Novel exponential fuzzy information measures

Abstract

Knowledge is basically a piece of information considered in a particular useful context under consideration. A knowledge measure as a dual of fuzzy entropy quantifies the knowledge associated with a fuzzy set. This communication is intended to introduce a novel exponential fuzzy knowledge measure. Furthermore, a new decision-making method using the VIKOR concept based on proposed knowledge measure is put forward. The working of proposed decision-making method is explained through two numerical examples. Considering the importance of criteria weights in decision-making, two ways have been discussed for their evaluation. Besides, three new measures, namely an exponential fuzzy accuracy measure, an exponential fuzzy entropy measure and an exponential fuzzy similarity measure, have been introduced.

Appendices

Appendix ‘A’

Proof of Theorem 3.2

• 1. Consider the function

  $$\begin{aligned} \digamma (\beta )=\frac{\beta e^{(1-\beta )}+(1-\beta ) e^\beta -\sqrt{e}}{1-\sqrt{e}}; ~ 0\le \beta \le 1. \end{aligned}$$

  (6.1)

  Since the function (6.1) is continuous on closed interval [0,1], therefore, it will attain its bounds on [0,1]. To determine the bounds, differentiating (6.1) with respect to $\beta$, we get

  $$\begin{aligned} \frac{d\digamma (\beta )}{d\beta }=\frac{(1-\beta ) e^{1-\beta }-\beta e^\beta }{1-\sqrt{e}}. \end{aligned}$$

  (6.2)

  Substituting (6.2) equals to zero and using the fact that $(1-\sqrt{e})\ne 0$, we have

  $$\begin{aligned} (1-\beta ) e^{(1-\beta )}=\beta e^\beta . \end{aligned}$$

  (6.3)

  Since (6.3) holds at  $\beta =\frac{1}{2}$, therefore,  $\beta =\frac{1}{2}$ is its stationary point. To check the nature of point  $\beta =\frac{1}{2}$, differentiating (6.2) with respect to $\beta$, we get

  $$\begin{aligned} \frac{d^2\digamma (\beta )}{d\beta ^2}=\frac{(\beta -2)e^{1-\beta }-e^\beta (1+\beta )}{1-\sqrt{e}}. \end{aligned}$$

  (6.4)

  At $\beta =\frac{1}{2}$, $\displaystyle \frac{d^2\digamma (\beta )}{d\beta ^2}>0$. This implies that the point $\beta =\frac{1}{2}$ is the point of minima. Also,

  $$\begin{aligned} \left[ \frac{d\digamma (\beta )}{d\beta }\right] _{\beta =0}<0 ~\text {and} ~ \left[ \frac{d\digamma (\beta )}{d\beta }\right] _{\beta =1}>0. \end{aligned}$$

  (6.5)

  Thus, the above discussion implies that the function defined by (6.1) is decreasing in the interval $[0, \frac{1}{2}]$ and increasing in the interval $[\frac{1}{2}, 1]$. Also, the function (6.1) being continuous attains its bounds in the interval [0,1]. Therefore, maximum value of (6.1) takes place at points $\beta =0, 1$ and minimum value at $\beta =\frac{1}{2}$. At $\beta =0, 1$, the value of (6.1) equals to 1 and at $\beta =\frac{1}{2}$, the value of (6.1) is zero. Therefore, $0\le {\mathbb {K}}(\Re )\le 1$.

• 2. Proof is obvious by taking $\mu _\Re ^c (\curlyvee _i)=1-\mu _\Re (\curlyvee _i)$ in (3.7).

• 3. Bifurcate the universe of discourse $\sqcup$ in two parts as follows:

  $$\begin{aligned} \sqcup _1=\{\curlyvee _i\in \sqcup |\mu _{\Re _1} (\curlyvee _i)\ge \mu _{\Re _2}(\curlyvee _i)\};\nonumber \\ \sqcup _2=\{\curlyvee _i\in \sqcup |\mu _{\Re _1} (\curlyvee _i)<\mu _{\Re _2}(\curlyvee _i)\}. \end{aligned}$$

  (6.6)

