Service outsourcing under different supply chain power structures

Abstract

This paper studies the impact of service outsourcing under three supply chain power structures: manufacturer-Stackelberg, vertical-Nash and retailer-Stackelberg supply chains. We first investigate price and service decisions by comparing the integrated channel and the decentralized channel with service outsourcing. It is found that a lower retail price or a higher service level could occur in the decentralized channel with service outsourcing compared to those in the integrated channel, but they never occur simultaneously. Next, we examine the manufacturer’s channel strategies, and demonstrate that service outsourcing can always help the manufacturer earn more profit from the decentralized channel than from the integrated channel in the manufacturer-Stackelberg and retailer-Stackelberg markets, as long as the retailer’s service investment is sufficiently efficient. However, in the vertical-Nash supply chain, this occurs only when the manufacturer is inefficient and the retailer is efficient enough in service provision. Finally, comparisons of channel efficiency and numerical analysis are provided.

Appendix: Proofs of Lemmas and Propositions

Proof of Lemma 1:

The Hessian matrix of the profit function $\Pi ^{C}$ with respect to $p^{C}$ and $s^{C}$ is given by $H_1 =\left[ {{\begin{array}{ll} {-2\beta }&{} \gamma \\ \gamma &{} {-\lambda _M } \\ \end{array} }} \right]$. Since $-2\beta <0$ and $-\lambda _M <0$, so if $2\lambda _M \beta -\gamma ^{2}>0$, profit function $\Pi ^{C}$ is strictly jointly concave with respect to $p^{C}$ and $s^{C}$.

Proof of Lemma 2:

Profit function $\Pi _R^M$, with respect to $u^{M}$ and $s^{M}$, has a Hessian matrix $H_2 =\left[ {{\begin{array}{ll} {-2\beta }&{} \gamma \\ \gamma &{} {-\lambda _R } \\ \end{array} }} \right]$. Similar to Lemma 1, it is easy to check if $2\lambda _R \beta -\gamma ^{2}>0$ holds, then $H_2$ is strictly jointly concave in $u^{M}$ and $s^{M}$.

Proof of Proposition 1:

• (a) Compare $s^{M^{{*}}}$and $s^{C^{{*}}}$, expressions of which are summarized in Table 1, we have

  $$\begin{aligned} \Delta s^{M^{{*}}}=s^{M^{{*}}}-s^{C^{{*}}}=\frac{\gamma \alpha }{2\left( {2\lambda _R \beta -\gamma ^{2}} \right) }-\frac{\gamma \alpha }{2\lambda _M \beta -\gamma ^{2}}=\frac{\left( {2\lambda _M \beta -4\lambda _R \beta +\gamma ^{2}} \right) ca}{2\left( {2\lambda _R \beta -\gamma ^{2}} \right) \left( {2\lambda _M \beta -\gamma ^{2}} \right) }. \end{aligned}$$

  $\Delta s^{M^{{*}}}\ge 0$ if $\frac{\gamma ^{2}}{2\beta }<\lambda _R \le \frac{2\lambda _M \beta +\gamma ^{2}}{4\beta }$; otherwise, $\Delta s^{M^{{*}}}<0$. Note that $\frac{2\lambda _M \beta +\gamma ^{2}}{4\beta }>\frac{\gamma ^{2}}{2\beta }$ is guaranteed by $\lambda _M >\frac{\gamma ^{2}}{2\beta }$.

• (b) First, comparing $s^{N^{{*}}}$ and $s^{C^{{*}}}$, we have

  $$\begin{aligned} \Delta s^{N^{{*}}}=s^{N^{{*}}}-s^{C^{{*}}}=\frac{\gamma \alpha }{3\lambda _R \beta -\gamma ^{2}}-\frac{\gamma \alpha }{2\lambda _M \beta -\gamma ^{2}}=\frac{\left( {2\lambda _M -3\lambda _R } \right) \beta \gamma \alpha }{\left( {3\lambda _R \beta -\gamma ^{2}} \right) \left( {2\lambda _M \beta -\gamma ^{2}} \right) }. \end{aligned}$$

