wardah aulia: progran matlab

Latihan 1

Metode bisection

n=0;

err=0.00000001;

b=2;

a=0;

disp('metode bisection')

disp('-------------------------------------------')

disp(' n a b m fx(m)')

disp('-------------------------------------------')

while abs (a-b)>=err

m=(a+b)/2;

if fx(a)*fx(m)<0

b=m;

else

a=m;

end

if fx(a)*fx(b)==0

exit

end

disp(sprintf('%3g %8.5f %8.5f %8.5f %8.5f',n,a,b,m,fx(m)))

n=n+1;

end

disp(sprintf('Hasil Akar=%7.5f',m))

xg=linspace(-0.5,2.5);

yg=fx(xg);

plot(xg,yg)

grid on

Fungsinya

function y=fx(x)

%y=x.^3-x.^2-1;

y=exp(-3*x).*cos(2*x-1)-sin(3*x);

Grafik fungsi bisection

Metode False Position

clear all

clc

n=0;

err=0.00000001;

b=1.05;

a=0.8;

disp('metode false position')

disp('-------------------------------------------')

disp(' n a b w fx(w)')

disp('-------------------------------------------')

selisih=1;

wl=1;

while selisih >=err

w=(abs(f(b)*a)-abs(f(a)*b))/(abs(f(b))-abs(f(a)));

if fx(a)*fx(b)<=0

b=w;

else

a=w;

end

selisih=abs(wl-w);

wl=w;

disp(sprintf('%3g %8.5f %8.5f %8.5f %8.5f',n,a,b,w,fx(w)))

n=n+1;

end

disp(sprintf('Hasil Akar=%7.5f',w))

xg=linspace(-0.5,2.5);

yg=fx(xg);

plot(xg,yg)

grid on

latihan 2

Metode Gauss Pivot

Fungsinya

function x = gauss_pivot(A,b)

[n,l] = size(A);

for i = 1:n-1

[pivot,k] = max(abs(A(i:n,i)));

if k > 1

temp1 = A(i,:);

temp2 = b(i,:);

A(i,:) = A(i+k-1,:);

b(i,:) = b(i+k-1,:);

A(i+k-1,:) = temp1;

b(i+k-1,:) = temp2;

end

for h = i+1:n

m = A(h,i)/A(i,i);

A(h,:) = A(h,:)-m*A(i,:);

b(h,:) = b(h,:)-m*b(i,:);

end

x(n,:) = b(n,:)/A(n,n);

for i = n-1:-1:1

x(1,:) = (b(i,:)-A(i,i+1:n)*x(i+1:n,:))/ A(i,i);

end

Program matlab’a

clc

clear all

AA = [2 1 3 ; 4 3 10 ; 2 4 17]

bb = [11 ; 28 ; 31]

x = gauss_pivot (AA,bb)

Metode Jacobi

Fungsinya

function x = jacobi(A,b,tol,max_i)

[n m] = size(A);

for i = 1:n

xlama(i) = b(i)/A(i,i);

end

xlama = xlama';

C = -A;

for i = 1:n

C(i,i) = 0.0;

C(i,:) = C(i,:)/A(i,i);

d(i,1) = xlama(i);

end

i = 1;

while (i<=max_i)

xbaru = C*xlama + d;

if norm(xbaru - xlama) <=tol;

x = xbaru;

disp('jacobi method konvergen')

return;

else

x = xbaru;

xlama = xbaru;

end

disp ([i xbaru']);

i = i + 1;

end

disp('jacobi method tidak konvergen')

program matlabnya

clc

clear all

AA = [2 1 3; 4 3 10; 2 4 17]

bb = [11; 28; 31]

err = 0.00000001;

maks = 1000;

xx = jacobi(AA,bb,err,maks)

latihjan 3

Metode Secant

Fungsinya

function y=f(x)

y = x.^3-3*x.^2-x+9;

Program matlabnya

clc

clear all

x(1) = 0;

x(2) =-1;

err = 0.00001;

i = 2;

disp(' metode secant ')

disp('------------------------------------')

disp(' i x f(x) ')

disp('------------------------------------')

disp(sprintf('%2g %14.10f %15.10f',i-1,x(i),f(x(i))))

while abs(x(i)-x(i-1))>err

x(i+1) = x(i)-f(x(i))*(x(i)-x(i-1))/(f(x(i))-f(x(i-1)));

disp(sprintf('%2g %14.10f %15.10f',i,x(i+1),f(x(i+1))))

i = i+1;

end

disp(sprintf('Hasil Akar=%7.5f',x(i)+1))

xg=linspace(-2,1);

yg=f(xg);

plot(xg,yg)

grid on

Grafiknya

Latihan 4

Metode Newton Raphson

Fungsinya

function [y,y1]= fungsi_x(x)

y= x^3-3*x-20;

y1=3*x^2-3;

Program matlab’a

clc

clear all

x0=5;

err=0.000001;

fprintf('err =%10.8f, x0=%8.6f\n',err,x0);

%y=x^3-3*x-20;

%y1=3*x^2-3;

