Neutrosophic Goal Geometric Programming Problem and Its Application to Multi-objective Reliability Optimization Model

Abstract

This paper presents the goal geometric programming method in neutrosophic environment. Neutrosophic set is one of the most useful tools to express uncertainty, impreciseness in a more general way compare to fuzzy set and intuitionistic fuzzy set. Thus, the proposed approach is described here as an extension of fuzzy goal geometric programming and intuitionistic fuzzy goal geometric programming. To demonstrate the methodology and applicability of the proposed approach, a multi-objective nonlinear reliability optimization model is taken here and it is evaluated comparing the result obtained by the proposed method with the solution obtained in intuitionistic fuzzy goal geometric programming technique at the end of this paper.

Appendices

Appendix 1

1.1 Proof of Lemma 3.1

From the Eq. (3.5) we have

$$\begin{array}{*{20}l} {} & {\mu_{i} \left( {f_{oi} } \right) \ge \sigma_{i} \left( {f_{oi} } \right)} \\ {\text{implies}} & {1 - \frac{{f_{oi} \left( x \right) - b_{0i} }}{{a_{0i} }} \ge 1 - \frac{{f_{oi} \left( x \right) - b_{0i} }}{{d_{0i} }}} \\ {{\text{or}},} & {\left( {f_{oi} \left( x \right) - b_{0i} } \right)\left( {\frac{1}{{d_{0i} }} - \frac{1}{{a_{0i} }}} \right) \ge 0} \\ \end{array}$$

(i)

In the above-mentioned neutrosophic goal programming problem, we consider each objective functions $f_{oi} \left( x \right)$ satisfying target achievement value $b_{0i}$ and also from the relation,

$$\begin{array}{*{20}l} {} & {\vartheta_{i} \left( {f_{oi} } \right) \ge 0} \\ {{\text{or}},} & {\frac{{f_{oi} \left( x \right) - b_{0i} }}{{r_{0i} }} \ge 0} \\ {{\text{or}},} & {\left( {f_{oi} \left( x \right) - b_{0i} } \right) \ge 0} \\ \end{array}$$

(ii)

Thus the relation (i) is true if $\left( {\frac{1}{{d_{0i} }} - \frac{1}{{a_{0i} }}} \right) \ge 0$

$${\text{i}} . {\text{e}} .\;\;\;a_{0i} > d_{0i}$$

(iii)

Hence from relation (iii), we have in neutrosophic goal geometric programming problem, acceptance tolerance $a_{0i}$ should be greater than indeterminacy tolerance $d_{0i}$.

Again from the relation $\mu_{i} \left( {f_{oi} } \right) \ge \vartheta_{i} \left( {f_{oi} } \right)$ and $\mu_{i} \left( {f_{oi} } \right) \ge \sigma_{i} \left( {f_{oi} } \right)$

$${\text{we have}},\;\;\;\;1 - \frac{{f_{oi} \left( x \right) - b_{0i} }}{{a_{0i} }} \ge \frac{{f_{oi} \left( x \right) - b_{0i} }}{{r_{0i} }}$$

(iv)

$${\text{and}}\;\;\;1 - \frac{{f_{oi} \left( x \right) - b_{0i} }}{{a_{0i} }} \ge 1 - \frac{{f_{oi} \left( x \right) - b_{0i} }}{{d_{0i} }}$$

(v)

Adding the above inequalities (iv) and (v), we get

$$1 - \frac{{f_{oi} \left( x \right) - b_{0i} }}{{a_{0i} }} \ge \frac{1}{2} + \frac{{\left( {f_{oi} \left( x \right) - b_{0i} } \right)}}{2}\left( {\frac{1}{{r_{0i} }} - \frac{1}{{d_{0i} }}} \right)$$

(vi)

Now from (3.5) using the relation

$$\begin{aligned} & \mu_{i} \left( {f_{oi} } \right) \ge \vartheta_{i} \left( {f_{oi} } \right) \ge 0\;\;\;{\text{and}} \\ & \mu_{i} \left( {f_{oi} } \right) + \vartheta_{i} \left( {f_{oi} } \right) + \sigma_{i} \left( {f_{oi} } \right) \le 3 \\ & {\text{we get}},\;\;\;\sigma_{i} \left( {f_{oi} } \right) \le 3 \\ & \quad {\text{or}},\;\;\;1 - \frac{{f_{oi} \left( x \right) - b_{0i} }}{{d_{0i} }} \le 3 \\ & \quad {\text{or}},\;\;\;f_{oi} \left( x \right) - b_{0i} \ge - 2d_{0i} \\ & \quad {\text{or}},\;\;\;\frac{1}{{f_{oi} \left( x \right) - b_{0i} }} \le - \frac{1}{{2d_{0i} }} \\ \end{aligned}$$

(vii)

Hence from $\mu_{i} \left( {f_{oi} } \right) + \vartheta_{i} \left( {f_{oi} } \right) + \sigma_{i} \left( {f_{oi} } \right) \le 3$ using (vi) and (vii)

$$\begin{aligned} & \frac{1}{2} + \frac{{\left( {f_{oi} \left( x \right) - b_{0i} } \right)}}{2}\left( {\frac{1}{{r_{0i} }} - \frac{1}{{d_{0i} }}} \right) + \frac{{f_{oi} \left( x \right) - b_{0i} }}{{r_{0i} }} \\ & \quad + 1 - \frac{{f_{oi} \left( x \right) - b_{0i} }}{{d_{0i} }} \le 3\;\;\;\;{\text{gives}}\;\;\;\;r_{0i} > 2d_{0i} . \\ \end{aligned}$$

Thus from the above relation, it is clear that in neutrosophic goal geometric programming problem half of the rejection tolerance $r_{0i}$ should be greater than the indeterminacy tolerance $d_{0i}$.

