A supply and demand economic order quantity inventory model under pythagorean fuzzy environment

Abstract

This article deals with a noble application of Pythagorean fuzzy set (PFS). In fuzzy set theory in particular for intuitionistic fuzzy set (IFS) we are quite familiar to pass through Attanassov’s concept. But to deal with non-standard fuzzy set, PFS plays an important role in many decision making problems which dominate Attanassov’s IFS. The main aim of this study is to show the effectiveness and applicability of the PFS in decision making under uncertain environment through inventory problems. Also, this works brings a solution to the situation when the sum of the membership and non-membership degree is greater than one i.e. when IFS fails. The primary goal of this study is to show how PFS has a wider space of membership degrees and it is more capable of handling uncertain information than any other fuzzy set. Here, we have considered an Economic Order Quantity (EOQ) model with customer’s income dependent demand of end-item and price-dependent supply of input component. We explicitly specify the demand as a function of customers’ income and unit selling price. We develop a crisp model for cost minimization problem first then we split the model into general fuzzy, intuitionistic fuzzy and Pythagorean fuzzy environment respectively. However, PFS lacks mathematical solution procedure. Employing the score function, we optimize the cost function using the Yager’s ranking index method. Moreover, our numerical results show the superiority of PFS with respect to the traditional model and other fuzzy environments. Finally, sensitivity analysis and graphical illustrations are made to justify the model.

Graphical abstract

Appendices

Appendix

A

The feasible region of PFS and IFS are shown in figure 5. [Here (.)/(.) stands for either or]

Figure 5

Feasible region of IFS and PFS.

Definition A.1

(Yager [34]) The set of Pythagorean membership grades is greater than the set of intuitionistic membership grades.

Example A.1

Let a point (a, b) is an intuitionistic membership grade. So, $a,b \in [0,1]$ implies $a+b \le 1$. For any $a,b \in [0,1]$, $a^{2}\le a$, $b^{2} \le b$ which implies $a^{2}+b^{2} \le 1$. So, any intuitionistic membership grade is a pythagorean membership grade. But, the point $(\frac{\sqrt{3}}{2},\frac{1}{2})$ we see that $(\frac{\sqrt{3}}{2})^{2}+(\frac{1}{2})^{2}=1$, thus this is a pythagorean membership grade. however since $\frac{\sqrt{3}}{2}+\frac{1}{2}>1$ this is not an intuitionistic membership grade.

From figure 5 we see that intuitionistic (non)membership grades are all points under the line ${\mu (x)}+{\nu (x)} = 1$ and Pythagorean (non)membership grades are all points with ${\mu ^{2}(x)}+{\nu ^{2}(x)} \le 1$. Hence Pythagorean fuzzy set represents a larger body of non-standard membership grades than intuitionistic membership grades.

B Particular case

If there is no expiration date i.e. if $m\rightarrow \infty$, then $\lim \limits _{m\rightarrow \infty } \bigg [\frac{1+m}{T}log\Big (\frac{1+m}{1+m-T}\Big )\bigg ]=\lim \limits _{m\rightarrow \infty } \frac{1+m}{1+m-T}=1$ (using L’Hospital rule). So, $Q_{2}=D(1+m)log\Big (\frac{1+m}{1+m-T}\Big )=DT ~\text{ as }~ m\rightarrow \infty$.

Now, $\lim \limits _{m\rightarrow \infty } \bigg [\frac{(1+m)^2}{2}log\Big (\frac{1+m}{1+m-T}\Big )-\frac{(1+m)T}{2}\bigg ]=\frac{1}{2}\lim \limits _{m\rightarrow \infty } \bigg [\frac{log\big (\frac{1+m}{1+m-T}\big )-\frac{T}{1+m}}{\frac{1}{(1+m)^2}}\bigg ]=\frac{T^2}{4}$ (using L’Hospital rule).

