An analysis of optimal pricing strategy and inventory scheduling policy for a non-instantaneous deteriorating item in a two-layer supply chain

Abstract

This paper presents a production inventory system with a manufacturer-retailer supply chain dealing with the non-instantaneous deteriorating products. The two-level supply chain model is analyzed with shortage and without shortage, considering the impact of business strategies in different sectors on the collaborating market system. Firstly, the integrated system and then the decentralized system under a retail fixed-mark-up strategy are studied. Further, we show that the retailer offers a fixed-mark-up policy as a signal to the manufacturer to resolve the gaming between channel members of the supply chain. This study’s prime objective is to determine the optimal retail price, wholesale price, and inventory schedules to maximize the overall supply chain’s profit. An analytical method is used to optimize the selling price and various time-length for maximum profit. The model is demonstrated through two numerical examples, and sensitivity analysis is conducted to study the behavior of parameters. It is observed from the numerical study that the supply chain system without shortage is beneficial compared to the shortage permitted supply chain. Manufacturer profit is improved after using RFM contract in contrast with integrate system.

Appendices

Appendix A

Proof of Lemma 1

Taking the first order and second order partial derivatives of TP_SC in (32) with respect to p gives

$$ \begin{array}{@{}rcl@{}} \frac{\partial TP_{SC} }{\partial p}&=& -bpT+(a-bp)T-\frac{b h_{r} {T_{3}^{2}}}{2} \\ && + \left[b h_{r} T_{3} -\frac{b (u-M D_{r})}{D_{r}} \right] \left[ T_{3} +\frac{e^{\theta (T-T_{3})}-1}{\theta} \right] \\ && -b \left( d+\frac{h_{r}}{\theta} \right) \left[ (T-T_{3})+ \frac{e^{\theta (T-T_{3})}}{\theta} \right] \\ && - \left( h_{r}-h_{m}+\frac{h_{m} D_{r}}{P} \right) \frac{b(a-bp)}{D_{r}} \\ && \times \left( T_{3} +\frac{e^{\theta (T-T_{3})}-1}{\theta} \right) \end{array} $$

(52)

and,

$$ \begin{array}{@{}rcl@{}} \frac{\partial^{2} TP_{SC} }{\partial p^{2}}&=& -2bT+\frac{b^{2}}{D_{r}} \left( h_{r}-h_{m}+\frac{h_{m} D_{r}}{P} \right) \\ && \times \left[ T_{3} +\frac{e^{\theta (T-T_{3})}-1}{\theta} \right]^{2} \end{array} $$

(53)

Now, $\frac {\partial ^{2} TP_{SC} }{\partial p^{2}}< 0$ if

$$ \begin{array}{@{}rcl@{}} 2bT> \frac{b^{2}}{D_{r}} \left( h_{r}-h_{m}+\frac{h_{m} D_{r}}{P} \right) \left[ T_{3} +\frac{e^{\theta (T-T_{3})}-1}{\theta} \right]^{2} \end{array} $$

(54)

holds. □

Proof of Lemma 2

Taking the first and second order partial derivatives of TP_SC in (32) with respect to T gives

$$ \begin{array}{@{}rcl@{}} \frac{\partial TP_{SC} }{\partial T}&=& p(a-bp)-u+ \left( \frac{u}{D_{r}}-M-h_{r} T_{3} \right) (a-bp) e^{\theta (T-T_{3})} \\ & &+ \frac{(a-bp)^{2} }{D_{r}} e^{\theta (T-T_{3})} \left( h_{r}-h_{m}+\frac{h_{m} D_{r}}{P} \right) \\ & &\times \left( T_{3} +\frac{e^{\theta (T-T_{3})}-1}{\theta} \right) \\ & & + (a-bp) \left( 1- e^{\theta (T-T_{3})}\right) \left( d+\frac{h_{r}}{\theta} \right) \end{array} $$

(55)

and,

$$ \begin{array}{@{}rcl@{}} \frac{\partial^{2} TP_{SC}}{\partial T^{2}} & =& \frac{(a-bp)^{2}}{D_{r}} e^{\theta (T-T_{3})} \left( h_{r}-h_{m}+\frac{D_{r}}{P} \right) \\ && \times \left[ \theta T_{3}+2 e^{\theta (T-T_{3})} -1 \right] \\ & & -\left( M+h_{r} T_{3}-\frac{u}{D_{r}} \right) (a-bp) \theta e^{\theta (T-T_{3})} \\ & & - \theta (a-bp) (d+\frac{h_{r}}{\theta}) e^{\theta (T-T_{3})} \end{array} $$

(56)

