mu _{2n}^*\in [\eta _1,\mu ^*]$. Define $\theta _1=\sup \{x|L(x;K)\le \mu _*\}$ and $\theta _2=\inf \{x|L(x;K)\ge \mu ^*\}$, in case of empty set $\emptyset$, let $\sup \emptyset =-\infty$ and $\inf \emptyset =+\infty$. Then $\theta _1\le \eta _1$, $\theta _2\ge \mu ^*$, and further we have $\mu _{1c}^*=L(\mu _{2n}^*;K)\in (\mu _*,\mu ^*)$ for all $\mu _{2n}^*\in (\theta _1,\theta _2)$, .

When $\mu _{1c}^*\in (\mu _*,\mu ^*)$, according to the Bayesian rule (6), we know that $m_n$ must be (F, f) conditional on $[\mu _*,\mu _{1c}^*]$ (i.e., $m_n=(F,f)_{[\mu _*,\mu _{1c}^*]}$). When $\mu _{2n}^*\in (\theta _1,\theta _2)$, we know that (4) is equivalent to

$$\begin{aligned} \begin{aligned} T=\int _{\mu _*}^{L(\mu _{2n}^*;K)} \left[ \pi _1(\mu _{2n}^*)-\pi _{0,0}(\mu _{2n}^*,t;K)\right] \frac{f(t)}{F(L(\mu _{2n}^*;K))}dt\doteq H_{n,1}(\mu _{2n}^*;K). \end{aligned} \end{aligned}$$

Note that

$$\begin{aligned} \begin{aligned} \text {if}\quad&\theta _1=-\infty , \quad \lim _{x\rightarrow \theta _1}H_{n,1}(x;K)=0<T\\ \text {if}\quad&\theta _1>-\infty , \quad \text {then} \quad L(\theta _1;K)=\mu _*,\\&\lim _{x\rightarrow \theta _1}H_{n,1}(x;K)=\lim _{x\rightarrow \theta _1}\int _{\mu _*}^{L(x;K)} \left[ \pi _1(x)-\pi _{0,0}(x,t;K)\right] \frac{f(t)}{F(L(x;K))}dt\\&=\pi _1(\theta _1)-\pi _{0,0}(\theta _1,\mu _*;K)\le \pi _1(\eta _1)-\pi _{0,0}(\eta _1,\mu _*;K)\\&=H(\mu _*,\eta _1;K)<T, \end{aligned} \end{aligned}$$

and

$$\begin{aligned} \begin{aligned} \text {if}\quad&\theta _2=+\infty , \quad \lim _{x\rightarrow \theta _2}H_{n,1}(x;K)=+\infty>T\\ \text {if}\quad&\theta _2>+\infty , \quad \text {then} \quad L(\theta _2;K)=\mu ^*\\&\lim _{x\rightarrow \theta _2}H_{n,1}(x;K)=\lim _{x\rightarrow \theta _2}\int _{\mu _*}^{L(x;K)} \left[ \pi _1(x)-\pi _{0,0}(x,t;K)\right] \frac{f(t)}{F(L(x;K))}dt\\&=\int _{\mu _*}^{\mu ^*} \left[ \pi _1(\theta _2)-\pi _{0,0}(\theta _2,t;K)\right] f(t)dt\ge \int _{\mu _*}^{\mu ^*} \left[ \pi _1(\mu ^*)-\pi _{0,0}(\mu ^*,t;K)\right] f(t)dt\\&=H(\mu ^*,\mu ^*;K)>T. \end{aligned} \end{aligned}$$

Hence, there exists a $\theta _n^*\in (\theta _1,\theta _2)$ such that $H_{n,1}(\theta _n^*;K)=T$. Let $\mu _{1c}^*=L(\theta _n^*;K)\in (\mu _*,\mu ^*)$, according to the Bayesian rule (6), we know that $m_b$ must be (F, f) conditional on $[L(\theta _n^*;K),\mu ^*]$ (i.e., $m_b=(F,f)_{[L(\theta _n^*;K),\mu ^*]}$). Therefore, (3) is equivalent to

$$\begin{aligned} \begin{aligned} T=\int _{L(\theta _n^*;K)}^{\mu ^*} \left[ \pi _1(\mu _{2b}^*)-\pi _{0,0}(\mu _{2b}^*,t;K)\right] \frac{f(t)}{F(L(\theta _n^*;K))}dt\doteq H_{b,1}(\mu _{2b}^*;K). \end{aligned} \end{aligned}$$

Note that $H_{b,1}(-\infty ;K)=0<T$ and $H_{b,1}(+\infty ;K)=+\infty >T$, there exists a $\theta _b^*$ such that $H_{b,1}(\theta _b^*;K)=T$.

