An improved inventory model with random review period and temporary price discount for deteriorating items

Abstract

This paper deals with a stochastic periodic review inventory system, wherein temporary price discount offer is taken into the account. The review period (time interval between two consecutive reviews) is considered as a random variable. In the real life, whereas, in one side, supplier stimulates the sale and/or rise cash flow by offering price discount/quantity discount, on the other side extra purchasing impels to deterioration of the product. To become the part of this, we translate some real-life situations such as deterioration, temporary price discount, partial backlogging into the mathematical model. This paper prudently studies the joint effect of deterioration and special sale offer. Furthermore, shortages are permissible in retailer’s inventory system and partially backlogged. The model is mathematically rigorously analyzed and concavity of saving function is also shown. Illustration of the proposed model is exposed through suitable numerical examples, sensitivity analysis and graphical representation.

Appendices

Appendix 1

$$\frac{d^{2}{T}_{p}}{{d{q_{s}}}^{2}}= \frac{-1}{r(x_{\max }-x_{\min })(\frac{\theta }{r}q_{s}+1)^{2}} \{ (a_{1}+a_{2})\theta (x_{\max }-\mu _{s})+a_{2} \}.$$

If $a_1\ge 0$ or $a_1 < 0$ and $a_2\ge -a_1$, then $d^2 T_p/dq_s^2 < 0$ if and only if $a_2 > -(a_1 + a_2)\theta (x_{\max } - \mu _s) \Rightarrow q_s < \left( e^{\frac{a_2}{a_1+a_2}+\theta x_{\max }}-1\right) r/\theta$. Moreover, if $a_1 < 0$ and $a_2<-a_1$, then $d^2 T_p/dq_s^2 < 0$ if and only if $a_2 > -(a_1 + a_2)\theta (x_{\max } - \mu _s) \Rightarrow q_s > \left( e^{\frac{a_2}{a_1+a_2}+\theta x_{\max }}-1\right) r/\theta$.

Appendix 2

$$\begin{aligned} \frac{d^{2}{T}_{p}}{{d{q_{s}}}^{2}} &= \frac{-1}{\sqrt{2\pi }r\sigma ({\varPhi }(x_{\max })-{\varPhi }(x_{\min })) (\frac{\theta }{r}q_{s}+1)^{2}}\left\{ a_{2} e^{-\frac{1}{2}(\frac{\rho -\mu _{s}}{\sigma })^{2}}\right. \\ &\quad\left. +\,\theta \sigma (a_{1}+a_{2})\sqrt{\frac{\pi }{2}}\left( Erf \left[ \frac{\rho -\mu _{s}}{\sqrt{2}\sigma }\right] +Erf \left[ \frac{x_{\max }-\rho }{\sqrt{2}\sigma }\right] \right) \right\} . \end{aligned}$$

If $a_1\ge 0$ or $a_1 < 0$ and $a_2\ge -a_1$, then $d^2 T_p/dq_s^2 < 0$ if $a_2 e^{-\frac{1}{2}(\frac{\rho -\mu _{s}}{\sigma })^{2}} +\theta \sigma (a_{1}+a_{2})\sqrt{\frac{\pi }{2}}\left( Erf \left[ \frac{\rho -\mu _{s}}{\sqrt{2}\sigma }\right] +Erf \left[ \frac{x_{\max }-\rho }{\sqrt{2}\sigma }\right] \right) >0,\Rightarrow \frac{a_{2}}{\theta \sigma (a_{1}+a_{2})} e^{-\frac{1}{2}(\frac{\rho -\mu _{s}}{\sigma })^{2}} +\sqrt{\frac{\pi }{2}}\left( Erf \left[ \frac{\rho -\mu _{s}}{\sqrt{2}\sigma }\right] +Erf \left[ \frac{x_{\max }-\rho }{\sqrt{2}\sigma }\right] \right) >0$. Moreover, if $a_1 < 0$ and $a_2<-a_1$, then $d^2 T_p/dq_s^2 < 0$ if and only if $\frac{a_{2}}{\theta \sigma (a_{1}+a_{2})} e^{-\frac{1}{2}(\frac{\rho -\mu _{s}}{\sigma })^{2}} +\sqrt{\frac{\pi }{2}}\left( Erf \left[ \frac{\rho -\mu _{s}}{\sqrt{2}\sigma }\right] +Erf \left[ \frac{x_{\max }-\rho }{\sqrt{2}\sigma }\right] \right) <0$.