Partial linked-to-order delayed payment and life time effects on decaying items ordering

Abstract

Trade-credit is an influential implementation in financial transactions. This paper proposes an inventory model for decaying items with lifetime under linked-to-order partial delay in payments. More precisely, the items are gradually deteriorating and also have a known maximum lifetime. The payment scheme is structured as follows: if the order quantity reaches a particular level, fully permissible trade credit is possible, otherwise the partial trade credit is offered. To the best of our knowledge, this is the first research incorporating the “Linked-to-order Trade Credit Financing” scheme for deteriorating items with lifetime. Selling price and purchasing costs are not considered equal and there is no need for the interest charged in stocks to be larger than the interest earned on investment. Theoretical results are developed to obtain the optimum solutions of the problem. The authenticity and pertinence of the model and solution procedure are illustrated through numerical results. Finally, sensitivity analysis and managerial insights are provided.

Appendices

Appendix 1

In order to prove Lemma 1 set

$$\begin{aligned} F_{1} (T) & = - A - hD\left[ {\frac{{(1 + G)^{2} }}{2}\ln \left( {\frac{1 + G}{1 + G - T}} \right) + \frac{{T^{2} }}{4} - \frac{(1 + G)T}{2}} \right] \\& \quad+ \frac{{I_{e} PDM^{2} }}{2} + hDT\left[ {\frac{{(1 + G)^{2} }}{2(1 + G - T)} + \frac{T}{2} - \frac{1 + G}{2}} \right] \\ & \quad + CDT\left( {\frac{1 + G}{1 + G - T}} \right) - CD(1 + G)\ln \left( {\frac{1 + G}{1 + G - T}} \right) + I_{k} CDT\left[ {\frac{{(1 + G - M)^{2} }}{2(1 + G - T)} + \frac{T + M}{2}} \right] \\ & \quad - I_{k} CD\left[ {\frac{{(1 + G - M)^{2} }}{2}\ln \left( {\frac{1 + G - M}{1 + G - T}} \right) + \frac{{T^{2} - M^{2} }}{4} + \frac{M(T + M) + M}{2}} \right] \\ \end{aligned}$$

(26)

$\frac{{dF_{1} (T)}}{dT}$ with respect to $T \in [M,\infty )$ yields:

$$\begin{aligned} \frac{{dF_{1} (T)}}{dT}& = hDT\left[ {\frac{{(1 + G)^{2} }}{{2(1 + G - T)^{2} }} + \frac{1}{2}} \right] + CD\frac{1 + G}{{(1 + G - T)^{2} }}\,\,\\&\quad + I_{k} CDT\left[ {\frac{{(1 + G - M)^{2} }}{{2(1 + G - T)^{2} }} + \frac{1}{2}} \right]\, > 0 \end{aligned}$$

(27)

Therefore,$F_{1} (T)$ is strictly increasing function of $T \in [M,\infty )$. From Eq. (26), $F_{1} (M) = \Delta_{1}$ and $\mathop {\lim }\nolimits_{T \to \infty } F_{1} (T) = \infty$. Thereafter, by using the Intermediate Value Theorem, we could claim that there exist a unique value of $T \in [M,\infty )$, say $T_{1}$, such that $F_{1} (T_{1} ) = 0$. Furthermore:

$$\frac{{d^{2} TRC_{1} (T)}}{{dT^{2} }}\mathop |\limits_{{T = T_{1} }} = \frac{D}{{T_{1} }}\left\{ {\frac{{(h + C)(1 + G)^{2} }}{{2(1 + G - T_{1} )}} + \frac{h}{2} + I_{k} C\left[ {\frac{{(1 + G - M)^{2} }}{{2(1 + G - T_{1} )^{2} }} + \frac{1}{2}} \right]\,\,} \right\} > 0\,$$

(28)

Then, apparently $T_{1} \in [M,\infty )$ is a unique optimal value for $TRC_{1} (T)$. If $\Delta_{1} > 0$, then for all $T \in [M,\infty ),F_{1} (T) > 0$ and $\frac{{dTRC_{1} (T)}}{dT} = \frac{{F_{1} (T)}}{{T^{2} }} > 0$. Consequently, $TRC_{1} (T)$ is a strictly increasing function of T in the interval $T \in [M,\infty )$. So, in this case, $TRC_{1} (T)$ has an optimal solution at $T = M$ and this completes the proof.

Appendix 2

In order to prove Lemma 3 set;

