The impact of cross-shareholding under different power structures considering green investment and green marketing

Abstract

Cross-shareholding has played an important role in strengthening strategic synergy among enterprises, but its impact on the green development of enterprises is unclear. In this paper, we construct an analytical framework that includes a manufacturer and a retailer to explore the impact of cross-shareholdings under different leaderships on green supply chain operational decisions and profits, in which the manufacturer invests in green technologies and the retailer conducts green marketing. By constructing and solving the game model of manufacturer-led and retailer-led before and after cross-shareholding, it is found that after cross-shareholding, the product’s green level and the retailers’ marketing level are improved. For supply chain leaders, cross-shareholdings always increase their profits. Only when the follower holds the leader’s shares no more than a certain value, cross-shareholding will increase the total profit of the supply chain. In addition, we obtain the optimal decisions and profits of the supply chain in the integrated situation and design two-part pricing contracts to achieve cross-shareholding supply chain coordination. The results of this paper can provide theoretical guidance and decision support for enterprises interested in using cross-shareholding to improve supply chain performance.

Appendices

Appendix

Proof of Lemma 1 We solve this model by backward induction and first consider the retailer’s decision. The Hessian matrix of ${\pi }_{r}^{DM}$ about $p,v$ is $[\begin{array}{cc}-2& k\\ k& -2\end{array}]$. Since $0<k<\sqrt{3}$, we can easily deduce that ${\pi }_{r}^{DM}$ is joint concave in $p,v$. Let $\frac{\partial {\pi }_{r}^{DM}}{\partial p}=0\frac{\partial {\pi }_{r}^{DM}}{\partial v}=0\frac{\partial {\pi }_{r}^{DM}}{\partial v}=0$, we have $p(w,e)=\frac{-w{k}^{2}+2a+2e+2w}{4-{k}^{2}}$, $v(w,e)=-\frac{ak+ek-kw}{4-{k}^{2}}$.

Substituting $p(w,e)$ and $v(w,e)$ into ${\pi }_{r}^{DM}$, the Hessian matrix of ${\pi }_{r}^{DM}$ about $w,e$ is $[\begin{array}{cc}-\frac{4}{4-{k}^{2}}& \frac{2}{4-{k}^{2}}\\ \frac{2}{4-{k}^{2}}& -2\end{array}]$. Obviously, its sequential principal minors are $|{H}_{1}|=-\frac{4}{4-{k}^{2}}<{0}$, $|{H}_{2}|=\frac{2(19-4{k}^{2})}{{(4-{k}^{2})}^{2}}>{0}$, respectively. Hence, we know that ${\pi }_{r}^{DM}$ is joint concave in $w$ and $e$. Let $\frac{\partial {\pi }_{m}^{DM}}{\partial w}=0\frac{\partial {\pi }_{m}^{DM}}{\partial e}=0$, we get ${w}^{DM}=\frac{4a+3c-a{k}^{2}-c{k}^{2}}{7-2{k}^{2}}$, ${e}^{MD}=\frac{a-c}{7-2{k}^{2}}$. From this, we can obtain ${p}^{DM}$, ${v}^{DM}$, ${\pi }_{m}^{DM}$, ${\pi }_{r}^{DM}$.

Proof of Lemma 2 We solve this model by backward induction and first consider the manufacturer’s decision. Let $p=w+y$, we derive that the Hessian matrix of ${\pi }_{m}^{DR}$ about $w,e$ is $\left[\begin{array}{cc}-2& 1\\ 1& -2\end{array}\right]$. Obviously, ${\pi }_{m}^{DR}$ is joint concave in $w$ and $e$. Let $\frac{\partial {\pi }_{m}^{DR}}{\partial w}=0\frac{\partial {\pi }_{m}^{DR}}{\partial e}=0$, we get $w\left(y,v\right)=\frac{2a}{3}+\frac{c}{3}-\frac{2y}{3}+\frac{2kv}{3}$, $e\left(y,v\right)=\frac{2a}{3}+\frac{c}{3}-\frac{2y}{3}+\frac{2kv}{3}$. Substituting $w\left(y,v\right)$ and $e\left(y,v\right)$ into ${\pi }_{r}^{DR}$, we get the Hessian matrix of ${\pi }_{r}^{DR}$ with respect to $y,v$ is $\left[\begin{array}{cc}-\frac{4}{3}& \frac{2k}{3}\\ \frac{2k}{3}& -2\end{array}\right]$. Since $0<k<\sqrt{3}$, we can easily induce that ${\pi }_{r}^{DR}$ is joint concave in $y,v$. Let $\frac{\partial {\pi }_{r}^{DR}}{\partial y}=0$, $\frac{\partial {\pi }_{r}^{DR}}{\partial v}=0$, we obtain ${y}^{DR}$, ${v}^{DR}$. From this, we get ${w}^{DR},{e}^{DR},{p}^{DR},{\pi }_{m}^{DR},{\pi }_{r}^{DR}$

The proofs of Lemmas 3 and 4 are similar to Lemmas 1 and 2. Hence, we omit them.