Now, $\frac {\partial ^{2} TP_{SC} }{\partial T^{2}}< 0$ if,

$$ \begin{array}{@{}rcl@{}} && \frac{(a-bp)^{2}}{D_{r}} e^{\theta (T-T_{3})} \left( h_{r}-h_{m}+\frac{D_{r}}{P} \right) \\ && \times \left[ \theta T_{3}+2 e^{\theta (T-T_{3})} -1 \right] \\ & &< \left( M+h_{r} T_{3}-\frac{u}{D_{r}} \right) (a-bp) \theta e^{\theta (T-T_{3})} \\ & & + \theta (a-bp) \left( d+\frac{h_{r}}{\theta}\right) e^{\theta (T-T_{3})} \end{array} $$

(57)

holds. □

Proof of Theorem 1

To verify the optimality of the solution obtained from (35) and (36), we have calculated the Hessian matrix (H_Model1−A) and show that the hessian matrix is negative definite. i.e., det(H_Model1−A) > 0.

Here,

$$ \mathbf{H_{Model 1-A}}=\left[\begin{array}{cc} \frac{\partial^{2} TP_{SC}}{\partial p^{2}}&\frac{\partial^{2} TP_{SC}}{\partial p\partial T}\\ \frac{\partial^{2} TP_{SC}}{\partial T\partial p }&\frac{\partial^{2} TP_{SC}}{\partial T^{2}} \end{array}\right] $$

Here, the expression

$$ \begin{array}{@{}rcl@{}} \frac{\partial^{2} TP_{SC}}{\partial p\partial T}&=&\frac{\partial^{2} TP_{SC}}{\partial T\partial p } \\ &=& a-2bp-b \left( d+\frac{h_{r}}{\theta} \right) \left( 1-e^{\theta (T-T_{3})} \right) \\ & &+ b e^{\theta (T-T_{3})} \left( h_{r} T_{3} +M-\frac{u}{D_{r}}\right) \\ & &+2\frac{(a-bp)}{D_{r}} b e^{\theta (T-T_{3})} \left( T_{3}+\frac{e^{\theta (T-T_{3})}-1}{\theta} \right) \\ & &\times \left[ h_{m}-h_{r}T_{3}-\frac{h_{m}D_{r}}{P} \right] \end{array} $$

(58)

The determinant of the hessian matrix are evaluated by

$$ \begin{array}{@{}rcl@{}} & det(H_{Model 1-A})= \frac{\partial^{2} TP_{SC} }{\partial p^{2}} \frac{\partial^{2} TP_{SC}}{\partial {T_{1}^{2}}} -\left\{ \frac{\partial^{2} TP_{SC}}{\partial p\partial T_{1}} \right\}^{2} \end{array} $$

(59)

Due to complexities of the expressions in Hessian matrix H_Model1−A, the concavity condition of TP_SC with respect to (p,T) is hardly verified by mathematical derivation, but in numerically conducted in Section (7) has shown its concavity (see Fig. (5a)). □

Appendix B

Proof of Lemma 3

Taking the first and second order partial derivatives of TP_SC in (33) with respect to p gives

$$ \begin{array}{@{}rcl@{}} \frac{\partial TP_{SC}}{\partial p} &= &(a-bp)T-bpT-\frac{h_{r} b {T_{3}^{2}}}{2} -\frac{(h_{r}+s)b q^{2}}{2(a-bp-D_{r})^{2}} \\ & -&b\left( \frac{u}{D_{r}} -M-h_{r} T_{3} \right) \left[ T_{3} + \frac{\left( e^{\theta (T-T_{3})}-1 \right)}{\theta} \right] \\ & -&\frac{b}{D_{r}} \left( h_{r}-h_{m}+\frac{h_{m} D_{r}}{P} \right) \left[ T_{3} + \frac{\left( e^{\theta (T-T_{3})}-1 \right)}{\theta} \right] \\ & \times& \left[ q+(a-bp)T_{3} + \frac{(a-bp) \left( e^{\theta (T-T_{3})}-1 \right)}{\theta} \right] \\ & -& b \left( d+\frac{h_{r}}{\theta} \right) \left[ (T-T_{3})-(e^{\theta (T-T_{3})}-1) \right] \end{array} $$

(60)

and,

$$ \begin{array}{@{}rcl@{}} \frac{\partial^{2} TP_{SC}}{\partial p^{2}} & = &-2bT - \frac{b^{2} q^{2} (h_{r}+s)}{(D_{r}-D_{c})^{3}} \\ & &- \frac{b}{D_{r}} \left( h_{r}-h_{m}+\frac{h_{m} D_{r}}{P} \right) \\ & &\times \left[ T_{3} +\frac{e^{\theta (T-T_{3})}-1}{\theta} \right]^{2} \end{array} $$

