# Another approach to formulate a fuzzy linear programming problem

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```					Another approach to formulate a fuzzy linear programming problem
Last Time, we introduced one approach to formulate a fuzzy linear programming problem.

 CT X  Z AX  b X 0 Where the constraint matrix A, the coefficients of the objective function C, and the right hand side vector b are crisp. The only fuzzy parts are the operators (  , can be read as essentially less than) used to formulate the fuzzy LP problem, and this operator can be implemented by defining fuzzy membership functions. To solve it, we transfer it into a crisp LP with one more variable, , which stands for the least degree of constrains and objectives satisfied. (The bigger, the better.) Maximize λ St. λpi+Bix<=di+pi i= 1, 2, ….M+1 X>=0 In the equations above, the pi is defined as the “lowest justifiable solution”, di is subjectively chosen constant of admissible violation, B is the matrix augmented by the rows of the objective functions. In fact, this approach formulates solves a max minimum problem. After transformation, we get a crisp LP with one more constraint and can solve it efficiently. After our discussion, we agreed that the same solution could be gotten by enlarging the searching area (relaxing some constraints) in certain cases.
Today, we introduce another approach to formulate a fuzzy linear programming problem where the constraint matrix A, the coefficients of the objective function C, the right hand side vector b are fuzzy, namely fuzzy numbers. The operators are fuzzy too. (The constraints and objective functions are not crisp, but can be violated to some extends.) 1. Classical definition of fuzzy number: A fuzzy number M is a convex normalized fuzzy set M of the real line such that 1. It exists exactly one x0  IR with  M ( x0 )  1 2.  M (x) is piecewise continuous. A triangle fuzzy number is shown here to give some insight.

1

c 0

c

c



c

Figure 1

(A triangle fuzzy number can be described by it’s center a and width c. The definition of fuzzy numbers is often modified to fit into different cases. ) 2. Definition of fuzzy set: If X is a collection of objects denoted generically by x, then a fuzzy set Ã in X is a set of ordered pairs:
~ A  {( x,  ( x)) | x  X }

A fuzzy set can be represented solely by stating its membership function.

3. Classical definition of fuzzy function: A fuzzy function is a generalization of a classical crisp function and it can be defined in many ways. Here, we just use a classical definition using fuzzy numbers to facilitate the understanding of “fuzzy function”. Suppose we have a classical function y=f (x, A) where x, y and A are crisp. Given a parameter A, this function maps x to y. Now, we replace the crisp parameter A ~ ~ ~ with a fuzzy number A . For a given x, we can define a fuzzy function Y  f ( x, A ) as follows:
~ ~ f : X   (Y ), Y  f ( x, A) ,  (Y ) is the set of all fuzzy subsets on Y. The fuzzy

~ set Y is defined by the membership function:

Y ( y )  max [  A (a)], {a | y  f ( x, a)}  ,
{a| y  f ( x , a )}

0,

otherwises

If f is monotonic with a, then the max operation would be over the degenerate case of just ~ one value. Note, for a different values of X, the fuzzy set Y will be different.

With all the definitions, we can get a statement:
~ Given the fuzzy parameters A  { , c} with center a and width c(Refer to Figure 1), the fuzzy linear function ~ ~ ~ Y  A1 X1  .... An X n can be defined as the following membership function: ( The proof is tedious and not so easy to understand, and we just use the conclusion here.)

| y - x | ,x  0 c|x| Y ( y )  1, x 0 and y  0 . 10,
So, a crisp LP:
 cT x   z

x 0 and y  0

Ax  b

x0

can be transformed into a fuzzy LP if we assume that the parameters( constraints matrix A, right-hand vector b, objectives Z and coefficients of objective function C) are fuzzy numbers.

Let The fuzzy LP will be:

~ z ~  c T  ~ B ~ ~   A b   

~ Bx  0
~

If we define the fuzzy relationship “almost positive’ denoted by  0 as  Y (0)  1   and the new variable , which stands for the degree of positive y is. (Degree of constrains and objectives satisfied too.) The bigger  is, the better. Since we normalize the fuzzy membership, we have 0<<1.) Thus, the fuzzy LP is transformed into Maximize λ St. ( i  ci ) x  0
~

x0

This problem is non-linear now, and we will give an example to show how the transformation works and how to solve the new problem. Begin from a classical LP problem (The objectives are treated equally with constraints, mathematically you can’t distinguish the equations coming from objectives and equations coming from constraints.):

6 X 1  5 X 2  30 2 X 1  9 X 2  45 11X 1  5 X 2  44  2 X 1  3 X 2  12
Replace all the coefficients with fuzzy numbers, where the width Cs are arbitrarily given.