  If $\curlyvee _i\in \sqcup _1$, then

  $$\begin{aligned} \mu _{\Re _1\cup \Re _2}(\curlyvee _i)=\max \{\mu _{\Re _1}(\curlyvee _i),\mu _{\Re _2}(\curlyvee _i)\}=\mu _{\Re _1}(\curlyvee _i);\nonumber \\ \mu _{\Re _1\cap \Re _2}(\curlyvee _i)=\min \{\mu _{\Re _1}(\curlyvee _i),\mu _{\Re _2}(\curlyvee _i)\}=\mu _{\Re _2}(\curlyvee _i). \end{aligned}$$

  (6.7)

  and if $\curlyvee _i\in \sqcup _2$, then

  $$\begin{aligned} \mu _{\Re _1\cup \Re _2}(\curlyvee _i)=\max \{\mu _{\Re _1}(\curlyvee _i),\mu _{\Re _2}(\curlyvee _i)\}=\mu _{\Re _2}(\curlyvee _i);\nonumber \\ \mu _{\Re _1\cap \Re _2}(\curlyvee _i)=\min \{\mu _{\Re _1}(\curlyvee _i),\mu _{\Re _2}(\curlyvee _i)\}=\mu _{\Re _1}(\curlyvee _i). \end{aligned}$$

  (6.8)

  Consider for all $\curlyvee _i\in \sqcup$,

  $$\begin{aligned} {\mathbb {K}}(\Re _1\cup \Re _2)+{\mathbb {K}}(\Re _1\cap \Re _2)&=\frac{1}{m}\sum _{i=1}^m\left( \frac{\mu _{\Re _1\cup \Re _2} (\curlyvee _i)e^{1-\mu _{\Re _1\cup \Re _2} (\curlyvee _i)}+(1-\mu _{\Re _1\cup \Re _2} (\curlyvee _i))e^{\mu _{\Re _1\cup \Re _2} (\curlyvee _i)}-\sqrt{e}}{1-\sqrt{e}}\right) \nonumber \\&\quad +\frac{1}{m}\sum _{i=1}^m\left( \frac{\mu _{\Re _1\cap \Re _2} (\curlyvee _i)e^{1-\mu _{\Re _1\cap \Re _2} (\curlyvee _i)}+(1-\mu _{\Re _1\cap \Re _2} (\curlyvee _i))e^{\mu _{\Re _1\cap \Re _2} (\curlyvee _i)}-\sqrt{e}}{1-\sqrt{e}}\right) \end{aligned}$$

  (6.9)

  This gives

  $$\begin{aligned}&{\mathbb {K}}(\Re _1\cup \Re _2)+{\mathbb {K}}(\Re _1\cap \Re _2)\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \nonumber \\&\quad =\frac{1}{m}\sum _{\sqcup _1}\left( \frac{\mu _{\Re _1} (\curlyvee _i)e^{1-\mu _{\Re _1} (\curlyvee _i)}+(1-\mu _{\Re _1} (\curlyvee _i))e^{\mu _{\Re _1} (\curlyvee _i)}-\sqrt{e}}{1-\sqrt{e}}\right) \nonumber \\&\quad +\frac{1}{m}\sum _{\sqcup _1}\left( \frac{\mu _{\Re _2} (\curlyvee _i)e^{1-\mu _{\Re _2} (\curlyvee _i)}+(1-\mu _{\Re _2} (\curlyvee _i))e^{\mu _{\Re _2} (\curlyvee _i)}-\sqrt{e}}{1-\sqrt{e}}\right) \nonumber \\&\quad +\frac{1}{m}\sum _{\sqcup _2}\left( \frac{\mu _{\Re _2} (\curlyvee _i)e^{1-\mu _{\Re _2} (\curlyvee _i)}+(1-\mu _{\Re _2} (\curlyvee _i))e^{\mu _{\Re _2} (\curlyvee _i)}-\sqrt{e}}{1-\sqrt{e}}\right) \nonumber \\&\quad +\frac{1}{m}\sum _{\sqcup _2}\left( \frac{\mu _{\Re _1} (\curlyvee _i)e^{1-\mu _{\Re _1} (\curlyvee _i)}+(1-\mu _{\Re _1} (\curlyvee _i))e^{\mu _{\Re _1} (\curlyvee _i)}-\sqrt{e}}{1-\sqrt{e}}\right) \end{aligned}$$

