Supplier’s cooperation strategy with two competing manufacturers under wholesale price discount contract considering technology investment

Abstract

Cooperation between upstream suppliers and downstream manufacturers in technology investment is a popular way to improve production technology for reducing suppliers’ production costs of key components. The suppliers’ cooperation strategies are mainly influenced by manufacturers’ technology investments and wholesale price discount contracts provided by suppliers. This paper explores whether a supplier should cooperate with two downstream competing manufacturers to accept their technology investments to reduce the supplier’s production cost of a key component. Specifically, we consider the following three cooperation strategies: The supplier does not accept manufacturers’ technology investments, only accepts one manufacturer’s technology investment and accepts both manufacturers’ technology investments. Our results demonstrate that the wholesale price discount contract and the technology investment can enhance the profits of the supplier and two manufacturers when the discount degree is low. Further, we conclude that when the discount degree is relatively low or when both the discount degree and the technology investment efficiency are relatively high, the supplier’s optimal cooperation strategy with two manufacturers is to accept both manufacturers’ technology investments and both manufacturers are also willing to invest simultaneously. At last, we extend the model to the asymmetric potential market size and show that our theoretical results are robust.

Appendix

Proof of Lemma 1

For any given w, we derive $\pi _{M_i}^{\mathrm{NN}}$ is concave in $p_i$ since $\frac{d^{2}\pi ^{\mathrm{NN}}_{M_i} }{d p_i^{2}}=-2<0$. By the first-order condition, that is $\frac{d \pi _{M_1}^{\mathrm{NN}}}{d p_1}=0$ and $\frac{d \pi _{M_2}^{\mathrm{NN}}}{d p_2}=0$, we conclude the unique optimal retail price (3).

By substituting (3) in the supplier’s profit function, we get $\pi _{S}^{\mathrm{NN}}(w)$. In order to make $\pi _{S}^{\mathrm{NN}}(w)$ to be concave in w, we require that $\frac{d^{2} \pi _{S}^{\mathrm{NN}}}{d w^{2}}=\frac{-4(b-1)}{b-2}<0$. From the first order condition, $\frac{d \pi _{S}^{\mathrm{NN}}}{d w}=0$, we obtain the unique optimal wholesale price (6). Then substituting Eq. (6) in Eq. (3), we get the optimal retail price (5).

Substituting Eqs. (5) and (6) in Eqs. (2) and (4), we can derive the optimal profit of the manufacturers (7) and the supplier’s optimal profit (8). $\square$

Proof of Lemma 2

For the given w, it is easy to show that $\pi _{M_1}^{\mathrm{TN}}$ is jointly concave in $p_1$ and $t_1$, because the Hessian matrix of $\pi _{M_1}^{\mathrm{TN}}$

$$\begin{aligned} H(\pi _{M_{1}}^{\mathrm{TN}})=\begin{pmatrix} \frac{\partial ^2 \pi _{M_{1}}^{TN}}{\partial p_{1}^2} &{} \frac{\partial ^2 \pi _{M_{1}}^{TN}}{\partial p_{1}\partial t_1} \\ \frac{\partial ^2 \pi _{M_{1}}^{TN}}{\partial t_1 \partial p_1} &{} \frac{\partial ^2 \pi _{M_{1}}^{TN}}{\partial t_{1}^2} \end{pmatrix}\\ =\begin{pmatrix} -2 &{}\alpha \\ -\alpha &{} -\mu \end{pmatrix} \end{aligned}$$

is negative definite when the leading principal minor $H_{1}^{\mathrm{TN}}=-2<0$, $H_2^{\mathrm{TN}}=2\mu -\alpha ^2>0$. Further, we conclude that $\pi _{M_2}^{\mathrm{TN}}$ is concave in $p_2$ when $\frac{d^{2} \pi ^{\mathrm{TN}}_{M_2}}{d p_2^{2}}=-2<0$.