Proof of Corollary 1 We can derive $\frac{\partial {e}^{DM}}{\partial k}\text{=}\frac{4k(a-c)}{{(2{k}^{2}-7)}^{2}}>0$, $\frac{\partial {e}^{DR}}{\partial k}\text{=}\frac{2k(a-c)}{{({k}^{2}-6)}^{2}}>0$, $\frac{\partial {v}^{DM}}{\partial k}=\frac{(a-c)(2{k}^{2}+7)}{{(2{k}^{2}-7)}^{2}}>0$, $\frac{\partial {v}^{DR}}{\partial k}=\frac{({k}^{2}+6)(a-c)}{{({k}^{2}-6)}^{2}}>0$.

Proof of Corollary 2 (1) In $ML$, we can derive

$$\begin{array}{c}\frac{\partial {e}^{CM}}{\partial {l}_{m}}=\frac{(4-{k}^{2})(a-c)(1-{l}_{r})}{{(4{l}_{m}+8{l}_{r}-4{l}_{m}{l}_{r}-{k}^{2}{l}_{m}-2{k}^{2}{l}_{r}+2{k}^{2}+{k}^{2}{l}_{m}{l}_{r}-7)}^{2}}>0,\\ \frac{\partial {e}^{CM}}{\partial {l}_{r}}=\frac{(4-{k}^{2})(a-c)({2}-{l}_{m})}{{(4{l}_{m}+8{l}_{r}-4{l}_{m}{l}_{r}-{k}^{2}{l}_{m}-2{k}^{2}{l}_{r}+2{k}^{2}+{k}^{2}{l}_{m}{l}_{r}-7)}^{2}}>0,\\ \begin{array}{c}\frac{\partial {v}^{CM}}{\partial {l}_{m}}=\frac{k(4-{k}^{2})(a-c){(1-{l}_{r})}^{2}}{{(4{l}_{m}+8{l}_{r}-4{l}_{m}{l}_{r}-{k}^{2}{l}_{m}-2{k}^{2}{l}_{r}+2{k}^{2}+{k}^{2}{l}_{m}{l}_{r}-7)}^{2}}>0,\\ \frac{\partial {v}^{CM}}{\partial {l}_{r}}=\frac{k(a-c)}{{(4{l}_{m}+8{l}_{r}-4{l}_{m}{l}_{r}-{k}^{2}{l}_{m}-2{k}^{2}{l}_{r}+2{k}^{2}+{k}^{2}{l}_{m}{l}_{r}-7)}^{2}}>0\end{array}\end{array}$$

(2) In RL, we can derive $\frac{\partial {e}^{CR}}{\partial {l}_{r}}=\frac{({3}-3{l}_{m})(a-c)(1-{l}_{m})}{{({k}^{2}+6{l}_{m}+3{l}_{r}-3{l}_{m}{l}_{r}-6)}^{2}}>0$, $\frac{\partial {e}^{CR}}{\partial {l}_{m}}=\frac{{k}^{2}(a-c)}{{({k}^{2}+6{l}_{m}+3{l}_{r}-3{l}_{m}{l}_{r}-6)}^{2}}>0$, $\frac{\partial {v}^{CR}}{\partial {l}_{m}}=\frac{k({6}-3{l}_{r})(a-c)}{{({k}^{2}+6{l}_{m}+3{l}_{r}-3{l}_{m}{l}_{r}-6)}^{2}}>0$, $\frac{\partial {v}^{CR}}{\partial {l}_{r}}=\frac{k({3}-3{l}_{m})(a-c)}{{({k}^{2}+6{l}_{m}+3{l}_{r}-3{l}_{m}{l}_{r}-6)}^{2}}>0$.