In summary, the above discussion indicates that,

$$\begin{aligned} \begin{aligned} (\mu _{1c}^*,\mu _{2b}^*,\mu _{2n}^*;m_b,m_n)= \left( L(\theta _n^*;K),\theta _b^*,\theta _n^*;(F,f)_{[L(\theta _n^*;K),\mu ^*]},(F,f)_{[\mu _*,L(\theta _n^*;K)]}\right) \end{aligned} \end{aligned}$$

forms a Bayesian Nash equilibrium when $T\in (H(\mu _*,\eta _1;K),H(\mu ^*,\mu ^*;K))$, and the threshold $L(\theta _n^*;K)\in (\mu _*,\mu ^*)$ is nontrivial.

Finally, we need to prove the relationships $\mu _{2n}^*\ge \mu _{1c}^*\ge \mu _{2b}^*$, i.e., $\theta _n^*\ge L(\theta _n^*;K)\ge \theta _b^*$.

According to (3), (4) and (6), we have

$$\begin{aligned} \begin{aligned} T=&\,\pi _1(\mu _{2b}^*)-\frac{1}{1-F(\mu _{1c}^*)}\int _{\mu _{1c}^*}^{\mu ^*}\pi _{0,1}(\mu _{2b}^*,t;K)f(t)dt\\ \ge&\,\pi _1(\mu _{2b}^*)-\int _{\mu _*}^{\mu ^*}\pi _{0,1}(\mu _{2b}^*,t;K)f(t)dt=H(\mu _{2b}^*,-\infty ;K)\\ \ge&\,H(\mu _{2b}^*,\mu _{2n}^*;K), \end{aligned} \end{aligned}$$

and

$$\begin{aligned} \begin{aligned} T=&\,\pi _1(\mu _{2n}^*)-\frac{1}{F(\mu _{1c}^*)}\int _{\mu _*}^{\mu _{1c}^*}\pi _{0,0}(\mu _{2n}^*,t;K)f(t)dt\\ \le&\,\pi _1(\mu _{2n}^*)-\int _{\mu _*}^{\mu ^*}\pi _{0,1}(\mu _{2n}^*,t;K)f(t)dt=H(\mu _{2n}^*,+\infty ;K)\\ \le&\,H(\mu _{2n}^*,\mu _{2n}^*;K). \end{aligned} \end{aligned}$$

Hence, we have $H(\mu _{2b}^*,\mu _{2n}^*;K)\le T=H(\mu _{1c}^*,\mu _{2n}^*;K)\le H(\mu _{2n}^*,\mu _{2n}^*;K)$. Further, we know that, $(\partial H/\partial x) (\mu _{1c}^*,\mu _{2b}^*;K)=(\partial H/\partial x) (L(\mu _{2b}^*;K),\mu _{2b}^*;K)>0$, therefore $\mu _{2n}^*\ge \mu _{1c}^*\ge \mu _{2b}^*$. $\square$

1.7 Proof of Theorem 4

If $\mu <\mu _{2b}^*$, then

$$\begin{aligned} \begin{aligned} \varPi _{R1}^{seq}(\mu ;K)-\varPi _{R2}^{seq}(\mu ;K)=\int _{\mu _{1c}^*\le t\le \mu _{2n}^*}[\pi _{0,0}(\mu ,t;K)-\pi _{0,1}(\mu ,t;K)]f(t)dt\ge 0. \end{aligned} \end{aligned}$$

If $\mu _{1c}^*\le \mu <\mu _{2n}^*$, then

$$\begin{aligned} \begin{aligned} \varPi _{R1}^{seq}(\mu ;K)-\varPi _{R2}^{seq}(\mu ;K)=\int _{t\le \mu _{1c}^*}[\pi _1(\mu )-\pi _{0,0}(\mu ,t;K)-T]f(t)dt. \end{aligned} \end{aligned}$$

According to (4) and (6), we have

$$\begin{aligned} \begin{aligned} T=\pi _1(\mu _{2n}^*)-\frac{1}{F(\mu _{1c}^*)}\int _{\mu _*}^{\mu _{1c}^*}\pi _{0,0}(\mu _{2n}^*,t;K)f(t)dt, \end{aligned} \end{aligned}$$

which is equivalent to

$$\begin{aligned} \begin{aligned} \int _{t\le \mu _{1c}^*}T f(t)dt=\int _{t\le \mu _{1c}^*}[\pi _1(\mu _{2n}^*)-\pi _{0,0}(\mu _{2n}^*,t;K)]f(t)dt. \end{aligned} \end{aligned}$$

