Bi-level Programming for Stackelberg Game with Intuitionistic Fuzzy Number: a Ranking Approach

Abstract

This paper introduces a ranking function procedure on a bi-level programming for Stackelberg game involving intuitionistic fuzzy parameters. Intuitionistic fuzzy number is considered in many real-life situations, so it makes perfect sense to address decision-making problem by using some specified intuitionistic fuzzy numbers. In this paper, intuitionistic fuzziness is characterized by a normal generalized triangular intuitionistic fuzzy number. A defuzzification method is introduced based on the proportional probability density function associated with the corresponding membership function, as well as the complement of non-membership function. Using the proposed ranking technique, a methodology is presented for solving bi-level programming for Stackelberg game. An application example is provided to demonstrate the applicability of the proposed methodology, and the achieved results are compared with the existing methods.

Appendices

Appendix

A detailed calculation to show $s_{1}=\frac{2}{(a_{2}-a_{1})}$ is given below:

$$\begin{aligned}&s_{1}\left[ \int _{a_{1}}^{a}\frac{(x-a_{1})}{(a-a_{1})}\hbox {d}x+\int _{a}^{a_{2}}\frac{(a_{2}-x)}{(a_{2}-a)}\hbox {d}x\right] =1,\\&\hbox {i.e.,}~~ ~s_{1}\left[ \frac{1}{(a-a_{1})}\int _{a_{1}}^{a}(x-a_{1})\hbox {d}x+\frac{1}{(a_{2}-a)}\int _{a}^{a_{2}}(a_{2}-x)\hbox {d}x\right] =1,\\&~~ ~s_{1}\left[ \frac{1}{(a-a_{1})}\left[ \frac{x^{2}}{2}-a_{1}x\right] _{a_{1}}^{a}+\frac{1}{(a_{2}-a)}\left[ a_{2}x-\frac{x^{2}}{2}\right] _{a}^{a_{2}}\right] =1,\\&~~ ~s_{1}\left[ \frac{1}{(a-a_{1})}\left[ \frac{a^{2}}{2}-aa_{1}+\frac{a_{1}^{2}}{2}\right] +\frac{1}{(a_{2}-a)}\left[ \frac{a_{2}^{2}}{2}-aa_{2}+\frac{a^{2}}{2}\right] \right] =1,\\&~~ ~s_{1}\left[ \frac{1}{2(a-a_{1})}\left[ a-a_{1}\right] ^{2}+\frac{1}{2(a_{2}-a)}\left[ a_{2}-a\right] ^{2}\right] =1,\\&~~ ~\frac{s_{1}}{2}\left[ \left[ a-a_{1}\right] +\left[ a_{2}-a\right] \right] =1,\\&\hbox {therefore}~~ s_{1}=\frac{2}{(a_{2}-a_{1})}.\\ \end{aligned}$$

Appendix

Another calculation is shown for $s_{2}$ as follows:

$$\begin{aligned}&s_{2}\left[ \int _{a_{3}}^{a}\frac{(x-a_{3})}{(a-a_{3})}\hbox {d}x+\int _{a}^{a_{4}}\frac{(a_{4}-x)}{(a_{4}-a)}\hbox {d}x\right] =1,\\&\hbox {i.e.,}~~ ~s_{2}\left[ \frac{1}{(a-a_{3})}\int _{a_{3}}^{a}(x-a_{3})\hbox {d}x+\frac{1}{(a_{4}-a)}\int _{a}^{a_{4}}(a_{4}-x)\hbox {d}x\right] =1,\\&~~ ~s_{2}\left[ \frac{1}{(a-a_{3})}\left[ \frac{x^{2}}{2}-a_{3}x\right] _{a_{3}}^{a}+\frac{1}{(a_{4}-a)}\left[ a_{4}x-\frac{x^{2}}{2}\right] _{a}^{a_{4}}\right] =1,\\&~~ ~s_{2}\left[ \frac{1}{(a-a_{3})}\left[ \frac{a^{2}}{2}-aa_{3}+\frac{a_{3}^{2}}{2}\right] +\frac{1}{(a_{4}-a)}\left[ \frac{a_{4}^{2}}{2}-aa_{4}+\frac{a^{2}}{2}\right] \right] =1,\\&~~ ~s_{2}\left[ \frac{1}{2(a-a_{3})}\left[ a-a_{3}\right] ^{2}+\frac{1}{2(a_{4}-a)}\left[ a_{4}-a\right] ^{2}\right] =1,\\&~~ ~\frac{s_{2}}{2}\left[ \left[ a-a_{3}\right] +\left[ a_{4}-a\right] \right] =1,\\&\hbox {therefore}~~ s_{2}=\frac{2}{(a_{4}-a_{3})}. \end{aligned}$$