[y,y1]= fungsi_x(x0);

fprintf('y(x0)=%11.8f, y1(x0)=%10.8f',y,y1);

i=1;

x=x0-y/y1;

fprintf(' maka x%g=%10.8f\n',i,x);

while abs((x-x0)/x)>err

x0=x;

[y,y1]=fungsi_x(x0);

fprintf('y(x%g)=%11.8f, y1(x%g)=%10.8f',i,y,i,y1);

i=i+1;

x=x0-y/y1;

fprintf(' maka x%g=%11.8f\n',i,x);

end

%fprintf('|(x-x0)/x|=%10.8f<=err=%10.8f\n',abs((x-x0)/x),err);

fprintf('akarnya=%8.6f, banyak iterasi=%g\n',x,i);

Newton raphson yang diperbaharui

Fungsinya

function [y,y1,y2]= fs(x)

y= x^3-3*x-20;

y1=3*x^2-3;

y2=6*x;

Program matlab’a

clc

clear all

x0=5;

err=0.000001;

fprintf('err =%5.8f, x0=%8.6f\n',err,x0);

%y=x^3-3*x-20;

%y1=3*x^2-3;

%y2=6*x

[y,y1,y2]= fs(x0);

fprintf('y(x0)=%10.8f, y1(x0)=%9.8f y2(x0)=%10.8',y,y1,y2);

i=1;

x=x0-y*y1/(((y1)^2)-(y*y2));

fprintf(' maka x%g=%10.8f\n',i,x);

while abs((x-x0)/x)>err

x0=x;

[y,y1,y2]=fs(x0);

fprintf('y(x%g)=%11.8f, y1(x%g)=%9.8f y2(x%g)=%10.8f',i,y,i,y1,i,y2);

i=i+1;

x=x0-y*y1/(((y1)^2)-(y*y2));

fprintf(' maka x%g=%11.8f\n',i,x);

end

%fprintf('|(x-x0)/x|=%10.8f<=err=%10.8f\n',abs((x-x0)/x),err);

Perbedaan “fprintf” dengan “sprintf”:

Jika menggunakan “fprintf”, maka harus menggunakan tanda “ \n” di akhir.

Jika menggunakan “sprintf” maka cukup tambahkan “disp”di depannya.

fprintf('akarnya=%8.6f, banyak iterasi=%g\n',x,i);

Latihan 5

Interpolasi

1. Interpolasi Linear

clear all

clc

x =15;

y1=12.5;

y2=14;

x2=20;

x1=10;

for i= 1:10;

y=y1 + ((y2-y1)*(x-x1)/(x2-x1));

disp(sprintf('%3g %4g %7.4f', i, x, y))

x=x+10;

end

2. Interpolasi Kuadrik

clear all

clc

x1=1;

x2=2;

x3=3;

y1=5;

y2=2;

y3=3;

x=5;

for i= 1:10;

y=y1 *(x-x2)*(x-x3)/((x1-x2)*(x1-x3))+y2*(x-x1)*(x-x3)/((x2-x1)*(x2-x3))+y3*(x-x1)*(x- x2)/((x3-x1)*(x3-x2));

disp(sprintf('%3g %4g %7.4f ', i, x, y))

x=x+10;

end

3. Interpolasi Polinomial

clear all

clc

x1=3.2; x2= 2.7; x3=1; x4=4.8; x5=5.6;

y1=22; y2=17.8; y3=14.2; y4=38.3; y5=51.7;

X = [x1^0 x1^1 x1^2 x1^3 x1^4 ;

x2^0 x2^1 x2^2 x2^3 x2^4 ;

x3^0 x3^1 x3^2 x3^3 x3^4 ;

x4^0 x4^1 x4^2 x4^3 x4^4 ;

x5^0 x5^1 x5^2 x5^3 x5^4 ];

Y = [22; 17.8; 14.2; 38.3; 51.7];

A = X\Y;

x = 3;

for i = 1:10;

y = A(1)*x^0 + A(2)*x^1 + A(3)*x^2 + A(4)*x^3 + A(5)*x^4;

disp(sprintf(' %4.2f %4.2f %3g %4.2f %4.2f', i, x, y))

x = x + 1;

Catatan :

· Pada saat display % untuk tempat desimal ada yang menggunakan” g” dan ada juga yang menggunakan “f”. Menggunakan “g” karena tidak ada pecahan atau decimal, jika ada decimal maka pake “f”.

· Untuk penulisan matriks harus menggunakan huruf capital kemudian menggunakan kurung siku

end

Latihan 6 :

Sigma dalam matlab menggunakan for,, karena batasnya telah diketahui.

Integrasi

1. Metode trapezoidal

L =

)+ 2

h = (b-a)/n

tentukan variable sigma, misal “sg”

contoh ;

Penyelesaian :

Dari soal di atas diketahui :

a = -2

b = 2

f(x)=

h = 1

Maka program matlabnya adalah :

Fungsinya:

function y = f(x)

y =x*exp(2*x);

programnya:

h = 1;

a = -2;

b = 2;

sg = 0;

x = a;

n = (b-a)/h;

for i = 1:n-1;

x = x + h;

sg = sg + f(x);

end

L = h*(f(a)+f(b)+2*sg)/2;

disp(sprintf('Luas = %7.4f',L))

2. Metode SIMPSON

Untuk “simpson” ini perbedaan terletak pada rumus luasnya. Dimana ada pemisahan antara sigma ganjil dengan sigma genap. Jadi ada 2 sigma yang digunakan.

