Backlogging EOQ model for promotional effort and selling price sensitive demand- an intuitionistic fuzzy approach

Abstract

An intuitionistic fuzzy economic order quantity (EOQ) inventory model with backlogging is investigated using the score functions for the member and non-membership functions. The demand rate is varying with selling price and promotional effort (PE). A crisp model is formulated first. Then, intuitionistic fuzzy set and score function (or net membership function) are applied in the proposed model, considering selling price and PE as fuzzy numbers. To obtain the best inventory policy, ranking index method has been adopted, showing that the score function can maintain the ranking rule also. Moreover, optimization is made under the general fuzzy optimal (GFO) and intuitionistic fuzzy optimal (IFO) policy. Finally, a graphical illustration, numerical examples with sensitivity analysis and conclusion is made to justify the model.

Appendix

Here, we shall show that the score (net membership) function follows Yager’s (1981) ranking index method.

We have

(i)

$$\begin{aligned} I ( \rho ) =& \frac{1}{2} \biggl( \delta_{1} + \delta_{3} + \frac{1}{2 \lambda_{1}} - \frac{1}{2 \lambda_{2}} \biggr) \\ =& \frac{1}{4} \biggl[ \frac{ ( \rho_{2} - \rho_{1} ) (\rho_{2} - \rho_{1} ' ) +2 ( \rho_{2}^{2} - \rho_{1} \rho_{1} ' )}{2 \rho_{2} - \rho_{1} - \rho_{1} '} + \frac{ ( \rho_{3} - \rho_{2} ) ( \rho_{3} ' - \rho_{2} ) +2 ( \rho_{3} \rho_{3} ' -\rho_{2}^{2} )}{\rho_{3} + \rho_{3} ' -2 \rho_{2}} \biggr] \end{aligned}$$

Letting $\rho_{1} ' = \rho_{1}$, $\rho_{3} ' = \rho_{3}$, in fuzzy sense, we get

$$\begin{aligned} I(\rho) =& \frac{1}{4} \biggl[ \frac{ ( \rho_{2} - \rho_{1} ) ( \rho_{2} - \rho_{1} ) +2 ( \rho_{2}^{2} - \rho_{1}^{2} )}{2 ( \rho_{2} - \rho_{1} )} + \frac{ ( \rho_{3} - \rho_{2} ) ( \rho_{3} - \rho_{2} ) +2 ( \rho_{3}^{2} -\rho_{2}^{2} )}{2 ( \rho_{3} - \rho_{2} )} \biggr] \\ =& \biggl( \frac{ ( \rho_{2} - \rho_{1} ) +2 ( \rho_{2} + \rho_{1} )}{8} + \frac{ ( \rho_{3} - \rho_{2} ) +2 ( \rho_{3} + \rho_{2} )}{8} \biggr) \\ =& \frac{3 \rho_{3} +4 \rho_{2} + \rho_{1}}{8} \xrightarrow{\mathit{yields}} \rho \quad \mbox{when }\rho_{1} = \rho_{2} = \rho_{3} =\rho,\ \mbox{in crisp sense.} \end{aligned}$$

Hence the proof.

Let $d_{2} = \frac{\tau\rho}{1+\rho}$ then $\rho= \frac{\tau}{\tau- d_{2}}$.

(ii) So, the net membership is given by

$$\zeta ( d_{2} ) = \left \{ \begin{array}{@{}l@{}} \frac{ [ \frac{\tau}{ ( \tau- d_{2} )} - ( 1+ \rho_{1} ) ]}{\rho_{2} - \rho_{1}} - \frac{ [ ( 1+ \rho_{2} ) - \frac{\tau}{ ( \tau- d_{2} )} ]}{\rho_{2} - \rho_{1} '} \quad \forall d_{2} ' < d_{2} < d_{2}^{''} \\ \frac{ [ ( 1+ \rho_{2} ) - \frac{\tau}{ ( \tau- d_{2} )} ]}{\rho_{3} - \rho_{2}} - \frac{ [ \frac{\tau}{ ( \tau- d_{2} )} - ( 1+ \rho_{3} ' ) ]}{\rho_{3} ' - \rho_{2}} \quad \forall d_{2}^{''} < d_{2} < d_{2}^{'''}\\ 0,\quad \mathrm{elsewhere} \end{array} \right \} $$

