Tutorial Matlab Komputasi

GOLDEN SEARCH METHOD

Mencari Nilai Minimum dr Fungsi berikut :

f(x)=x^2/10-2*sin(x)

Kode Matlab :

%Golden Search Method

syms x

A = (x^2/10)-(2*sin(x)); %fungsi yg dicari nilai maksimum

a = 0; %nilai awal xl

b = pi/2;

e = 0.5;

t = (sqrt(5)-1/2);

x1 = a+t*(b-a);

x2 = b-t*(b-a);

n = 100;

N = [];

for i = 1:n

fa = subs (A,x,x1);

fb = subs (A,x,x2);

if fa>fb

a = x2;

x2 = x1;

x1 = a+t*(b-a);

else

b = x1;

x1 = x2;

x2 = b-t*(b-a);

end

N = [N; i a b x1 x2 fa fb];

if abs (b-a)<= e

break

end

nilai_x = (b+a)/2

f_minimum = subs (A,x,nilai_x)

plot (N(:,1),N(:,7),'-ro','linewidth',3,...

'MarkerEdgeColor','y',...

'MarkerFaceColor','w',...

'MarkerSize',1.5)

title('Golden Seacrh Method','fontname','algerian')

xlabel('Sumbu Y','fontname','algerian')

ylabel('Sumbu X','fontname','algerian')

grid

INTERVAL EQUATION METHOD

Mencari Nilai Maksimum dr Fungsi berikut :

f(x)=x^2/10-2*sin(x)

Kode Matlab :

%interval Method

syms x

A = (x^2/10)-(2*sin(x));

a = 0;

b = pi/2;

e = 0.2;

n = 100;

N = []

for i = i:n

s = ((a+b)/2+(e/2));

t = ((a+b)/2-(e/2));

fs = subs (A,x,s);

ft = subs (A,x,t);

if fs>=ft

a=t;

else

b=s;

end

N = [N; i a b fs ft]

if abs (b-a)<=0.02

break

end

nilai_x = (a+b)/2

f_maximum = subs (A,x,nilai_x)

plot (N(:,1),N(:,5),'-ro','linewidth',3,...

'MarkerEdgeColor','y',...

'MarkerFaceColor','w',...

'MarkerSize',1.5)

title('Interval Equation Method','fontname','algerian')

xlabel('Fs','fontname','algerian')

ylabel('Ft','fontname','algerian')
grid

SECHANT METHOD

SECHANT METHOD

4 akar dari Fungsi Persamaan Berikut :

(f = x^4+2*x^3-7*x^2+3)

KOde MAtlab :

%SECANT METOHD

syms x

f = x^4-2*x^3+7*x^2+3;

a =0.1;

b=0.3;

n = 100;

ea = 0.000001;

N = [];

for i = 1:n

fa = subs (f,x,a);

fb = subs (f,x,b);

c = b - (((b-a)/(fb-fa))*fb);

es = abs((c-a)/c);

fc= subs(f,x,c);

d=c-(((c-b)/(fc-fb))*fb);

es1= abs((d-b)/d);

if es <= ea;

break

end

a = b;

b = c;

N=[N; i a b c d es es1]

end

plot (N(:,1),N(:,7),'--gs','LineWidth',3,...

'MarkerEdgeColor','k',...

'MarkerFaceColor','g',...

'MarkerSize',5)

title ('SECANT METHOD')

xlabel ('Iterasi')

ylabel ('%error')

grid

Newton Raphson METHOD

4 akar dari fungsi persamaan berikut :

(f = x^4+2*x^3-7*x^2+3)

KOde Matlab :

%NEWTON METHOD
syms x
f = x^4+2*x^3-7*x^2+3;
a =0.1;
n = 100;
ea = 0.000001;
N = [];
for i = 1:n;
fa = subs (f,x,a);
fb = diff (f,x);
fc = subs (fb,x,a);
b = a - (fa/fc);
es = abs ((b-a)/b);
fd= subs (f,x,b);
fe= subs(fb,x,b);
c= b-(fd/fe);
es2 = abs ((c-b)/c);
ff=subs(f,x,c);
fg=subs(fb,x,c);
d= c-(ff/fg);
es3= abs ((d-c)/d);
if es <= ea
break
end
a = b;
N = [N; i a b c d es es2 es3]
end
plot (N(:,1),N(:,8),'--rs','LineWidth',3,...
'MarkerEdgeColor','k',...
'MarkerFaceColor','g',...
'MarkerSize',5)
title ('Newton METHOD')
xlabel ('Iterasi')
ylabel ('%error')
grid

Grafik :

Falsi Method

Falsing Method

Sekarang kita gunakan metode yang lain yaitu regulas falsi. Fungsi yang digunakan tetap sama hanya saja perhitungannya yang berbeda.

Kode MATLAB :

%Falsepoint

a=0;

b=1;

n=100;

es= 0.00001;

ea= 1;

syms x

f= x^3-3*x+1;

N= [a b ea];

for i=1:n;

fa= subs (f,x,a);

fb= subs (f,x,b);

c= ((b*fa)-(a*fb))/(fa-fb);

fc= subs (f,x,c);

if (fa*fc)<0

ea= (b-c)^2/b;

b=c;

else

ea=(a-c)^2/b;

a=c;

end

if ea<=es

break

end

N=[N;a b ea]

end

Tampak bahwa hanya rumus perhitungannya saja yang diubah. Jika program dijalankan maka hasilnya seperti berikut:

Hasil :

N =

[ 0, 1, 1]

[ 0, 1/2, 1/4]

N =

[ 0, 1, 1]

[ 0, 1/2, 1/4]

[ 0, 4/11, 9/242]

N =

[ 0, 1, 1]

[ 0, 1/2, 1/4]

[ 0, 4/11, 9/242]

[ 0, 121/347, 3249/5297996]

Pada iterasi ke-3 perhitungan dihentikan karena nilai absolut dari fx₀ lebih kecil dari nilai toleransi. Berdasarkan percobaan, tampak bahwa metode regula falsi lebih cepat dalam mendapatkan hasil.

Grafik :

Bisection Method

BISECTION METHOD

Kode MATLAB :

a = 0; %tebakan awal 1 (x1)

b = 1; %tebakan awal 2 (x2)

c = (a+b)/2; %rata-rata (xm)

n = 100; %jumlah iterasi

es = 0.001; %batas eror

ea = 1; %eror sebenarnya

syms x %mendeklarasikan variabel x

f= x^3-3*x+1; %fungsi yang dihitung akarnya

N = [i a b c ea];

for i = 1:n

c = (a+b)/2;

fa = subs (f,x,a);

fb = subs (f,x,b);

fc = subs (f,x,c);

if (fa*fc)<0

ea = (b-c)^2/b;

b = c;

else

ea = (a-c)^2/a;

a = c;

end

if ea<=es

break

end

N = [N; i a b c ea]

End

Plot (N(:,1),N(:,5))

Grafik :