Assessing Potential of Organizations with Fuzzy Entropy

Abstract

Assessing the performance of organizations in the near future has been a challenging problem because of the factors of subjective assessment of the uncertainty. To cater to such applications, a novel form of fuzzy entropy is proposed to unify the concepts of fuzzy entropy and the conventional Shannon’s entropy. The proposed entropy provides a direct and convenient framework to quantify the uncertainty associated with a firm in the near future. It is also shown that the proposed form is amenable in extending the fuzzy entropy functions to the probabilistic-fuzzy domain. Thus, the proposed form of fuzzy entropy finds a special significance in the human decision-making problems. A case study illustrates the applicability of the proposed functions in assessment of future performance of a firm.

Appendices

Appendix 1

1.1 Proofs for the Properties of R-LT Entropy

Proof of $P_1$

From (6), we know that $H_{R-Sh, d} = \frac{1}{n}\sum _{i = 1}^{n} (1 - \Upsilon _i)\mathcal {G}_i$ . Since $\mu _i \in [0, 1]$, $\Upsilon _i \in [0, 1]$ [see (3)]. Hence $(1 - \Upsilon _i) \ge 0$, and $\mathcal {G}_i = -\log (\Upsilon _i) \ge 0$. Since both $(1 - \Upsilon _i) \ge 0$, and $\mathcal {G}_i \ge 0$, $H_{{R-Sh}, d} \ge 0$.

Proof of $P_2$

The proof follows trivially. When $\forall x_i \in X = 0$ or 1, $\Upsilon _i = 1$. Therefore, $\log \left( \Upsilon _i\right) = 0$, $\forall i$, and hence $H_{{R-Sh}, d} = 0$.

Proof of $P_3$

At $\mu _i = 0.5$, $\Upsilon _i = 0$, i.e., the minimum. As $\mu _i$ moves towards 0.5, $\Upsilon _i$ decreases. As $\Upsilon _i$ decreases, $\mathcal {G}_i$ increases. Therefore, the maximum value for $\mathcal {G}_i$ is obtained at $\mu _i = 0.5$. As $\log$ is undefined at 0, we consider $\mu _i \rightarrow 0.5$, or $\Upsilon _i \rightarrow 0$. Similarly, the weight $(1 - \Upsilon _i)$ attains its maximum value at $\mu _i = 0.5$. Therefore, $H_{{R-Sh}, d}$ is maximum at $\mu _i \rightarrow 0.5, \forall x_i \in U$.

Proof of $P_4$

We know that $\Upsilon _i$ decreases as $\mu _i$ increases in the interval [0, 0.5], and $\Upsilon _i$ increases as $\mu _i$ increases in the interval [0.5, 1]. Therefore, both $(1 - \Upsilon _i)$ and $\mathcal {G}_i = \log (\frac{1}{\Upsilon _i})$ are monotonically increasing for $\mu _i \in [0, 0.5]$, and is monotonically decreasing for $\mu _i \in [0.5, 1]$, and attains the maximum value at $\mu _i = 0.5$. Resultantly, H monotonically increases in $\mu _i \in [0, 0.5]$, and monotonically decreasing in $\mu _i \in [0.5, 1]$.

Proof of $P_5$

$H(X,Y) = H(X) + H(Y)$. Let $\mathcal {M}^X$ and $\mathcal {M}^Y$ denote the sets for relative membership grades for the fuzzy sets X and Y defined on U. Then from (5), we can write:

$$\begin{aligned} \begin{aligned} H(X)&= -\frac{1}{n}\sum _{i = 1}^{n} (1 - \Upsilon _i^X)\log ({\Upsilon _i^X}) \\ H(Y)&= -\frac{1}{n}\sum _{i = 1}^{n} (1 - \Upsilon _i^Y)\log ({\Upsilon _i}^Y) \end{aligned} \end{aligned}$$

(16)

Since $\Upsilon _i^X$ and $\Upsilon _i^Y$ are independent of each other, we can write:

$$\begin{aligned} \begin{aligned} H^{X, Y}&= -\frac{1}{n}\sum _{i = 1}^{n} \left( (1 - \Upsilon _i^X)\log ({\Upsilon _i^X}) + (1 - \Upsilon _i^Y)\log ({\Upsilon _i^Y})\right) \\&= -\frac{1}{n}\sum _{i = 1}^{n} (1 - \Upsilon _i^X)\log ({\Upsilon _i^X}) - \sum _{i = 1}^{n} (1 - \Upsilon _i^Y)\log ({\Upsilon _i}^Y)\\&= H(X) + H(Y) \end{aligned} \end{aligned}$$

(17)

Proof of $P_6$

The proof follows straightforward from P3 and P4.

Proof of $P_7$

From (3) : $\Upsilon _i = 2*|0.5 - \mu _i|$. For a completment set $\mu _i = 1 - \mu _i, \, \forall i$. Hence, $\Upsilon _i$ for the complement set: $\Upsilon _i^{'} = 2*|0.5 - (1 - \mu _i)| = \Upsilon _i$. Therefore $H = H^{'}$.