Here, $\frac{ [ \frac{\tau}{ ( \tau- d_{2} )} - ( 1+ \rho_{1} ) ]}{\rho_{2} - \rho_{1}} - \frac{ [ ( 1+ \rho_{2} ) - \frac{\tau}{ ( \tau- d_{2} )} ]}{ \rho_{2} - \rho_{1} '} \geq\alpha$ and $\frac{ [ ( 1+ \rho_{2} ) - \frac{\tau}{ ( \tau- d_{2} )} ]}{\rho_{3} - \rho_{2}} - \frac{ [ \frac{\tau}{ ( \tau- d_{2} )} - ( 1+ \rho_{3} ' ) ]}{\rho_{3} ' - \rho_{2}} \geq\alpha$, after a little bit calculation, we have

$$d_{2} \geq\tau- \frac{\tau ( \frac{1}{\rho_{2} - \rho_{1}} + \frac{1}{\rho_{2} - \rho_{1} '} )}{\alpha+ ( \frac{1+ \rho_{1}}{\rho_{2} - \rho_{1}} + \frac{1+ \rho_{2}}{\rho_{2} - \rho_{1} '} )}\quad \mbox{and}\quad d_{2} \leq\tau- \frac{\tau ( \frac{1}{\rho_{3} - \rho_{2}} + \frac{1}{\rho_{3} ' - \rho_{2}} )}{ ( \frac{1+ \rho_{1}}{\rho_{3} - \rho_{2}} + \frac{1+ \rho_{2}}{\rho_{3} ' - \rho_{2}} ) -\alpha}, \mathrm{respectively}. $$

Therefore,

$$\begin{aligned} \bigl[ L_{d 2}^{-1} ( \alpha ), R_{d 2}^{-1} (\alpha) \bigr] =& \biggl[ \tau- \frac{\tau ( \frac{1}{\rho_{2} - \rho_{1}} + \frac{1}{\rho_{2} - \rho_{1} '} )}{\alpha+ ( \frac{1+ \rho_{1}}{\rho_{2} - \rho_{1}} + \frac{1+ \rho_{2}}{\rho_{2} - \rho_{1} '} )},\tau- \frac{\tau ( \frac{1}{\rho_{3} - \rho_{2}} + \frac{1}{\rho_{3} ' - \rho_{2}} )}{ ( \frac{1+ \rho_{1}}{\rho_{3} - \rho_{2}} + \frac{1+ \rho_{2}}{\rho_{3} ' - \rho_{2}} ) -\alpha} \biggr]\\ =& \biggl[ \tau- \frac{k_{1}}{\alpha+ k_{2}},\tau- \frac{k_{3}}{ k_{4} -\alpha} \biggr] \end{aligned}$$

where

$$\begin{aligned} &k_{1} =\tau \biggl( \frac{1}{\rho_{2} - \rho_{1}} + \frac{1}{\rho_{2} - \rho_{1} '} \biggr), \qquad k_{2} = \biggl( \frac{1+ \rho_{1}}{\rho_{2} - \rho_{1}} + \frac{1+ \rho_{2}}{\rho_{2} - \rho_{1} '} \biggr),\\ & k_{3} =\tau \biggl( \frac{1}{\rho_{3} - \rho_{2}} + \frac{1}{\rho_{3} ' - \rho_{2}} \biggr)\quad \mbox{and}\quad k_{4} = \biggl( \frac{1+ \rho_{1}}{\rho_{3} - \rho_{2}} + \frac{1+ \rho_{2}}{\rho_{3} ' -\rho_{2}} \biggr) \end{aligned}$$

Now, the indexed value is

$$\begin{aligned} I ( d_{2} ) =& \frac{1}{2} \int_{0}^{1} \bigl[ L_{d 2}^{-1} ( \alpha ) + R_{d 2}^{-1} ( \alpha ) \bigr] d\alpha \\ =& \frac{1}{2} \int_{0}^{1} \biggl[ \tau- \frac{k_{1}}{\alpha+ k_{2}} +\tau- \frac{k_{3}}{k_{4} -\alpha} \biggr] d\alpha \\ =& \frac{1}{2} \biggl[ 2\tau- k_{1} \operatorname{Log}\biggl( \frac{1+ k_{2}}{k_{2}} \biggr) + k_{3} \operatorname{Log}\biggl( \frac{k_{4} -1}{k_{4}} \biggr) \biggr]. \end{aligned}$$

To have a fuzzy value, we have $\rho_{1} ' \rightarrow \rho_{1}$ and $\rho_{3} ' \rightarrow \rho_{3}$ those provide

$$\begin{aligned} &k_{1} \rightarrow \biggl( \frac{2\tau}{\rho_{2} - \rho_{1}} \biggr),\qquad k_{2} \rightarrow \biggl( \frac{2+ \rho_{1} + \rho_{2}}{\rho_{2} - \rho_{1}} \biggr),\qquad k_{3} \rightarrow \biggl( \frac{2\tau}{\rho_{3} - \rho_{2}} \biggr)\quad \mbox{and}\\ & k_{4} \rightarrow \biggl( \frac{2+ \rho_{1} + \rho_{2}}{\rho_{3} - \rho_{2}} \biggr). \end{aligned}$$