Therefore, $\lim \limits _{m\rightarrow \infty } h_{e}D \bigg [\frac{(1+m)^2}{2}log\Big (\frac{1+m}{1+m-T}\Big )-\frac{(1+m)T}{2}+\frac{T^2}{4}\bigg ]=\frac{h_{e}DT^{2}}{2}$

C Selection of $\alpha$-cut intervals for IFS

We have the score function of the fuzzy parameters $d_{i}$ developed in section 5

$$\begin{aligned} \omega (d_{i})= \left\{ \begin{array}{ll} \sqrt{\frac{d_{i}-d_{i1}}{d_{i2}-d_{i1}}}-\sqrt{\frac{d_{i2}-d_{i}}{d_{i2}-{d_{i1}}^{'}}} &{} \sigma _{1i} \leqslant d_{i} \leqslant \sigma _{2i} \\ \sqrt{\frac{d_{i3}-d_{i}}{d_{i3}-d_{i2}}}-\sqrt{\frac{d_{i}-d_{i2}}{{d_{i3}}^{'}-d_{i2}}} &{} \sigma _{2i} \leqslant d_{i}\leqslant \sigma _{3i} \\ 0 &{} \hbox {otherwise} \end{array} \right. \end{aligned}$$

(C.1)

Now, to get $\alpha$-cut intervals we write

$$\begin{aligned}&\sqrt{\frac{d_{i}-d_{i1}}{d_{i2}-d_{i1}}}-\sqrt{\frac{d_{i2}-d_{i}}{d_{i2}-{d_{i1}}^{'}}} \ge \alpha \nonumber \\&\quad \Rightarrow \sqrt{(d_{i}-d_{i1})(d_{i2}-{d_{i1}}^{'})}-\sqrt{(d_{i2}-d_{i})(d_{i2}-d_{i1})}\nonumber \\&\qquad \ge \alpha \sqrt{(d_{i2}-d_{i1})(d_{i2}-{d_{i1}}^{'})} \nonumber \\&\quad \Rightarrow Ad_{i}^{2}-2Bd_{i}+C \ge 0, \end{aligned}$$

(C.2)

where

$$\begin{aligned} \begin{aligned} A=&\{({d_{i1}}+{d_{i1}}^{'})^{2}+4d_{i2}(d_{i2}-d_{i1}-d_{i1}^{'})\},\\ B=&2(d_{i2}-d_{i1}^{'})(d_{i2}-d_{i1}^{'})(d_{i2}+d_{i1})\\&-(d_{i1}-d_{i1}^{'}) \{(d_{i2}^{2}-2d_{i1}d_{i2}+d_{i1}d_{i1}^{'})\\&-\alpha ^{2}(d_{i2}-d_{i1})(d_{i2}-d_{i1}^{'})\}\\ C=&\{(d_{i2}^{2}-2d_{i1}d_{i2}+d_{i1}d_{i1}^{'})-\alpha ^{2}(d_{i2}-d_{i1})(d_{i2} -d_{i1}^{'})\}^{2}\\&+4(d_{i2}-d_{i1}^{'})(d_{i2}-d_{i1}^{'})\\ \end{aligned} \end{aligned}$$

From (C.2), we get

$$\begin{aligned} \Rightarrow d_{i} \ge \frac{B}{A} + \frac{\sqrt{B^{2}-AC}}{A} ~\text{ or }~ d_{i} \le \frac{B}{A} - \frac{\sqrt{B^{2}-AC}}{A} \end{aligned}$$

(C.3)

Again,

$$\begin{aligned} \sqrt{\frac{d_{i3}-d_{i}}{d_{i3}-d_{i2}}}-\sqrt{\frac{d_{i}-d_{i2}}{{d_{i3}}^{'}-d_{i2}}} \ge \alpha \end{aligned}$$

implies

$$\begin{aligned}&\sqrt{(d_{i3}-d_{i})(d_{i3}-d_{i2})}-\sqrt{(d_{i}-d_{i2})({d_{i3}}^{'}-d_{i2})}\nonumber \\&\qquad \ge \alpha \sqrt{(d_{i3}-d_{i2})({d_{i3}}^{'}-d_{i2})} \nonumber \\&\quad \Rightarrow A^{'}d_{i}^{2}-2B^{'}d_{i}+C^{'} \ge 0, \end{aligned}$$

