Logistics cost sharing in supply chains involving a third-party logistics provider

Abstract

In many cases, end customers are sensitive to a product’s logistics service level which is provided by a third-party logistics (3PL) provider, therefore, the continuous improvement of the logistics service is imperative and valuable. However, the problem is that improving the logistics service benefits all of them, but is costly to only the 3PL provider. The 3PL provider is not willing to do this. Sharing the logistics cost is one solution to this problem. This study investigates cost sharing in two kinds of supply chains, i.e., one manufacturer-one 3PL provider-one retailer supply chain and two manufacturers-one 3PL provider-one retailer supply chain. Two types of cost sharing mechanisms, i.e., decentralized cost sharing mechanisms and centralized cost sharing mechanisms, are explored. Decentralized cost sharing mechanisms are proposed as contracts that chain members separately decide their cost sharing portions to optimize their own profits, ignoring the collective impacts of their decisions on the channel as a whole. Centralized cost sharing mechanisms are in the situation that chain members negotiate their cost sharing portions so that their profits are the shares of the entire supply chain’s profit, implying that the supply chain is coordinated perfectly. This study aims to analyse how cost sharing mechanisms affect supply chain performance and under what conditions chain members are willing to engage in cost sharing mechanisms. Conditions necessary for cost sharing mechanisms to achieve win-win outcomes are identified.

Appendix

Proof of Proposition 1

Because 2$m_{R}\ge m_{L},4m_{M}\ge 2m_{R} + m_{L}$, we have

• (i) $\begin{array}{ll} _{DCS}\Pi _{M}^{*} - \Pi _{M}^{*} &{}= \frac{\alpha ^{2}[(4m_M +2m_R +m_L )^{2}-32m_M m_L ]}{64k}\ge \frac{\alpha ^{2}[(4m_M +2m_L )^{2}-32m_M m_L ]}{64k}\\ &{}=\frac{\alpha ^{2}(2m_M -m_L )^{2}}{16k}\ge 0. \end{array}$

• (ii) $_{DCS}\Pi _L^*-\Pi _L^*=\frac{\alpha ^{2}m_L (4m_M +2m_R -3m_L )}{16k}\ge 0$.

• (iii) $\begin{array}{ll} _{DCS}\Pi _R^*-\Pi _R^*&{}=\frac{\alpha ^{2}[4m_M (2m_R +m_L )+(2m_R +m_L )^{2}-16m_R m_L ]}{32k}\\ &{}\ge \frac{\alpha ^{2}[2(2m_R +m_L )^{2}-16m_R m_L ]}{32k}\\ &{}=\frac{\alpha ^{2}(2m_R -m_L )^{2}}{16k}\ge 0. \end{array}$

Proof of Proposition 2

The proof is already in the text and therefore the detail is omitted.

Proof of Proposition 3

• (i) $\begin{array}{l} _{CCS}\Pi _R^*-_{DCS}\Pi _R^*\\ \quad =\frac{\alpha ^{2}[8m_R (m_M +m_R +m_L )-(2m_R +m_L )(4m_M +2m_R +m_L )]}{32k}\\ \quad =\frac{\alpha ^{2}(4m_R^2 +4m_R m_L -4m_M m_L -m_L^2 )}{32k}. \end{array}$ If 4$m_{R}^{2} + 4m_{R}m_{L} - 4m_{M}m_{L} - m_{L}^{2}\ge$ 0, then $_{CCS}\Pi _{R}^{*}\ge _{DCS}\Pi _{R}^{*}$

• (ii) $\begin{array}{ll} _{CCS}\Pi _M^*-_{DCS}\Pi _M^*&{}=\frac{\alpha ^{2}[16m_M (m_M +m_R +m_L )-(4m_M +2m_R +m_L )^{2}]}{64k}\\ &{}=\frac{\alpha ^{2}[8m_M m_L -(2m_R +m_L )^{2}]}{64k}. \end{array}$ If $8m_{M}m_{L} - (2m_{R} + m_{L})^{2}\ge$ 0, then $_{CCS}\Pi _{M}^{*}\ge _{DCS}\Pi _{M}^{*}$.

• (iii) $_{CCS}\Pi _L^*-_{DCS}\Pi _L^*=\frac{\alpha ^{2}m_L (2m_R +3m_L )}{16k}>0$.

