---m file---
%check equation
clear;
clc;
f = @(x) (x-1).^3;
%initial guess
x1 = -5;
x2 = -2;
x3 = 5;
ep = 0.001;
iter = 1;
while abs(x1-x3) >ep
y = [f(x1) f(x2) f(x3)]';
x = [1 x1 x1^2
1 x2 x2^2
1 x3 x3^2];
cc = inv(x)*y;
b = cc(2);
c = cc(3);
xr = -b/2/c;f
iter = iter + 1;
disp([iter x1 x2 x3 xr])
if (xr > x1) & (xr < x2)
x3 = x2;
x1 = x1;
x2 = xr;
else
x3 = x3;
x1 = x2;
x2 = xr;
end
end
%condition to make stop is not good so it take much calculate loops to get
%the result
--- จบ ---
หน้า CM จะแสดงคำตอบ
f =
@(x)(x-1).^3
1.0e+004 *
1.5905 0.0001 0.0001 0.0004 0.0001
ใช้ ฟังก์ชั่น Newton ในการหา optimize solution
--- m file ---
clear;
clc;
fx = @(x) (x-1).^3;
df = @(x) 3*(x-1).^2;
ddf = @(x) 6*(x-1);
ep = 0.001;
x0 = 5;
xn = x0 - df(x0)/ddf(x0);
while abs(xn-x0) >ep
x0 = xn;
xn = x0 - df(x0)/ddf(x0);
disp([x0 xn]);
end
%check min or max
if ddf(xn) > 0
disp('min point');
else
disp('max point');
end
%optimize solution
[xn fx(xn)]
--- end ---
ในหน้า CM จะแสดงคำตอบ
3 2
2.0000 1.5000
1.5000 1.2500
1.2500 1.1250
1.1250 1.0625
1.0625 1.0313
1.0313 1.0156
1.0156 1.0078
1.0078 1.0039
1.0039 1.0020
1.0020 1.0010
min point
ans =
1.0010 0.0000
การใช้ Grid search เพื่อหาคำตอบ
--- m file ---
clear;
clc;
f = @(x) (x-3).^2;
al = 0.01; %result is up to alpha change all result will be change too
%alpha is range if it's lare it's not good, if it's small it's good
xcold = -100; %initial guess
xc = 1;
iter = 1;
while al > 10^-4 %cal until alpha > 0.0001 i will OK!
xcold = xc;
xl = xc + al*(-1);
xr = xc + al*(1);
if f(xl) < f(xc) & f(xl) < f(xr)
xc = xl;
end
if f(xr) < f(xc) & f(xr) < f(xl)
xc = xr;
end
if abs(xc-xcold) > 0 %condition for change alpha to prove result
al = al*2;
else
al = al/2;
end
iter = iter +1;
disp([iter xc])
end
--- end ---
ในหน้า CM จะแสดง
... 32 3
33 3
34 3
โจทย์อีกข้อนึง (Grid search)
--- m file ---
clear;clc ;
f = @(x)(x-3).^2;
al = 4;
xc_old = -100;
xc = 1;
iter = 1;
while al > 10^-4 % abs(xc - xc_old) > 0
xc_old = xc;
xl = xc + al*(-1);
xr = xc + al*(1);
if f(xl) < f(xc) & f(xl) < f(xr)
xc = xl;
end
if f(xr) < f(xc) & f(xr) < f(xl)
xc = xr;
end
if abs(xc-xc_old) > 0
al = al*2;
else
al = al/2;
end
disp([iter xc al])
iter = iter +1;
end
xc
--- end ---
หน้า CM จะแสดง
. . .