Appendix

Again detailed calculation is presented for $M_{X}(2)$

$$\begin{aligned} M_{X}(t)= & {} \int _{0}^{\infty }x^{t-1}\left[ \lambda h_{1}(x)+(1-\lambda )h_{2}(x)\right] \hbox {d}x\\= & {} 2\lambda \left[ \int _{a_{1}}^{a}x^{t-1}\frac{(x-a_{1})}{(a_{2}-a_{1})(a-a_{1})}\hbox {d}x+ \int _{a}^{a_{2}}x^{t-1}\frac{(a_{2}-x)}{(a_{2}-a_{1})(a_{2}-a)}\hbox {d}x\right] \\&+2(1-\lambda )\left[ \int _{a_{3}}^{a}x^{t-1}\times \frac{(x-a_{3})}{(a_{4}-a_{3})(a-a_{3})}\hbox {d}x+\int _{a}^{a_{4}}x^{t-1} \frac{(a_{4}-x)}{(a_{4}-a_{3})(a_{4}-a)}\hbox {d}x\right] \\= & {} 2\lambda \left[ \frac{1}{(a_{2}-a_{1})(a-a_{1})}\left[ \frac{x^{t+1}}{(t+1)}- \frac{a_{1}x^{t}}{t}\right] _{a_{1}}^{a}+\frac{1}{(a_{2}-a_{1})(a_{2}-a)}\times \left[ \frac{a_{2}x^{t}}{t}-\frac{x^{t+1}}{(t+1)}\right] _{a}^{a_{2}}\right] \\&+2(1-\lambda )\times \left[ \frac{1}{(a_{4}-a_{3})(a-a_{3})} \left[ \frac{x^{t+1}}{(t+1)}-\frac{a_{3}x^{t}}{t}\right] _{a_{1}}^{a} \right. \\&\left. +\frac{1}{(a_{4}-a_{3})(a_{4}-a)} \left[ \frac{a_{4}x^{t}}{t}-\frac{x^{t+1}}{(t+1)}\right] _{a}^{a_{4}}\right] \\= & {} 2\lambda \left[ \frac{1}{(a_{2}-a_{1})(a-a_{1})} \left[ \frac{\left[ a^{t+1}-a_{1}^{t+1}\right] }{(t+1)}- \frac{a_{1}\left[ a^{t}-a_{1}^{t}\right] }{t}\right] \right. \\&\left. + \frac{1}{(a_{2}-a_{1})(a_{2}-a)}\left[ \frac{a_{2} \left[ a_{2}^{t}-a^{t}\right] }{t} -\frac{\left[ a_{2}^{t+1}-a^{t+1}\right] }{(t+1)}\right] \right] \\&+2(1-\lambda )\left[ \frac{1}{(a_{4}-a_{3})(a-a_{3})} \left[ \frac{\left[ a^{t+1}-a_{3}^{t+1}\right] }{(t+1)}- \frac{a_{3}\left[ a^{t}-a_{3}^{t}\right] }{t}\right] \right. \\&\left. + \frac{1}{(a_{4}-a_{3})(a_{4}-a)}\left[ \frac{a_{4}\left[ a_{4}^{t}-a^{t}\right] }{t}- \frac{\left[ a_{4}^{t+1}-a^{t+1}\right] }{(t+1)}\right] \right] . \end{aligned}$$

$$\begin{aligned} \hbox {Therefore}\\ M_{X}(2)= & {} 2\lambda \left[ \frac{1}{(a_{2}-a_{1})(a-a_{1})} \left[ \frac{\left[ a^{3}-a_{1}^{3}\right] }{3}-\frac{a_{1} \left[ a^{2}-a_{1}^{2}\right] }{2}\right] \right. \\&\left. +\frac{1}{(a_{2}-a_{1})(a_{2}-a)} \left[ \frac{a_{2}\left[ a_{2}^{2}-a^{2}\right] }{2} -\frac{\left[ a_{2}^{3}-a^{3}\right] }{3}\right] \right] \\&+2(1-\lambda )\left[ \frac{1}{(a_{4}-a_{3})(a-a_{3})} \left[ \frac{\left[ a^{3}-a_{3}^{3}\right] }{3}-\frac{a_{3} \left[ a^{2}-a_{3}^{2}\right] }{2}\right] \right. \\&\left. +\frac{1}{(a_{4}-a_{3})(a_{4}-a)}\left[ \frac{a_{4}\left[ a_{4}^{2}-a^{2}\right] }{2}-\frac{\left[ a_{4}^{3}-a^{3}\right] }{3}\right] \right] \\= & {} 2\lambda \left[ \frac{1}{(a_{2}-a_{1})}\left[ \frac{(2a^{2}-aa_{1}-a_{1}^{2})}{6}+\frac{a_{2}^{2}+aa_{2}-2a^{2}}{6}\right] \right] \\&+2(1-\lambda )\left[ \frac{1}{(a_{4}-a_{3})}\left[ \frac{(2a^{2}-aa_{3}-a_{3}^{2})}{6}+ \frac{(a_{4}^{2}+aa_{4}-2a^{2})}{6}\right] \right] \\= & {} 2\lambda \left[ \frac{(a_{2}-a_{1})(a+a_{1}+a_{2})}{6(a_{2}-a_{1})}\right] +2(1-\lambda )\left[ \frac{(a_{4}-a_{3})(a+a_{3}+a_{4})}{6(a_{4}-a_{3})}\right] \\= & {} \frac{(a+a_{3}+a_{4})+\lambda (a_{1}+a_{2}-a_{3}-a_{4})}{3}. \end{aligned}$$