When $\mu >\frac{\alpha ^2}{2}$, by the first order condition, that is $\frac{\partial \pi _{M_1}^{\mathrm{TN}}}{\partial p_1}=0,$ $\frac{\partial \pi _{M_1}^{\mathrm{TN}}}{\partial t_1}=0,$ and $\frac{d \pi _{M_2}^{\mathrm{TN}}}{d p_2}=0,$ we can conclude (11), (12) and (13). By substituting Eqs. (11), (12) and (13) in the supplier’s profit (14), we obtain $\pi _{S}^{\mathrm{TN}}(w)$. To ensure that $\pi _{S}^{\mathrm{TN}}(w)$ is concave in w, $\frac{d^{2} \pi _{S}^{\mathrm{TN}}}{d w^{2}}<0$ is required, that is $\mu >\mu _{u}^{\mathrm{TN}}={\frac{ \left( b-1 \right) \left( b+2 \right) \alpha + \left( b-2 \right) \left( b+1 \right) {\alpha }^{2}-\alpha \,\sqrt{(\alpha ^2(b^2-b-2)-b^2-b+2)(1-b)(b+2)}}{2\left( b-2 \right) \left( b+2 \right) }}$. From $\frac{d \pi _{S}^{\mathrm{TN}}}{d w}=0,$ we can obtain the unique optimal wholesale price (18). Then substituting Eq. (18) in Eqs. (11), (12) and (13), we have the optimal retail prices (15), (16) and the optimal technology investment level (17).

Substituting Eqs. (15), (16), (17) and (18) in Eqs. (9), (10) and (14), we can get the optimal profits of the manufacturers (19), (20) and the supplier’s optimal profit (21). $\square$

Proof of Lemma 3

For any given w, the Hessian matrix of $\pi _{M_i}^{\mathrm{TT}}, i=1,2$ are as follows:

$$\begin{aligned} H(\pi _{M_{i}}^{\mathrm{TT}})=\begin{pmatrix} \frac{\partial ^2 \pi _{M_{i}}^{\mathrm{TT}}}{\partial p_{i}^2} &{} \frac{\partial ^2 \pi _{M_{i}}^{\mathrm{TT}}}{\partial p_{i}\partial t_i} \\ \frac{\partial ^2 \pi _{M_{i}}^{\mathrm{TT}}}{\partial t_i \partial p_i} &{} \frac{\partial ^2 \pi _{M_{i}}^{\mathrm{TT}}}{\partial t_{i}^2} \end{pmatrix}\\ =\begin{pmatrix} -2 &{}-\alpha \\ -\alpha &{} -\mu \end{pmatrix}. \end{aligned}$$

To ensure that $\pi _{M_i}^{\mathrm{TT}}$ is jointly concave in $p_i$ and $t_{i}$, we require that the leading principal minor $H_{1}^{\mathrm{TT}}=-2<0$, $H_2^{\mathrm{TT}}=2\mu -\alpha ^2>0$. By the first order condition, that is $\frac{\partial \pi _{M_i}^{\mathrm{TT}}}{\partial p_i}=0,$ $\frac{\partial \pi _{M_i}^{\mathrm{TT}}}{\partial t_i}=0,$ we can obtain the unique optimal retail price (23) and the optimal technology investment level (24).

Substituting Eqs. (23) and (24) in Eq. (25), we get the supplier’s profit $\pi _{S}^{\mathrm{TT}}(w)$. Because $\frac{d^{2} \pi _{S}^{\mathrm{TT}}}{d w^{2}} \!=\!{\frac{4\mu (1\!-\!b)((b\!-\!2)\mu \!-\!2(b\!-\!1)\alpha ) }{ \left( \left( \!-\!{\alpha }^{2}\!+\!\mu \right) b\!+\!{\alpha }^{2}\!-\!2\,\mu \right) ^{2}}} <0$, that is $\mu >\frac{2(1-b)\alpha }{(2-b)}$, we have $\pi _{S}^{\mathrm{TT}}(w)$ is concave in w. By the first order condition, we get the unique optimal wholesale price (28). By substituting Eq. (28) in Eqs. (23) and (24), we obtain the unique optimal retail price (26) and the technology investment level (27).