(C.4)

where

$$\begin{aligned} \begin{aligned} A^{'}=&\{({d_{i3}}+{d_{i3}}^{'})^{2}+4d_{i2}(d_{i2}-d_{i3}-d_{i3}^{'})\}\\ B^{'}=&2(d_{i2}-d_{i3})(d_{i2}-{d_{i3}}^{'})(d_{i2}+d_{i3})\\&-(d_{i3}-d_{i3}^{'})\{(d_{i2}^{2}-2d_{i3}d_{i2}+d_{i3}d_{i3}^{'})\\&-\alpha ^{2}(d_{i2}-d_{i3})(d_{i2}-d_{i3}^{'})\}\\ C^{'}=&\{(d_{i2}^{2}-2d_{i3}d_{i2}+d_{i3}d_{i3}^{'})-\alpha ^{2}(d_{i2}-d_{i3})(d_{i2}-d_{i3}^{'})\}^{2}\\&+4(d_{i2}-d_3^{'})(d_{i2}-d_{i3}^{'}) \end{aligned} \end{aligned}$$

From (C.4), we get

$$\begin{aligned} \Rightarrow d_{i} \ge \frac{B^{'}}{A^{'}} + \frac{\sqrt{{B^{'}}^{2}-A^{'}C^{'}}}{A^{'}} ~\text{ or }~ d_{i} \le \frac{B^{'}}{A^{'}} - \frac{\sqrt{{B^{'}}^{2}-A^{'}C^{'}}}{A^{'}} \end{aligned}$$

Let $\sigma _{1i}=\frac{B}{A} - \frac{\sqrt{B^{2}-AC}}{A}$, $\sigma _{2i}=\frac{B}{A} + \frac{\sqrt{B^{2}-AC}}{A}$, $\sigma _{3i}=\frac{B^{'}}{A^{'}} - \frac{\sqrt{{B^{'}}^{2}-A^{'}C^{'}}}{A^{'}}$ and $\sigma _{4i}=\frac{B^{'}}{A^{'}} + \frac{\sqrt{{B^{'}}^{2}-A^{'}C^{'}}}{A^{'}}$

Figure 6

(a), (b), (c), (d), (e), (f) All possible intervals of $d_{i}$’s.

Therefore from above we can write $\sigma _{1i}\ge d_{i} \ge \sigma _{2i}$ and $\sigma _{3i} \ge d_{i} \ge \sigma _{4i}$. The possible intervals are

• (a) ${d_{i1}}^{'} \le d_{i} \le \sigma _{1i}$ and $\sigma _{2i} \le d_{i} \le d_{i2}$    (b) ${d_{i1}}^{'} \le d_{i} \le \sigma _{1i}$ and $d_{i2} \le d_{i} \le \sigma _{3i}$

• (c) ${d_{i1}}^{'} \le d_{i} \le \sigma _{1i}$ and $\sigma _{4i} \le d_{i} \le {d_{i3}}^{'}$   (d) $\sigma _{2i} \le d_{i} \le d_{i2}$ and $d_{i2} \le d_{i} \le \sigma _{3i}$ i.e. $\sigma _{2i} \le d_{i} \le \sigma _{3i}$

• (e) $\sigma _{2i} \le d_{i} \le d_{i2}$ and $\sigma _{4i} \le d_{i} \le {d_{i3}}^{'}$    (f) $d_{i2} \le d_{i} \le \sigma _{3i}$ and $\sigma _{4i} \le d_{i} \le {d_{i3}}^{'}$

Figures 6(a), 6(b), 6(c), 6(d), 6(e), 6(f) show the graphical illustrations of possible intervals of $d_{i}$’s.

Therefore, the possible defuzzifying integrals are

• (a) $I(d_{i}) = \frac{1}{2} \int _{0}^{1}[{d_{i1}}^{'} + \sigma _{1i}]d\alpha + \frac{1}{2} \int _{0}^{1}[\sigma _{2i} + d_{i2}]d\alpha$;  (b) $I(d_{i}) = \frac{1}{2} \int _{0}^{1}[{d_{i1}}^{'} + \sigma _{1i}]d\alpha + \frac{1}{2}\int _{0}^{1}[d_{i2}+ \sigma _{3i}]d\alpha$

• (c) $I(d_{i}) = \frac{1}{2} \int _{0}^{1}[{d_{i1}}^{'} + \sigma _{1i}]d\alpha + \frac{1}{2}\int _{0}^{1}[\sigma _{4i} + {d_{i3}}^{'}]d\alpha$   (d) $I(d_{i}) = \frac{1}{2} \int _{0}^{1}[\sigma _{2i} + d_{i2}]d\alpha + \frac{1}{2}\int _{0}^{1}[d_{i2} + \sigma _{3i}]d\alpha$

