Page 324 - MATLAB an introduction with applications

Optimization ——— 309

p1=a1+s*b1;p2=a2+s*b2;
z=90*(p2–p1^2)^2+(1–p1)^2;

The output is as follows:
Optimum point after 30 cycles is
1.003135
1.000263

2
2
Example E5.27: Use the Fletcher-Reeves method to locate the minimum of f (x) = 9x + 3x – 8x x + 2x .
1 2
2
1
1
Start with X =[0 0] .
T
o
Solution:
This method is also called conjugate gradient method. It is modification of Cauchy’s method of gradient
decent algorithm. The complete algorithm is given below:
0
Given x perform the following steps:
∇
1
0
1. Compute fx∇ () and set u = – fx 0
().
2. For i = 1, 2, ..., n do:
i
i
2.1 Set x = x i− 1 + λ i u where λ such that
i
( fx i− 1 +λ i u i ) = min (fx i –1 +λ u i ) (line search)
λ
2.2 Compute fx∇ (),
i
2.3 If convergence criteria satisfied, then STOP and * x ≅ x i , else go to Step 2.4
i
≤≤
x
2.4 If 1 i n − 1, u i+ 1 = −∇ f ( ) + β i u i
0
n
3. Set x = x and go to Step 2 (restart).
∇ f 2
Here β= i .
i
∇ f i− 1 2
 18x − 8x + 2 0 
1
2
In this problem ∇f =   and X = 
0
0
 6x − 8x 1  
2
Iteration 1:
2 
1
0
Thus U = –∇f (X ) =  .
0

 2λ 1 
For i = 1, find X = X + λ U =   0  and substituting in function f and find the minimum value of f and
1
1
0
1

1
corresponding λ and X .
1
Compute ∇f (X )
1