Finally, substituting Eqs. (26), (27) and (28) in Eqs. (22) and (25), we derive the manufacturers’ optimal profits (29) and the supplier’s optimal profit (30). $\square$

Proof of Corollary 1

In order to make the decision variables to be positive and the uniqueness of optimal solutions in case $\mathrm{TT}$, $a+(b-1)c>0$, $\mu >\max \left\{ \frac{\alpha ^2}{2}, \mu _{p}^{\mathrm{TT}}\right\}$ are needed. With resect to $\max \left\{ \frac{\alpha ^2}{2}, \mu _{p}^{\mathrm{TT}}\right\}$, we can simplify it as follows. When $0<\alpha <\frac{8a(1-b)}{2a(1-b)+(a-bc+c)}$, we have $\frac{\alpha ^2}{2}<\mu _{p}^{\mathrm{TT}}$, then $\mu _{p}^{\mathrm{TT}}=\max \left\{ \frac{\alpha ^2}{2}, \mu _{p}^{\mathrm{TT}}\right\}$, when $\alpha >\frac{8a(1-b)}{2a(1-b)+(a-bc+c)}$, we have $\mu _{p}^{\mathrm{TT}}<\frac{\alpha ^2}{2}$, then $\frac{\alpha ^2}{2}=\max \left\{ \frac{\alpha ^2}{2}, \mu _{p}^{\mathrm{TT}}\right\}$.

For any given $\alpha$, we have

$$\begin{aligned} \frac{\partial t_{i}^{\mathrm{TT}}}{\partial \alpha }= & {} -{\frac{\mu \, \left[ \left( b-1 \right) c+a \right] \left( b-2\right) }{ 2\left[ \mu \, \left( b-2 \right) -2\,\alpha \, \left( b-1\right) \right] ^{2}}},\nonumber \\ \frac{\partial w^{\mathrm{TT}}}{\partial \alpha }= & {} -{\frac{ \left[ \left( 1-b \right) {\alpha }^{2}+\mu \, \left( b-2\right) \alpha -\mu \, \left( b-2 \right) \right] \left[ \left( b-1\right) c+a \right] }{ \left[ \mu \, \left( b-2 \right) -2\,\alpha \,\left( b-1 \right) \right] ^{2}}},\nonumber \\ \frac{\partial p_{i}^{\mathrm{TT}}}{\partial \alpha }= & {} -{\frac{\mu \, \left[ \left( b-1 \right) c+a \right] }{\left[ \mu \,\left( b-2 \right) -2\,\alpha \, \left( b-1 \right) \right] ^{2}}},\nonumber \\ \frac{\partial \pi _{M_{i}}^{\mathrm{TT}}}{\partial \alpha }= & {} -{\frac{{\mu }^{2} \left[ \left( b-1 \right) c+a \right] ^{2}\left[ 4(1-b)-\alpha \, \left( 2-b \right) \right] }{4 \left[ \mu \,\left( b-2 \right) -2\,\alpha \, \left( b-1 \right) \right] ^{3}}},\nonumber \\ \frac{\partial \pi _{s}^{\mathrm{TT}}}{\partial \alpha }= & {} {\frac{ \left[ \left( b-1 \right) c+a \right] ^{2}\mu }{ \left[ \mu \,\left( b-2 \right) -2\,\alpha \, \left( b-1 \right) \right] ^{2}}}. \end{aligned}$$

(31)

From (31), we have the following results.

• (1) Apparently, when $\mu >\max \{\frac{\alpha ^2}{2}, \mu _{p}^{\mathrm{TT}}\}$, we have $\frac{\partial t^{\mathrm{TT}}}{\partial \alpha }>0$ and $\frac{\partial p_i^{\mathrm{TT}}}{\partial \alpha }<0$. Therefore, results (i) and (iii) hold.