L =

)+ 2

contoh ;

Penyelesaian :

Dari soal di atas diketahui :

a = -2

b = 2

f(x)=

h = 1

Maka program matlabnya adalah :

Fungsinya:

function y = f(x)

y =x*exp(2*x);

Programnya matlab

h = 1;

a = -2;

b = 2;

sg1 = 0;

sg2 = 0;

x = a;

n = (b-a)/h;

for i = 1:2:n-1;

x = x + h;

sg1 = sg + f(x);

end

for i = 2:2:n-1;

x = x + h;

sg2 = sg + f(x);

end

L = h*(f(a)+f(b)+2*sg1+4*sg2)/3;

disp(sprintf('Luas = %7.4f',L))

untuk membuat grafiknya tambahkan pada program matlab :

xm=linspace(-2,2);

ym=f(xm);

plot(xm,ym)

grid on

Penulisan % pada matlab.. angka depan harus lebih besar dari angka belakang, misal %8.4f.

Latihan 7

Differensial

1. Metode Selisih Maju

Fungsinya:

function y=f(x)

y=3*x*exp(-1*x);

Program Matlabnya:

clc

clear all

disp(' diferensial selisih maju ')

disp(' i h f1 ')

disp('--------------------------- ')

h=0.1;

x=2;

for i=1:10;

f1=(f(x+h)-f(x))/h;

disp(sprintf('%3g %10.8f %10.8f', i,h,f1))

h=h/10;

end

2. Metode Titik Terpusat

Fungsinya :

function y=f(x)

y=3*x*exp(-1*x);

Program Matlabnya :

clc

clear all

disp(' Metode Titik Terpusat ')

disp(' i h f1 ')

disp('--------------------------- ')

h=0.1;

x=2;

for i=1:10;

f1=(f(x+h)-f(x-h))/(2*h);

disp(sprintf('%3g %10.8f %10.8f', i,h,f1))

h=h/10;

end

3. Formula Titik Tiga

Fungsinya :

function y=f(x)

y=3*x*exp(-1*x);

Program Matlabnya :

clc

clear all

disp(' Formula Titik Tiga ')

disp(' i h f1 ')

disp('--------------------------- ')

h=0.1;

x=2;

for i=1:10;

f1=((-3*f(x))+(4*f(x+h))-(f(x+(2*h))))/(2*h);

disp(sprintf('%3g %10.8f %10.8f', i,h,f1))

h=h/10;

end

4. Formula Titik Lima

Fungsinya :

function y=f(x)

y=3*x*exp(-1*x);

Program Matlabnya 1:

clc

clear all

disp(' Formula Lima Titik Part 1 ')

disp(' i h f1 ')

disp('--------------------------- ')

h=0.1;

x=2;

for i=1:10;

f1=((8*f(x+h))-(8*f(x-h))+(f(x-(2*h)))-(f(x+(2*h))))/(12*h);

disp(sprintf('%3g %10.8f %10.8f', i,h,f1))

h=h/10;

end

Program Matlab 2:

clc

clear all

disp(' Formula Lima Titik part 2 ')

disp(' i h f1 ')

disp('--------------------------- ')

h=0.1;

x=2;

for i=1:10;

f1=((-25*f(x))+(48*f(x+h))-(36*f(x+(2*h)))+(16*f(x+(3*h)))-(3*f(x+(4*h))))/(12*h);

disp(sprintf('%3g %10.8f %10.8f', i,h,f1))

h=h/10;

end

Grafik

1. Jika fungsinya f(x)=ln x, yang dalam matlab ditulis f(x)=log(x)

Latihan 8

Differensial Tingkat Dua dan Romberg

1. Differensial Tingkat Dua

Program Matlabnya

disp('Diferensial Tingkat Dua')

disp('------------------------------')

disp(' i h f11 ')

disp('===================================')

h = 0.1;

x = 2;

for i = 1:3;

f11 = (f(x-2*h)-(2*f(x))+f(x+2*h))/(4*(h^2));

disp(sprintf('%3g %12.3f %15.5f', i,h,f11))

h = h/10;

end

Fungsinya

function y=f(x)

%y =3*x*exp(-1*x);

%y =log(x);

y =x^2*exp(1*x);

Atau

function f2 = turun2(fstr,x,h,r)

disp('Diferensial Tingkat Dua')

disp('------------------------------')

disp(' i h f2 ')

disp('================================== ')

f = inline(fstr);

for i = 1:r;

f2 = (f(x-2*h)-(2*f(x))+f(x+2*h))/(4*(h^2));

disp(sprintf('%3g %12.3f %15.5f', i,h,f2))

h = h/10;

end

2. Romberg

function L = rmb(fstr,a,b,n)

f = inline(fstr);