17.0000 3.0000 0.0001
18.0000 3.0000 0.0001
xc =
3
การใช้ วิธี??? ในการหาค่า
clear; clc; format compact;format short g ;
f = @(x)(x(1)-3).^2+(x(2)-4).^2;
al = @(x1,x2)(x1.^2+x2.^2-6*x1-8*x2+25)/(2*(x1-3).^2+2*(x2-4).^2);
df1 = @(x1,x2)2*(x1-3);
df2 = @(x1,x2)2*(x2-4);
x0 = [-10 -10]';
xn = [1 1]';
ep = 10^-2;
iter = 1;
while min(abs(xn-x0)) > ep
x0 = xn;
s = -[df1(x0(1),x0(2))
df2(x0(1),x0(2))];
%s = -[df1; df2];
iter = iter +1;
xn = x0 + (al(x0(1),x0(2)))*s;
disp(x0)
end
--- จบ ---
ในหน้า CM จะแสดง
1
1
3
4
การใช้ ฟังก์ชั่น Newton ในการแก้ปัญหา multi condition
--- m file ---
clear;
clc;
x0 = [10 10]';
xn = [5 5]'; % ' is transpost
iter = 1; maxiter = 50;
ep = 0.01;
while abs(min(xn-x0)) > ep
x0 = xn;
x1 = x0(1);
x2 = x0(2);
df = [8*x1 - 2*x2
2*x2 - 2*x1];
H = [ 8 -2
-2 2 ];
xn = x0 - inv(H)*df;
iter = iter + 1;
disp([x0 xn]);
end
--- end ---
ในหน้า CM จะแสดง
5 0
5 0
0 0
0 0
หรือสามารถเขียน code ได้อีกแบบนึง ซึ่งก็คือ
--- m file---
clear;
clc;
f = @(x1,x2) 4*x1^2 + x2^2 - 2*x1*x2;
x0 = [10 10]';
xn = [5 5]'; % ' is transpost
iter = 1; maxiter = 50;
epx = 0.01;
epf = 0.01;
cond1 = 1;
cond2 = 1;
while cond1 | cond2
x0 = xn;
x1 = x0(1);
x2 = x0(2);
df = [8*x1 - 2*x2
2*x2 - 2*x1];
H = [ 8 -2
-2 2 ];
xn = x0 - inv(H)*df;
iter = iter + 1;
disp([iter x0' xn']);
cond1 = abs(min(xn-x0)) > epx;
cond2 = abs(min(f(x0(1),x0(2))...
- f(xn(1),xn(2)))) > epf
disp([cond1 cond2]);
end
--- end ---
ในหน้า CM จะแสดง
cond2 =
1
1 1
3 0 0 0 0
cond2 =
0
0 0
การใช้ function buildin ของ Matlab
--- m file ---
function fx = fobj(x)
x1 = x(1);
x2 = x(2);
fx = 4*x1^2 + x2^2 - 2*x1*x2;
--- end ---
ในหน้า CM พิม...
>> [x,fval,exitflag]=fminsearch(@fobj,[1 1])
x =
-2.7148e-006 -3.6605e-005
fval =
1.1707e-009
exitflag =
1
หรือ
>> [x,fval,exitflag]=fminunc(@fobj,[1 1])
Warning: Gradient must be provided for trust-region method;
using line-search method instead.
> In fminunc at 354
Optimization terminated: relative infinity-norm of gradient less than options.TolFun.
x =
-8.056e-008 -1.6227e-009
fval =
2.5701e-014
exitflag =
1
การนำไปใช้ในการทำ Regression
--- m file ---
function sr = fobj1(x)
a0 = x(1);
a1 = x(2);
x = [ 10 20 30 40 50 60 70 80];
y = [ 25 70 380 550 610 1220 830 1450];
sr = 0;
n = length(y);
for i = 1:n
sr = sr + (y(i)-a0-a1*x(i))^2;
end
%sr = sum((y-a0-a1*x).^2 can take place in line 7 to line 11
--- end ---
ในหน้า CM พิม...
>> [x,fv,ex] = fminsearch(@fobj1,[1 1])
x =
-234.29 19.47
fv =
2.1612e+005
ex =
1
ถ้าโจทย์มี 3 ตัวแปร
--- m file ---
function sr = fobj2(x)
a0 = x(1);
a1 = x(2);
a2 = x(3);
x = [ 10 20 30 40 50 60 70 80];
y = [ 25 70 380 550 610 1220 830 1450];
sr = 0;
n = length(y);
for i = 1:n
sr = sr + (y(i)-a0-a1*x(i)-a2*x(i)^2)^2;
end
--- end ---
ในหน้า CM พิม...
>> [x,fv,ex] = fminsearch(@fobj2,[1 1 1])
x =
-178.48 16.122 0.037202
fv =
2.1379e+005
ex =
1
จบ บบ บบบ เย่!!!