• (e) $I(d_{i}) = \frac{1}{2} \int _{0}^{1}[\sigma _{2i} + d_{i2}]d\alpha + \frac{1}{2}\int _{0}^{1}[\sigma _{4i} + {d_{i3}}^{'}]d\alpha$    (f) $I(d_{i}) = \frac{1}{2} \int _{0}^{1}[d_{i2} + \sigma _{3i}]d\alpha + \frac{1}{2}\int _{0}^{1}[\sigma _{4i} + {d_{i3}}^{'}]d\alpha$

Let us take as limiting case: ${d_{i1}}^{'} \rightarrow {d_{i1}}$ and ${d_{i3}}^{'}\rightarrow {d_{i3}}$. We get

$$\begin{aligned} \begin{aligned} A^{'}&=\{(2{d_{i3}})^{2}+4d_{i2}(d_{i2}-2d_{i3})\} \rightarrow 4(d_{i3}-d_{i2})^{2},~\\&\quad B^{'}=2(d_{i2}-d_{i3})^{2}(d_{i2}+d_{i3}) \rightarrow 2(d_{i3}-d_{i2})^{2}(d_{i3}+d_{i2})\\ C^{'}&=\{(d_{i2}^{2}-2d_{i3}d_{i2}+d_{i3}^{2})-\alpha ^{2}(d_{i2}-d_{i3})(d_{i2}-d_{i3})\}^{2}\\&\quad +4(d_{i2}-d_{3})(d_{i2}-d_{i3})\\&\rightarrow (d_{i3}-d_{i2})^{2}[(1-{\alpha ^{2}}^{2})^{2}(d_{i3}-d_{i2})^{2}+4d_{i2}d_{i3}]\\ \end{aligned} \end{aligned}$$

D Crisp converging test for the proposed IFS

We take the limits of $d_{i1} \rightarrow d_{i2}$ and $d_{i3} \rightarrow d_{i2}$. So, $\frac{B^{'}}{A^{'}}\rightarrow \frac{2(d_{i3}-d_{i2})^{2}(d_{i3}+d_{i2})}{4(d_{i3}-d_{i2})^{2}} \rightarrow d_{i2}$

$$\begin{aligned} \begin{aligned}&\frac{\sqrt{{B^{'}}^{2}-A^{'}C^{'}}}{A^{'}} \rightarrow \frac{\sqrt{4(d_{i3}-d_{i2})^{4}(d_{i3}+d_{i2})^{2}}-4(d_{i3}-d_{i2})^{2}(d_{i3}-d_{i2})^{2} [(1-{\alpha ^{2}}^{2})^{2}(d_{i3}-d_{i2})^{2}+4d_{i2}d_{i3}}{4(d_{i3}-d_{i2})^{2}}\\&\quad \rightarrow \frac{\sqrt{(d_{i3}+d_{i2})^{2}}-[(1-{\alpha ^{2}}^{2})^{2}(d_{i3}-d_{i2})^{2}+4d_{i2}d_{i3}}{2} \rightarrow \frac{\sqrt{(2d_{i2})^{2}-(4d_{i2}d_{i2})}}{2} \rightarrow 0 \end{aligned} \end{aligned}$$

Similarly, $\frac{B}{A} \rightarrow d_{i2}, ~\text{ and }~\frac{\sqrt{{B}^{2}-AC}}{A} \rightarrow \frac{\sqrt{(2d_{i2})^{2}-(4d_{i2}d_{i2})}}{2} \rightarrow 0$

Therefore, $\sigma _{1i} \rightarrow \sigma _{2i} \rightarrow \sigma _{3i} \rightarrow \sigma _{4i} \rightarrow d_{i2}$. Hence, the possible defuzzifying limiting integrals are

• 1. $I(d_{i}) = \frac{1}{2} \int _{0}^{1}[{d_{i1}}^{'} + \sigma _{1i}]d\alpha + \frac{1}{2} \int _{0}^{1}[\sigma _{2i} + d_{i2}]d\alpha \rightarrow 2d_{i2}$

• 2. $I(d_{i}) = \frac{1}{2} \int _{0}^{1}[{d_{i1}}^{'} + \sigma _{1i}]d\alpha + \frac{1}{2}\int _{0}^{1}[d_{i2}+ \sigma _{3i}]d\alpha \rightarrow 2d_{i2}$