Hence, we have

$$\begin{aligned} \begin{aligned}&\varPi _{R1}^{seq}(\mu ;K)-\varPi _{R2}^{seq}(\mu ;K)\\&\quad =\int _{t\le \mu _{1c}^*}[\pi _1(\mu )-\pi _{0,0}(\mu ,t;K)-(\pi _1(\mu _{2n}^*)-\pi _{0,0}(\mu _{2n}^*,t;K))]f(t)dt. \end{aligned} \end{aligned}$$

Note that, $\pi _1(x)-\pi _{0,0}(x,t;K)$ is increasing in x (see the proof of Theorem 1). Therefore, $\varPi _{R1}^{seq}(\mu ;K)-\varPi _{R2}^{seq}(\mu ;K)\le 0$.

If $\mu _{2n}^*\le \mu$, then $\varPi _{R1}^{seq}(\mu ;K)-\varPi _{R2}^{seq}(\mu ;K)=0$.

Finally, if $\mu _{2b}^*\le \mu <\mu _{1c}^*$, then

$$\begin{aligned} \begin{aligned} \varPi _{R1}^{seq}(\mu ;K)-\varPi _{R2}^{seq}(\mu ;K)=&\int _{\mu _{1c}^*\le t\le \mu _{2n}^*}[T+\pi _{0,0}(\mu ,t;K)-\pi _1(\mu )]f(t)dt\\&+\int _{\mu _{2n}^*\le t}[T+\pi _{0,1}(\mu ,t;K)-\pi _1(\mu )]f(t)dt. \end{aligned} \end{aligned}$$

Note that $\pi _1(x)-\pi _{0,1}(x,t;K)$ and $\pi _1(x)-\pi _{0,0}(x,t;K)$ are both increasing in x [see the proof of Theorem 1)]. Hence, $\varPi _{R1}^{seq}(\mu ;K)-\varPi _{R2}^{seq}(\mu ;K)$ is decreasing in $\mu$.

According to (3), (4), (5)and (6), we can check that $\varPi _{R1}^{seq}(\mu ;K)$ and $\varPi _{R2}^{seq}(\mu ;K)$ are continuous functions, and we know that $\varPi _{R1}^{seq}(\mu _{2b}^*;K)-\varPi _{R2}^{seq}(\mu _{2b}^*;K)\ge 0$ and $\varPi _{R1}^{seq}(\mu _{1c}^*;K)-\varPi _{R2}^{seq}(\mu _{1c}^*;K)\le 0$. Therefore, there exists a $\mu _{12}\in [\mu _{2b}^*,\mu _{1c}^*]$ such that $\varPi _{R1}^{seq}(\mu ;K)-\varPi _{R2}^{seq}(\mu ;K)\ge 0$ for $\mu _{2b}^*\le \mu \le \mu _{12}$, and $\varPi _{R1}^{seq}(\mu ;K)-\varPi _{R2}^{seq}(\mu ;K)\le 0$ for $\mu _{12}\le \mu \le \mu _{1c}^*$.

In summary, we have proved the theorem. $\square$

1.8 Proof of Theorem 5

Since H(x, y) is decreasing in y, $\pi _{0,0}(x,y)$ is decreasing in y and $\pi _{0,0}(x,y)\ge \pi _{0,1}(x,y)$, we have

$$\begin{aligned} \begin{aligned} H(\mu _0^*,\mu _0^*;K)=&\,T=\pi _1(\mu _{2n}^*)-\frac{1}{F(\mu _{1c}^*)}\int _{\mu _*}^{\mu _{1c}^*}\pi _{0,0}(\mu _{2n}^*,t;K)f(t)dt\\ \le&\,\pi _1(\mu _{2n}^*)-\int _{\mu _*}^{\mu ^*}\pi _{0,0}(\mu _{2n}^*,t;K)f(t)dt\\ \le&\,\pi _1(\mu _{2n}^*)-\int _{\mu _*}^{\mu ^*}\pi _{0,1}(\mu _{2n}^*,t;K)f(t)dt=H(\mu _{2n}^*,+\infty ;K)\\ \le&\,H(\mu _{2n}^*,\mu _0^*;K), \end{aligned} \end{aligned}$$

If $\mu _{2n}^*<\mu _0^*$, we have $H(\mu _{2n}^*,\mu _0^*;K)=H(\mu _0^*,\mu _0^*;K)=T$ (since H(x, y) is increasing in x). Therefore $\mu _0^*=L(\mu _0^*;K)=\mu _{2n}^*<\mu _0^*$, which is a contradiction! Hence it must hold $\mu _0^*\le \mu _{2n}^*$.