  (6.10)

  On solving, we have

  $$\begin{aligned} {\mathbb {K}}(\Re _1\cup \Re _2)+{\mathbb {K}}(\Re _1\cap \Re _2)={\mathbb {K}}(\Re _1)+{\mathbb {K}}(\Re _2). \end{aligned}$$

  (6.11)

$\square$

Appendix ‘B’

• 1. Let $\Re _1, \Re _2$ be two FSs such that  $\Re _1=\Re _2$. Therefore, (5.6) becomes

  $$\begin{aligned} {\mathbb {K}}(\Re _1,\Re _1)&=\frac{1}{m}\sum _{i=1}^m\frac{\mu _{\Re _1}(\curlyvee _i) e^{1-\mu _{\Re _1}(\curlyvee _i)}+(1-\mu _{\Re _1}(\curlyvee _i))e^{\mu _{\Re _1}(\curlyvee _i)}-\sqrt{e}}{2(1-\sqrt{e})}\nonumber \\&\quad +\frac{1}{m}\sum _{i=1}^m\frac{\sqrt{\mu _{\Re _1}(\curlyvee _i)\mu _{\Re _1}(\curlyvee _i)} e^{1-\sqrt{\mu _{\Re _1}(\curlyvee _i)\mu _{\Re _1}(\curlyvee _i)}}+(1-\sqrt{\mu _{\Re _1}(\curlyvee _i)\mu _{\Re _1}(\curlyvee _i)}) e^{\sqrt{\mu _{\Re _1}(\curlyvee _i)\mu _{\Re _1}(\curlyvee _i)}}-\sqrt{e}}{2(1-\sqrt{e})}\nonumber \\ \end{aligned}$$

  (6.12)

  If we consider  $\Re _1,\Re _2$ to be crisp sets, then $\mu _{\Re _1} (\curlyvee _i)=\mu _{\Re _2}(\curlyvee _i)=0 ~\text {or}~ 1$. Therefore, (6.12) gives

  $$\begin{aligned} {\mathbb {K}}^{acu} (\Re _1,\Re _1)={\mathbb {K}}^{acu} (\Re _1)=1; \end{aligned}$$

  (6.13)

  which is the maximum value of fuzzy accuracy measure (5.6). Therefore, ${\mathbb {K}}^{acu}(\Re _1,\Re _2)=1$ if $\mu _{\Re _1} (\curlyvee _i)=\mu _{\Re _2} (\curlyvee _i)=0~ \text {or}~1$ for all $\curlyvee _i\in U$.

• 2. Consider a function

  $$\begin{aligned}&{\mathbb {K}}^{acu}(\tau _1,\tau _2)\nonumber \\ {}&\quad =\frac{\tau _1e^{1-\tau _1}+(1-\tau _1)e^{\tau _1}-\sqrt{e}}{2(1-\sqrt{e})}\nonumber \\ {}&\qquad +\frac{(\sqrt{\tau _1\tau _2}) e^{(1-\sqrt{\tau _1\tau _2})}+(1-\sqrt{\tau _1\tau _2})e^{\sqrt{\tau _1\tau _2}}-\sqrt{e}}{2(1-\sqrt{e})} \end{aligned}$$

  (6.14)

  where $0\le \tau _1,\tau _2\le 1$. Differentiating (6.14) partially with respect to $\tau _1$ and solving, we get

  $$\begin{aligned}&\frac{\partial {\mathbb {K}}^{acu}(\tau _1,\tau _2)}{\partial \tau _1}=\frac{(1-\tau _1)e^{(1-\tau _1)}-\tau _1e^{\tau _1}}{2(1-\sqrt{e})}\nonumber \\&\quad +\left( \frac{\tau _2}{\tau _1}\right) \left( \frac{(1-\sqrt{\tau _1\tau _2})e^{(1-\sqrt{\tau _1\tau _2})}-(\sqrt{\tau _1\tau _2})e^{\sqrt{\tau _1\tau _2}}}{4(1-\sqrt{e})}\right) . \end{aligned}$$

  (6.15)