L = 0;

h =(b-a)/n;

for i = 0:n;

if i==0|i==n

L=L+f(a+i*h)*h/2;

else

L=L+f(a+i*h)*h;

end

%fprintf('L=%12.9f\n',L)

Atau

clear all

disp('Integral Romberg')

disp('------------------------------')

disp(' j LL ')

disp('================================== ')

for j=1:10;

LL = rmb('1/(1+x)',0,1,j);

disp(sprintf('%3g %15.9f',j,LL));

end

Latihan 9

Persamaan Differensial

1. Difeuler

Program Matlabnya

function y=difeuler(fs,a,b,y0,n)

f=inline(fs);

h=(b-a)/n;

y(1)=y0;

x(1)=a;

x(2)=x(1)+h;

for i=1:n;

y(i+1)=y(i)+h*(f(x(i)));

x(i+1)=x(i)+h;

end

[x' y']

plot(x,y)

2. Metode Poligon

Programnya

function y=poligon(fs,a,b,y0,n)

f=inline(fs);

h=(b-a)/n;

y(1)=y0;

x(1)=a;

x(2)=x(1)+h;

for i=1:n;

x(i+1)=x(i)+h;

y_st=y(i)+f(x(i),y(i))*h/2;

x_st=f(x(i)+h/2,y_st);

y(i+1)=y(i)+x_st*h;

end

[x' y']

plot(x,y)

3. Metode Raphson

Program matlabnya

function y=raphson(fs,a,b,y0,n)

f=inline(fs,'x','y');

h=(b-a)/n;

y(1)=y0;

x(1)=a;

x(2)=x(1)+h;

for i=1:n;

x(i+1)=x(i)+h;

k1=f(x(i),y(i));

k2=f(x(i)+h*3/4,y(i)+h*k1*3/4);

y(i+1)=y(i)+h*(k1*1/3+2/3*k2);

end

[x' y']

plot(x,y)

4. Metode Rungekutta 4

Program matlabnya

function y=rungekutta(fs,a,b,y0,n)

f=inline(fs,'x','y');

h=(b-a)/n;

y(1)=y0;

x(1)=a;

x(2)=x(1)+h;

for i=1:n;

x(i+1)=x(i)+h;

k1=h*f(x(i),y(i));

k2=h*f(x(i)+h/2,y(i)+k1/2);

k3=h*f(x(i)+h/2,y(i)+k2/2);

k4=h*f(x(i)+h/2,y(i)+k3);

y(i+1)=y(i)+(k1+2*k2+2*k3+k4)/6;

end

[x' y']

plot(x,y)

5. Metode Rungekutte 3

Program Matlabnya

function y=rungekutta3(fs,a,b,y0,n)

f=inline(fs,'x','y');

h=(b-a)/n;

y(1)=y0;

x(1)=a;

x(2)=x(1)+h;

for i=1:n;

x(i+1)=x(i)+h;

k1=f(x(i),y(i));

k2=f(x(i)+h/2,y(i)+k1*h/2);

k3=f(x(i)+h,y(i)-k1*h+2*h*k2);

y(i+1)=y(i)+(k1+4*k2+k3)/6;

end

[x' y']

plot(x,y)

6. Metode Heun

Program Matlabnya

function y=heun(fs,a,b,y0,n)

f=inline(fs,'x','y');

h=(b-a)/n;

y(1)=y0;

x(1)=a;

x(2)=x(1)+h;

for i=1:n;

x(i+1)=x(i)+h;

g(i)=f(x(i),y(i));

y(i+1)=y(i)+h*f(x(i),y(i));

g(i+1)=f(x(i+1),y(i+1));

y(i+1)=y(i)+h*(g(i)+g(i+1))/2;

end

[x' y']

plot(x,y)

Latihan 10

Lanjutan Dari Latihan 9

1. Dengan Metode Euler (PD tingkat 2)

function pd2=feuler(sf,sg,x0,y0,z0,h)

f=inline(sf);

g=inline(sg);

n=11;

x(1)=x0 ;

y(1)=y0 ;

z(1)=z0 ;

disp(' i-1 x(i) y(i) z(i)')

for i=1:n;

fprintf('%3g |%8.5f |%8.5f |%8.5f\n',i-1,x(i),y(i),z(i))

y(i+1)=y(i)+h*f(z(i));

z(i+1)=z(i)+h*g(x(i),y(i),z(i));

x(i+1)=x(i)+h;

end

pd2=z(n);

Cara memanggil feuler =feuler('z','1-2*x*y-3*z',0,0,0,0.2)

Hasil command window:

i-1 x(i) y(i) z(i)