If $\mu _{2n}^*\ge \mu _0^*>\mu _{1c}^*$, then $H(\mu _0^*,\mu _0^*;K)=T=H(\mu _{1c}^*,\mu _{2n}^*;K)\le H(\mu _{1c}^*,\mu _0^*;K)$. Note that H(x, y) is increasing in x, we have $H(\mu _0^*,\mu _0^*;K)=H(\mu _{1c}^*,\mu _0^*;K)=T$, i.e., $\mu _0^*=L(\mu _0^*;K)=\mu _{1c}^*<\mu _0^*$, which is a contradiction! Hence, we have proved $\mu _0^*\le \mu _{1c}^*$.

Finally, since H(x, y) is decreasing in y and $\pi _{0,1}(x,y)$ is decreasing in y, we have

$$\begin{aligned} \begin{aligned} H(\mu _0^*,\mu _0^*;K)=&\,T=\pi _1(\mu _{2b}^*)-\frac{1}{1-F(\mu _{1c}^*)}\int _{\mu _{1c}^*}^{\mu ^*}\pi _{0,1}(\mu _{2b}^*,t;K)f(t)dt\\ \ge&\,\pi _1(\mu _{2b}^*)-\int _{\mu _*}^{\mu ^*}\pi _{0,1}(\mu _{2b}^*,t;K)f(t)dt=H(\mu _{2b}^*,-\infty ;K)\\ \ge&\,H(\mu _{2b}^*,\mu _0^*;K). \end{aligned} \end{aligned}$$

If $\mu _0^*<\mu _{2b}^*$, the $H(\mu _0^*,\mu _0^*;K)=H(\mu _{2b}^*,\mu _0^*;K)=T$ (since H(x, y) is increasing in x). Therefore $\mu _0^*=L(\mu _0^*;K)=\mu _{2b}^*>\mu _0^*$, which is a contradiction! Hence it must hold $\mu _0^*\ge \mu _{2b}^*$.

In summary, we have proved $\mu _{2n}^*\ge \mu _{1c}^*\ge \mu _0^*\ge \mu _{2b}^*$.

For the profit comparison, we prove it by two steps: 1. prove $\varPi _{R1}^{seq}(\mu ;K)\ge \varPi _{Ri}^{sim}(\mu ;K)$; 2. prove $\varPi _{R2}^{seq}(\mu ;K)\ge \varPi _{Ri}^{sim}(\mu ;K)$.

Step 1

If $\mu <\mu _0^*$, then

$$\begin{aligned} \begin{aligned} \varPi _{R1}^{seq}(\mu ;K)-\varPi _{Ri}^{sim}(\mu ;K)=\int _{\mu _0^*\le t\le \mu _{2n}^*}[\pi _{0,0}(\mu ,t;K)-\pi _{0,1}(\mu ,t;K)]f(t)dt\ge 0. \end{aligned} \end{aligned}$$

If $\mu _0^*\le \mu <\mu _{1c}^*$, then

$$\begin{aligned} \begin{aligned}&\varPi _{R1}^{seq}(\mu ;K)-\varPi _{Ri}^{sim}(\mu ;K)\\&\quad =T-\left[ \pi _1(\mu )-\int _{t\le \mu _{2n}^*} \pi _{0,0}(\mu ,t;K)f(t)dt-\int _{t\ge \mu _{2n}^*} \pi _{0,1}(\mu ,t;K)f(t)dt\right] \\&\quad =H(\mu _{1c}^*,\mu _{2n}^*;K)-H(\mu ,\mu _{2n}^*;K)\ge 0. \end{aligned} \end{aligned}$$

If $\mu _{1c}^*\le \mu$, then R1 with average demand $\mu$ in the sequential scenario and Ri with average demand $\mu$ in the simultaneous scenario both purchase the guaranteed supply, hence $\varPi _{R1}^{seq}(\mu ;K)-\varPi _{Ri}^{sim}(\mu ;K)=0$.

Therefore, the first step is completed.