  $\Delta s^{N^{{*}}}\ge 0$ if $\frac{\gamma ^{2}}{2\beta }<\lambda _R \le \frac{2}{3}\lambda _M$, and $\Delta s^{N^{{*}}}<0$ otherwise. Based on the range of $\lambda _M$, we divide the discuss into two cases below:

• (i) if $\lambda _M \le \frac{3\gamma ^{2}}{4\beta }$, then $\lambda _R >\frac{2}{3}\lambda _M$, which indicates $\Delta s^{N^{{*}}}<0$;

• (ii) if $\lambda _M >\frac{3\gamma ^{2}}{4\beta }$, then either $\frac{\gamma ^{2}}{2\beta }<\lambda _R \le \frac{2}{3}\lambda _M$ or $\lambda _R >\frac{2}{3}\lambda _M$ holds, corresponding to $\Delta s^{N^{{*}}}\ge 0$ or $\Delta s^{N^{{*}}}<0$, respectively.

• (c) Similar to the proof of Proposition 1(a) and therefore omitted.

Proof of Proposition 2:

• (a) Comparing $p^{C^{{*}}}$ and $p^{M^{{*}}}$ in Sect. 3, we have

  $$\begin{aligned} \Delta p^{M^{{*}}}= & {} p^{M^{{*}}}-p^{C^{{*}}}=\frac{\left( {3\lambda _R \beta -\gamma ^{2}} \right) \alpha }{2\beta \left( {2\lambda _R \beta -\gamma ^{2}} \right) }-\frac{\lambda _M \alpha }{2\lambda _M \beta -\gamma ^{2}}\\= & {} \frac{\left( {2\lambda _M \lambda _R \beta ^{2}-3\lambda _R \beta \gamma ^{2}+\gamma ^{4}} \right) a}{2\beta \left( {2\lambda _R \beta -\gamma ^{2}} \right) \left( {2\lambda _M \beta -\gamma ^{2}} \right) }. \end{aligned}$$

  $\Delta p^{M^{{*}}}>0$ if $\beta \left( {3\gamma ^{2}-2\lambda _M \beta } \right) \lambda _R <\gamma ^{4}$, which always holds when $\lambda _M \ge \frac{3\gamma ^{2}}{2\beta }$ (case (ii)). We next consider the case in which $\lambda _M <\frac{3\gamma ^{2}}{2\beta }$ (case (i)). $\Delta p^{M^{{*}}}>0$ holds when $\frac{\gamma ^{2}}{2\beta }<\lambda _R <\frac{\gamma ^{4}}{\beta \left( {3\gamma ^{2}-2\lambda _M \beta } \right) }$, which entails $\frac{\gamma ^{2}}{2\beta }<\frac{\gamma ^{4}}{\beta \left( {3\gamma ^{2}-2\lambda _M \beta } \right) }$, equivalent to $\lambda _M >\frac{\gamma ^{2}}{2\beta }$.

• (b) Next, we prove the case for vertical-Nash market. First, comparing $p^{N^{{*}}}$ and $p^{C^{{*}}}$, we have $\Delta p^{N^{{*}}}=p^{N^{{*}}}-p^{C^{{*}}}=\frac{2\lambda _R \alpha }{3\lambda _R \beta -\gamma ^{2}}-\frac{\lambda _M \alpha }{2\lambda _M \beta -\gamma ^{2}}=\frac{\left( {\lambda _M \lambda _R \beta -2\lambda _R \gamma ^{2}+\lambda _M \gamma ^{2}} \right) \alpha }{\left( {2\lambda _M \beta -\gamma ^{2}} \right) \left( {3\lambda _R \beta -\gamma ^{2}} \right) }$. $\Delta p^{N^{{*}}}>0$ if $\left( {2\gamma ^{2}-\lambda _M \beta } \right) \lambda _R <\lambda _M \gamma ^{2}$, which is always satisfied when $\lambda _M \ge \frac{2\gamma ^{2}}{\beta }$. Note that inequality$\lambda _R >\frac{\gamma ^{2}}{2\beta }$ must hold to guarantee the retailer’s participation. If $\frac{\gamma ^{2}}{2\beta }<\lambda _M \le \frac{2\gamma ^{2}}{3\beta }$, then $\frac{\lambda _M \gamma ^{2}}{2\gamma ^{2}-\lambda _M \beta }\le \frac{\gamma ^{2}}{2\beta }<\lambda _R$, which leads to $p^{N^{{*}}}<p^{C^{{*}}}$. Finally, if $\frac{2\gamma ^{2}}{3\beta }<\lambda _M <\frac{2\gamma ^{2}}{\beta }$, $p^{N^{{*}}}$ can either be higher or lower than $p^{C^{{*}}}$, depending on whether $\frac{\gamma ^{2}}{2\beta }<\lambda _R <\frac{\lambda _M \gamma ^{2}}{2\gamma ^{2}-\lambda _M \beta }$ or $\lambda _R \ge \frac{\lambda _M \gamma ^{2}}{2\gamma ^{2}-\lambda _M \beta }$ holds, i.e., $p^{N^{{*}}}\left\{ {\begin{array}{l} \le \\ > \\ \end{array}} \right\} p^{C^{{*}}}$ for $\left\{ {\begin{array}{l} \lambda _R \ge \frac{\lambda _M \gamma ^{2}}{2\gamma ^{2}-\lambda _M \beta } \\ \frac{\gamma ^{2}}{2\beta }<\lambda _R <\frac{\lambda _M \gamma ^{2}}{2\gamma ^{2}-\lambda _M \beta } \\ \end{array}} \right.$.