0 | 0.00000 | 0.00000 | 0.00000

1 | 0.20000 | 0.00000 | 0.20000

2 | 0.40000 | 0.04000 | 0.28000

3 | 0.60000 | 0.09600 | 0.30560

4 | 0.80000 | 0.15712 | 0.29920

5 | 1.00000 | 0.21696 | 0.26940

6 | 1.20000 | 0.27084 | 0.22098

7 | 1.40000 | 0.31504 | 0.15839

8 | 1.60000 | 0.34671 | 0.08693

9 | 1.80000 | 0.36410 | 0.01288

10 | 2.00000 | 0.36668 |-0.05700

ans =

-0.0570

2. Dengan Metode Runge Kutta

function pd2=frungekutta2(sf,sg,x0,y0,z0,h)

f=inline(sf);

g=inline(sg);

n=11;

x(1)=x0;

y(1)=y0;

z(1)=z0;

disp(' i-1 x(i) k1 l1 k2 l2 y(i) z(i)')

k1=0;

l1=0;

k2=0;

l2=0;

for i=1:n;

fprintf('%3g |%8.5f |%8.5f |%8.5f |%8.5f |%8.5f |%8.5f|%8.5f\n',i-1,k1,l1,k2,l2,x(i),y(i),z(i))

l1=h*g(x(i),y(i),z(i));

k1=h*f(z(i));

k2=h*f(z(i)+l1);

l2=h*g(x(i)+h,y(i)+k1,z(i)+l1);

y(i+1)=y(i)+(k1+k2)/2;

z(i+1)=z(i)+(l1+l2)/2;

x(i+1)=x(i)+h;

end

pd2=z(n);

plot(x,y)

hasilnya:

i-1 x(i) k1 l1 k2 l2 y(i) z(i)

0 | 0.00000 | 0.00000 | 0.00000 | 0.00000 | 0.00000 | 0.00000| 0.00000

1 | 0.00000 | 0.20000 | 0.04000 | 0.08000 | 0.20000 | 0.02000| 0.14000

2 | 0.02800 | 0.11440 | 0.05088 | 0.03968 | 0.40000 | 0.05944| 0.21704

3 | 0.04341 | 0.06027 | 0.05546 | 0.00893 | 0.60000 | 0.10887| 0.25164

4 | 0.05033 | 0.02289 | 0.05491 |-0.01566 | 0.80000 | 0.16149| 0.25525

5 | 0.05105 |-0.00483 | 0.05008 |-0.03527 | 1.00000 | 0.21206| 0.23520

6 | 0.04704 |-0.02595 | 0.04185 |-0.04992 | 1.20000 | 0.25650| 0.19727

7 | 0.03945 |-0.04148 | 0.03116 |-0.05921 | 1.40000 | 0.29181| 0.14692

8 | 0.02938 |-0.05157 | 0.01907 |-0.06278 | 1.60000 | 0.31604| 0.08975

9 | 0.01795 |-0.05611 | 0.00673 |-0.06065 | 1.80000 | 0.32838| 0.03137

10 | 0.00627 |-0.05525 |-0.00478 |-0.05339 | 2.00000 | 0.32912|-0.02295

ans =

-0.0230

Grafiknya:

Latihan 11

Regresi Linear

Diperoleh data penelitian adalah sbb:

BB	50	70	56	64	66	73	74	78	83	85
TB	115	130	130	125	135	134	140	138	145	145

Dari data di atas bagaimana jika diketahui BB= 73,5 berapakah TB nya??

Jawaban :

Misalkan X= BB

Y=TB

S= sigma

Program Matlabnya :

X = [50 70 56 64 66 73 74 78 83 85];

Y = [115 130 130 125 135 134 140 138 145 145];

n = 10;

SXY = 0;

SX = 0;

SY = 0;

SX2= 0;

for i= 1:n

SXY = SXY + X(i)*Y(i);

SX = SX +X(i);

SY = SY +Y(i);

SX2= SX2 + X(i)^2;

end

a = (n*SXY-(SX*SY))/(n*SX2-SX^2);

b = SY/n- a*SX/n;

sprintf('Y = %7.4f x+ %7.4f',a,b)

hasilnya

Y = 0.7513 x+ 81.1807

Catatan : tinggal ganti x dengan yang ditanyakan

Atau :

function TB = regresi (BB);

X = [50 70 56 64 66 73 74 78 83 85];

Y = [115 130 130 125 135 134 140 138 145 145];

n = 10;

SXY = 0;

SX = 0;

SY = 0;

SX2= 0;

for i= 1:n

SXY = SXY + X(i)*Y(i);

SX = SX +X(i);

SY = SY +Y(i);

SX2= SX2 + X(i)^2;

end

a = (n*SXY-(SX*SY))/(n*SX2-SX^2);

b = SY/n- a*SX/n;

sprintf('Y = %7.4f x+ %7.4f',a,b)

TB = a*BB+b;

sprintf('berat badan = %3g\n',BB)

sprintf('tinggi badan = %8.4f\n',TB)

catatan: simpan sesuai nama funsi jangan diganti

cara memanggil :

misalkan berapa tinggi badan dari berat badan 40??