• (ii) If $0<\alpha <1$, we have $\mu _{p}^{\mathrm{TT}}> \frac{\alpha ^2}{2}$, when $\mu >\mu _{p}^{\mathrm{TT}}$, then $\frac{\partial w^{\mathrm{TT}}}{\partial \alpha }<0$. If $1<\alpha <\frac{4a(2-b)}{2a(2-b)+(a+bc-c)}$, we have $\frac{\alpha ^2}{2}<\mu _{p}^{\mathrm{TT}}<\frac{(1-b)\alpha ^2}{(2-b)(\alpha -1)}$, when $\mu _{p}^{\mathrm{TT}}<\mu <\frac{(1-b)\alpha ^2}{(2-b)(\alpha -1)}$, then $\frac{\partial w^{\mathrm{TT}}}{\partial \alpha }<0$, when $\mu >\frac{(1-b)\alpha ^2}{(2-b)(\alpha -1)}$, then $\frac{\partial w^{\mathrm{TT}}}{\partial \alpha }>0$. If $\frac{4a(2-b)}{2a(2-b)+(a+bc-c)}<\alpha <\frac{8a(1-b)}{2a(1-b)+(a-bc+c)}$, we have $\max \{\frac{\alpha ^2}{2}, \frac{(1-b)\alpha ^2}{(2-b)(\alpha -1)}\}<\mu _{p}^{\mathrm{TT}}$, when $\mu >\mu _{p}^{\mathrm{TT}}$, then $\frac{\partial w^{\mathrm{TT}}}{\partial \alpha }>0$. If $\alpha >\frac{8a(1-b)}{2a(1-b)+(a-bc+c)}$, we have $\frac{\alpha ^2}{2}>\max \{\frac{(1-b)\alpha ^2}{(2-b)(\alpha -1)}, \mu _{p}^{\mathrm{TT}}\}$, when $\mu >\frac{\alpha ^2}{2}$, then $\frac{\partial w^{\mathrm{TT}}}{\partial \alpha }>0$. That is, if $\alpha >\frac{4a(2-b)}{2a(2-b)+(a+bc-c)}$, when $\mu >\max \{\frac{\alpha ^2}{2}, \mu _{p}^{\mathrm{TT}}\}$, then $\frac{\partial w^{\mathrm{TT}}}{\partial \alpha }>0$. Therefore, result (ii) holds.

• (2) (i) From $\frac{4(1-b)}{2-b}<\frac{8a(1-b)}{2a(1-b)+(a-bc+c)}$, if $0<\alpha <\frac{4(1-b)}{2-b}$, when $\mu >\mu _{p}^{\mathrm{TT}}$, we have $\frac{\partial \pi _{M_{i}}^{\mathrm{TT}}}{\partial \alpha }>0$. If $\frac{4(1-b)}{2-b}<\alpha <\frac{8a(1-b)}{2a(1-b)+(a-bc+c)}$, when $\mu >\mu _{p}^{\mathrm{TT}}$, we have $\frac{\partial \pi _{M_{i}}^{\mathrm{TT}}}{\partial \alpha }<0$, if $\alpha >\frac{8a(1-b)}{2a(1-b)+(a-bc+c)}$, when $\mu > \frac{\alpha ^2}{2}$, we have $\frac{\partial \pi _{M_{i}}^{\mathrm{TT}}}{\partial \alpha }<0$. That is, if $\alpha >\frac{4(1-b)}{2-b}$, when $\mu >\max \{\frac{\alpha ^2}{2}, \mu _{p}^{\mathrm{TT}}\}$, then $\frac{\partial \pi _{M_{i}}^{\mathrm{TT}}}{\partial \alpha }<0$. Therefore, result (i) holds.

