Optimal ordering policy for an inventory system with linearly increasing demand and allowable shortages under two levels trade credit financing

Abstract

To reduce default risk, a retailer may offer a partial down-stream trade credit to its credit-risk customers who should pay a portion of their purchasing costs at the time of receiving items as a collateral deposit, and then receive a permissible trade credit on the rest of the outstanding amount. To reflect this fact, we consider an inventory model with linear time dependent demand under two levels of trade credit and allowable shortages. Depending on the relationship between up-stream and down-stream trade credit periods, several cases are considered and the necessary and sufficient conditions are derived for finding the optimal solution. We also present a simple algorithm to determine the optimal solution. Numerical examples are provided to illustrate the solution procedure. Sensitivity analysis of important model-parameters is performed and some relevant managerial implications are discussed.

Appendix

Proof of Proposition 1

For maximization of $TP_1(T_1,T)$, the necessary conditions are $\frac{\partial TP_1(T_1,T)}{\partial T_1}=0$ and $\frac{\partial TP_1(T_1,T)}{\partial T}=0$ and the sufficient conditions are $\frac{\partial ^2 TP_1(T_1,T)}{\partial T_1^2}<0$, $\frac{\partial ^2 TP_1(T_1,T)}{\partial T^2}<0$ and $\left| \begin{array}{cc} \frac{\partial ^2 TP_1(T_1,T)}{\partial T_1^2} &{} \frac{\partial ^2 TP_1(T_1,T)}{\partial T_1\partial T}\\ \frac{\partial ^2 TP_1(T_1,T)}{\partial T\partial T_1} &{} \frac{\partial ^2 TP_1(T_1,T)}{\partial T^2} \end{array}\right| >0$.

Now,

$$\begin{aligned} \frac{\partial TP_1(T_1,T)}{\partial T_1}= & {} \frac{1}{2T}\bigg\{ 2a\big[ sT-(h+cI_p+s)T_1-I_ep(M-N(1-\alpha ))\big ]+b\big[ 2T_1(sT-(h+s)T_1 \\&-\,\left. I_ep\big (M-N(1-\alpha ))\big )+cI_p\left( (M-N)^2-2T_1^2+(2M-N)N\alpha \right) \right] \bigg\} \\ \frac{\partial TP_1(T_1,T)}{\partial T}= & {} \frac{1}{6T^2}\big [6A+3aX_{11}+b(X_{12}+cX_{13}\big )\big ] \end{aligned}$$

Therefore, $\frac{\partial TP_1(T_1,T)}{\partial T_1}=0$ and $\frac{\partial TP_1(T_1,T)}{\partial T}=0$ give $2a\{sT-(h+cI_p+s)T_1-I_ep[M-N(1-\alpha) ] \}+b \{2T_1 [sT-(h+s)T_1-I_ep (M-N(1-\alpha ))]+cI_p [(M-N)^2-2T_1^2+(2M-N)N\alpha ] \}=0$ and $6A+3aX_{11}+b(X_{12}+cX_{13})=0$.

The optimal values of $T$ and $T_1$ which maximize $TP_1(T_1,T)$ are obtained by solving the above pair of equations, provided that they satisfy the sufficient conditions.

The sufficient conditions are given by

$$\begin{aligned} \frac{\partial ^2 TP_1(T_1,T)}{\partial T_1^2}=-\frac{1}{T}\big [aX_{31}+bX_{32}\big ]<0 \end{aligned}$$

which gives $aX_{31}+bX_{32}>0$,

$$\begin{aligned} \frac{\partial ^2 TP_1(T_1,T)}{\partial T^2}=\frac{1}{3T^3}\big [-6A+bX_{41}+3a\big (X_{42}+cX_{43}\big )\big ]<0 \end{aligned}$$

which gives $-6A+bX_{41}+3a\big (X_{42}+cX_{43}\big )<0$.

and $\left| \begin{array}{cc} \frac{\partial ^2 TP_1(T_1,T)}{\partial T_1^2} &{} \frac{\partial ^2 TP_1(T_1,T)}{\partial T_1\partial T}\\ \frac{\partial ^2 TP_1(T_1,T)}{\partial T\partial T_1} &{} \frac{\partial ^2 TP_1(T_1,T)}{\partial T^2} \end{array}\right| =-\frac{1}{12T^4}\big \{3\big [2aX_{21}+b(X_{22}-cX_{23})\big ]^2+4(aX_{31}+bX_{32})(-6A +bX_{41}+3a(X_{42}+cX_{43}))\big \}>0,$ which gives $\{3[2aX_{21}+b(X_{22}-cX_{23})]^2+4 (aX_{31}+bX_{32} ) (-6A +bX_{41}+3a(X_{42}+cX_{43}) ) \}<0$. This completes the proof.