A neural network to solve quadratic programming problems with fuzzy parameters

Abstract

In this paper, a representation of a recurrent neural network to solve quadratic programming problems with fuzzy parameters (FQP) is given. The motivation of the paper is to design a new effective one-layer structure neural network model for solving the FQP. As far as we know, there is not a study for the neural network on the FQP. Here, we change the FQP to a bi-objective problem. Furthermore, the bi-objective problem is reduced to a weighting problem and then the Lagrangian dual is constructed. In addition, we consider a neural network model to solve the FQP. Finally, some illustrative examples are given to show the effectiveness of our proposed approach.

Appendices

Appendix 1: Some results on fuzzy calculus

Lemma 7.1

Let ${\tilde{f}}, {\tilde{g}}$ be convex fuzzy mappings defined on $C\subseteq {\varOmega }$, and $int\ C\ne \emptyset$, then,

$$\begin{aligned} {\tilde{\partial }}(\lambda {\tilde{f}})(x)=\lambda {\tilde{\partial }} {\tilde{f}}(x),\quad (\lambda >0),\qquad {\tilde{\partial }}({\tilde{f}}+{\tilde{g}})(x)={\tilde{\partial }} {\tilde{f}}(x)+{\tilde{\partial }} {\tilde{g}}(x). \end{aligned}$$

Proof

From Theorem 2.14, $\lambda {\tilde{f}}$ and ${\tilde{f}}+{\tilde{g}}$ are convex fuzzy mapping. The completeness of the proof follows from theorem 23.8 in Zhong and Shi (2002). $\square$

Theorem 7.2

Let ${\tilde{f}}, {\tilde{g}}$ be convex fuzzy mappings defined on $C\subseteq {\varOmega }$, and $int\ C\ne \emptyset$, if ${\tilde{f}}, {\tilde{g}}$ are differential at $x^*$, then $\lambda {\tilde{f}}$ ($\lambda >0$) and ${\tilde{f}}+{\tilde{g}}$ are also differential at $x^*$, i. e.,

$$\begin{aligned} {\tilde{\nabla }} (\lambda {\tilde{f}})(x^*)=\lambda {\tilde{\nabla }} {\tilde{f}}(x^*),\quad {\tilde{\nabla }}({\tilde{f}}+{\tilde{g}})(x^*)={\tilde{\nabla }} {\tilde{f}}(x^*)+{\tilde{\nabla }} {\tilde{g}}(x^*). \end{aligned}$$

Proof

From Theorem 2.14, $\lambda {\tilde{f}}$ and ${\tilde{f}}+{\tilde{g}}$ are convex fuzzy mapping. Using Definition 2.11 and Lemma 7.1, the proof is trivial. $\square$

Appendix 2: Some results on FQP

Lemma 7.3

If fuzzy matrix ${\tilde{H}}$, is positive semi-definite and symmetric, $x\in {\mathbb {R}}^n_+$, then $x^T{\tilde{H}}x$ is a fuzzy number and,

$$\begin{aligned} x^T{\tilde{H}}x&=\left( \left\{ \left( \sum _{i,j=1}^nx_ix_j\underline{h_{ij}}(\alpha ), \sum _{i,j=1}^nx_ix_j\overline{h_{ij}}(\alpha ), \alpha \right) : \alpha \in [0,1]\right\} \right) _{n\times n}\\&=(\{(x^T{\underline{H}}(\alpha )x, x^T{\overline{H}}(\alpha )x, \alpha ): \alpha \in [0,1])\})_{n\times n}, \end{aligned}$$

(24)

where $x=(x_1,x_2,\ldots ,x_n)^T, {\tilde{H}}=(\{(\underline{h_{ij}}(\alpha ), \overline{h_{ij}}(\alpha ), \alpha ): \alpha \in [0,1]\})_{n\times n}$.

Proof

The proof follows from Definition 3.1. $\square$

Lemma 7.4

Let fuzzy matrix ${\tilde{H}}$ be positive semi-definite and symmetric, $x\in {\mathbb {R}}^n_+$, then ${\tilde{h}}(x)=x^T{\tilde{H}}x$ is a convex fuzzy mapping.

Proof

The proof follows from Definition 3.1, Theorem 2.13, and Lemma 7.3. $\square$

Now, consider the FQP defined in (1). Here, we are going to prove some results for the FQP.

Lemma 7.5

Let fuzzy matrix ${\tilde{H}}$ be a positive semi-definite and symmetric, then ${\tilde{f}}(x)={\tilde{c}}^Tx+\frac{1}{2}x^T{\tilde{H}}x$ in (1) is a convex fuzzy mapping.

Proof

Since $x\ge 0, {\tilde{c}}^Tx$ is a convex fuzzy mapping. Using Lemma 7.4 and Theorem 2.14, the proof is complete. $\square$

Remark 7.6

Since in FQP (1), ${\tilde{f}}(x)$ is a convex fuzzy mapping and $T=\{x:\ x\ge 0, {\tilde{A}}x\le {\tilde{b}}\}$ is a convex feasible set, so FQP (1) is a convex fuzzy programming.