• (ii) Apparently, if $\mu >\max \{\frac{\alpha ^2}{2}, \mu _{p}^{\mathrm{TT}}\}$, then $\frac{\partial \pi _{s}^{\mathrm{TT}}}{\partial \alpha }>0$. Therefore, result (ii) holds. $\square$

Proof of Corollary 2

For any given $\mu$, we have

$$\begin{aligned} \frac{\partial t_{i}^{\mathrm{TT}}}{\partial \mu }= & {} {\frac{ \alpha \,\left[ \left( b-1 \right) c+a \right] \left( b-2 \right) }{ 2\left[ \mu \, \left( b-2 \right) -2\,\alpha \,\left( b-1\right) \right] ^{2}}},\nonumber \\ \frac{\partial w^{\mathrm{TT}}}{\partial \mu }= & {} {\frac{ \left( \alpha -2 \right) \left( b-2 \right) \left[ \left( b-1 \right) c+a \right] \alpha }{ 2\left[ \mu \, \left( b-2\right) -2\,\alpha \, \left( b-1 \right) \right] ^{2}}},\nonumber \\ \frac{\partial p_{i}^{\mathrm{TT}}}{\partial \mu }= & {} {\frac{ \left[ \left( b-1 \right) c+a \right] \alpha }{ \left[ \mu \,\left( b-2 \right) -2\,\alpha \, \left( b-1 \right) \right] ^{2}}},\nonumber \\ \frac{\partial \pi _{M_{i}}^{\mathrm{TT}}}{\partial \mu }\!=\! & {} {\frac{\alpha \, \left[ \left( b\!-\!1 \right) c\!+\!a \right] ^{2}\left[ 2\, \left( b\!-\!1 \right) {\alpha }^{2}\!+\!\mu \, \left( b\!-\!2 \right) \alpha \!-\!8\,\mu \, \left( b\!-\!1 \right) \right] }{ 8\left[ \mu \, \left( b\!-\!2\right) \!-\!2\,\alpha \, \left( b\!-\!1 \right) \right] ^{3}}},\nonumber \\ \frac{\partial \pi _{s}^{\mathrm{TT}}}{\partial \mu }= & {} -{\frac{\alpha \, \left[ \left( b-1 \right) c+a \right] ^{2}}{\left[ \mu \, \left( b-2 \right) -2\,\alpha \, \left( b-1 \right) \right] ^{2}}}. \end{aligned}$$

(32)

From (32), we have the following results.

• (1) Apparently, if $\mu >\max \{\frac{\alpha ^2}{2}, \mu _{p}^{\mathrm{TT}}\}$, we can obtain $\frac{\partial t^{\mathrm{TT}}}{\partial \mu }<0$ and $\frac{\partial p_i^{\mathrm{TT}}}{\partial \mu }>0$. Therefore, results (i) and (iii) hold.

• (ii) From $\frac{8a(1-b)}{2a(1-b)+(a-bc+c)}\le 2$, that is $b>\frac{2(a-c)}{4a-2c}$, if $0<\alpha <2$, when $\mu >\max \{\frac{\alpha ^2}{2}, \mu _{p}^{\mathrm{TT}}\}$, then $\frac{\partial w^{\mathrm{TT}}}{\partial \mu }>0$. If $\alpha >2$, when $\frac{\mu >\alpha ^2}{2}$, then $\frac{\partial w^{\mathrm{TT}}}{\partial \mu }<0$. From $\frac{8a(1-b)}{2a(1-b)+(a-bc+c)}>2$, that is $0<b<\frac{2(a-c)}{4a-2c}$, if $0<\alpha <2$, when $\mu >\mu _{p}^{\mathrm{TT}}$, then $\frac{\partial w^{\mathrm{TT}}}{\partial \mu }>0$. If $\alpha >2$, when $\mu >\max \{\frac{\alpha ^2}{2}, \mu _{p}^{\mathrm{TT}}\}$, then $\frac{\partial w^{\mathrm{TT}}}{\partial \mu }<0$. Therefore, result (ii) holds.

