Partial up-stream advanced payment and partial down-stream delayed payment in a three-level supply chain

Abstract

In a competitive market, the retailers, in order to encourage the customers to increase their orders, give them the opportunity to pay a fraction of the purchasing cost after delivery of the ordered items (i.e., down-stream partial delayed payment). On the other hand, the suppliers, in order to reduce the risk of cancellations of orders from buyers, may ask the retailers to pay a portion of the purchasing cost before delivery of products (i.e., up-stream partial prepayment). In this paper, an EOQ model with down-stream partial delayed payment and up-stream partial prepayment under three different scenarios (without shortage, with full backordering and with partial backordering) is presented. In order to find the optimal solutions of the models developed for different scenarios, the convexity of the objective functions (i.e., total cost functions) are proved and then closed-form optimal solutions are derived. Also, a solution algorithm is proposed for the model of the third scenario. To demonstrate the applicability of the framework, some numerical examples are presented. Finally, sensitivity analyses are made on several key parameters, in order to gain some managerial insight.

Appendices

Appendix A: Convexity of equation (10) and finding its roots

The first partial derivative of $\Delta (K,T)$ respect to T is as below;

$$\begin{aligned} \frac{\partial \Delta _1 (K,T)}{\partial T}=-\frac{\psi _1}{T^{2}}+\gamma (K) \end{aligned}$$

(61)

Setting the above equation equal to zero gives:

$$\begin{aligned} T^*=T^*\left( K \right) =\sqrt{\frac{\psi _1 }{\gamma (K)}} \end{aligned}$$

(62)

Discriminant of $\gamma (K)=\psi _2 K^{2}-2\psi _3 K+\psi _4 , \Delta =b^{2}-4ac$ is shown in Eq. (63).

$$\begin{aligned} \Delta =4\psi ^{2}_3 -4\psi _2 \psi _4 =C^{2}_b D^{2}-\left( {D(C_b +C_h +i_1 C_p )} \right) \left( {C_b D} \right) =-D^{2}C_b \left( {C_h +i_1 C_p } \right) <0\nonumber \\ \end{aligned}$$

(63)

Since $\Delta$ is always negative, $\gamma (K)$ does not have any root, and so its value is always either positive or negative. On the other hand, regarding $\gamma (0)=\frac{C_b D}{2}>0$, it is obvious that $\gamma (K)$ takes always a positive value in the range of [0, 1]. Hence, for any given value ofK, a unique value is obtained for $T^{*}$, using Eq. (62). If Eq. (62) is substituted into Eq. (26), we will have:

$$\begin{aligned} \Delta (K)=\frac{\psi _1 }{\sqrt{\frac{\psi _1 }{\gamma (K)}}}+\sqrt{\frac{\psi _1 }{\gamma (K)}}\gamma (K)+\psi _5 =2\sqrt{\psi _1 \gamma (K)}+\psi _5 \end{aligned}$$

(64)

The above Equation gives a unique minimal cost for any given value of K. It should be noticed that $\Delta (K)$is continuous on the compact interval [0, 1]. Therefore, it may have more than one local minima, the smallest of which will be the global minimum. In order to find the global minimum solution, the first and the second derivatives of $\Delta (K)$ with respect to K are taken, as provided in Eqs. (65) and (66), respectively.

$$\begin{aligned} \frac{d\Delta (K)}{dK}= & {} \sqrt{\psi _1 }\frac{\gamma ^{{\prime }}(K)}{\sqrt{\gamma (K)}}\end{aligned}$$

(65)

$$\begin{aligned} \frac{d^{2}\Delta (K)}{dK^{2}}= & {} \frac{\sqrt{\psi _1 }\left[ {2\gamma ^{{\prime }{\prime }}(K)\gamma (K)-\left( {\gamma ^{{\prime }}(K)} \right) ^{2}} \right] }{2\left( {\gamma (K)} \right) ^{\frac{3}{2}}} \end{aligned}$$

(66)

For all amounts of K, we have:

$$\begin{aligned} \frac{d^{2}\Delta (K)}{dK^{2}}= & {} \frac{\sqrt{\psi _1 }\left[ {2\gamma ^{{\prime }{\prime }}(K)\gamma (K)-\left( {\gamma ^{{\prime }}(K)} \right) ^{2}} \right] }{2\left( {\gamma (K)} \right) ^{\frac{3}{2}}}\nonumber \\= & {} \frac{\sqrt{\psi _1 }\left[ {\left( {2\psi _2 } \right) \left( {\psi _2 K^{2}-2\psi _3 K+\psi _4 } \right) -\left( {\psi _2 K-\psi _3 } \right) ^{2}} \right] }{\left( {\gamma (K)} \right) ^{\frac{3}{2}}} \nonumber \\= & {} \frac{\sqrt{\psi _1 }\left[ {2\left( {\psi ^{2}_2 K^{2}-2\psi _2 \psi _3 K+\psi _2 \psi _4 } \right) -\left( {\psi ^{2}_2 K-2\psi _2 \psi _3 K+\psi ^{2}_3 } \right) } \right] }{\left( {\gamma (K)} \right) ^{\frac{3}{2}}} \nonumber \\= & {} \frac{\sqrt{\psi _1 }\left[ {\left( {\psi ^{2}_2 K^{2}-2\psi _2 \psi _3 K+\psi _2 \psi _4 } \right) +\psi _2 \psi _3 -\psi ^{2}_3 } \right] }{\left( {\gamma (K)} \right) ^{\frac{3}{2}}}\nonumber \\= & {} \frac{\sqrt{\psi _1 }\left[ {\psi _2 \left( {\psi _2 K^{2}-2\psi _3 K+\psi _4 } \right) +(\psi _2 -\psi _3 )\psi _3 } \right] }{\left( {\gamma (K)} \right) ^{\frac{3}{2}}} \nonumber \\= & {} \frac{\sqrt{\psi _1 }\left[ {\psi _2 \gamma (K)+(\psi _2 -\psi _3 )\psi _3 } \right] }{\left( {\gamma (K)} \right) ^{\frac{3}{2}}}>0 \end{aligned}$$

(67)

Considering that $\psi _1 ,\psi _2 ,\psi _3 ,\gamma (K)$ are positive and also $(\psi _2 -\psi _3 )=\frac{D(C_h +i_1 C_p )}{2}>0$, function $\Delta _1 (K)$ is convex and so has a global minimum. In order to obtain this optimal point, the first derivative of $\mathop \Delta \limits ^\wedge _1 (K)$ with respect to K is set equal to zero:

$$\begin{aligned} \frac{d\Delta _1 (K)}{dK}=\sqrt{\psi _1 }\frac{\gamma ^{{\prime }}(K)}{\sqrt{\gamma (K)}}=0 \end{aligned}$$

(68)

Since $\sqrt{\psi _1 }$ and $\gamma (K)$ are positive, it can be concluded from Eq. (34) that $\gamma ^{{\prime }}(K)$ is equal to zero, i.e.,

$$\begin{aligned} \gamma ^{{\prime }}(K)=2\psi _2 K-2\psi _3 =0 \end{aligned}$$

(69)

Equation (69) gives:

$$\begin{aligned} K^{*}=\frac{\psi _3 }{\psi _2 }=\frac{C_b }{(C_b +C_h +i_1 C_p )} \end{aligned}$$

(70)

After Substituting $K^{*}$ into Eq. (62) and simplifying it, the optimal period length is obtained as follows:

$$\begin{aligned} T^{*}=T^{*}\left( {K^{*}} \right) =\sqrt{\frac{2A(C_b +C_h +i_1 C_p )}{DC_b \left( {(C_h +i_1 C_p )} \right) }} \end{aligned}$$

(71)

Appendix B: Finding the roots of $\varphi (K,T)$ with respect to K and T

From Eq. (51), we have:

$$\begin{aligned} \varphi (K,T)=\psi _1 K^{2}T-\psi _2 KT-\psi _3 K+\psi _4 T+\frac{\psi _5 }{T}+\psi _6 \end{aligned}$$

Taking thefirstderivativesof $\varphi (K,T)$ with respectto K and T gives:

$$\begin{aligned} \frac{\varphi (K,T)}{dT}= & {} \psi _1 K^{2}-\psi _2 K+\psi _4 -\frac{\psi _5 }{T^{2}} \rightarrow T=\frac{\psi _5 }{\psi _1 K^{2}-\psi _2 K+\psi _4}\end{aligned}$$

(72)

$$\begin{aligned} \frac{\varphi (K,T)}{dK}= & {} 2\psi _1 KT-\psi _2 T-\psi _3 \rightarrow K=\frac{\psi _2 T+\psi _3 }{2\psi _1 T} \end{aligned}$$

(73)

After some algebra we have:

$$\begin{aligned} T= & {} \sqrt{\frac{4\psi _1 \psi _5 -\psi _3^{2}}{4\psi _1 \psi _4 -\psi _2^{2}}}\end{aligned}$$

(74)

$$\begin{aligned} K= & {} \frac{\psi _2 }{2\psi _1 }+\frac{\psi _3 }{2\psi _1 }\sqrt{\frac{4\psi _1 \psi _4 -\psi _2^{2}}{4\psi _1 \psi _5 -\psi _3^{2}}} \end{aligned}$$

(75)