• (c) Similar to case (a) above and, therefore, omitted.

Proof of Proposition 3:

Based on the equilibrium results of price and service in Sect. 3, we have the price-service ratios in different channel structures as follows:

$$\begin{aligned} p_S^{C*}= & {} \frac{p^{C^{{*}}}}{s^{C^{{*}}}}=\frac{\lambda _M }{\gamma },\hbox { with} \ \lambda _M>\frac{\gamma ^{2}}{2\beta };\\ p_S^{M*}= & {} \frac{p^{M^{{*}}}}{s^{M^{{*}}}}=\frac{3\lambda _R \beta -\gamma ^{2}}{\beta \gamma }, \hbox { with}\ \lambda _R>\frac{\gamma ^{2}}{2\beta }\\ p_S^{N*}= & {} \frac{p^{N^{{*}}}}{s^{N^{{*}}}}=\frac{2\lambda _R }{\gamma } ,\hbox { with }\ \lambda _R>\frac{\gamma ^{2}}{2\beta }\\ p_S^{R*}= & {} \frac{p^{R^{{*}}}}{s^{R^{{*}}}}=\frac{3\lambda _R }{\gamma },\hbox { with} \ \lambda _R >\frac{\gamma ^{2}}{4\beta } \end{aligned}$$

Comparing the price-service ratios, we have

• (a) In the manufacturer-Stackelberg supply chain, $p_S^{M*} -p_S^{C*} =\frac{3\beta \lambda _R -\beta \lambda _M -\gamma ^{2}}{\gamma \beta }\le 0$ if $\frac{\gamma ^{2}}{2\beta }<\lambda _R \le \frac{\beta \lambda _M +\gamma ^{2}}{3\beta }$, which implies $\lambda _M >\frac{\gamma ^{2}}{2\beta }$.

• (b) In the vertical-Nash supply chain,

  $p_S^{N*} -p_S^{C*} =\frac{2\lambda _R -\lambda _M }{\gamma }\le 0$ if $\frac{\gamma ^{2}}{2\beta }<\lambda _R \le \frac{1}{2}\lambda _M$, which holds if and only if $\lambda _M >\frac{\gamma ^{2}}{\beta }$; otherwise, $p_S^{N*} -p_S^{C*} >0$ for $\frac{\gamma ^{2}}{2\beta }<\lambda _M \le \frac{\gamma ^{2}}{\beta }$.

• (c) In the retailer-Stackelberg supply chain,

  $p_S^{R*} -p_S^{C*} =\frac{3\lambda _R -\lambda _M }{\gamma }\le 0$ if $\frac{\gamma ^{2}}{4\beta }<\lambda _R \le \frac{1}{3}\lambda _M$, which holds if and only if $\lambda _M >\frac{3\gamma ^{2}}{4\beta }$; otherwise, $p_S^{R*} -p_S^{C*} >0$ for $\frac{\gamma ^{2}}{2\beta }<\lambda _M \le \frac{3\gamma ^{2}}{4\beta }$.