• (2) (i) If $0<\alpha <\frac{16a(1-b)}{2a(3-2b)+a-bc+c}$, we have $\frac{2(1-b)\alpha ^2}{(b-2)\alpha +8(1-b)}<\mu _{p}^{\mathrm{TT}}$, when $\mu >\mu _{p}^{\mathrm{TT}}$, then $\frac{\partial \pi _{M_{i}}^{\mathrm{TT}}}{\partial \mu }<0$. If $\frac{16a(1-b)}{2a(3-2b)+a-bc+c}<\alpha <\frac{8a(1-b)}{2a(1-b)+(a-bc+c)}$, we have $\mu _{p}^{\mathrm{TT}}<\frac{2(1-b)\alpha ^2}{(b-2)\alpha +8(1-b)}$, when $\mu _{p}^{\mathrm{TT}}<\mu <\frac{2(1-b)\alpha ^2}{(b-2)\alpha +8(1-b)}$, then $\frac{\partial \pi _{M_{i}}^{\mathrm{TT}}}{\partial \mu }>0$, when $\mu >\frac{2(1-b)\alpha ^2}{(b-2)\alpha +8(1-b)}$, then $\frac{\partial \pi _{M_{i}}^{\mathrm{TT}}}{\partial \mu }<0$. If $\frac{8a(1-b)}{2a(1-b)+(a-bc+c)}<\alpha < \frac{8(1-b)}{2-b}$, we have $\frac{\alpha ^2}{2}<\frac{2(1-b)\alpha ^2}{(b-2)\alpha +8(1-b)}$, when $\frac{\alpha ^2}{2}<\mu <\frac{2(1-b)\alpha ^2}{(b-2)\alpha +8(1-b)}$, then $\frac{\partial \pi _{M_{i}}^{\mathrm{TT}}}{\partial \mu }>0$, when $\mu >\frac{2(1-b)\alpha ^2}{(b-2)\alpha +8(1-b)}$, then $\frac{\partial \pi _{M_{i}}^{\mathrm{TT}}}{\partial \mu }<0$. That is, if $\frac{16a(1-b)}{2a(3-2b)+a-bc+c}<\alpha <\frac{8(1-b)}{2-b}$, when $\max \{\frac{\alpha ^2}{2}, \mu _{p}^{\mathrm{TT}}\}<\mu <\frac{2(1-b)\alpha ^2}{(b-2)\alpha +8(1-b)}$, then $\frac{\partial \pi _{M_{i}}^{\mathrm{TT}}}{\partial \mu }>0$, when $\mu >\frac{2(1-b)\alpha ^2}{(b-2)\alpha +8(1-b)}$, then $\frac{\partial \pi _{M_{i}}^{\mathrm{TT}}}{\partial \mu }<0$. If $\alpha >\frac{8(1-b)}{2-b}$, we have $\frac{2(1-b)\alpha ^2}{(b-2)\alpha +8(1-b)}<\frac{\alpha ^2}{2}$, when $\mu >\frac{\alpha ^2}{2}$, then $\frac{\partial \pi _{M_{i}}^{\mathrm{TT}}}{\partial \mu }>0$. Therefore, result (i) holds.

• (ii) Apparently, when $0<\alpha <\frac{8a(1-b)}{2a(1-b)+(a-bc+c)}$ and $\mu >\mu _{p}^{\mathrm{TT}}$, or when $\alpha >\frac{8a(1-b)}{2a(1-b)+(a-bc+c)}$ and $\mu > \frac{\alpha ^2}{2}$,$\frac{\partial \pi _{s}^{\mathrm{TT}}}{\partial \mu }<0$. Therefore, result (ii) holds.

