Solving linear optimization problems with max-star composition equation constraints

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Abstract

In this paper an optimization model with a linear objective function subject to a system of fuzzy relation composition equations is presented. Since the non-empty feasible solution set of the fuzzy relation equations is generally a non-convex set, the conventional linear programming method will not be suitable for solving such a problem. Therefore, an efficient solution procedure for such problems appears to be necessary. In this paper, firstly, we characterize the feasible solution set, and then introduce two efficient procedures for solving the problem. Finally, two concrete examples are included for more details.

Introduction

Let A = (aij), 0 ⩽ aij ⩽ 1, be an (m × n) dimensional fuzzy matrix and b = (b1, … , bn), 0 ⩽ bj ⩽ 1, be an n-dimensional vector, then, the following system of fuzzy relation max-star composition equations is defined by A and b:xo∗A=b,where x = (x1, x2, … , xm)t with 0 ⩽ xi ⩽ 1, is a solution vector that we attempt to find, and “o∗” ismaxi=1,2,…,m[xi+aij-xiaij]=bjforj=1,2,…,n.

The resolution of fuzzy relation Eq. (1) is interesting and ongoing research topic [1], [2], [3], [4], [5], [8], [9], [10], [11], [12], [14], [15], [16]. In this paper, we learn a variant of such problem. We are interested in solving the following linear programming model with fuzzy relation constraints:∑i=1mcixixo∗A=b,0⩽xi⩽1,where c = (c1, c2, … , cn)t is an m-dimensional vector, and ci represents the cost associated with variable xi for i = 1, 2, … , m.

The regular linear programming problems [6], [13] have a completely different nature with this linear optimization problem subject to fuzzy composition equations. The non-empty feasible solution set of fuzzy relation Eq. (1) is generally a non-convex set which can be completely determined by one maximum solution and a finite number of minimum solutions [3], [8]. The traditional linear programming method, such as the simplex and interior-point algorithm is useless, because the solution set is non-convex. Recently, Fang and Li [7] have studied (3) subject to max–min relation equations and presented a branch and bounded method to discover an optimal solution, also Khorram and Ghodousian [17] have considered (3) regarding to max–av relation equations and have achieved a tabular form method to access the minimum points by using the single maximum point. Fuzzy relation equation plays an important role in fuzzy modeling, fuzzy diagnosis, and fuzzy control and also applications in fields such as psychology, medicine, economics, and sociology [4], [18].

In this paper in Section 2 the feasible solution and properties of system (1) are studied, in Section 3 the tabular form of the present relation is studied in order to find minimum solutions by applying the single maximum point. In Section 4, we aim to use an alternative technique by using some concepts from Section 3 in order to find the real minimum point. In Section 5, two numerical examples with two mentioned approaches are solved, and finally a conclusion is derived in Section 6.

Section snippets

Characterization of a feasible solution set

Let X = {x ∈ Rm : 0 ⩽ xi ⩽ 1, i ∈ I}, I = {1, 2, … , m}, J = {1, 2, … , n}. The feasible solution of (3) is denoted by X[A,b]={x∈X:xo∗A=b}. To characterize X[A, b], we remind that for x1, x2 ∈ X, it is said to be x1 ⩽ x2 if and only if xi1⩽xi2,∀i∈I. In this way, “⩽” forms a partial order relation on X. Moreover, we call x∧ ∈ X[A, b], is said to be a maximum solution, if x ⩽ x∧, ∀x ∈ X[A, b]. Similarly, x∨ ∈ X[A, b] is called a minimum solution, if x ⩽ x∨ implies x = x∨, ∀x ∈ X[A, b].

Now, we bring in two lemmas that are required in the

First procedure

Definition 2

For each x∈jX[A,b], I(x) is definedI(x)=I1(x)×I2(x)×⋯×In(x),whereIj(x)=i∈I:xi=bj-aij1-aij,j∈Jand for f = (f(1), f(2), … , f(n)) ∈ I(x) the ith component of f[x] is defined byf[x]i=maxj∈Jf(i)bj-aij1-aij,Jf(i)≠∅,0,Jf(i)=∅,whereJf(i)={j∈J:f(j)=i}.

Definition 3

Suppose jx∧ be the maximum solution of (2). Definexi∧=minj∈Jjxi∧,f[x]=maxj∈Jjx∨(f(i)).

Remark 1

It is quite noticeable that when, x ∈ X[A, b], f ∈ I(x) then it can be said that there exists i ∈ I ∋ : Jf(i) ≠ ∅. In order to prove it, suppose for each i ∈ I, Jf(i) = ∅ then xi≠bj-aij1-aij,∀j

Second procedure

At this moment we consider a second method. Some required points are studied in the following theorem:

Theorem 9

X0[A, b] = F0[x∧] where F0[x∧] is the set of minimal of F[x∧].

Proof

According to Theorem 7 for each x ∈ X[A, b] ∃ x′ ∈ X[A, b] ∋ : x′ ⩽ x, and since X0[A, b] ⊆ F[x∧], for each x ∈ X[A, b] there exists x′ ∈ f[x∧] ∋ : x′ ⩽ x. Now, it is proved by using Lemma 2 and Theorem 6. □

Lemma 3

F0[x∧] ⊆ X∗ ⊆  F[x∧] where X∗ = {f[x∧]:f ∈ E} and E = {(i1, i2, … , in) ∈ I(x∧):{i1, i2, … , ij−1} ⋂  Ij(x∧) = ∅ or ij = max({i1, i2, … , ij−1} ⋂ Ij(x∧))}.

Proof

[3]. □

At the moment, we consider an

Numerical result

In this section, we consider two examples and after studying feasible region, they are solved with the two procedures.

Example 1

Considerminz=4x1+7x2+5x3xo∗A=b,xi∈[0,1],whereA=.2.4.3.4.3.4.4.55.4andb=(.6,.7,.6)we can easily find x∧=37,13,13. Here, I1(x∧) = {2, 3}, I2(x∧) = {3}, I3(x∧) = {1, 2, 3}.

Now, Table 1 is formed.

But min{12/7, 7/3, 5/3} = 5/3, hence, the 3rd row that corresponds to f = (3, 3, 3) is chosen, then x∨ = (0, 0, 1/3). Here S = {(3/7, 1/3, 1/3), (0, 1/3, 1/3), (0, 0, 1/3), (3/7, 0, 1/3)}.

Now we consider this example with

Conclusion

In this paper, after considering the feasible solutions space, we considered two methods. In the first one, we used the maximum points, the minimum points achieved by the tabular technique, and in the second one through an algorithm set E, which shows the number of choices is smaller. Finally, the performance of two numerical examples showed that the second method is preferable.

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