Proof Proposition 1 In ML, we can rewrite ${w}^{CM}$ and ${w}^{DM}$ as ${w}^{CM}=c+\frac{(a-c)(4-{k}^{2}){(1-{l}_{m})}^{2}(1-{l}_{r})}{(1-{l}_{m}-{l}_{r})[(4-{k}^{2})(2-{l}_{m})(1-{l}_{r})-1]}$, ${w}^{DM}=c+\frac{(4-{k}^{2})(a-c)}{7-2{k}^{2}}$. So, to compare ${w}^{CM}$ and ${w}^{DM}$, we only need to compare their second part. Let ${\theta }_{1}=\frac{\frac{(a-c)(4-{k}^{2}){(1-{l}_{m})}^{2}(1-{l}_{r})}{(1-{l}_{m}-{l}_{r})[(4-{k}^{2})(2-{l}_{m})(1-{l}_{r})-1]}}{\frac{(4-{k}^{2})(a-c)}{7-2{k}^{2}}}$ and $\delta =\frac{1}{4-{k}^{2}}$, we can get ${\theta }_{1}=\frac{(2-\delta ){(1-{l}_{m})}^{2}(1-{l}_{r})}{(1-{l}_{m}-{l}_{r})[(2-{l}_{m})(1-{l}_{r})-\delta ]}$. If ${l}_{r1}=\frac{2+{l}_{m}{}^{2}\delta -{l}_{m}{}^{2}-2{l}_{m}\delta -\sqrt{{l}_{m}{}^{2}{(2-{l}_{m})}^{2}{(1-\delta )}^{2}+4{(1-{l}_{m})}^{2}}}{2(2-{l}_{m})}<{l}_{r}<\frac{2+{l}_{m}{}^{2}\delta -{l}_{m}{}^{2}-2{l}_{m}\delta \text{+}\sqrt{{l}_{m}{}^{2}{(2-{l}_{m})}^{2}{(1-\delta )}^{2}+4{(1-{l}_{m})}^{2}}}{2(2-{l}_{m})}$, we have ${\theta }_{1}>1$, that is, ${w}^{CM}>{w}^{DM}$. However, we can derive $\frac{2+{l}_{m}{}^{2}\delta -{l}_{m}{}^{2}-2{l}_{m}\delta \text{+}\sqrt{{l}_{m}{}^{2}{(2-{l}_{m})}^{2}{(1-\delta )}^{2}+4{(1-{l}_{m})}^{2}}}{2(2-{l}_{m})}-\frac{1}{2}>\frac{2-(2-{l}_{m})+{l}_{m}{}^{2}\delta -{l}_{m}{}^{2}-2{l}_{m}\delta +2(1-{l}_{m})}{2(2-{l}_{m})}>0$. Therefore, if ${l}_{r1}<{l}_{r}\le 50\%$, then ${w}^{CM}>{w}^{DM}$ and ${w}^{CM}\le {w}^{DM}$ otherwise.

(2) In RL, we can rewrite ${w}^{CM}$ and ${w}^{DR}$ as ${w}^{CR}=c+\frac{(a-c)(1-{l}_{m})\{2[(2{l}_{r}-3){l}_{m}-{l}_{r}+1]+{l}_{m}(1-{l}_{r})\}}{(1-{l}_{m}-{l}_{r})[3(1-{l}_{m})(2-{l}_{r})-{k}^{2}]}$, ${w}^{DR}=c+\frac{2(a-c)}{6-{k}^{2}}$. To compare ${w}^{CR}$ and ${w}^{DR}$, we only need to compare their second part.

Let ${\theta }_{2}=\frac{\frac{(a-c)(1-{l}_{m})\{2[(2{l}_{r}-3){l}_{m}-{l}_{r}+1]+{l}_{m}(1-{l}_{r})\}}{(1-{l}_{m}-{l}_{r})[3(1-{l}_{m})(2-{l}_{r})-{k}^{2}]}}{\frac{2(a-c)}{6-{k}^{2}}}$. By ${\theta }_{2}=1$, we have ${l}_{m1}=\frac{\frac{3}{2}{l}_{m}{}^{2}+[\frac{1}{2}+\frac{1}{2}(\frac{1}{2}{k}^{2}-6)-\frac{3}{2}{k}^{2}+4]{l}_{r}-\frac{1}{2}+(8-\frac{1}{2}{k}^{2})\frac{1}{2}+\frac{3}{2}{k}^{2}-8-\sqrt{\Delta }}{\frac{3}{2}{l}_{r}({k}^{2}-4)-1+(16-{k}^{2})\frac{1}{2}+4({k}^{2}-4)-{k}^{2}}$, ${l}_{m2}=\frac{\frac{3}{2}{l}_{m}{}^{2}+[\frac{1}{2}+\frac{1}{2}(\frac{1}{2}{k}^{2}-6)-\frac{3}{2}{k}^{2}+4]{l}_{r}-\frac{1}{2}+(8-\frac{1}{2}{k}^{2})\frac{1}{2}+\frac{3}{2}{k}^{2}-8+\sqrt{\Delta }}{\frac{3}{2}{l}_{r}({k}^{2}-4)-1+(16-{k}^{2})\frac{1}{2}+4({k}^{2}-4)-{k}^{2}}$, where $\Delta ={(\frac{1}{2}-3)}^{2}{(1-{l}_{r})}^{2}\frac{1}{4}{k}^{2}+\frac{3}{2}(1-{l}_{m})(-\frac{1}{2}{l}_{r}{}^{2}+\frac{15}{2}{l}_{r}-\frac{15}{2}){k}^{2}+\frac{9}{4}{({l}_{r}{}^{2}-3{l}_{r}+4)}^{2}$. Therefore, if $\mathrm{max}(\mathrm{min}({l}_{m1},{l}_{m2}),0)<{l}_{m}\le \mathrm{min}(\mathrm{max}({l}_{m1},{l}_{m2}),50\%)$, then ${w}^{CR}>{w}^{DR}$ and ${w}^{CR}\le {w}^{DR}$ otherwise.

