Transcript Pertemuan_5

Mata Praktikum:
Metode Numerik
& FORTRAN
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Pertemuan 5
Metode Sekan
… … 2008
Copyright © 2008.
This presentation is dedicated to Laboratorium Informatika Universitas Gunadarma.
This presentation is for education purpose only.
Daftar Isi
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VII.
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Pendahuluan
Metode Sekan
Algoritma Sekan
Contoh Soal
Contoh Program
Laporan Akhir
Laporan Pendahuluan Pertemuan 6
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This presentation is for education purpose only.
VII
Daftar
Isi
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PENDAHULUAN
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PENDAHULUAN
Dalam komputasi terkadang sering dihadapkan dengan
permasalahan yang berkaiatan dengan analisa terhadap
numeril. Salah satunya mencari nilai akar sautu persamaan.
Dan untuk mencari akar dari suatu persamaan nonlinier
dapat digunakan beberapa metode. Secara metode
numerik, dapat digunakan 2 cara :
1.Tanpa menggunakan derivatif (turunan)
- Metode Biseksi
- Metode Regulafalsi
- Metode Sekan
- Metode Iterasi titik tetap
2.Menggunakan derivatif (turunan)
- Metode Newton-Raphson
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METODE SEKAN
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METODE SEKAN
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Metoda Sekan disebut juga metoda interpolasi linier.
Dalam prosesnya tidak dilakukan penjepitan akar
sehingga [x0,x1] tidak harus mengandung akar, serta f(x0)
dan f(x1) bisa bertanda sama.
f(x)
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x0
x1
x2
x
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Daftar
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Tarik garis lurus melalui (x0, f(x0)) dan (x1, f(x1)) dan
memotong sumbu x di (x2,0)
x 2  x1 
x 1  x 0 
* f x 
f  x 1   f x 0 
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METODE SEKAN
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Iterasi berikutnya dengan pergeseran :
x0  x1
x1  x2
Iterasi berlangsung sampai batas maksimum iterasi atau
sampai
x1  x 2
T
x1
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Daftar
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Daftar
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III
ALGORITMA SEKAN
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Algoritma Sekan
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a) Tentukan x0, x1, T, iterasi maksimum dan F(x)
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b) Hitung x2 = x1 – f(x1) (x1 - x0) / [f(x1) – f(x0)]
c) Jika nilai |(x1-x2) / x1| < T, tulis x2 sebagai akar dan akhiri
program. Jika tidak, lanjutkan ke langkah berikutnya.
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d) Jika jumlah iterasi > iterasi maksimum, akhiri program.
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e) x0 = x1
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f) x1 = x2
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g) Kembali ke b
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IV
CONTOH SOAL
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1/4
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Contoh Soal
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1. Cari akar dari f(x) = x3 – 2x – 5, dimana :
- x0 = 1
- x1 = 2
- Toleransi (T) = 0,001 atau 10-3
Jawab :
Iterasi 1 :
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Daftar
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f x  x  x 
0
 1
x x 
2 1 f  x   f  x 
 1
 0
 2,2
 2  f 22 1  2  -11
-1 - 6
f 2 f 1
f 2,2  1,248
2  2,2  0,1
2
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2/4
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Contoh Soal
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Iterasi 2 :
x  2  f(2)  -1
0
x  2,2  f(2,2)  1,248
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x  2,2  1,2482,2  2  2,089
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1,248  -1
f 2,089  0,062
2,2  2,089  0,051
2,2
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3/4
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Contoh Soal
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Iterasi 3 :
x  2,2  f(2,2) 1,248
0
x  2,089  f(2,089)  -0,062
1
- 0,0622,089  2,2  2,094