• 3. $I(d_{i}) = \frac{1}{2} \int _{0}^{1}[{d_{i1}}^{'} + \sigma _{1i}]d\alpha + \frac{1}{2}\int _{0}^{1}[\sigma _{4i} + {d_{i3}}^{'}]d\alpha \rightarrow 2d_{i2}$

• 4. $I(d_{i}) = \frac{1}{2} \int _{0}^{1}[\sigma _{2i} + d_{i2}]d\alpha + \frac{1}{2}\int _{0}^{1}[d_{i2} + \sigma _{3i}]d\alpha \rightarrow d_{i2}$

• 5. $I(d_{i}) = \frac{1}{2} \int _{0}^{1}[\sigma _{2i} + d_{i2}]d\alpha + \frac{1}{2}\int _{0}^{1}[\sigma _{4i} + {d_{i3}}^{'}]d\alpha \rightarrow 2d_{i2}$

• 6. $I(d_{i}) = \frac{1}{2} \int _{0}^{1}[d_{i2} + \sigma _{3i}]d\alpha + \frac{1}{2}\int _{0}^{1}[\sigma _{4i} + {d_{i3}}^{'}]d\alpha \rightarrow 2d_{i2}$

Thus we accept the integral $I(d_{i}) = \frac{1}{2} \int _{0}^{1}[\sigma _{2i} + d_{i2}]d\alpha + \frac{1}{2}\int _{0}^{1}[d_{i2} + \sigma _{3i}]d\alpha \rightarrow d_{i2}$

E Crisp converging test for PFS

Here we shall take the index values of the proposed PFS parameters in terms of $\delta _{ij}$’s and $\lambda _{ij}$’s defined in section (8). Let us take the index value for the fuzzy parameters ${\tilde{K}}$ as

$$\begin{aligned} \begin{aligned} I(K)&=\frac{\lambda ^{'}}{2}\int _{0}^{1}\Big [\left( \delta _{16}+\frac{\alpha }{\lambda _{16}}\right) ^{\lambda } +\left( \delta _{36}-\frac{\alpha }{\lambda _{26}}\right) ^{\lambda }\Big ]d\alpha \\&=\frac{\lambda ^{'}}{2(1+\lambda )}\bigg [\lambda _{16}\Big \{\left( \delta _{16}+\frac{1}{\lambda _{16}}\right) ^{1+\lambda } -\delta _{16}^{1+\lambda }\Big \}\\&\quad -\lambda _{26}\Big \{(\delta _{36} -\frac{1}{\lambda _{26}})^{1+\lambda }-\delta _{36}^{1+\lambda }\Big \}\bigg ] \end{aligned} \end{aligned}$$

Now, as $p_{1}^{'}\rightarrow p_{1}$ and $p_{3}^{'}\rightarrow p_{3}$ we have

$$\begin{aligned} \begin{aligned} I(K)&\rightarrow \frac{\lambda ^{'}}{2(1+\lambda )}\bigg [\frac{2}{p_{2}-p_{1}}\Big \{p_{2}^{1+\lambda } -\left( \frac{p_{1}+p_{2}}{2}\right) ^{1+\lambda }\Big \}\\&\quad -\frac{2}{p_{3}-p_{2}}\Big \{p_{2}^{1+\lambda } -\left( \frac{p_{2}+p_{3}}{2}\right) ^{1+\lambda }\Big \}\bigg ] \end{aligned} \end{aligned}$$

Also for $p_{1}\rightarrow p_{2}$ and $p_{3}\rightarrow p_{2}$, we get $I(K) = \lambda ^{'}p_{2}^{\lambda } \Rightarrow$ crisp supply rate Again, let us take the index value of the fuzzy demand rate as

$$\begin{aligned} \begin{aligned} I(D)&=\frac{d_{0}E^{\delta }}{2}\int _{0}^{1}\Big [\left( \delta _{37}-\frac{\alpha }{\lambda _{27}}\right) ^{-\lambda } +\left( \delta _{17}+\frac{\alpha }{\lambda _{17}}\right) ^{-\lambda }\Big ]d\alpha \\&=\frac{d_{0}E^{\delta }}{2(1-\lambda )}\bigg [\lambda _{17}\Big \{\left( \delta _{17} +\frac{1}{\lambda _{17}}\right) ^{1-\lambda }-\delta _{17}^{1-\lambda }\Big \}\\&\quad -\lambda _{27}\Big \{\left( \delta _{37}-\frac{1}{\lambda _{27}}\right) ^{1-\lambda }-\delta _{37}^{1-\lambda }\Big \}\bigg ] \end{aligned} \end{aligned}$$