The proof of Proposition 1 (3) and (4) is similar to Proposition 1 (1) and (2), we omit it.

Proof Proposition 2 In ML, we have ${e}^{CM}-{e}^{DM}=\frac{(a-c)((4-{k}^{2})[2-(2-{l}_{m})(1-{l}_{r})])}{(7-2{k}^{2})[(4-{k}^{2})(2-{l}_{m})(1-{l}_{r})-1]}$.

Since $(4-{k}^{2})[2-(2-{l}_{m})(1-{l}_{r})]>0$, we have ${e}^{CM}-{e}^{DM}> \text{0}$. Similarly, it can be proved that ${e}^{CR}-{e}^{DR} \, \text{>} \, {0}$.

In RL, we have ${v}^{CM}-{v}^{DM}=\frac{k(a-c)((4-{k}^{2})(1-{l}_{r}){l}_{m}+{l}_{r})}{(7-2{k}^{2})[(4-{k}^{2})(2-{l}_{m})(1-{l}_{r})-1]}$.

Since $(4-{k}^{2})(1-{l}_{r}){l}_{m}+{l}_{r}>0$, we have ${v}^{CM}-{v}^{DM}> \text{0}$. Similarly, it can be proved that ${v}^{CR}-{v}^{DR}\text{>}{0}$.

Proof Proposition 3 In ML, let $\delta =\frac{1}{4-{k}^{2}}$, by ${\pi }_{sc}^{CM}>{\pi }_{sc}^{DM}$, we have $0<{l}_{r}<\frac{(1-{l}_{m}){\delta }^{2}+({l}_{m}{}^{2}+{l}_{m}-2)\delta -3{l}_{m}{}^{2}+4{l}_{m}+\sqrt{[{(1-{l}_{m})}^{2}\delta -3{l}_{m}{}^{2}+4{l}_{m}]{(2-\delta )}^{2}\delta }}{(3-2{l}_{m}){\delta }^{2}+({l}_{m}{}^{2}+4{l}_{m}-8)\delta -3{l}_{m}{}^{2}+4{l}_{m}}={l}_{r2}$.

In RL, let $\gamma =\frac{{k}^{2}}{3}$, by ${\pi }_{sc}^{CR}>{\pi }_{sc}^{DR}$, we have $0<{l}_{m}<\frac{(1-{l}_{r}){\gamma }^{2}+({l}_{r}{}^{2}+{l}_{r}-2)\gamma -3{l}_{r}{}^{2}+4{l}_{r}+\sqrt{[{(1-{l}_{r})}^{2}\gamma -3{l}_{r}{}^{2}+4{l}_{r}]{(2-\gamma )}^{2}\gamma }}{(3-2{l}_{r}){\gamma }^{2}+({l}_{r}{}^{2}+4{l}_{r}-8)\gamma -3{l}_{r}{}^{2}+4{l}_{r}}={l}_{m3}$.

Proof of Proposition 4 We have ${\pi }_{m}^{CM}-{\pi }_{m}^{DM}=\frac{{(a-c)}^{2}((4-{k}^{2})(1-{l}_{r}){l}_{m}+{l}_{r})}{(7-2{k}^{2})[(4-{k}^{2})(2-{l}_{m})(1-{l}_{r})-1]}>0$, that is, ${\pi }_{m}^{CM}>{\pi }_{m}^{DM}$ for any ${l}_{m}$ and ${l}_{r}$.

