1. Begin

Algoritma untuk menyelesaikan persamaan f(X)=0 dengan
Metode Regula Falsi:
1.
2.
3.
4.
5.
6.
Begin;
Definition …Y=f(X0)
Read(..X0,..X1,..E);
Y0=..(f(X0); Y1=..f(X1);
D  Y0.Y1;
If D> 0..then write ..(‘Akar bukan pada interval’)goto 14;...
7. X2 …
…
Yo=… 1-10+11 = 2..
Y1=… 8-20+11 = -1..
D = …-2..
X2 =
..
Y2 = …-1,0369..
P = …-2,0738..
Xtetap = Xt = X0 = …1.. dan Ytetap = Yt = Y0 = 2
8. Y2  ..f(X2)
9. P  ..Y0.Y2
10.If P = 0 then..(‘Akar =’,..X2);
11.If P < 0 then
Begin
Xt  ..X0; Yt  ..Y0;
i  2;
Repeat
Begin
Xi+1 =…
Xi+1 =…
..
..
Yi+1  f(Xi+1)
Yi+1 = …(1,4390)3 – 10(1,4390) + 11 = -0,4102…
Xi+1 
Yi+1 = …-0,4102…
Yi+1  f(Xi+1)
i i+1
R Abs(Y i+1)
End;
Until R
Perhitungan:
Persamaan X3 – 10X + 11 = 0
Dengan X0 = 0, X1 = 2 dan E = 0,1
R Abs(Y i+1) =…0,4102…
Diulangi terus sampai ketemu R
E;
E
..End;
12.If P > 0 then;
Begin
Xt  Xi ; Yt  Y1;
i  2;
Repeat
Begin
Xi+1 
Yi+1  f(Xi+1)
i i+1
R Abs(Y i+1)
End;
Until R
E;
End;
13.Write (“Akar = X i “);
14. End.
Sehingga jika dijalankan dalam table hasilnya….
Xi+1 
Yi+1  f(Xi+1)
R Abs(Y i+1)
-1,0369
-0,4102
0,4102
1,4390
-0,4102
-0,1036
0,1036
1,3643
-0,1036
-0,0228
0,0228
Iterasi ke-i
Xi
Yi
2
1,6666
3
4
Algoritma untuk menyelesaikan persamaan f(X)=0 dengan
Metode Newton Rhapson
1.
2.
3.
4.
5.
Begin;
Definition …Y=f(X)…dan Ft(X);
Read(..X0,..E);
i  … 0;..
Repeat
Begin
Yi  F(Xi);
Yti  Ft(Xi);
D  ...Abs(Ft(Xi))…
If D …
..goto 7;…
Else
Begin;
Xi+1 …
Perhitungan:
Dengan X0 = 1 dan E = 0,1
Persamaan X3 – 10X + 11 = 0  Ft(X) = 3X2 - 10
F(X) = …(1)3 – 10(1) + 11 = 2…
Ft(X) = …3(1)2 – 10 = -7…
X0 =…1= 1,2857
F(X0) = …(1,2857)3 – 10(1,2857) + 11 = 0,2683…
Yt0 = Ft(X0) = …3(-7)2 – 10 = -1,0390…
D = …1,039..
X1 = …1,2857…
Xi+1 = …1,2857 –
= 1,5439…
;...
R Abs
i  i+1
6.
7.
End;
End;
Until R E;
Write (“Akar = “, Xi);
End.
Sehingga jika dijalankan dalam table hasilnya….
Iterasi ke-i
Xi
Yi
Yti
0
1
2
1
1,2857
1,5439
2
0,2683
-0,7589
-7
-1,0390
-13,3649
Xi+1 …
1,2857
1,5439
1,6007
...
R Abs
0,2857
0,2008
0,0368
Algoritma menyelesaikan persamaan Linier dengan 3
peubah:
Missal persamaan linier:
2X1 + 3X2 + 5X3 = 17
4X1 + 8X2 + 9X3 = 37
6X1 + 7X2 + 8X3 = 40
Bila dijabarkan dalam matriks:
Sehingga didapat:
a11 = 2 ; a12 = 3 ; a13 = 5 ; a14 = 17
a21 = 4 ; a22 = 8 ; a23 = 9 ; a24 = 37
a31 = 6 ; a32 = 7 ; a33 = 8 ; a34 = 40
4.
Xn 
=
= -1
;
n=3; X3 
i = 2; s = 0
j=n  s = s + aij.Xj
j=3  s = s + a23.X3 = 0 + (-1).1 = -1
Xn 
 Xi 
X2 
i = 1; s = 0
j=n  s = s + aij.Xj
j=3  s = s + a13.X3 = 0 + 5.1 = 5
j=2  s = s + a12.X2 = 5 + 3.2 = 11
End;
5.
Xn 
6.
For i  (n-1) down to 1 do
Begin
s0
For j=n down to 1 do
Begin
s  s + aij + Xj
End;
8.
= =3
j=k  aij  aij-Uik*akj
j=2  a32  a32-U32*a22 = -2 – (-1).2 = 0
j=3  a33  a33-U32*a23 = -7 – (-1).(-1) = -8
j=4  a34  a34-U32*a24 = -11 – (-1).3 = -8
End;
End;
For I = 1 to n do
Begin
Write (‘Xi’);
End;
End.
j=k  aij  aij-Uik*akj
j=1  a21  a21-U21*a11 = 4 – 2.2 = 0
j=2  a22  a22-U21*a12 = 8 – 2.3 = 2
j=3  a23  a23-U21*a13 = 9 – 2.5 = -1
j=4  a24  a24-U21*a14 = 37 – 2.17 = 3
i=3
For j = k to (n+1) do
Begin
aij  aij-Uik*akj
End;
7.
= =2
j=k  aij  aij-Uik*akj
j=1  a31  a31-U31*a11 = 6 – 3.2 = 0
j=2  a32  a32-U31*a12 = 7 – 3.3 = -2
j=3  a33  a33-U31*a13 = 8 – 3.5 = -7
j=4  a34  a34-U31*a14 = 40 – 3.17 = -11
Begin;
Read(..n..);
For i = 1 to n;
Begin
For j=1 to (n+1) do
Begin
Read (aij)
End;
End;
For k=1 to (n-1) do
Begin;
For i = (k+1) to n do
Begin
Xi 
i=2
i=3
Algoritmanya:
1.
2.
3.
Perhitungan persamaannya:
“{dimana
X1 
Catatan:
Pada pers: Xn 
 Xi 
(down to) hanya n, sedangkan (n+1) tetap.
(index)}”
, yang berubah