(61)

Clearly, $\frac {\partial ^{2} TP_{SC} }{\partial p^{2}} < 0$. Since, we have considered a positive inventory of the retailer for D_r > D_c. □

Proof of Lemma 4

Taking the first and second order derivative of TP_SC in (33) with respect to T, we have

$$ \begin{array}{@{}rcl@{}} \frac{\partial TP_{SC} }{\partial T} &=& p(a-bp)-u + (a-bp) \left( e^{\theta (T-T_{3})}-1\right) \\ & &\times \left( \frac{u}{D_{r}} -M-h_{r} T_{3} \right)\\ & &- (a-bp) \left( 1-e^{\theta (T-T_{3})}\right) \left( d+\frac{h_{r}}{\theta} \right) \\ & &+ \frac{(a-bp)}{D_{r}} \left( h_{r}-h_{m}+\frac{h_{m} D_{r}}{P} \right) e^{\theta (T-T_{3})} \\ & &\times \left[ q + (a - bp)T_{3} + \frac{(a - bp) \left( e^{\theta (T-T_{3})} - 1 \right)}{\theta} \right] \\ \end{array} $$

(62)

and,

$$ \begin{array}{@{}rcl@{}} \frac{\partial^{2} TP_{SC}}{\partial T^{2}} & =& -\theta (a-bp) e^{\theta (T-T_{3})} \\ & &\times \left[ (M+h_{r} T_{3}-\frac{u}{D_{r}} )+(d+\frac{h_{r}}{\theta}) \right] \\ & &+ (a-bp)^{2} e^{2 \theta (T-T_{3})} \left( h_{r}-h_{m}+\frac{h_{m} D_{r}}{P} \right) \\ & &+ (a-bp) e^{\theta (T-T_{3})} \left( 1h_{r}-h_{m}+\frac{h_{m} D_{r}}{P} \right) \\ & &\times \left[ q+(a-bp)T_{3} + \frac{(a-bp) \left( e^{\theta (T-T_{3})}-1 \right)}{\theta} \right] \\ \end{array} $$

(63)

Now, $\frac {\partial ^{2} TP_{SC}}{\partial T^{2}}<0$ if

$$ \begin{array}{@{}rcl@{}}&&\theta (a-bp) e^{\theta (T-T_{3})} \times \left[ (M+h_{r} T_{3}-\frac{u}{D_{r}} )+(d+\frac{h_{r}}{\theta}) \right] \\ & &> (a-bp)^{2} e^{2 \theta (T-T_{3})} \left( h_{r}-h_{m}+\frac{h_{m} D_{r}}{P} \right) \\ & &+ (a-bp) e^{\theta (T-T_{3})} \left( h_{r}-h_{m}+\frac{h_{m} D_{r}}{P} \right) \\ && \times \left[ q+(a-bp)T_{3} + \frac{(a-bp) \left( e^{\theta (T-T_{3})}-1 \right)}{\theta} \right] \end{array} $$

(64)

holds. □

Proof of Theoram 2

To verify the optimality of the solution obtained from (37), (38), we have calculated the hessian matrix of the profit functions is negative definite i.e. if $\frac {\partial ^{2} TP_{SC} }{\partial p^{2}}\frac {\partial ^{2} TP_{SC} }{\partial T^{2}}- \{ \frac {\partial ^{2} TP_{SC}}{\partial p\partial T} \}^{2}>0$

Here,

$$ \begin{array}{@{}rcl@{}} \frac{\partial^{2} TP_{SC}}{\partial p\partial T}&=&\frac{\partial^{2} TP_{SC}}{\partial T\partial p} \\ &= &a-2bp -b e^{\theta (T-T_{3})} \left( \frac{u}{D_{r}} -M-h_{r} T_{3} \right) \\ & &-\frac{b}{D_{r}} (a-bp) e^{\theta (T-T_{3})} \left( h_{r}-h_{m}+\frac{h_{m} D_{r}}{P} \right) \\ & &\times \left[ T_{3} + \frac{\left( e^{\theta (T-T_{3})}-1 \right)}{\theta} \right] \\ & &-\frac{b}{D_{r}} e^{\theta (T-T_{3})} \left( h_{r}-h_{m}+\frac{h_{m} D_{r}}{P} \right) \\ & &\times \left[ q+(a-bp)T_{3} + \frac{(a-bp) \left( e^{\theta (T-T_{3})}-1 \right)}{\theta} \right] \\ \end{array} $$

(65)

High complexity of the expressions in Hessian matrix H_Model1−B, the concavity condition of TP_SC with respect to (p,T) is difficult to verify by mathematical derivation, but numerically in Section (7) has shown its concavity (see Fig. (5b)). □