$\square$

Proof of Proposition 1

In order to make the decision variables to be positive and the uniqueness of optimal solutions in cases $\mathrm{TN}$ and $\mathrm{TT}$, we require $a+(b-1)c>0$, $B_1+B_2<0$, $\mu >\max \left\{ \frac{\alpha ^2}{2}, \mu _{u}^{\mathrm{TN}}, \mu _{p}^{\mathrm{TT}}\right\}$=$\max \left\{ \frac{\alpha ^2}{2}, \mu _{p}^{\mathrm{TT}}\right\}$, because $\mu _{u}^{\mathrm{TN}}<\max \left\{ \frac{\alpha ^2}{2}, \mu _{p}^{\mathrm{TT}}\right\}$. With resect to $\max \left\{ \frac{\alpha ^2}{2}, \mu _{p}^{\mathrm{TT}}\right\}$, we can simplify it as follows. If $0<\alpha <\frac{8a(1-b)}{2a(1-b)+(a-bc+c)}$, then $\max \left\{ \frac{\alpha ^2}{2}, \mu _{p}^{\mathrm{TT}}\right\} =\mu _{p}^{\mathrm{TT}}$. If $\alpha >\frac{8a(1-b)}{2a(1-b)+(a-bc+c)}$, then $\max \left\{ \frac{\alpha ^2}{2}, \mu _{p}^{\mathrm{TT}}\right\} =\frac{\alpha ^2}{2}$.

• (1) $\pi _{S}^{\mathrm{TT}}-\pi _{S}^{\mathrm{TN}}= {\frac{ f_{s}(\mu )\left( a+bc-c \right) ^{2}\alpha }{4\left[ 2\, \left( 1-b\right) \alpha + \left( b-2 \right) \mu \right] (B_1+B_2) }}$, where $f_{s}(\mu )= 4(b+2)\mu ^2+(b+1)(b+2)\alpha ^2(\alpha -6)\mu +2(b+1)^2\alpha ^4$. For $f_{s}(\mu )$, from $\Delta =(b+1)^2(b+2)^2\alpha ^4[(\alpha -6)^2-32]$, we have $f_{s}(\mu )>0$ when $\Delta <0$ if $6-4\sqrt{2}<\alpha <6+4\sqrt{2}$. Clearly, if $0<\alpha <6-4\sqrt{2}$ and $\alpha >6+4\sqrt{2}$, then $\Delta >0$. For $f_{s}(\mu )=0$, we have two different real roots $\mu _i=\frac{\alpha ^2(b+1)(6-\alpha \pm \sqrt{\alpha ^2-12\alpha +4})}{8(b+2)}$, i=1,2. Since $\min \left\{ \frac{\alpha ^2}{2}, \mu _{p}^{\mathrm{TT}}\right\}>\mu _2>\mu _1$, when $\mu >\max \left\{ \frac{\alpha ^2}{2}, \mu _{p}^{\mathrm{TT}}\right\}$, we have $f_{s}(\mu )>0$. In summary, when $\mu >\max \left\{ \frac{\alpha ^2}{2}, \mu _{p}^{\mathrm{TT}}\right\}$, we have $f_{s}(\mu )>0$. Consequently, $\pi _{S}^{\mathrm{TT}}>\pi _{S}^{\mathrm{TN}}$.

• (2) $\pi _{S}^{\mathrm{TN}}-\pi _{S}^{\mathrm{NN}}={\frac{ \left( b+2 \right) \left( a+bc-c \right) ^{2}\alpha f_{s1}(u)}{4\left( b-2 \right) (B_1+B_2)}}$, where $f_{s1}(u)=4(b+2)\mu -(b+1)(\alpha +2)\alpha ^2$. Clearly, when $0<\mu <\frac{(b+1)(\alpha +2)\alpha ^2}{4(b+2)}$, we get $f_{s1}(u)<0$, when $\mu >\frac{(b+1)(\alpha +2)\alpha ^2}{4(b+2)}$, we get $f_{s1}(u)>0$.