Maka ketik pada comand window!

regresi(40)

hasil:

berat badan = 40

tinggi badan = 111.2347

function Yt = regresi2 (d);

t = linspace (0,pi/4,10);

Y = sin (2*t);

n = 10;

StY = 0;

St = 0;

SY = 0;

St2= 0;

for i= 1:n ;

StY = StY + t(i)*Y(i);

St = St +t(i);

SY = SY +Y(i);

St2= St2 + t(i)^2;

end

a = (n*StY-(St*SY))/(n*St2-St^2);

b = SY/n- a*St/n;

sprintf('Y = %7.4f x+ %7.4f',a,b)

Y = a*d+b;

sprintf('t = %3g\n',d)

sprintf('Y = %8.4f\n',Y)

hasil:

regresi2(30)

Y = 1.3117 x+ 0.1064

t = 30

Y = 39.4588

Regresi Eksponensial

function TB= eksponensial (BB)

X = [50 70 56 64 66 73 74 78 83 85];

Y = [115 130 130 125 135 134 140 138 145 145];

n = 10;

SXY = 0;

SX = 0;

SY = 0;

SX2= 0;

for i= 1:n

SXY = SXY + X(i)*Y(i);

SX = SX +X(i);

SY = SY +Y(i);

SX2= SX2 + X(i)^2;

end

a = (n*SXY-(SX*SY))/(n*SX2-SX^2);

b = log(SY/n)-a*SX/n;

sprintf('Y = %7.4f x+ %7.4f',a,b)

TB = a*BB+b;

sprintf('berat badan = %3g\n',BB)

sprintf('tinggi badan = %8.4f\n',TB)

hasil:

eksponensial(65)

Y = 0.7513 x+ -47.6237

berat badan = 65

tinggi badan = 1.2140

1.2140

Latihan 1

Metode bisection

n=0;

err=0.00000001;

b=2;

a=0;

disp('metode bisection')

disp('-------------------------------------------')

disp(' n a b m fx(m)')

disp('-------------------------------------------')

while abs (a-b)>=err

m=(a+b)/2;

if fx(a)*fx(m)<0

b=m;

else

a=m;

end

if fx(a)*fx(b)==0

exit

end

disp(sprintf('%3g %8.5f %8.5f %8.5f %8.5f',n,a,b,m,fx(m)))

n=n+1;

end

disp(sprintf('Hasil Akar=%7.5f',m))

xg=linspace(-0.5,2.5);

yg=fx(xg);

plot(xg,yg)

grid on

Fungsinya

function y=fx(x)

%y=x.^3-x.^2-1;

y=exp(-3*x).*cos(2*x-1)-sin(3*x);

Grafik fungsi bisection

Metode False Position

clear all

clc

n=0;

err=0.00000001;

b=1.05;

a=0.8;

disp('metode false position')

disp('-------------------------------------------')

disp(' n a b w fx(w)')

disp('-------------------------------------------')

selisih=1;

wl=1;

while selisih >=err

w=(abs(f(b)*a)-abs(f(a)*b))/(abs(f(b))-abs(f(a)));

if fx(a)*fx(b)<=0

b=w;

else

a=w;

end

selisih=abs(wl-w);

wl=w;

disp(sprintf('%3g %8.5f %8.5f %8.5f %8.5f',n,a,b,w,fx(w)))

n=n+1;

end

disp(sprintf('Hasil Akar=%7.5f',w))

xg=linspace(-0.5,2.5);

yg=fx(xg);

plot(xg,yg)

grid on

latihan 2

Metode Gauss Pivot

Fungsinya

function x = gauss_pivot(A,b)

[n,l] = size(A);

for i = 1:n-1

[pivot,k] = max(abs(A(i:n,i)));

if k > 1

temp1 = A(i,:);

temp2 = b(i,:);

A(i,:) = A(i+k-1,:);

b(i,:) = b(i+k-1,:);

A(i+k-1,:) = temp1;

b(i+k-1,:) = temp2;

end

for h = i+1:n

m = A(h,i)/A(i,i);

A(h,:) = A(h,:)-m*A(i,:);

b(h,:) = b(h,:)-m*b(i,:);

end

x(n,:) = b(n,:)/A(n,n);

for i = n-1:-1:1

x(1,:) = (b(i,:)-A(i,i+1:n)*x(i+1:n,:))/ A(i,i);

end

Program matlab’a

clc

clear all

AA = [2 1 3 ; 4 3 10 ; 2 4 17]

bb = [11 ; 28 ; 31]

x = gauss_pivot (AA,bb)

Metode Jacobi

Fungsinya

function x = jacobi(A,b,tol,max_i)

[n m] = size(A);

for i = 1:n

xlama(i) = b(i)/A(i,i);

end

xlama = xlama';

C = -A;

for i = 1:n

C(i,i) = 0.0;

C(i,:) = C(i,:)/A(i,i);

d(i,1) = xlama(i);

end

i = 1;

while (i<=max_i)

xbaru = C*xlama + d;

if norm(xbaru - xlama) <=tol;

x = xbaru;

disp('jacobi method konvergen')

return;

else

x = xbaru;

xlama = xbaru;

end

disp ([i xbaru']);

i = i + 1;

end

disp('jacobi method tidak konvergen')

program matlabnya

clc

clear all

AA = [2 1 3; 4 3 10; 2 4 17]

bb = [11; 28; 31]

err = 0.00000001;

maks = 1000;

xx = jacobi(AA,bb,err,maks)

latihjan 3

Metode Secant

Fungsinya

function y=f(x)

y = x.^3-3*x.^2-x+9;