Thus for $s_{1}^{'}\rightarrow s_{1}$ and $s_{3}^{'}\rightarrow s_{3}$ we have

$$\begin{aligned} \begin{aligned} I(D) =&\frac{d_{0}E^{\delta }}{(1-\lambda )}\bigg [\frac{1}{s_{2}-s_{1}}\Big \{1-\left( \frac{s_{1}+s_{2}}{2s_{2}}\right) ^{1-\lambda }\Big \}\\&-\frac{1}{s_{3}-s_{2}}\Big \{1-\left( \frac{s_{2}+s_{3}}{2s_{2}}\right) ^{1-\lambda }\Big \}\bigg ] \end{aligned} \end{aligned}$$

Also for $s_{1}\rightarrow s_{2}$ and $s_{3}\rightarrow s_{2}$, $I(D) \Rightarrow d_{0}E^{\delta }s_{2}^{-\lambda } \Rightarrow$ crisp demand rate. However, for the objective function the index value is given by

$$\begin{aligned} \begin{aligned} I(z)&=\frac{1}{2T}\Big [\delta _{11}+\delta _{31}+\frac{1}{2}\big (\frac{1}{\lambda _{11}}-\frac{1}{\lambda _{21}}\big )\Big ]\\&\quad +\frac{d_{0}E^{\delta }}{2T}\int _{0}^{1}[\Big (\frac{\alpha }{\lambda _{17}} +\delta _{17}\Big )^{-\lambda } \left( \delta _{35}-\frac{\alpha }{\lambda _{25}}\right) \\&\quad +\Big (\delta _{37}-\frac{\alpha }{\lambda _{27}}\Big )^{-\lambda }\left( \frac{\alpha }{\lambda _{15}}+\delta _{15}\right) \Big ]d\alpha \\&\quad +\frac{d_{0}E^{\delta }}{T}\Big \{\frac{(1+m)^2}{2T}log\left( \frac{1+m}{1+m-T}\right) +\frac{T}{4} - \frac{(1+m)}{2}\Big \}\\&\quad \int _{0}^{1}\Big [\left( \frac{\alpha }{\lambda _{12}}+\delta _{12}\right) \Big (\delta _{37}-\frac{\alpha }{\lambda _{27}}\Big )^{-\lambda } +\left( \delta _{32}-\frac{\alpha }{\lambda _{22}}\right) \Big (\frac{\alpha }{\lambda _{17}}\\&\quad +\delta _{17}\Big )^{-\lambda } \Big ]d\alpha +\frac{\lambda ^{'}}{2T}\Big [\left( \frac{\alpha }{\lambda _{16}}+\delta _{16}\right) ^{\lambda +1}\\&\quad +\left( \delta _{36}-\frac{\alpha }{\lambda _{26}}\right) ^{\lambda +1}\Big ] +\frac{\lambda ^{'}T}{4}\int _{0}^{1}\Big [\left( \frac{\alpha }{\lambda _{13}} +\delta _{13}\right) \left( \frac{\alpha }{\lambda _{16}}+\delta _{16}\right) ^{\lambda }\\&\quad +\left( \delta _{33}-\frac{\alpha }{\lambda _{23}}\right) \left( \delta _{36}-\frac{\alpha }{\lambda _{26}}\right) ^{\lambda }\Big ]d\alpha \\&\quad +\frac{d_{0}E^{\delta }}{2}\Big \{\frac{1+m}{T} log\left( \frac{1+m}{1+m-T}\right) -1 \Big \}\\&\quad \int _{0}^{1}\Big [\left( \frac{\alpha }{\lambda _{14}}+\delta _{14}\right) \Big (\delta _{37} -\frac{\alpha }{\lambda _{27}}\Big )^{-\lambda }\\&\quad +\left( \delta _{34}-\frac{\alpha }{\lambda _{24}}\right) \Big (\frac{\alpha }{\lambda _{17}}+\delta _{17}\Big )^{-\lambda } \Big ]d\alpha \end{aligned} \end{aligned}$$

