NAFIPS 2005 - 2005 Annual Meeting of the North American Fuzzy Information Processing Society

Geometric Programming with Max-Product FuzzyRelation Equation ConstraintsJi-hui Yang

Institute of MathematicsShantou University

Guangdong, ZIP 515063, PR.Chinayangjihui@ 1 63.com

Abstract- An optimization model is presented with a posyn-omial objective function subject to max-product fuzzy relationEquation. Then the structure of solution set and its solutionmethod are related with max-product fuzzy relation Equation.Next the optimal solution is discussed, based on exponent ofmonomial among objective function, a solution procedure isproposed. And finally, two practical examples are given forillustration purpose.

I. INTRODUCTIONWe call

Aox = b (1)

a max-product fuzzy relation Equation, where A = (aij) is(n x m)-dimensional fuzzy matrix, and x (x1, X2 .X*, Xm)T,0 < xi < 1 (1 < i < m) is m-dimensional vector,composition operator " o " is " V- "., that is,

V (aij xj)=bi (1<i<n).j2=1

Recent year, the programming problem with Equation(1) constraints is being paid more attention to by people,the linear programming with max-product fuzzy relationEquation constraints has been discussed by J.Loetamonphongand S.C.Fang [1][7]. However, the case with nonlinear objectfunction has been developing very slowly all the time. Itis difficult to receive ideal result by traditional nonlinearoptimization method since the feasible domain of this kind ofprogramming is generally nonconvex[10][12]. Besides owingto nonlinear objective function, it is also very difficult toprovide the general algorithm to this kind of optimizationquestion, and we can only make corresponding discussion onsome concrete nonlinear objective. Reference [2] has provideda solution method to such a problem by genetic algorithm.However, when the scale of variable enlarges, it is difficultto solve premature convergence problem of genetic algorithm.Under the circumstances, we have proposed the fuzzy relationgeometric programming based on "V - A" operator [3]. The

Supported by National Natural Science Foundation of China (No.70271047 and No. 79670012 ).

Supported by " 211" Project Foundation of Shantou University and Li Jia-cheng Science Development Foundation of Shantou University P.R.China.

Bing-yuan CaoDept. and Institute of Math., Shantou University

Guangdong, ZIP 515063, P.R.ChinaSchool of Math. and Inf: Sci., Guangzhou University

bycao@stu.edu.cn

practice has proved, sometimes operator " V - A " may easilylose a lot of important information, and it is disadvantageto delineate a practical problem exactly, to some problems,operator "V -7" can overcome the above-mentioned shortage,and it is irreplaceable to solve some practical problem.

In this paper, we propose the following fuzzy relationgeometric programming model.

min f(x) = (Cl .x) V (c2 _x22) V ... V (cm. xv )s.t Aox= b,

O< xj < 1,(2)

where cj > 0, rj (1 < j < m) is an arbitrary real number.We call the Model (2) a max-product fuzzy relation

geometric programming.Structure of solution set on the max-product fuzzy re-

lation Equation (1) is introduced in the second part, thencompatibility and comparison on (1) are explained, withthe maximal solution formula given. And how to find aminimum solutions by simplification matrix is expatiated. Inthe third part, solution method is described on Model (2),an optimal solution is given under assumption of negativeor nonnegative or both positive and negative exponent. Thefourth part provides the practical solution procedure to Model(2). In the fifth part, two practical examples are given forillustration purpose. Finally, further research direction is putforward on (2).

II. STRUCTURE OF SOLUTION SET ON EQUATION (1)Since the feasible domain of Model (2) is a solution

to (1). Solving Equation (1) is very important in order tooptimize Model (2), so that we make some exposition tostructure of solution set of Model (1) as follows.

Definition 2.1 [4] If there exists a solution in Equation (1),it is called compatible.

SupposeX(A,b) = {x = (X1,X2,...,Xm)T E RmlAox-b, 0 < xj < 1} is the solution set of Equation (1). We defineVxl,x2 E X(A,b), x1 < x2 X xl < x2 (1 < j < m), sucha definition " < " is a partial order relation on X(A, b).

Definition 2.2 If 3x E X(A, b), such that x < x,, Vx EX(A, b), then x is called a maximal solution to Equation

0-7803-91 87-X/05/$20.00 ©2005 IEEE. 650

(1). If 3x E X(A,b), such that x < x, Vx E X(A,b),then x is called a minimal solution to Equation (1). And if3x E X(A, b), such that x < x, then x = x, x is called aminimum solution to Equation (1).

Let

(3)xj = A (aija-'bi) (1 < j < m),i=-1

where

-l= a3'b aej > bi,

Vai7bi E [0,11], < i < n, 1 < j < m.

If x = (x1,x2,...,±m) is a solution to Equation (1), wecan easily prove that x must be a maximal solution to (1).For maximal solution to Equation (1), we have

Lemma 2.1 [5] A o x = b is compatible, if and only ifA o x = b and xi is a maximal solution.Proof: The sufficiency is evident, and now we prove thenecessity.