Proof of Proposition 4

Because $4\alpha m_M \ge (\alpha -\beta )(2m_R +m_L ),2m_R \ge m_L$, we have

• (i) ${{\begin{array}{ll} {}_{PCS,MM}\Pi _{M1}^*\!-\!{}_{MM}\Pi _{M1}^*&{}=\frac{16\alpha ^{2}m_M^2 \!+\!16\alpha (\alpha \!-\!\beta )m_M m_R -24\alpha (\alpha \!-\!\beta )m_M m_L +(\alpha -\beta )^{2}(2m_R +m_L )^{2}}{64k}.\\ &{}\ge \frac{16\alpha ^{2}m_M^2 -16\alpha (\alpha -\beta )m_M m_L +(\alpha -\beta )^{2}(2m_R +m_L )^{2}}{64k}\\ &{}\ge \frac{16\alpha ^{2}m_M^2 -16\alpha (\alpha -\beta )m_M m_L +4(\alpha -\beta )^{2}m_L ^{2}}{64k}\\ &{}=\frac{[4\alpha m_M -2(\alpha -\beta )m_L ]^{2}}{64k}\ge 0. \end{array}}}$

• (ii) $\begin{array}{ll} {}_{PCS,MM}\Pi _L^*-{}_{MM}\Pi _L^*&{}=\frac{(\alpha -\beta )m_L [4\alpha m_M +(\alpha -\beta )(2m_R +9m_L )-12(\alpha -\beta )m_{L}}{16k}\\ &{}\ge \frac{(\alpha -\beta )m_{L} [(\alpha -\beta )(2m_R +m_L )+(\alpha -\beta )(2m_R +9m_L )-12(\alpha -\beta )m_{L}}{16k}\\ &{}\ge \frac{(\alpha -\beta )m_L [2(\alpha -\beta )m_L +10(\alpha -\beta )m_L -12(\alpha -\beta )m_L ]}{16k}=0. \end{array}$

• (iii) $\begin{array}{ll} _{PCS,MM}\Pi _R^*-_{MM}\Pi _R^*&{}=\frac{(\alpha -\beta )m_R [4\alpha m_M +(\alpha -\beta )(2m_R -7m_L )]}{16k}\\ &{}\quad +\frac{(\alpha -\beta )m_L [4\alpha m_M +(\alpha -\beta )(2m_R +m_L )]}{32k}\\ &{}\ge \frac{(\alpha -\beta )m_L [8\alpha m_M +(\alpha -\beta )(4m_R -6m_L )]}{32k}\\ &{}\ge \frac{(\alpha -\beta )m_L [2(\alpha -\beta )(2m_R +m_L )+(\alpha -\beta )(4m_R -6m_L )]}{32k}\\ &{}=\frac{(\alpha -\beta )^{2}m_L (2m_R -m_L )}{8k}\ge 0. \end{array}$

• (iv) $\begin{array}{ll} _{PCS,MM}\Pi _{M2}^*-_{MM}\Pi _{M2}^*&{}=\frac{\beta m_M [3(\alpha -\beta )m_L -4\alpha m_M -2(\alpha -\beta )m_R ]}{8k}\\ &{}\le \frac{\beta m_M [3(\alpha -\beta )m_L -(\alpha -\beta )(2m_R +m_L )-2(\alpha -\beta )m_R ]}{8k}\\ &{}=\frac{\beta (\alpha -\beta )m_M (m_L -2m_R )}{4k}\le 0. \end{array}$

Proof of Proposition 5

Because 4$m_{M}\ge 2m_{R} + m_{L}, 2m_{R}\ge m_{L}$, we have

• (i) ${{\begin{array}{ll} _{BCS,MM}\Pi _{M1}^*-_{MM}\Pi _{M1}^*&{}=\frac{(\alpha -\beta )^{2}[(4m_M +2m_R +m_L )^{2}-32m_M m_L ]}{64k}\\ &{}\ge \frac{(\alpha -\beta )^{2}[(4m_M +2m_L )^{2}-32m_M m_L ]}{64k}\!=\!\frac{(\alpha -\beta )^{2}(2m_M -m_L )^{2}}{16k}\!\ge \! 0.\\ \end{array}}}$

• (ii) $_{BCS,MM}\Pi _L^*-_{MM}\Pi _L^*=\frac{(\alpha -\beta )^{2}m_L (4m_M +2m_R -3m_L )}{8k}\ge 0$.

• (iii) $\begin{array}{ll} _{BCS,MM}\Pi _R^*-_{MM}\Pi _R^*&{}=\frac{(\alpha -\beta )^{2}[(4m_M +2m_R +m_L )(2m_R +m_L )-16m_R m_L ]}{16k}\\ &{}\ge \frac{(\alpha -\beta )^{2}[(2m_R +m_L )^{2}-8m_R m_L ]}{8k}=\frac{(\alpha -\beta )^{2}(2m_R -m_L )^{2}}{8k}\ge 0. \end{array}$

Proof of Proposition 6

The proof is already in the text and therefore the detail is omitted.

Proof of Proposition 7

The proof is already in the text and therefore the detail is omitted.

Proof of Proposition 8

The proof is similar to that in Proposition 3 and therefore the detail is omitted.