Integration of pricing and inventory decisions of deteriorating item in a decentralized supply chain: a Stackelberg-game approach

Abstract

In the present worldwide highly competitive markets, competition occurs among the supply chain members on behalf of organizations. In this way, partners of the supply chain try to apply effective coordination to increase market shares. Because of the significance and utilization of inventory decisions and pricing strategies on accepting a product in the current business scenario, in this study, decentralized channel coordination is generalized to increase the profitability of a two-echelon supply chain. Here, a one-manufacturer–one-retailer supply chain mechanism for the deteriorating product with a leader–follower relationship under price-dependent demand is developed. Up-stream member-manufacturer sells the on-hand item to the downstream member-retailer in which the retailer faces off the customer. This model considers the effect of the deterioration of the product, selling price of both the channel members, cycle duration, idle time, ordering lot size in a decentralized supply chain system. The Stackelberg game method has been used considering the retailer as a leader and manufacturer as a follower to optimize the sales price of channel members and time–length up to zero manufacturer inventory for maximum profit. Finally, a numerical example and sensitivity analysis are given to demonstrate the model. The result shows that manufacturer profit is far better than the retailer profit though the retailer’s selling price is higher than that of the manufacturer’s selling price. Also, a little change in the manufacturing cost is highly sensitive for the profit of the channel members, which encourages the manufacturer to reduce the manufacturing cost and increases supply chain profit as well as channel members profit.

Appendix 1

Proof of Lemma 3

Second order principal minor follows

$$\begin{aligned} \bigtriangleup _2=&\frac{e^{\theta t_1}}{\theta t_1^3}(-1+e^{\theta t_1}) \Big (\frac{k h_m}{\theta } + \frac{h_m^2}{2 \theta ^2 }+ \frac{k^2}{2} \Big ) \nonumber \\&+ \Big (\frac{-1}{2}+e^{\theta t_1} -\frac{e^{2\theta t_1}}{4} \Big ) \frac{1}{2 \theta ^2 t_1^4} \Big (\frac{k h_m}{\theta } + \frac{h_m^2}{2 \theta ^2 } + \frac{k^2}{2} \Big ) \nonumber \\&+ \frac{e^{2 \theta t_1 }}{2 t_1^2} \Big (-\frac{c^2}{2}-\frac{k^2}{2}-kc-\frac{k h_m}{\theta } +\frac{h_m^2}{2 \theta ^2} \Big ) \nonumber \\&\times \Big ( -e^{\theta t_1}+ e^{2 \theta t_1}-\frac{1}{2 \theta }+\frac{e^{\theta t_1}}{\theta }-\frac{e^{2\theta t_1}}{2\theta } \Big ) \nonumber \\&\times \Big (k+\frac{c}{2}+\frac{h_m}{2} \Big ) \frac{c}{\theta t_1^3} + (a_1-b_1 p_m) \nonumber \\&\frac{2 e^{\theta t_1}b_1 t_1}{T^2} (k\theta +a_1 +c)-\frac{e^{2\theta t_1} h_m c}{2 \theta t_1^2} > 0 \end{aligned}$$

(A.1)

If,

$$\begin{aligned}&\Big ( \frac{k h_m}{\theta } + \frac{h_m^2}{2 \theta ^2 }+ \frac{k^2}{2} \Big ) \Big ( \frac{e^{\theta t_1}}{\theta t_1^3}+\frac{e^{\theta t_1}}{2 \theta ^2 t_1^4} \Big ) +\frac{h_m^2 e^{2 \theta t_1}}{4 \theta ^2 t_1^2} \nonumber \\&\qquad + \Big (k+\frac{c}{2}+\frac{h_m}{\theta } \Big )\frac{c}{\theta t_1^3} \Big (e^{2\theta t_1}+ \frac{e^{\theta t_1}}{\theta } \Big ) \nonumber \\&\qquad \times (a_1-b_1 p_m) \frac{2 e^{\theta t_1 }b_1 t_1}{k \theta +a_1+c} \nonumber \\&\quad > \Big ( \frac{k h_m}{\theta } + \frac{h_m^2}{2 \theta ^2 }+ \frac{k^2}{2} \Big ) \Big (\frac{1}{4 \theta ^2 t_1^4}+ \frac{e^{2 \theta t_1}}{8\theta ^2 t_1^4 }+ \frac{e^{\theta t_1}}{\theta t_1^3} \Big ) \nonumber \\&\qquad +\frac{e^{2 \theta t_1}h_m c}{2 \theta t_1^2} + \Big (\frac{c^2}{2}+\frac{k^2}{2}+kc+{\frac{k h_m}{\theta }} \Big ) \frac{e^{2 \theta t_1}}{2 t_1^2} \nonumber \\&\qquad + \Big ({e^{\theta t_1}}+\frac{1}{2\theta }+{2 \theta } \Big ) \frac{c}{\theta t_1^2} \Big (k+\frac{c}{2}+\frac{h_m}{\theta } \Big ) \end{aligned}$$