Appendix 2

1.1 Proofs for the Properties of R-PP fuzzy entropy

Proof of $P_1$

We know that the entropy is minimum when $\mu _i = 0$ or 1, or $\Upsilon _i = 1$, $\forall x_i \in X$. Recalling (8): $H = \frac{1}{n}\sum _{i = 1}^n (1 - \Upsilon _i)e^{1 - \Upsilon _i}$, we know that H is minimum when both $(1 - \Upsilon _i)$ and $e^{1 - \Upsilon _i}$ are at their respective minimum values, for all $x_i \in X$. Since $\mu _i \in [0, 1]$, we have $\Upsilon _i \in [0, 1]$. At the upper bound $\Upsilon _i = 1$, or $\mu _i = 0$ or 1, both $(1 - \Upsilon _i)$ and $e^{1 - \Upsilon _i}$ are at their minimums. The minimum H is thus $\frac{1}{n}\left( n(1 - 1)e^{1 - 1}\right) = 0$.

Proof of $P_2$

From (8), we know that H is maximum when $\mathcal {G}_i$ is maximum, and its weight $(1 - \Upsilon _i)$ is maximum, for all $x_i \in X$. Since $\mathcal {G}_i = e^{1 - \Upsilon _i}$. Hence $\mathcal {G}_i$ attains its maxima at $\Upsilon _i = 0$. The weight of $\mathcal {G}_i$, i.e., $(1 - \Upsilon _i)$ is also maximum at $\Upsilon _i = 0$, or $\mu _i = 0.5$. Hence, H attains the maximum value, when $\mu _i = 0.5$ (or $\Upsilon _i = 0$), $\forall x_i \in X$. Replacing $\Upsilon _i = 0$, $\forall x_i \in X$ in (8), we obtain $H_{max} = e$.

Proof of $P_3$

We know that $\Upsilon _i$ decreases as $\mu _i$ increases in the interval [0, 0.5], and $\Upsilon _i$ increases as $\mu _i$ increases in the interval [0.5, 1]. Therefore, both $(1 - \Upsilon _i)$ and $\mathcal {G}_i = e^{(1 - \Upsilon _i)}$ are monotonically increasing for $\mu _i \in [0, 0.5]$, and is monotonically decreasing for $\mu _i \in [0.5, 1]$, and attains the maximum value at $\mu _i = 0.5$. Resultantly, $H = \frac{1}{n}\sum _{i = 1}^n(1 - \Upsilon _i)(e^{(1 - \Upsilon _i)})$ has the same trend.

Proof of $P_4$

The proof follows directly from that for Property 3.

Proof of $P_5$

The proof is same as for that of Property 7 of R-Sh entropy.

Appendix 3

1.1 Proofs for the Properties of the Proposed Fuzzy Entropy

Proof of $P_1$

If $\mu _i = 0$ or 1, then $\Upsilon _i = 1$. Hence $\mathcal {G}_i = e - e^{\Upsilon _i} = 0$. Therefore, if $\mu _i = 0$ or 1, for all $x_i \in U$, then $H = \frac{1}{n}\sum (1 - 1)0 = 0$, the minimum value for H (Since $\Upsilon _i \in [0, 1]$, $e - e^{\Upsilon _i}$ has a lower bound at 0.).

Proof of $P_2$

Since $\mu _i \in [0, 1]$, $\Upsilon _i \in [0, 1]$. At $\mu _i = 0$ or 1, $\Upsilon _i = 1$. As $\mu _i$ moves towards 0.5, $\Upsilon _i$ decreases towards 0, and $\mathcal {G}_i$ increases towards $e - 1$, the upper bound for $\mathcal {G}_i$. If $\mu _i = 0.5$, for all $x_i \in U$, then $H = \frac{1}{n}\sum _{i = 1}^n(1 - 0)(e - 1) = (e - 1)$, the maximum possible value for H.

Proof of $P_3$

We know that $\Upsilon _i$ decreases as $\mu _i$ increases in the interval [0, 0.5], and $\Upsilon _i$ increases as $\mu _i$ increases in the interval [0.5, 1]. For a given a, therefore, both $(1 - \Upsilon _i)$ and $\mathcal {G}_i = (e - e^{(\Upsilon _i)^a})$ are monotonically increasing for $\mu _i \in [0, 0.5]$, and is monotonically decreasing for $\mu _i \in [0.5, 1]$, and attains the maximum value at $\mu _i = 0.5$. Resultantly, $H = \frac{1}{n}\sum _{i = 1}^n(1 - \Upsilon _i)(e - e^{(\Upsilon _i)^a})$ varies in the same way.

Proof of $P_4$

The proof follows trivially from that for Property 3.

Proof of $P_5$

We know that $\Upsilon _i = 2*|0.5 - \mu _i|$. Therefore, $\overline{\Upsilon }_i = 2*|0.5 - (1 - \mu _i)| = |-(0.5 - \mu _i)| = |0.5 - \mu _i| = \Upsilon _i$. Hence $H = {H}^{'}$.