From $g_{s1}^{\mu }=\mu _{p}^{\mathrm{TT}}-\frac{(b+1)(\alpha +2)\alpha ^2}{4(b+2)}=\frac{16a(1-b)(b+2)\alpha -(1+b)(\alpha ^2+2\alpha )[2a(1-b)+(a-bc+c)]\alpha }{4(b+2)[2a(1-b)+(a-bc+c)]}$, when $0<\alpha <\alpha _{s1}$, where $\alpha _{s1}=\frac{2(1+b)[2a(1-b)+(a-bc+c)]+\sqrt{\Delta _{s1}}}{2(1+b)[2a(1-b)+(a-bc+c)]}$, $\Delta _{s1}=4(b+1)^2[2a(1-b)+(a-bc+c)]^2+64a(1-b^2)(b+2)[2a(1-b)+(a-bc+c)]$, we have $g_{s1}^{u}>0$, when $\alpha >\alpha _{s1}$, we have $g_{s1}^{u}<0$. From $g_{s2}^{\mu }=\frac{\alpha ^2}{2}-\frac{(b+1)(\alpha +2)\alpha ^2}{4(b+2)}=\frac{\alpha ^2[2-(b+1)\alpha ]}{4(b+2)}$, we derive that $g_{s2}^{\mu }>0$ if $0<\alpha <\frac{2}{b+1}$ and $g_{s2}^{\mu }<0$ if $\alpha >\frac{2}{b+1}$.

When $\max \{\alpha _{s1},\frac{2}{b+1}\}<\frac{8a(1-b)}{2a(1-b)+(a-bc+c)}$, where $\max \{\alpha _{s1},\frac{2}{b+1}\}=\alpha _{s1}$, we can conclude that if $0<\alpha <\max \{\alpha _{s1},\frac{2}{b+1}\}=\alpha _{s1}$, when $\mu >\mu _{p}^{\mathrm{TT}}$, we have $f_{s1}(u)>0$, that is $\pi _{S}^{\mathrm{TN}}>\pi _{S}^{\mathrm{NN}}$. If $\alpha >\alpha _{s1}$, when $\max \left\{ \frac{\alpha ^2}{2}, \mu _{p}^{\mathrm{TT}}\right\}<\mu <\frac{(b+1)(\alpha +2)\alpha ^2}{4(b+2)}$, we have $f_{s1}(u)<0$, that is $\pi _{S}^{\mathrm{TN}}<\pi _{S}^{\mathrm{NN}}$, when $\mu >\frac{(b+1)(\alpha +2)\alpha ^2}{4(b+2)}$, we get $f_{s1}(u)>0$, that is $\pi _{S}^{\mathrm{TN}}>\pi _{S}^{\mathrm{NN}}$.

When $\min \{\alpha _{s1},\frac{2}{b+1}\}>\frac{8a(1-b)}{2a(1-b)+(a-bc+c)}$, where $\min \{\alpha _{s1},\frac{2}{b+1}\}=\alpha _{s1}$, we can conclude that if $0<\alpha <\max \{\alpha _{s1},\frac{2}{b+1}\}=\frac{2}{b+1}$, when $\mu >\max \left\{ \frac{\alpha ^2}{2}, \mu _{p}^{\mathrm{TT}}\right\}$, we have $f_{s1}(u)>0$, that is $\pi _{S}^{\mathrm{TN}}>\pi _{S}^{\mathrm{NN}}$. If $\alpha >\frac{2}{b+1}$, when $\frac{\alpha ^2}{2}<\mu <\frac{(b+1)(\alpha +2)\alpha ^2}{4(b+2)}$, we have $f_{s1}(u)<0$, that is $\pi _{S}^{\mathrm{TN}}<\pi _{S}^{\mathrm{NN}}$, when $\mu >\frac{(b+1)(\alpha +2)\alpha ^2}{4(b+2)}$, we get $f_{s1}(u)>0$, that is $\pi _{S}^{\mathrm{TN}}>\pi _{S}^{\mathrm{NN}}$.

• (3) When $\mu >\max \left\{ \frac{\alpha ^2}{2}, \mu _{p}^{\mathrm{TT}}\right\}$, $\pi _{S}^{\mathrm{TT}}-\pi _{S}^{\mathrm{NN}}= {\frac{\alpha \, \left( \left( b-1 \right) c+a \right) ^{2}}{ \left( b-2 \right) (2\alpha (1-b)+\mu (b-2))}}>0$.

From results (1), (2) and (3), we conclude that Proposition 1 holds. $\square$