Remark 7.7

As in crisp programming problem, we say a fuzzy programming is convex if both objective function and the region solution are convex.

Lemma 7.8

The fuzzy mapping ${\tilde{f}}(x)={\tilde{c}}^Tx+\frac{1}{2}x^T{\tilde{H}}x$ is differentiable on $int\ {\mathbb {R}}^n_+$ and,

$$\begin{aligned} {\tilde{\nabla }}{\tilde{f}}(x)={\tilde{c}}+{\tilde{H}}{x}. \end{aligned}$$

(25)

Proof

The proof follows from Theorem 7.2 and Lemma 7.5. $\square$

Appendix 3: Proof of Theorem 3.2

Proof

Since ${\bar{x}}$ is a local optimal solution, there exists a neighborhood $N({\bar{x}})$ around ${\bar{x}}$, such that:

$$\begin{aligned} {\tilde{f}}({\bar{x}})\le {\tilde{f}}(x),\quad \forall \ x\in N({\bar{x}}). \end{aligned}$$

i.e., according to the Definition 2.3, we get,

$$\begin{aligned} {\underline{f}}({\bar{x}})(\alpha )\le {\underline{f}}(x)(\alpha ),\qquad {\overline{f}}({\bar{x}})(\alpha )\le {\overline{f}}(x)(\alpha ),\qquad 0\le \alpha \le 1. \end{aligned}$$

(26)

By contradiction, suppose that ${\bar{x}}$ is not a global optimal solution, so that ${\tilde{f}}(x^*)<{\tilde{f}}({\bar{x}})$ for some $x^*\in T$, where $T=\{x:\ x\ge 0, {\tilde{A}}x\le {\tilde{b}}\}$ is the feasible set. In other words, we have:

$$\begin{aligned} {\underline{f}}({x}^*)(\alpha )\le {\underline{f}}({\bar{x}})(\alpha ),\qquad {\overline{f}}({x}^*)(\alpha )\le {\overline{f}}({\bar{x}})(\alpha ),\qquad 0\le \alpha \le 1. \end{aligned}$$

From the convexity of ${\tilde{f}}$ for all $\lambda \in (0,1)$, we have:

$$\begin{aligned} \underline{f}(\lambda x^*+(1-\lambda )\bar{x})(\alpha )&\le \lambda \underline{f}(x^*)(\alpha )+ (1-\lambda )\underline{f}(\bar{x})(\alpha )\\&\le \lambda \underline{f}(\bar{x})(\alpha )+(1-\lambda )\underline{f}(\bar{x})(\alpha )=\underline{f}(\bar{x})(\alpha ),\\ \overline{f}(\lambda x^*+(1-\lambda )\bar{x})(\alpha )&\le \lambda \overline{f}(x^*)(\alpha )+(1-\lambda ) \overline{f}(\bar{x})(\alpha )\\ {}&\le \lambda \overline{f}(\bar{x})(\alpha )+(1-\lambda )\overline{f}(\bar{x})(\alpha )=\overline{f}(\bar{x})(\alpha ). \end{aligned}$$

But for $\lambda >0$ and sufficiently small, $\lambda x^*+(1-\lambda ){\bar{x}}\in N({\bar{x}})$. Hence, the above inequalities contradict with (26), this leads to the conclusion that ${\bar{x}}$ is a global optimal solution. Suppose that ${\bar{x}}$ is not the unique global optimal solution, so that there exists a ${\hat{x}}\in T, {\hat{x}}\ne {\bar{x}}$, such that ${\tilde{f}}({\hat{x}})={\tilde{f}}({\bar{x}})$, i. e.,

$$\begin{aligned} {\underline{f}}({\hat{x}})(\alpha )={\underline{f}}({\bar{x}})(\alpha ),\qquad {\overline{f}}({\hat{x}})(\alpha )={\overline{f}}({\bar{x}})(\alpha ),\qquad 0\le \alpha \le 1. \end{aligned}$$

By the strict convexity,

$$\begin{aligned}&{\underline{f}}(\frac{1}{2}{\hat{x}}+\frac{1}{2}{\bar{x}})(\alpha )<\frac{1}{2}{\underline{f}}({\hat{x}})(\alpha )+\frac{1}{2}{\underline{f}}({\bar{x}})(\alpha )={\underline{f}}({\bar{x}})(\alpha ),\\&{\overline{f}}(\frac{1}{2}{\hat{x}}+\frac{1}{2}{\bar{x}})(\alpha )<\frac{1}{2}{\overline{f}}({\hat{x}})(\alpha )+\frac{1}{2}{\overline{f}}({\bar{x}})(\alpha )={\overline{f}}({\bar{x}})(\alpha ). \end{aligned}$$

By the convexity of $T, \frac{1}{2}{\hat{x}}+\frac{1}{2}{\bar{x}}\in T$, and the above inequalities violate global optimality of ${\bar{x}}$. Hence, ${\bar{x}}$ is the unique global minimum. $\square$