Appendix 2

2.1 Proof of Theorem 3.1

If $x^{*}$ be a Pareto optimal solution of the FGGPP (3.10) then there does not exist any x such that $f_{oi} \left( x \right) \le f_{oi} \left( {x^{*} } \right)$ for all i = 1, 2,…, p. and $f_{oi} \left( {x^{*} } \right) \ne f_{oi} \left( x \right)$ for at least one i.

Then we have for all $x = \left( {x_{1} ,x_{2} , \ldots ,x_{m} } \right)$,

$$f_{oi} \left( x \right) \le f_{oi} \left( {x^{*} } \right)$$

(A)

with strict inequality hold for at least one i.

$$\begin{aligned} & {\text{i}} . {\text{e}} .\;\;\;f_{oi} \left( x \right) - b_{0i} \le f_{oi} \left( {x^{*} } \right) - b_{0i} \\ & {\text{or}},\;\;\;\frac{{f_{oi} \left( x \right) - b_{0i} }}{{a_{0i} }} \le \frac{{f_{oi} \left( {x^{*} } \right) - b_{0i} }}{{a_{0i} }} \\ & {\text{or}},\;\;\;1 - \frac{{f_{oi} \left( x \right) - b_{0i} }}{{a_{0i} }} \ge 1 - \frac{{f_{oi} \left( {x^{*} } \right) - b_{0i} }}{{a_{0i} }}\;\;\;{\text{implies}}\,\;\;\mu_{i} \left( {f_{oi} \left( x \right)} \right) \ge \mu_{i} \left( {f_{oi} \left( {x^{*} } \right)} \right). \\ \end{aligned}$$

Similarly from (A) we have $\frac{{f_{oi} \left( x \right) - b_{0i} }}{{r_{0i} }} \le \frac{{f_{oi} \left( {x^{*} } \right) - b_{0i} }}{{r_{0i} }}$ which implies $\vartheta_{i} \left( {f_{oi} \left( x \right)} \right) \le \vartheta_{i} \left( {f_{oi} \left( {x^{*} } \right)} \right)$

$$\begin{aligned} & {\text{and also}}\;\;\;\frac{{f_{oi} \left( x \right) - b_{0i} }}{{d_{0i} }} \le \frac{{f_{oi} \left( {x^{*} } \right) - b_{0i} }}{{d_{0i} }} \\ & {\text{or}},\;\;\;1 - \frac{{f_{oi} \left( x \right) - b_{0i} }}{{d_{0i} }} \ge 1 - \frac{{f_{oi} \left( {x^{*} } \right) - b_{0i} }}{{d_{0i} }} \\ \end{aligned}$$

or, $\sigma_{i} \left( {f_{oi} \left( x \right)} \right) \ge \sigma_{i} \left( {f_{oi} \left( {x^{*} } \right)} \right)$. Hence from the definition of Pareto optimal solution to the NGGPP, we have $x^{*}$ is the Pareto optimal solution of (3.1).

Conversely, let $x^{*}$ is a Pareto optimal solution to NGGPP (3.1), then from the expression of membership function given in (3.2) we get

$$\begin{aligned} & 1 - \frac{{f_{oi} \left( x \right) - b_{0i} }}{{a_{0i} }} \ge 1 - \frac{{f_{oi} \left( {x^{*} } \right) - b_{0i} }}{{a_{0i} }} \\ & {\text{i}} . {\text{e}} .\;\;\; f_{oi} \left( x \right) \le f_{oi} \left( {x^{*} } \right). \\ \end{aligned}$$

Again using (3.3) we have $\frac{{f_{oi} \left( x \right) - b_{0i} }}{{d_{0i} }} \le \frac{{f_{oi} \left( {x^{*} } \right) - b_{0i} }}{{d_{0i} }}$ which implies $f_{oi} \left( x \right) \le f_{oi} \left( {x^{*} } \right).$

Similarly, using (3.4), $1 - \frac{{f_{oi} \left( x \right) - b_{0i} }}{{d_{0i} }} \ge 1 - \frac{{f_{oi} \left( {x^{*} } \right) - b_{0i} }}{{d_{0i} }}$ gives $f_{oi} \left( x \right) \le f_{oi} \left( {x^{*} } \right).$

Thus we have $f_{oi} \left( x \right) \le f_{oi} \left( {x^{*} } \right)$ with strict inequality hold for at least one i, $i \in \left\{ {1,2, \ldots ,p} \right\}$ and which shows that $x^{*}$ is a Pareto optimal solution of (3.10).