  Again, differentiating (6.14) partially with respect to $\tau _2$, we get

  $$\begin{aligned}&\frac{\partial {\mathbb {K}}^{acu}(\tau _1,\tau _2)}{\partial \tau _2}\nonumber \\ {}&\quad =\left( \frac{\tau _2}{\tau _1}\right) \!\! \left( \frac{(1-\sqrt{\tau _1\tau _2})e^{(1-\sqrt{\tau _1\tau _2})}-(\sqrt{\tau _1\tau _2})e^{\sqrt{\tau _1\tau _2}}}{4(1-\sqrt{e})}\right) . \end{aligned}$$

  (6.16)

  We start with discussing the nature of (6.14) in intervals [.5, 1] and [0,.5]. For this, differentiating (6.14) with respect to $\tau _1$ and $\tau _2$, we obtain (6.15) and (6.16). First, we analyze the sign of (6.15) in the interval [.5, 1]. Now, (6.15) may be written as

  $$\begin{aligned}&\frac{\partial {\mathbb {K}}^{acu}(\tau _1,\tau _2)}{\partial \tau _1}=A+B ~~~\text {where} \end{aligned}$$

  (6.17)
  $$\begin{aligned}&A=\frac{(1-\tau _1)e^{(1-\tau _1)}-\tau _1e^{\tau _1}}{2(1-\sqrt{e})} ~\text {and} \end{aligned}$$

  (6.18)
  $$\begin{aligned}&B=\left( \frac{\tau _2}{\tau _1}\right) \!\! \left( \frac{(1-\sqrt{\tau _1\tau _2})e^{(1-\sqrt{\tau _1\tau _2})}-(\sqrt{\tau _1\tau _2})e^{\sqrt{\tau _1\tau _2}}}{4(1-\sqrt{e})}\right) . \end{aligned}$$

  (6.19)

  In interval [.5, 1], $\tau _1\ge (1-\tau _1)$. This implies $\tau _1e^{\tau _1}\ge (1-\tau _1)e^{(1-\tau _1)}$. Also, $(1-\sqrt{e})<0$. From these arguments, we may conclude that $A\ge 0$. Similarly, for $.5\le \tau _1,\tau _2\le 1$, $(1-\sqrt{\tau _1\tau _2})\le (\sqrt{\tau _1\tau _2})$,   which gives   $(1-\sqrt{\tau _1\tau _2})e^{(1-\sqrt{\tau _1\tau _2})}\le (\sqrt{\tau _1\tau _2})e^{\sqrt{\tau _1\tau _2}}$. This assertion together with $(1-\sqrt{e})<0$ implies that  $B\ge 0$. Thus, we may conclude that  $\displaystyle \frac{\partial {\mathbb {K}}^{acu}(\tau _1,\tau _2)}{\partial \tau _1}\ge 0$ in the interval [.5, 1]. From the fact  $B\ge 0$, we may also infer that  $\displaystyle \frac{\partial {\mathbb {K}}^{acu}(\tau _1,\tau _2)}{\partial \tau _2}\ge 0$. From the above discussion, it may be concluded that the function defined by (6.14) is increasing in interval [.5, 1]. Similarly, it can be established that the function (6.14) is decreasing in the interval [0,.5]. Now, we prove the required property. Since, $\Re _1\subseteq \Re _2\subseteq \Re _3$, this gives  $0\le \mu _{\Re _1}(\curlyvee _i)\le \mu _{\Re _2}(\curlyvee _i)\le \mu _{\Re _3}(\curlyvee _i)\le 1$. Therefore,  $\mu _{\Re _1}(\curlyvee _i)\ge .5$,  $(\mu _{\Re _2}(\curlyvee _i)-\mu _{\Re _1}(\curlyvee _i))\le (\mu _{\Re _3}(\curlyvee _i)-\mu _{\Re _1}(\curlyvee _i))$ and the fact that function defined by (6.14) is increasing in the interval [.5, 1] together implies that  ${\mathbb {K}}^{acu} (\Re _1, \Re _2)\le {\mathbb {K}}^{acu}(\Re _1, \Re _3)$. Similarly.second part of theorem may be proved.

• 3. The proof of third property is straightforward.