x  2,089 
2
- 0,062 1,248
f 2,094  -0,006
2,089  2,094
 0,002
2,089
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4/4
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Contoh Soal
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Iterasi 4 :
x  2,089  f(2,089)  -0,062
0
x  2,094  f(2,094)  -0,006
1
- 0,0062,094  2,089  2,095
x  2,094 
2
- 0,006 - 0,062
f 2,095  0,005
2,094  2,095
 0,0005
2,094
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Karena Tolerasi (T) yang didapat = 0,0005 < 10-3
Jadi, akarnya adalah = 2,095
Daftar
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Representasi dalam bentuk tabel
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Iterasi
x0
x1
x2
|(x1-x2) / x1|
1
1
2
2,2
0,1
2
2
2,2
2,089
0,051
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3
2,2
2,089
2,094
0,002
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4
2,089
2,094
2,095
0,0005
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Karena Tolerasi (T) yang didapat = 0,0005 < 10-3
Jadi, akarnya adalah = 2,095
Daftar
Isi
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CONTOH PROGRAM
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CONTOH PROGRAM
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$ title: Sekan
c Contoh program Sekan
REAL X0,X1,X2,T
INTEGER ORDO, ITER
DIMENSION KOEF(20)
WRITE (*,’(24(/))’)
WRITE (*,’(30X,A)’) ‘Input Persamaan’
WRITE (*,’(30X,A)’) ‘===============’
WRITE (*,*)
WRITE (*,’(a,\)’) ‘Orde/Derajat : ’
READ (*,’(I2)’) ORDO
1/11
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CONTOH PROGRAM
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WRITE (*,*)
DO 10 I=ORDO+1, 1, -1
WRITE (*,’(A,I2,A,\)’) ‘Koefisien X^’,I-1,‘ = ’
READ (*,’(I3)’) KOEF(I)
10 CONTINUE
WRITE (*,*)
WRITE (*,*) ‘Persamaan yang diinput : ’
WRITE (*,*)
CALL OUTPUT (ORDO,KOEF)
PAUSE
2/11
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CONTOH PROGRAM
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WRITE (*,’(24(/))’)
WRITE (*,’(20X,A)’) ‘Pencarian Akar Menggunakan
Metode Sekan’
WRITE (*,’(20X,A)’) ‘==========================’
WRITE (*,*)
CALL OUTPUT(ORDE,KOEF)
WRITE (*,*)
WRITE (*,’(A,\)’) ‘X0(Batas Bawah) = ’
READ (*,*) X0
WRITE (*,’(A,\)’) ‘X1(Batas Atas) = ’
READ (*,*) X1
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3/11
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CONTOH PROGRAM
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WRITE (*,’(A,\)’) ‘Toleransi Kesalahan = ’
READ (*,*) T
WRITE (*,*)
X2 = X1 – (FNG(ORDO,KOEF,X1)*(X1-X0))/
(FNG(ORDO,KOEF,X1)-FNG(ORDO,KOEF,X0))
ITER = 1
WRITE (*,*) ‘ÚÄÄÄÄÄÄÄÄÄÄÄÄÄÄÂÄÄÄÄÄÄÄÄÄÄÄÄÄÄÂÄÄÄÄÄ
ÄÄÄÄÄÄÄÄÄÂÄÄÄÄÄÄÄÄÄÄÄÄÄÄÂÄÄÄÄÄÄÄÄÄÄÄÄÄÄ¿’
WRITE (*,'(2(A,3X),7(A,6X),A,3X,A)') '³',
' ITERASI','³','X0','³','X1','³','X2','³',
'F(X2)','³’
WRITE (*,*) ‘ÃÄÄÄÄÄÄÄÄÄÄÄÄÄÄÅÄÄÄÄÄÄÄÄÄÄÄÄÄÄÅÄÄÄÄÄ
ÄÄÄÄÄÄÄÄÄÅÄÄÄÄÄÄÄÄÄÄÄÄÄÄÅÄÄÄÄÄÄÄÄÄÄÄÄÄÄ´’
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4/11
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Copyright © 2008.
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CONTOH PROGRAM
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WHILE (( ABS((X1-X2)/X1) .GT. T) .AND.
( FNG(ORDO,KOEF,X2) .NE. 0 )) DO
WRITE (*,'(A,5X,I3,6X,A,4(1X,F12.7,1X,A))')
'³',ITER,'³',X0,'³',X1,'³',X2,'³',
FNG(ORDO,KOEF,X2),'³'
X0 = X1
X1 = X2
X2 = X1 - (FNG(ORDO,KOEF,X1)*(X1-X0))/
(FNG(ORDO,KOEF,X1)-FNG(ORDO,KOEF,X0))
ITER = ITER + 1
PAUSE
ENDWHILE
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5/11
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Copyright © 2008.
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CONTOH PROGRAM
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WRITE (*,'(A,5X,I3,6X,A,4(1X,F12.7,1X,A))')
'³',ITER,'³',X0,'³',X1,'³',X2,'³',
FNG(ORDO,KOEF,X2),'³'
WRITE (*,*) 'ÀÄÄÄÄÄÄÄÄÄÄÄÄÄÄÁÄÄÄÄÄÄÄÄÄÄÄÄÄÄÁÄÄÄÄ
ÄÄÄÄÄÄÄÄÄÄÁÄÄÄÄÄÄÄÄÄÄÄÄÄÄÁÄÄÄÄÄÄÄÄÄÄÄÄÄÄÙ'
WRITE (*,*)
WRITE (*,*) 'SOLUSINYA = ',X2
END
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SUBROUTINE OUTPUT (ORDO,KOEF)
DIMENSION KOEF(20)
INTEGER ORDO
WRITE (*,'(A,\)') 'F(X) = '
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6/11
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CONTOH PROGRAM
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DO 20 I = ORDO + 1, 1, -1
IF (I .GT. 2) THEN
IF (KOEF(I) .EQ. (-1)) THEN
WRITE (*,'(A,I2,\)') 'X^',I – 1
ELSEIF ((KOEF(I) .NE. 1) .AND.