Therefore, from above, we have

$$I ( d_{2} ) = \frac{1}{2} \biggl[ 2\tau- \frac{2\tau}{\rho_{2} - \rho_{1}} \operatorname{Log}\biggl( 1+ \frac{\rho_{2} - \rho_{1}}{2+ \rho_{2} + \rho_{1}} \biggr) + \frac{2\tau}{\rho_{3} - \rho_{2}} \operatorname{Log}\biggl( 1- \frac{\rho_{3} - \rho_{2}}{2+ \rho_{2} + \rho_{3}} \biggr) \biggr]. $$

The crisp value of d ₂ is

$$\begin{aligned} d_{2} =& \lim_{ \substack{ \rho_{1} \rightarrow \rho_{2}\\ \rho_{3} \rightarrow \rho_{2} } } I ( d_{2} ) \\ =& \lim_{\substack{ \rho_{1} \rightarrow \rho_{2}\\ \rho_{3} \rightarrow \rho_{2} } } \biggl[ \tau- \frac{\tau}{\rho_{2} - \rho_{1}} \operatorname{Log}\biggl( 1+ \frac{\rho_{2} - \rho_{1}}{2+ \rho_{1} + \rho_{2}} \biggr) + \frac{\tau}{ \rho_{3} - \rho_{2}} \operatorname{Log}\biggl( 1+ \frac{\rho_{3} - \rho_{2}}{2+ \rho_{2} + \rho_{3}} \biggr) \biggr] \end{aligned}$$

where

$$\begin{aligned} &\lim_{\rho 1 \rightarrow \rho 2} \biggl[ \frac{1}{\rho_{2} - \rho_{1}} \operatorname{Log}\biggl( 1+ \frac{\rho_{2} - \rho_{1}}{2+ \rho_{1} + \rho_{2}} \biggr) \biggr] \\ &\quad= \lim_{\rho 1 \rightarrow \rho 2} \biggl( \frac{1}{2+ \rho_{1} + \rho_{2}} \biggr) \times \lim _{\rho 1 \rightarrow \rho 2} \biggl[ \frac{\operatorname{Log}( 1+ \frac{\rho_{2} - \rho_{1}}{2+ \rho_{1} + \rho_{2}} )}{ ( \rho_{2} - \rho_{1} ) / ( 2+ \rho_{1} + \rho_{2} )} \biggr] = \frac{1}{2 ( 1+ \rho_{2} )} \end{aligned}$$

and

$$\begin{aligned} &\lim_{\rho 3 \rightarrow \rho 2} \biggl[ \frac{1}{\rho_{3} - \rho_{2}} \operatorname{Log}\biggl( 1- \frac{\rho_{3} - \rho_{2}}{2+ \rho_{2} + \rho_{3}} \biggr) \biggr] \\ &\quad = \lim_{\rho 3 \rightarrow \rho 2} \biggl( \frac{-1}{2+ \rho_{2} + \rho_{3}} \biggr) \times \lim _{\rho 3 \rightarrow \rho 2} \biggl[ \frac{\operatorname{Log}( 1- \frac{\rho_{3} - \rho_{2}}{2+ \rho_{2} + \rho_{3}} )}{- ( \rho_{3} - \rho_{2} ) / ( 2+ \rho_{2} + \rho_{3} )} \biggr] = \frac{1}{2 ( 1+ \rho_{2} )}. \end{aligned}$$

Thus

$$\begin{aligned} d_{2} =& \lim_{\substack{ \rho_{1} \rightarrow \rho_{2}\\ \rho_{3} \rightarrow \rho_{2} } } I ( d_{2} ) \\ =& \biggl[ \tau- \frac{\tau}{2 ( 1+ \rho_{2} )} - \frac{\tau}{2 ( 1+ \rho_{2} )} \biggr] \\ =&\tau \biggl[ 1- \frac{1}{1+ \rho_{2}} \biggr] = \frac{\tau \rho_{2}}{1+ \rho_{2}} \rightarrow \frac{\tau\rho}{ 1+\rho} \xrightarrow{\mathit{yields}} \mbox{Crisp value}. \end{aligned}$$

Proceeding this way we can prove the other IFS as well. Hence the Yager’s Ranking method can be applied on net membership function also. This completes the proof.