Program matlabnya

clc

clear all

x(1) = 0;

x(2) =-1;

err = 0.00001;

i = 2;

disp(' metode secant ')

disp('------------------------------------')

disp(' i x f(x) ')

disp('------------------------------------')

disp(sprintf('%2g %14.10f %15.10f',i-1,x(i),f(x(i))))

while abs(x(i)-x(i-1))>err

x(i+1) = x(i)-f(x(i))*(x(i)-x(i-1))/(f(x(i))-f(x(i-1)));

disp(sprintf('%2g %14.10f %15.10f',i,x(i+1),f(x(i+1))))

i = i+1;

end

disp(sprintf('Hasil Akar=%7.5f',x(i)+1))

xg=linspace(-2,1);

yg=f(xg);

plot(xg,yg)

grid on

Grafiknya

Latihan 4

Metode Newton Raphson

Fungsinya

function [y,y1]= fungsi_x(x)

y= x^3-3*x-20;

y1=3*x^2-3;

Program matlab’a

clc

clear all

x0=5;

err=0.000001;

fprintf('err =%10.8f, x0=%8.6f\n',err,x0);

%y=x^3-3*x-20;

%y1=3*x^2-3;

[y,y1]= fungsi_x(x0);

fprintf('y(x0)=%11.8f, y1(x0)=%10.8f',y,y1);

i=1;

x=x0-y/y1;

fprintf(' maka x%g=%10.8f\n',i,x);

while abs((x-x0)/x)>err

x0=x;

[y,y1]=fungsi_x(x0);

fprintf('y(x%g)=%11.8f, y1(x%g)=%10.8f',i,y,i,y1);

i=i+1;

x=x0-y/y1;

fprintf(' maka x%g=%11.8f\n',i,x);

end

%fprintf('|(x-x0)/x|=%10.8f<=err=%10.8f\n',abs((x-x0)/x),err);

fprintf('akarnya=%8.6f, banyak iterasi=%g\n',x,i);

Newton raphson yang diperbaharui

Fungsinya

function [y,y1,y2]= fs(x)

y= x^3-3*x-20;

y1=3*x^2-3;

y2=6*x;

Program matlab’a

clc

clear all

x0=5;

err=0.000001;

fprintf('err =%5.8f, x0=%8.6f\n',err,x0);

%y=x^3-3*x-20;

%y1=3*x^2-3;

%y2=6*x

[y,y1,y2]= fs(x0);

fprintf('y(x0)=%10.8f, y1(x0)=%9.8f y2(x0)=%10.8',y,y1,y2);

i=1;

x=x0-y*y1/(((y1)^2)-(y*y2));

fprintf(' maka x%g=%10.8f\n',i,x);

while abs((x-x0)/x)>err

x0=x;

[y,y1,y2]=fs(x0);

fprintf('y(x%g)=%11.8f, y1(x%g)=%9.8f y2(x%g)=%10.8f',i,y,i,y1,i,y2);

i=i+1;

x=x0-y*y1/(((y1)^2)-(y*y2));

fprintf(' maka x%g=%11.8f\n',i,x);

end

%fprintf('|(x-x0)/x|=%10.8f<=err=%10.8f\n',abs((x-x0)/x),err);

Perbedaan “fprintf” dengan “sprintf”:

Jika menggunakan “fprintf”, maka harus menggunakan tanda “ \n” di akhir.

Jika menggunakan “sprintf” maka cukup tambahkan “disp”di depannya.

fprintf('akarnya=%8.6f, banyak iterasi=%g\n',x,i);

Latihan 5

Interpolasi

1. Interpolasi Linear

clear all

clc

x =15;

y1=12.5;

y2=14;

x2=20;

x1=10;

for i= 1:10;

y=y1 + ((y2-y1)*(x-x1)/(x2-x1));

disp(sprintf('%3g %4g %7.4f', i, x, y))

x=x+10;

end

2. Interpolasi Kuadrik

clear all

clc

x1=1;

x2=2;

x3=3;

y1=5;

y2=2;

y3=3;

x=5;

for i= 1:10;

y=y1 *(x-x2)*(x-x3)/((x1-x2)*(x1-x3))+y2*(x-x1)*(x-x3)/((x2-x1)*(x2-x3))+y3*(x-x1)*(x- x2)/((x3-x1)*(x3-x2));

disp(sprintf('%3g %4g %7.4f ', i, x, y))

x=x+10;

end

3. Interpolasi Polinomial

clear all

clc

x1=3.2; x2= 2.7; x3=1; x4=4.8; x5=5.6;

y1=22; y2=17.8; y3=14.2; y4=38.3; y5=51.7;

X = [x1^0 x1^1 x1^2 x1^3 x1^4 ;

x2^0 x2^1 x2^2 x2^3 x2^4 ;

x3^0 x3^1 x3^2 x3^3 x3^4 ;

x4^0 x4^1 x4^2 x4^3 x4^4 ;

x5^0 x5^1 x5^2 x5^3 x5^4 ];

Y = [22; 17.8; 14.2; 38.3; 51.7];

A = X\Y;

x = 3;

for i = 1:10;

y = A(1)*x^0 + A(2)*x^1 + A(3)*x^2 + A(4)*x^3 + A(5)*x^4;

disp(sprintf(' %4.2f %4.2f %3g %4.2f %4.2f', i, x, y))

x = x + 1;

Catatan :

· Untuk penulisan matriks harus menggunakan huruf capital kemudian menggunakan kurung siku

end

Latihan 6 :

Sigma dalam matlab menggunakan for,, karena batasnya telah diketahui.