$$\begin{aligned} F_{3} (T) & = - A - hD\left[ {\frac{{(1 + G)^{2} }}{2}\ln \left( {\frac{1 + G}{1 + G - T}} \right) + \frac{{T^{2} }}{4} - \frac{(1 + G)T}{2}} \right] \\& \quad + hDT\left[ {\frac{{(1 + G)^{2} }}{2(1 + G - T)} + \frac{T}{2} - \frac{1 + G}{2}} \right] - CD(1 + G)\ln \left( {\frac{1 + G}{1 + G - T}} \right) \\ & \quad + CDT\left( {\frac{1 + G}{1 + G - T}} \right) + \frac{{I_{e} PDT^{2} }}{2} - I_{k} CD(1 + G)^{2} \left\{ { - \frac{1}{2}\ln \left( {\frac{1 + G}{1 + G - T}} \right)}\right.\\& \quad \times \left.{\left[ {\alpha \left( {\frac{1 + G}{1 + G - T}} \right)^{2(\alpha - 1)} } \right] + \frac{1}{4}\left[ {\left( {\frac{1 + G}{1 + G - T}} \right)^{2(\alpha - 1)} - 1} \right]} \right\} \\ & \quad + I_{k} CD(1 + G)^{2} T\left\{ { - \frac{\alpha }{2}\frac{{(1 + G)^{{^{2(\alpha - 1)} }} }}{{(1 + G - T)^{{^{2\alpha - 1} }} }} - \frac{\alpha - 1}{2}\frac{{(1 + G)^{{^{2(\alpha - 1)} }} }}{{(1 + G - T)^{{^{2\alpha - 1} }} }} }\right.\\&\quad \left.{- \alpha (\alpha - 1)\ln \left( {\frac{1 + G}{1 + G - T}} \right)\frac{{(1 + G)^{{^{2(\alpha - 1)} }} }}{{(1 + G - T)^{{^{2\alpha - 1} }} }} + \frac{1}{2(1 + G - T)}} \right\} \\ & \quad + I_{k} CD\alpha (1 + G)^{2} \left\{ {\ln \left( {\frac{1 + G}{1 + G - T}} \right)\left[ {1 - \left( {\frac{1 + G}{1 + G - T}} \right)^{\alpha - 1} } \right]} \right\} \\ & \quad - I_{k} CD\alpha (1 + G)^{2} T\left\{ { - \frac{{(1 + G)^{{^{\alpha - 1} }} }}{{(1 + G - T)^{\alpha } }} }\right.\\&\quad \left.{- \frac{{\alpha - 1(1 + G)^{{^{2(\alpha - 1)} }} }}{{2(1 + G - T)^{{^{2\alpha - 1} }} }} - (\alpha - 1)\ln \left( {\frac{1 + G}{1 + G - T}} \right)\frac{{(1 + G)^{{^{\alpha - 1} }} }}{{(1 + G - T)^{\alpha } }} + \frac{1}{(1 + G - T)}} \right\} \\ \end{aligned}$$

(29)

$\frac{{dF_{3} (T)}}{dT}$ with respect to $T \in (0,T_{w} )$, yields:

$$\begin{aligned} \frac{{dF_{3} (T)}}{dT}& = hDT\left[ {\frac{{(1 + G)^{2} }}{{2(1 + G - T)^{2} }} + \frac{1}{2}} \right] + \frac{CD(1 + G)}{{(1 + G - T)^{2} }}\, + I_{e} PD \\& \quad + \,I_{k} CD(1 + G)^{2} T\left\{ {\frac{1}{{2(1 + G - T)^{2} }}\left[ {1 - (\frac{1 + G}{1 + G - T})^{2\alpha - 1} \frac{(\alpha - 1)(2\alpha - 1)}{1 + G}} \right]} \right\} \end{aligned}$$

(30)

Since $\alpha < 1$ we have:

$$1 - \left( {\frac{1 + G}{1 + G - T}} \right)^{2\alpha - 1} \frac{(\alpha - 1)(2\alpha - 1)}{1 + G} > 1 - \frac{(\alpha - 1)(2\alpha - 1)}{1 + G - T}$$

(31)

As mentioned in assumptions, $\theta (t) = \frac{1}{1 + G - t}$ for $t \in [0,T]$ and since $\theta (t) < 1$ we have $\theta (T) = \frac{1}{1 + G - T} < 1$, then $\frac{(\alpha - 1)(2\alpha - 1)}{1 + G - T} < 1$ and consequently $1 - \frac{(\alpha - 1)(2\alpha - 1)}{1 + G - T} > 0$ which gives $\frac{{dF_{3} (T)}}{dT} > 0$. Therefore, $F_{3} (T)$ is a strictly increasing function of T in the interval $(0,T_{w} )$. From Eq. (29) we know that $\mathop {\lim }\nolimits_{{T \to T_{w}^{ - } }} F_{3} (T) = \Delta_{3}$ and $\mathop {\lim }\nolimits_{T \to \infty } F_{1} (T) < 0$. Thus if $\mathop {\lim }\nolimits_{{T \to T_{w}^{ - } }} F_{3} (T) = \Delta_{3} > 0$, by using the Intermediate Value Theorem, there is a unique $T_{3} \in (0,T_{w} )$ which gives $F_{3} (T) = 0$. Moreover, taking the second derivative of $TRC_{3} (T)$ with respect to T at the point $T_{3}$ yields:

$$\begin{aligned} \frac{{d^{2} TRC_{3} (T)}}{{dT^{2} }}\mathop |\limits_{{T_{3} }} &= \frac{D}{{T_{3} }}\left\{ {\frac{{(h + C)(1 + G)^{2} }}{{2(1 + G - T_{3} )}} + \frac{h}{2} + I_{e} P + I_{k} C(1 + G)^{2} }\right.\\&\quad \times\left.{\left\{ {\frac{1}{{2(1 + G - T_{3} )^{2} }}\left[ {1 - \left( {\frac{1 + G}{{1 + G - T_{3} }}} \right)^{2\alpha - 1} \frac{(\alpha - 1)(2\alpha - 1)}{1 + G}} \right]} \right\}} \right\} > 0 \end{aligned}$$

(32)

Therefore, $T_{3} \in (0,T_{w} )$ is the unique optimal solution for $TRC_{3} (T)$. However, if $\mathop {\lim }\nolimits_{{T \to T_{w}^{ - } }} F_{3} (T) = \Delta_{3} < 0$, then for all $T \in (0,T_{w} )$, we have $F_{3} (T) < 0$. As a result for all $T \in (0,T_{w} )$ also $\frac{{dTRC_{3} (T)}}{dT} = \frac{{F_{3} (T)}}{{T^{2} }} < 0$. Then we can claim that $TRC_{3} (T)$ is a strictly decreasing function of T in the interval $(0,T_{w} )$. Subsequently, there is no T in the open interval $(0,T_{w} )$ under which $TRC_{3} (T)$ is minimized and this completes the proof.