Now, 1st term$= \frac{1}{2T}\Big [\delta _{11}+\delta _{31}+\frac{1}{2}\big (\frac{1}{\lambda _{11}}-\frac{1}{\lambda _{21}}\big )\Big ]$. As $Cs_{1}^{'}\rightarrow Cs_{1}$ and $Cs_{3}^{'}\rightarrow Cs_{3}$, 1st term $\rightarrow \frac{Cs_{1}+6Cs_{2}+Cs_{3}}{8T} \rightarrow \frac{Cs_{2}}{T}$ as $Cs_{1}\rightarrow Cs_{2}$, $Cs_{3}\rightarrow Cs_{2}$. 2nd term$=\frac{d_{0}E^{\delta }}{2T}\int _{0}^{1}[\Big (\frac{\alpha }{\lambda _{17}}+\delta _{17}\Big )^{-\lambda } (\delta _{35}-\frac{\alpha }{\lambda _{25}})+\Big (\delta _{37}-\frac{\alpha }{\lambda _{27}}\Big )^{-\lambda } (\frac{\alpha }{\lambda _{15}}+\delta _{15})\Big ]d\alpha$

$$\begin{aligned} \begin{aligned}&=\frac{d_{0}E^{\delta \lambda _{27}}}{2T}\bigg [-\frac{\delta _{15}}{1-\lambda }\{\Big (\delta _{37} -\frac{1}{\lambda _{27}}\Big )^{1-\lambda }-\delta _{37}^{1-\lambda }\}\\&\quad +\frac{\lambda _{27}}{\lambda _{15}} \{\frac{\delta _{37}}{1-\lambda }\left[ \Big (\delta _{37}-\frac{1}{\lambda _{27}}\Big )^{1-\lambda }-\delta _{37}^{1-\lambda }\right] \\&\quad -\frac{1}{2-\lambda }\left[ \Big (\delta _{37}-\frac{1}{\lambda _{27}}\Big )^{2-\lambda }\right. \\&\quad \left. -\delta _{37}^{2-\lambda }\right] \}\bigg ]+\frac{d_{0}E^{\delta \lambda _{17}}}{2T}\bigg [\frac{\delta _{35}}{1-\lambda } \Big \{(\Big (\frac{1}{\lambda _{17}}\\&\quad +\delta _{17}\Big )^{1-\lambda } -\delta _{17}^{1-\lambda }\Big \}-\frac{\lambda _{17}}{\lambda _{25}(2-\lambda )}\Big \{\Big (\frac{1}{\lambda _{17}} +\delta _{17}\Big )^{2-\lambda }-\delta _{17}^{2-\lambda }\Big \}\\&\quad +\frac{\lambda _{17}\delta _{17}}{\lambda _{25}(1-\lambda )} \Big \{\Big (\frac{1}{\lambda _{17}}+\delta _{17}\Big )^{1-\lambda }-\delta _{17}^{1-\lambda }\Big \} \bigg ] \end{aligned} \end{aligned}$$

As $c_{11}^{'}\rightarrow c_{11}$, $c_{13}^{'}\rightarrow c_{13}$, $s_{1}^{'}\rightarrow s_{1}$ and $s_{3}^{'}\rightarrow s_{3}$ 2nd terms becomes

$$\begin{aligned} \begin{aligned}&\frac{d_{0}E^{\delta }}{(s_{3}-s_{2})T}\bigg [\frac{c_{12}-c_{11}}{s_{3}-s_{2}}\Big \{\frac{s_{3}+s_{2}}{2(1-\lambda )}\Big (s_{2}^{1-\lambda }-(\frac{s_{2}+s_{3}}{2})^{1-\lambda }\Big )\\&\quad -\frac{1}{2-\lambda } \Big (s_{2}^{2-\lambda }-\left( \frac{s_{2}+s_{3}}{2}\right) ^{2-\lambda }\Big )\Big \}\\&\quad -\frac{c_{11}+c_{12}}{2(1-\lambda )}\Big (s_{2}^{1-\lambda }-\left( \frac{s_{2}+s_{3}}{2}\right) ^{1-\lambda }\Big )\bigg ]\\&\quad +\frac{d_{0}E^{\delta }}{(s_{2}-s_{1})T}\bigg [\frac{c_{13}+c_{12}}{2(1-\lambda )} \Big \{\Big (s_{2}^{1-\lambda }-\left( \frac{s_{2}+s_{1}}{2}\right) ^{1-\lambda }\Big )\\&\quad -\frac{c_{13}-c_{12}}{(s_{2}-s_{1})(2-\lambda )}\Big (s_{2}^{2-\lambda } -\left( \frac{s_{2}+s_{1}}{2}\right) ^{2-\lambda }\Big )\Big \}\\&\quad -\frac{(c_{13}-c_{12})(s_{2}+s_{1})}{(1-\lambda )(s_{2}-s_{1})} \Big (s_{2}^{1-\lambda }-\left( \frac{s_{2}+s_{1}}{2}\right) ^{1-\lambda }\Big )\bigg ] \end{aligned} \end{aligned}$$