Proof Proposition 5 We have ${\pi }_{r}^{CR}-{\pi }_{r}^{DR}=\frac{12(a-c)(1-{l}_{m}){l}_{r}+{l}_{m}{k}^{2}{(a-c)}^{2}}{4[3(1-{l}_{m})(2-{l}_{r})-{k}^{2}](6-{k}^{2})}>0$, that is, ${\pi }_{r}^{CR}>{\pi }_{r}^{DR}$ for any ${l}_{m}$ and ${l}_{r}$.

Proof of Lemma 5 Let $\frac{\partial {\pi }_{sc}^{C}}{\partial p}=0$, $\frac{\partial {\pi }_{sc}^{C}}{\partial v}=0$, $\frac{\partial {\pi }_{sc}^{C}}{\partial e}=0$, we have ${p}_{sc}^{C}=c+\frac{2(a-c)}{3-{k}^{2}}{e}_{sc}^{C}=\frac{a-c}{3-{k}^{2}}{v}_{sc}^{C}=\frac{k(a-c)}{3-{k}^{2}}$. From this, we get ${\pi }_{sc}^{C}=\frac{{(a-c)}^{2}}{3-{k}^{2}}$.

Proof Proposition 6 (1) We can derive $\frac{\partial {e}_{sc}^{C}}{\partial k}=\frac{2k(a-c)}{{(3-{k}^{2})}^{2}}>0$, $\frac{\partial {v}_{sc}^{C}}{\partial k}=\frac{({k}^{2}+3)(a-c)}{{(3-{k}^{2})}^{2}}>0$.

(2) We can derive ${v}_{sc}^{C}-{v}^{CM}=\frac{k(a-c)((4-{k}^{2})(1-{l}_{m})(1-{l}_{r})+{l}_{r})}{(3-{k}^{2})[(4-{k}^{2})(2-{l}_{m})(1-{l}_{r})-1]}>0$, ${e}_{sc}^{C}-{e}^{CM}=\frac{(a-c)((4-{k}^{2})(2+{l}_{m}-2{l}_{r}-{l}_{m}{l}_{r}))}{(3-{k}^{2})[(4-{k}^{2})(2-{l}_{m})(1-{l}_{r})-1]}>0$, that is, ${v}_{sc}^{C}>{v}^{CM}$ and ${e}_{sc}^{C}>{e}^{CM}$.

Similarly, it can be proved that ${v}_{sc}^{C}>{v}^{CM}$ and ${e}_{sc}^{C}>{e}^{CR}$.

Proof Proposition 7 (1) In ML, for the manufacturer is in the dominant position, we first consider the retailer’s decisions. Let $\frac{\partial {\pi }_{r}^{TC}}{\partial p}=0$ $\frac{\partial {\pi }_{r}^{TC}}{\partial v}=0$, we have $p(w,e)=\frac{-(2(a+e)-(-2+{k}^{2})w)(-1+{l}_{\text{m}})+(-2+{k}^{2})(-c+w){l}_{\text{r}}}{(-4+{k}^{2})(-1+{l}_{\text{m}})}$, $v(w,e)=\frac{-k(a+e-w)(-1+{l}_{\text{m}})+k(-c+w){l}_{\text{r}}}{(-4+{k}^{2})(-1+{l}_{\text{m}})}$.

To reach the level of profit in the integration situation, there should be $p(w,e)={p}_{sc}^{C},v(w,e)={v}_{sc}^{C}$. From this, we get ${e}^{TC}={e}^{C},{w}^{TC}=c$. Since the manufacturer is dominant, it can have all the profits obtained by the coordination, and the retailer can have the reserved profits. By ${\pi }_{r}^{TC}-S={\pi }_{r}^{CM}$, we obtain ${S}^{TC}$.

(2) In RL, for the retailer is in the dominant position, we first consider the manufacturer’s decisions. Let $\frac{\partial {\pi }_{m}^{TC}}{\partial e}=0$, we have $e(p,v)=\frac{{l}_{m}(p-w)+(c-w)({l}_{r}-1)}{2(1-l){r}}$. To reach the level of profit in the case of integration, there should be $p={p}_{sc}^{C},v={v}_{sc}^{C},e(p,v)={e}_{sc}^{C}$. From this, we get ${w}^{TC}=\frac{-c{k}^{2}+2a+c}{3-{k}^{2}}$. Since the retailer is dominant, it can have all the profits obtained by the coordination, and the manufacturer can have the reserved profits. By ${\pi }_{m}^{TC}-S={\pi }_{m}^{CR}$, we obtain ${S}^{TC}$.