Step 2

If $\mu <\mu _{2b}^*$, then

$$\begin{aligned} \begin{aligned} \varPi _{R2}^{seq}(\mu ;K)-\varPi _{Ri}^{sim}(\mu ;K)=\int _{\mu _0^*\le t\le \mu _{1c}^*}[\pi _{0,0}(\mu ,t;K)-\pi _{0,1}(\mu ,t;K)]f(t)dt\ge 0. \end{aligned} \end{aligned}$$

If $\mu _{2b}^*\le \mu <\mu _0^*$, then

$$\begin{aligned} \begin{aligned}&\varPi _{R2}^{seq}(\mu ;K)-\varPi _{Ri}^{sim}(\mu ;K)\\&\quad =\int _{\mu _0^*\le t\le \mu _{1c}^*}[\pi _{0,0}(\mu ,t;K)-\pi _{0,1}(\mu ,t;K)]f(t)dt\\&\qquad +\int _{\mu _{1c}^*\le t}[\pi _1(\mu )-\pi _{0,1}(\mu ,t;K)-T]f(t)dt\\&\quad \ge \int _{\mu _{1c}^*\le t}[\pi _1(\mu )-\pi _{0,1}(\mu ,t;K)-T]f(t)dt. \end{aligned} \end{aligned}$$

According to (3) and (6), we have

$$\begin{aligned} \begin{aligned} T=\pi _1(\mu _{2b}^*)-\frac{1}{1-F(\mu _{1c}^*)}\int _{\mu _{1c}^*}^{\mu ^*}\pi _{0,1}(\mu _{2b}^*,t;K)f(t)dt, \end{aligned} \end{aligned}$$

which is equivalent to

$$\begin{aligned} \begin{aligned} \int _{\mu _{1c}^*\le t}T f(t) dt=\int _{\mu _{1c}^*}^{\mu ^*}[\pi _1(\mu _{2b}^*)-\pi _{0,1}(\mu _{2b}^*,t;K)]f(t)dt. \end{aligned} \end{aligned}$$

Hence, we have

$$\begin{aligned} \begin{aligned}&\varPi _{R2}^{seq}(\mu ;K)-\varPi _{Ri}^{sim}(\mu ;K)\\&\quad \ge \int _{\mu _{1c}^*\le t}[\pi _1(\mu )-\pi _{0,1}(\mu ,t;K)-T]f(t)dt.\\&\quad =\int _{\mu _{1c}^*\le t}\left\{ \pi _1(\mu )-\pi _{0,1}(\mu ,t;K)-[\pi _1(\mu _{2b}^*)-\pi _{0,1}(\mu _{2b}^*,t;K)]\right\} f(t)dt. \end{aligned} \end{aligned}$$

Note that $\pi _1(x)-\pi _{0,1}(x,t;K$ is increasing in x (see the proof of Theorem 1). Therefore, $\varPi _{R2}^{seq}(\mu ;K)-\varPi _{Ri}^{sim}(\mu ;K)\ge 0$.

If $\mu _0^*\le \mu <\mu _{2n}^*$, then

$$\begin{aligned} \begin{aligned} \varPi _{R2}^{seq}(\mu ;K)-\varPi _{Ri}^{sim}(\mu ;K)=\int _{t\le \mu _{1c}^*}[T-(\pi _1(\mu )-\pi _{0,0}(\mu ,t;K))]f(t)dt. \end{aligned} \end{aligned}$$

According to (4) and (6), we have

$$\begin{aligned} \begin{aligned} T=\pi _1(\mu _{2n}^*)-\frac{1}{F(\mu _{1c}^*)}\int _{\mu _*}^{\mu _{1c}^*}\pi _{0,0}(\mu _{2n}^*,t;K)f(t)dt, \end{aligned} \end{aligned}$$

which is equivalent to

$$\begin{aligned} \begin{aligned} \int _{t\le \mu _{1c}^*}T f(t)dt=\int _{t\le \mu _{1c}^*}[\pi _1(\mu _{2n}^*)-\pi _{0,0}(\mu _{2n}^*,t;K)]f(t)dt. \end{aligned} \end{aligned}$$

Hence, we have

$$\begin{aligned} \begin{aligned}&\varPi _{R2}^{seq}(\mu ;K)-\varPi _{Ri}^{sim}(\mu ;K)\\&\quad =\int _{t\le \mu _{1c}^*}[\pi _1(\mu _{2n}^*)-\pi _{0,0}(\mu _{2n}^*,t;K)-(\pi _1(\mu )-\pi _{0,0}(\mu ,t;K))]f(t)dt. \end{aligned} \end{aligned}$$

Note that, $\pi _1(x)-\pi _{0,0}(x,t;K)$ is increasing in x (see the proof of Theorem 1). Therefore, $\varPi _{R2}^{seq}(\mu ;K)-\varPi _{Ri}^{sim}(\mu ;K)\ge 0$.

If $\mu _{2n}^*\le \mu$, then R2 with average demand $\mu$ in the sequential scenario and Ri with average demand $\mu$ in the simultaneous scenario both purchase the guaranteed supply, hence $\varPi _{R2}^{seq}(\mu ;K)-\varPi _{Ri}^{sim}(\mu ;K)=0.$

Hence, the second step is proved, and the theorem is proved. $\square$