Then $c_{11}\rightarrow c_{12}$, $c_{13}\rightarrow c_{12}$, $s_{1}\rightarrow s_{2}$ and $s_{3}\rightarrow s_{2}$, using L-Hospital rule we get the 2nd term $\rightarrow \frac{d_{0}E^{\delta }}{T} c_{12}s_{2}^{-\lambda }$.

4th term$=\frac{\lambda ^{'}}{2T}\Big [(\frac{\alpha }{\lambda _{16}}+\delta _{16})^{\lambda +1}+(\delta _{36} -\frac{\alpha }{\lambda _{26}})^{\lambda +1}\Big ]$. As $p_{1}^{'}\rightarrow p_{1}$ and $p_{3}^{'}\rightarrow p_{3}$,

$$\begin{aligned} \begin{aligned} \text{4th } \text{ term } =&\frac{\lambda ^{'}p_{2}^{2+\lambda }}{T(2+\lambda )}\bigg [\frac{1}{p_{2} -p_{1}}\Big \{1-\left( \frac{p_{1}+p_{2}}{2p_{2}}\right) ^{2+\lambda }\Big \}\\&-\frac{1}{p_{3} -p_{2}}\Big \{1-\left( \frac{p_{2}+p_{3}}{2p_{2}}\right) ^{2+\lambda }\Big \}\bigg ] \end{aligned} \end{aligned}$$

4th term $\rightarrow \frac{\lambda ^{'}}{T}p_{2}^{1+\lambda }$ = crisp value, as $p_{1}\rightarrow p_{2}$, $p_{3}\rightarrow p_{2}$ by using L-Hospital rule.

5th term$=\frac{\lambda ^{'}T}{4}\int _{0}^{1}\Big [(\frac{\alpha }{\lambda _{13}} +\delta _{13})(\frac{\alpha }{\lambda _{16}}+\delta _{16})^{\lambda }+(\delta _{33} -\frac{\alpha }{\lambda _{23}})(\delta _{36}-\frac{\alpha }{\lambda _{26}})^{\lambda }\Big ]d\alpha$ . As $p_{1}^{'}\rightarrow p_{1}$, $p_{3}^{'}\rightarrow p_{3}$, $h_{s1}^{'}\rightarrow h_{s1}$ and $h_{s3}^{'}\rightarrow h_{s3}$

$$\begin{aligned} \begin{aligned} \text{5th } \text{ term }=&\frac{\lambda ^{'}Th_{s2}p_{2}^{1+\lambda }}{2(1+\lambda )}\bigg [\frac{1}{p_{2}-p_{1}} \Big \{1-\left( \frac{p_{1}+p_{2}}{2p_{2}}\right) ^{1+\lambda }\Big \}\\&-\frac{1}{p_{3}-p_{2}}\Big \{1 -\left( \frac{p_{2}+p_{3}}{2p_{2}}\right) ^{1+\lambda }\Big \}\bigg ] \end{aligned} \end{aligned}$$

5th term $\rightarrow \lambda ^{'}Th_{s2}p_{2}^{1+\lambda }$ = crisp value, as $p_{1}\rightarrow p_{2}$, $p_{3}\rightarrow p_{2}$, $h_{s1}\rightarrow h_{s2}$ and $h_{s3}\rightarrow h_{s2}$ by using L-Hospital rule.

Similarly, the 3rd and 6th term tends to a crisp value. Therefore, I(z) becomes a crisp value.