Integrasi

1. Metode trapezoidal

L =

)+ 2

h = (b-a)/n

tentukan variable sigma, misal “sg”

contoh ;

Penyelesaian :

Dari soal di atas diketahui :

a = -2

b = 2

f(x)=

h = 1

Maka program matlabnya adalah :

Fungsinya:

function y = f(x)

y =x*exp(2*x);

programnya:

h = 1;

a = -2;

b = 2;

sg = 0;

x = a;

n = (b-a)/h;

for i = 1:n-1;

x = x + h;

sg = sg + f(x);

end

L = h*(f(a)+f(b)+2*sg)/2;

disp(sprintf('Luas = %7.4f',L))

2. Metode SIMPSON

Untuk “simpson” ini perbedaan terletak pada rumus luasnya. Dimana ada pemisahan antara sigma ganjil dengan sigma genap. Jadi ada 2 sigma yang digunakan.

L =

)+ 2

contoh ;

Penyelesaian :

Dari soal di atas diketahui :

a = -2

b = 2

f(x)=

h = 1

Maka program matlabnya adalah :

Fungsinya:

function y = f(x)

y =x*exp(2*x);

Programnya matlab

h = 1;

a = -2;

b = 2;

sg1 = 0;

sg2 = 0;

x = a;

n = (b-a)/h;

for i = 1:2:n-1;

x = x + h;

sg1 = sg + f(x);

end

for i = 2:2:n-1;

x = x + h;

sg2 = sg + f(x);

end

L = h*(f(a)+f(b)+2*sg1+4*sg2)/3;

disp(sprintf('Luas = %7.4f',L))

untuk membuat grafiknya tambahkan pada program matlab :

xm=linspace(-2,2);

ym=f(xm);

plot(xm,ym)

grid on

Penulisan % pada matlab.. angka depan harus lebih besar dari angka belakang, misal %8.4f.

Latihan 7

Differensial

1. Metode Selisih Maju

Fungsinya:

function y=f(x)

y=3*x*exp(-1*x);

Program Matlabnya:

clc

clear all

disp(' diferensial selisih maju ')

disp(' i h f1 ')

disp('--------------------------- ')

h=0.1;

x=2;

for i=1:10;

f1=(f(x+h)-f(x))/h;

disp(sprintf('%3g %10.8f %10.8f', i,h,f1))

h=h/10;

end

2. Metode Titik Terpusat

Fungsinya :

function y=f(x)

y=3*x*exp(-1*x);

Program Matlabnya :

clc

clear all

disp(' Metode Titik Terpusat ')

disp(' i h f1 ')

disp('--------------------------- ')

h=0.1;

x=2;

for i=1:10;

f1=(f(x+h)-f(x-h))/(2*h);

disp(sprintf('%3g %10.8f %10.8f', i,h,f1))

h=h/10;

end

3. Formula Titik Tiga

Fungsinya :

function y=f(x)

y=3*x*exp(-1*x);

Program Matlabnya :

clc

clear all

disp(' Formula Titik Tiga ')

disp(' i h f1 ')

disp('--------------------------- ')

h=0.1;

x=2;

for i=1:10;

f1=((-3*f(x))+(4*f(x+h))-(f(x+(2*h))))/(2*h);

disp(sprintf('%3g %10.8f %10.8f', i,h,f1))

h=h/10;

end

4. Formula Titik Lima

Fungsinya :

function y=f(x)

y=3*x*exp(-1*x);

Program Matlabnya 1:

clc

clear all

disp(' Formula Lima Titik Part 1 ')

disp(' i h f1 ')

disp('--------------------------- ')

h=0.1;

x=2;

for i=1:10;

f1=((8*f(x+h))-(8*f(x-h))+(f(x-(2*h)))-(f(x+(2*h))))/(12*h);

disp(sprintf('%3g %10.8f %10.8f', i,h,f1))

h=h/10;

end

Program Matlab 2:

clc

clear all

disp(' Formula Lima Titik part 2 ')

disp(' i h f1 ')

disp('--------------------------- ')

h=0.1;

x=2;

for i=1:10;

f1=((-25*f(x))+(48*f(x+h))-(36*f(x+(2*h)))+(16*f(x+(3*h)))-(3*f(x+(4*h))))/(12*h);

disp(sprintf('%3g %10.8f %10.8f', i,h,f1))

h=h/10;

end

Grafik

1. Jika fungsinya f(x)=ln x, yang dalam matlab ditulis f(x)=log(x)