(A.2)

which implying the concavity of $\varPi _m$ with respect to $p_m$ and $t_1$. □

Proof of Lemma 4

Taking the first order derivative of retailer profit function, we get

$$\begin{aligned} \frac{\partial \varPi _r}{\partial p_r}=&\frac{1}{B_1} \Big [ \frac{(1-e^{-\theta t_1})b_2 (a_1-b_1p_m)p_r}{\theta (a_2-b_2 p_r)(1+(1-e^{-\theta t_1})(-1+\frac{a_1-b_1p_m}{a_2-b_2 p_r}))} \nonumber \\&+(k+\frac{h_r}{\theta })(a_2-b_2 p_r) \Big \{-\frac{(1-e^{-\theta t_1})(a_1-b_1p_m)b_2}{\theta (a_2-b_2 p_r)^2} \nonumber \\&+\frac{(1-e^{-\theta t_1})(a_1-b_1p_m)b_2}{\theta (a_2-b_2 p_r)^2 \Big ( 1+(1-e^{-\theta t_1})(-1+\frac{a_1-b_1p_m}{a_2-b_2 p_r}) \Big ) } \Big \} \nonumber \\&- b_2 \Big (k+\frac{h_r}{\theta } \Big ) \Big [ \Big (\frac{1}{\theta }+\frac{B_2}{\theta } -\frac{e^{B_2}}{\theta } \Big ) - \Big (\frac{-1}{\theta }+\frac{e^{-\theta t_1}}{\theta }+ t_1 \Big )\Big ] \nonumber \\&-b_2 p_r \Big (\frac{B_2}{\theta }+t_1 \Big ) +(a_2-b_2 p_r) \Big (\frac{B_2}{\theta }+t_1 \Big ) \Big ] \nonumber \\&-\frac{A_1}{B_3}-\frac{A_2}{B_3}+\frac{A_3}{B_3} \end{aligned}$$

(A.3)

where,

$$\begin{aligned} A_1=&(1-e^{-\theta t_1})b_2(a_1-b_1 p_m)(a_2-b_2p_r) \Big (-A_r+\frac{k+h_r}{\theta } \Big ) \\&\times \Big ( \frac{1+log[1+(1-e^{-\theta t_1})(-1+\frac{a_1-b_1p_m}{a_2-b_2 p_r})]}{\theta } \Big ) \\ A_2=&p_m(a_1-b_1p_m)t_1+ (k+\frac{h_r}{\theta })(a_1-a_2-b_1p_m+b_2 p_r) \\&\times \Big (-\frac{1}{\theta }+\frac{e^{-\theta t_1}}{\theta }+t_1 \Big ) \\ A_3=&p_r(a_2-b_2p_r)log[1+(1-e^{-\theta t_1})] \Big (-1+\frac{a_1-b_1p_m}{a_2-b_2p_r} \Big ) \\ B_1=&t_1+\frac{log[1+(1-e^{-\theta t_1})(-1+\frac{a_1-b_1p_m}{a_2-b_2 p_r})]}{\theta } \\ B_2=&log \Big [1+(1-e^{-\theta t_1}) \Big (-1+\frac{a_1-b_1p_m}{a_2-b_2 p_r} \Big ) \Big ] \\ B_3=&\theta (a_2-b_2p_r)^2 \Big [1+(1-e^{-\theta t_1})(-1+\frac{a_1-b_1p_m}{a_2-b_2 p_r} ) \Big ] \\&\times (\frac{log[1+(-1+\frac{a_1-b_1p_m}{a_2-b_2 p_r})]}{\theta }+t_1)^2 \end{aligned}$$

Due to complexity of (A.3), we can not analytically obtained the value of $p_r$ but the numerical simulation is done in Sect. 6 has also shown the numeric value of $\frac{\partial ^2 \varPi _r}{\partial p_r^2}< 0$ . □