If x is a solution to A o x = b, then

V(aij.xj)=bi (1<i<n),j=1

so Vi, j, there exists aij .xj < bi, let j be fixed, when aij < bi,then 0 < xj <1 ,when aij > bi, then 0 < xj <k , and,hence, we have

x A (aija-1bi) = xji=1

which is x < x.Step forward, suppose that bi < aij, since

n

¾ = A(aija1bz),i=1

then aij - xj < bi; suppose that bi > aij then aij -j < aij <bi so we have

m

V(azj j) < bi,that is, A o K< b. Since x <Kx then b= A o x < A o K< b.Hence, A o x = b, and x is a maximal solution.

Corollary 2.1 If X(A, b) . 0, then x E X(A, b).In terms of minimal solution to Equation (1), Reference

[6] has provided a sufficient and necessary condition, butit is difficult to be satisfied. Generally speaking, minimalsolution to Equation (1) may not exist in X(A, b). SinceX(A, b) is a partial order set to " < ", its minimumelement exists, as minimum solution is often considered inmany practical problems, we usually pay more attention tominimum solution to Equation (1). For minimum element ofX(A, b), we have

Lemma 2.2 If X(A, b) # 0, then a minimum element mustexist in X(A, b).

If X(A, b) has a minimum element, its numbers usuallyare not unique. If we denote all minimum element byX(A, b), then solution set of Equation (1) can be denoted asfollows.

X(A,b) = U {xlx< x < x, x E X}. (4)xEX(A,b)

We can clearly see by Formula (4), the solution setstructure of Equation (1) can be ascertained by X(A,b),solving X(A, b) means X(A, b) is known. Now we introducethe method to find the minimum solution to a max-productfuzzy relation Equation.

Definition 2.3 Matrix D = (dij)nXtn is called a discriminatematrix of A, where

dij _ aij, aij xj =bi,l O aij*x,j bi.

We can easily prove by Definition 2.3, Equation (1) hasa solution if and only if discriminate matrix D of A containsat least a nonzero entry in each row.

Definition 2.4 Matrix GC (gij)nxm is called a simplifica-tion matrix of A, where

_i aJjX; -ii=bi9ij -o{ aij &j ={ bi

Based on matrix G, X(A, b) can be filtrated as thefollowing rules.

Rule 2.1 (Filtration rule of minimum solution)1) If bi = 0, then delete the ith row of G.2)If bi> 0,and 3 kE{1,2,..., n},suchthat k>i,Vj=

1,2, ...,m, Ckjz3 0 =cij #0 , then delete ith row of G.3) The matrix gained by 1) and 2) can be denoted by C.

To each row of G, the only nonzero entry is selected with allentries of the rest seen as zero, perhaps all of matrices aredenoted by C1, 02,..., Cp. To each column of Gk(1 < k < p),the maximal entry is selected, a minimum solution xi canbe obtained through such a method. Set composed of allxi is called a quasi-minimum solution one, and it includesall minimum solution to Equation (1). If repeat solution isdeleted, and according to Definition 2.2, minimum solutionset X(A,b) can be got[8][11][15].

III. SOLVING SOLUTION ON MODEL (2)Let us consider the function as follows .

f(x) = (Cl * X") V (C2 * X22) V ... V (Cm * XI-). (5)

The optimal value of f(x) is related to exponent rj of xj (1 <i< m).Now we discuss Model (2) through the following three

cases.

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Lemma 3.1 If rj < 0 (1 < j < m), then maximal solutionx to Equation (1) is an optimal one to (2).Proof: Since rj < 0 (1 < j < m), then

d(xrj)- rx -1dxj 3z~x <0

for each xj with 0 < xj < 1. Then xrj is a monotonedecreasing function about xj, so it is easy to know that cj-xis also a monotone function about xj.

Moreover, Vx E X(A, b), when x < x, then

cAXj > C*A3ri (1< j < m),I -J-

such that f(x) > f(x), so x is an optimal solution to Model(2).Lemma 3.2 If ri > 0 (1 < j < m), then a minimum solutionx to Equation (1) is an optimal one to (2).Proof: Since rj > 0 (1 < j < rm), then

d(Xrj)dx xrjx

1> 0

for each xj with O < Xj < 1. Then xrjJ iS a monotoneincreasing function about xj, so is Cj . x? about xj.

Moreover, Vx E X(A, b), according to Formula (4), thenthere exists x E X(A, b), such that x > x, that is, xj > xj.Therefore,

c Xrj >c xj rj (1 j m),

then f(x) > fbx), that is, the optimal solution to Model (2)must exist in X(A, b). Let

f(x)= min{f (x) |x E X (A, b)},here x* E X(A,b). Then Vx E X(A,b), there is f(x) >f(x*). So x* is an optimal solution to Model (2).