Proof of Proposition 5:

$$\begin{aligned} \Pi _M^{M^{{*}}} -\Pi ^{C^{{*}}}=\frac{\lambda _R \alpha ^{2}}{4\left( {2\lambda _R \beta -\gamma ^{2}} \right) }-\frac{\lambda _M \alpha ^{2}}{2\left( {2\lambda _M \beta -\gamma ^{2}} \right) }=\frac{\left( {2\lambda _M \gamma ^{2}-2\lambda _M \lambda _R \beta -\lambda _R \gamma ^{2}} \right) \alpha ^{2}}{4\left( {2\lambda _M \beta -\gamma ^{2}} \right) \left( {2\lambda _R \beta -\gamma ^{2}} \right) }. \end{aligned}$$

Combining with the concavity constraints in Lemma 1 and 2, we can easily see that $\Pi _M^{M^{{*}}} -\Pi ^{C^{{*}}}>0$ if $\lambda _M >\frac{\gamma ^{2}}{2\beta }$ and $\frac{\gamma ^{2}}{2\beta }<\lambda _R <\frac{2\lambda _M \gamma ^{2}}{2\lambda _M \beta +\gamma ^{2}}$.

Proof of Proposition 6:

$$\begin{aligned} \Pi _M^{N^{{*}}} -\Pi ^{C^{{*}}}=\frac{\lambda _R^2 \beta \alpha ^{2}}{\left( {3\lambda _R \beta -\gamma ^{2}} \right) ^{2}}-\frac{\lambda _M \alpha ^{2}}{2\left( {2\lambda _M \beta -\gamma ^{2}} \right) }=\frac{\Phi \alpha ^{2}}{2\left( {2\lambda _M \beta -\gamma ^{2}} \right) \left( {3\lambda _R \beta -\gamma ^{2}} \right) ^{2}}, \end{aligned}$$

where $\Phi =-\beta \left( {5\lambda _M \beta +2\gamma ^{2}} \right) \lambda _R^2 +6\lambda _M \beta \gamma ^{2}\lambda _R -\lambda _M \gamma ^{4}$. Solving the equation $\Phi =0$ for $\lambda _R$, we obtain $\lambda _R^{*} =\frac{3\lambda _M \beta \gamma ^{2}\pm \gamma ^{2}\sqrt{2\lambda _M \beta \left( {2\lambda _M \beta -\gamma ^{2}} \right) }}{\beta \left( {5\lambda _M \beta +2\gamma ^{2}} \right) }$. Combining the concavities constraints of the respective profit functions, we conclude that $\Pi ^{C^{{*}}}<\Pi _M^{N^{{*}}}$ if and only if $\lambda _M \ge \frac{2\gamma ^{2}}{3\beta }$ and $\frac{\gamma ^{2}}{2\beta }<\lambda _R <\bar{{\lambda }}_R$, where $\bar{{\lambda }}_R$ is the larger value of the root of the quadratic function $\Phi =0$.

Proof of Proposition 7:

We, again, compare profits from the integrated channel and the indirect one in which service is provided by the retailer. Then we have $\Pi _M^{R^{{*}}} -\Pi ^{C^{{*}}}=\frac{\lambda _R^2 \beta \alpha ^{2}}{\left( {4\lambda _R \beta -\gamma ^{2}} \right) ^{2}}-\frac{\lambda _M \alpha ^{2}}{2\left( {2\lambda _M \beta -\gamma ^{2}} \right) }=\frac{a^{2}\Gamma }{2\left( {2\lambda _M \beta -\gamma ^{2}} \right) \left( {4\lambda _R \beta -\gamma ^{2}} \right) ^{2}},$ where $\Gamma$ is given by $\Gamma =-2\beta \left( {6\lambda _M \beta +\gamma ^{2}} \right) \lambda _R^2 +8\lambda _M \lambda _R \beta \gamma ^{2}-\lambda _M \gamma ^{4}$. It can be verified that $\Gamma$ is non-negative if $\lambda _M \ge \frac{\gamma ^{2}}{2\beta }$ and $\frac{\gamma ^{2}}{4\beta }<\lambda _R <\mathop {\lambda _R }\limits ^=$.