(KOEF(I) .NE. 0)) THEN
WRITE (*,'(I3,A,I2,\)') KOEF(I),'X^',I – 1
ELSEIF (KOEF(I) .NE. 0) THEN
WRITE (*,'(A,I2,\)') 'X^',I – 1
ENDIF
IF (KOEF(I-1) .GT. 0) THEN
WRITE (*,'(A,\)') ' + ‘
ENDIF
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7/11
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CONTOH PROGRAM
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8/11
ELSEIF (I .EQ. 2) THEN
IF (KOEF(I) .EQ. (-1)) THEN
WRITE (*,'(A,\)') '-X ‘
ELSEIF ((KOEF(I) .NE. 1) .AND.
(KOEF(I) .NE. 0)) THEN
WRITE (*,'(I3,A,\)') KOEF(I),' X ‘
ELSEIF (KOEF(I) .NE. 0) THEN
WRITE (*,'(A,\)') ' X ‘
ENDIF
IF (KOEF(I-1) .GT. 0) THEN
WRITE (*,'(A,\)') ' + ‘
ENDIF
ELSEIF (KOEF(I) .NE. 0) THEN
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CONTOH PROGRAM
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WRITE (*,'(I3)') KOEF(I)
ENDIF
20 CONTINUE
END
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REAL FUNCTION FNG(ORDO,KOEF,MX)
INTEGER ORDO
DIMENSION KOEF(20)
REAL MX
FNG = 0
DO 30 I = ORDO + 1, 1, -1
IF (MX .NE. 0) FNG = FNG + (KOEF(I)*MX**(I-1))
30 CONTINUE
RETURN
9/11 END
Daftar
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I
CONTOH OUTPUT PROGRAM
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Input Persamaan
===============
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Orde/Derajat : 3
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Koefisien
Koefisien
Koefisien
Koefisien
X^
X^
X^
X^
3
2
1
0
=
=
=
=
1
0
-2
-5
Persamaan yang diinput :
F(X) = X^ 3 -2 X -5
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10/11
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Copyright © 2008.
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CONTOH OUTPUT PROGRAM
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Pencarian Akar menggunakan Metode Sekan
=======================================
F(X) = X^ 3 -2 X -5
X0(Batas Bawah) = 1
X1(Batas Atas) = 2
Toleransi Kesalahan = 0.001
┌──────────┬───────────┬───────────┬───────────┬───────────────┐
│ ITERASI │
X0
│
X1
│
X2
│
F(X2)
│
├──────────┼───────────┼───────────┼───────────┼───────────────┤
│
1
│ 1.0000000 │ 2.0000000 │ 2.2000000 │
1.2480010
│
│
2
│ 2.0000000 │ 2.2000000 │ 2.0889680 │ -0.0621233
│
│
3
│ 2.2000000 │ 2.0889680 │ 2.0942330 │ -0.0035534
│
│
4
│ 2.0889680 │ 2.0942330 │ 2.0945530 │
0.0000114
│
└──────────┴───────────┴───────────┴───────────┴───────────────┘
Solusinya = 2.0945530
11/11
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VI
LAPORAN AKHIR
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LAPORAN AKHIR
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1. Tuliskan Logika untuk program yang telah
ditulis.
2. Logika tidak boleh sama.
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Daftar
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VII
LAPORAN PENDAHULUAN
PERTEMUAN 6
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LAPORAN PENDAHULUAN PERTEMUAN 6
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2.
Sebutkan dan jelaskan metode yang dapat digunakan
untuk mencari invers suatu matriks !
Diketahui sebuah matriks :
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│
│
│
│
│
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Tentukan : a. Matriks adjoin ?
b. Eliminasi Gauus Jordan ?
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2
4
8
16
32
3
6
12
24
48
4
8
16
32
64
5
10
20
40
80
6
12
24
48
96
│
│
│
│
│
Daftar
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Sampai bertemu lagi di
Pertemuan ke 6
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