~ A1  { 1  (30,6,5), c1  (6,2,1)} ~ A2  { 2  (45,2,9), c 2  (8,2,3)} ~ A3  { 3  (44,11,5), c3  (4,2,1)} ~ A4  { 4  (12,2,3), c 4  (4,1,2)}
The transformed problem will be

Max  (30  6 )  (6  2 ) X 1  (5  1 ) X 2  0 (45  8 )  (2  2 ) X 1  (9  3 ) X 2  0 (44  4 )  (11  2 ) X 1  (5  1 ) X 2  0 (12  4 )  (2   ) X 1  (3  2 ) X 2  0
Since the new problem is nonlinear, we need an algorithm to solve it. 1.Find an initial number of  so that there exits a vector of X satisfy ( i  ci ) x  0 2. Increase  until not X exists to meet ( i  ci ) x  0 Then, we get the maximum . There maybe infinity X exits, which can give the optimum . (The solution is not necessarily unique.) To find a crisp solution, we can ask for the decision maker’s preference or more information. Conclusion: This approach can use fuzzy parameters and fuzzy operators. Since the objective is to maximize the , we can think it as some kind of max-min techniques. (In the step 2, where the magnitude of  is restricted by the most demanding requirement for

decision variables x.) During the solving procedure, for each , the problem changes to be a LP. Thus, it’s possible to use standard or steamed algorithm to solve it once we can find an optimum algorithm to speed the search of optimum . Discussion: What’s the difference between this approach with parametric analysis? Second approach to fuzzy LP Changes of Constraint matrix A, Cost parameters equation C, Right Hand Vector b and the objectives Z Degree of requirements met Meaning of  Results A fuzzy set (solutions not unique.) Parametric analysis Cost equation C, and Right Hand Vector b The degree of perturbation A crisp one

The most importance is that the fuzzy LP can be implemented use different kinds of fuzzy sets. Thus, the final mathematical formulation will be different with parametric analysis.

A simple example of GENCO: Suppose a GENCO had 3 units available:

H ( p1 )  510  7.2 * p1  0.00142 * p1

2 2

( Coal - fired, fuel price Pc  1.1) 600  p1  150MW ( Oil - fired, fuel price Po  1.0) 400  p 2  100MW ( Oil - fired, fuel price Pc  1.0) 200  p3  50MW

H ( p 2 )  510  7.2 * p 2  0.00142 * p 2 H ( p3 )  510  7.2 * p3  0.00142 * p3

2

Before re-regulation, the GENCO has obligation to meet load of 850MW. The problem is to minimize the production cost (ignoring the forced outage rate). The solution to the regular LP:

min H(p 1 ) * Pc  H(p 2 ) * Po  H(p 3 ) * Po st .P1  P2  P3  850 600  p1  150 400  p 2  100 200  p3  50 Solution : P1  393.2, P2  334.6, P3  122.2,   9.148 The total cost is 8194.36

Now, we use fuzzy numbers to represent the fuel prices and load demand. Given the new parameter, we try to minimize our production cost. 1. Linearize the input-output curve of different units. To use linear programming, we need to linearize the input-output curve of all three units. Choose the original solution as the base points and extends to 10%. Unit Unit 1 Unit 2 Unit 3 90% 360 3286.0 300 2839.6 110 1013.0 base 392.2 3560.6 334.6 3153.8 122.2 1123.9 110% 420 3784.5 360 3387.4 135 1241.8

The new curves are: H ( p1 )  8.038 * p1  295 ( Coal - fired, fuel price Pc  1.1) 420  p1 360MW

H ( p2 )  9.13 * p2  100.6 ( Oil - fired, fuel price Po  1.0) 360  p2  300MW H ( p3 )  9.152 * p3  6.28 ( Oil - fired, fuel price Pc  1.0) 135  p3  110MW
If we multiply the fuel prices with the new input-output curves, we get the new LP problem:

min 9.1388P1  9.13P2  9.152P3  431.38 st .P1  P2  P3  850  0 420  p1  360 360  p 2  300 135  p3  110
If we assume 1% fluctuation on fuel prices and load demand, allow 5% deviation from the lowest cost gotten from the regular LP, and represent them as fuzzy numbers. Namely, we assume:

A1  (7768.62,9.1388,9.13,9.152) C1  (388.43,0.0914,0.0913,0.0915) A2  (1,1,1,850) C1  (0,0,0,8.5)

Let  be the degree of requirements met, we can get our transformed non-linear problem.

Max st .(7768.62  388.43 )  (9.1388  0.0914 ) P1  (9.13  0.0913 ) P2  (9.152  0.0915 ) P3  0 P1  P2  P3  (850  8.5 )  0 420  p1  360 360  p 2  300 135  p3  110
Use the algorithm to solve it and we get 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9  cost 7.7663 7.7818 7.7974 7.8129 7.8285 7.8441 7.8598 7.8754 7.8910 7.9067

[P1,P2,P3]’= 380.0000 380.8500 381.7000 382.5501 383.4000 384.2500 385.1000 360.0000 360.0000 360.0000 359.9999 360.0000 360.0000 360.0000 110.0000 110.0000 110.0000 110.0000 110.0000 110.0000 110.0000 385.95 386.8000 387.6500 388.5000 360.00 360.0000 360.0000 360.0000 110.00 110.0000 110.0000 110.0000

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Jun Wang Dr
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