As for the general situation, that is, in function (5), theexponent rj (1 < j < m) of xj is either a positive number ora negative one. Let

R, = {Jrj <0,1 < .m},R2 = {jlrj > 0,1 < j < m}.

Then R1 n R2 =0, R1 u R2 = J, here J ={1, 2,...,m}.Let

fl(x) = V {(cj *x)})jeR1

f2(X) = V {(Cj *Xj )jER2

Then f(x) = fi (x) Vf2(x). Therefore, we have the followingtwo optimization models based on the above.

min fi (x)s.t A o x = b, (6)

0 < xj < 1,

andmin f2 (x)s.t Aox = b,

0 < xj < 1.(7)

By Lemma 3.1, x is an optimal solution to Model (6). ByLemma 3.2, 3x* E X(A,b), x* is an optimal solution toModel (7). Let

X* _ jfI j E R1,j j ER2,

We have the following theorem.

Theorem 3.1 If exponent rj (1 < j < m) of xj is eithera positive number or a negative one, then x* is an optimalsolution to Model (2).Proof: Vx E X(A, b). According to (4), 3x E X(A, b), suchthat x < x <ix. By Lemma 3.1 and 3.2, we have

f(X) = fl (X) V f2 (X). f1(d) Vf2 (t) > flW()vf2(*) f (X*).

So x* is an optimal solution to Model (2).

IV. ALGORITHM TO MODEL (2)Algorithm 4.1

Step 1: At first, x is found by Formula (3). If x is not asolution to Equation (1), then turn to Step 9. Otherwise, turnto Step 2.

Step 2: Check the sign of rj (1 < j < m). If rj < 0 (1 <j < in), then turn to Step 8. Otherwise, turn to Step 3.

Step 3: Solving simplification matrix G of A. The mini-mum solution set X(A, b) of Equation (1) is filtrated by Rule2.1.

Step 4: If rj > 0 (1 < j < m), we obtain x* by Lemma3.2. Turn to Step 7. Otherwise, turn to Step 5.

Step 5: Gain x* by Theorem 3.1.Step 6: Print f(x*), stop.Step 7: Print f(x*), stop.Step 8: Print f(x), stop.Step 9: Print "have no solution", stop.

V. EXAMPLESExample 1 We now consider the following max-productfuzzy relation geometric programming.

min f(x) = (0.3 xix-2) V (1.8 3

V (1.4 X3- 2) V (0.45. X42)s.t Aox=b,

0<xi<1 (.<j<4),where b = (0.4,0.2,0.2)T,

(0.5A= 0.5

0 0.6 0.8\0.2 0 041.0.1 0.3 0.2 /

By Formula (3), we can solve x~= (0.4, 1, 2, 0.5)T. SinceA o x= b, then x is a maximal solution to A o x = b. It

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is easy to see rj < 0 (1 < j < 4). By Lemma 3.1, x isan optimal solution to Example 1, and the optimal value isf(x) = 1.875.

Example 2 Consider finding

min f(x) = (0.4. x1-) V (0.7 X22)V(0.6 - x3 1) V (0.1 X42)

s.t Aox=b,0.<xxj1, (1<j<4)

where A, b is the same as Example 1.The discriminate matrix of A is

0 0

D= 0.5 0.2O 0

0.6 0.80 0.40.3 0 /

Since each row ofD contains at least a nonzero entry, solutionexists in max-product fuzzy relation Equation A o x = b. Theoutcome consists with Example 1. Because the exponent rjis either positive or negative, we solve simplification matrixo of A by Algorithm 4.1, then

/ 0 0 o0.51 ~~~31G= 0.4 1 0 0.5

0 0 2 0

the matrix G is dealt with by Rule 2.1. So we can get

0.4 1 00 0 2

3

0.5).O

Step forward, we have

0.4 0 0 0)

~0 _

GI- 0O O -0 2 X

So all of minimum solutions to A o x-b is

2= 120 0

v3 (0 0 2 05)T

3

Notice thatf (x) = (0.4* x1- ) V (0.1 *

f2(x) = (0.7. X22) V (0.6. X32).By Lemma 3.1, x is an optimal solution to fi(x).By Lemma 3,2, X and X are optimal solutions to f2(x).By Theorem 3.1, x* = (0.4, 0, 2, 0.5)T is an optimal solutionto f(x), and the optimal value is f(x*) = 0.632.

VI. CONCLUSIONIn the above, we research the max-product fuzzy relation

geometric programming similar with Model (2), and giveits finding optimal solution procedure. Numerical experimentproves that, when the variable scale of Model (2) is not verylarge, we can smoothly reach the optimal point by applyingthis algorithm and that, however, when the variable scale ofModel (2) is very large, the number of element among theminimum solution set X(A, b) of Equation (1) will increasesignificantly. It is worthy of considering a problem whetherthere is a more rapid optimization method than Algorithm 4.1to quickly reach the optimal point under the circumstanceswithout solving X(A, b).

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