Lapsem Modul 5
Lapsem Modul 5
Lapsem Modul 5
FAKULTAS TEKNIK
UPN “VETERAN” JAWA TIMUR
Nama : CLARISSA PUTRI M
Praktikum : MATEMATIKA TEKNIK NPM/Semester : 21031010041/IV
Percobaan : METODE JACOBI & METODE Sesi : A2
Paralel :A
GAUSS SIEDEL
Tanggal : 16 MARET 2023
Pembimbing : IR. KETUT SUMADA.M.S LAPORAN SEMENTARA
Iterasi 1
A =0
B =0
C =0
D =0
4−(0)−3(0)
A = = 0,8
5
3+(0)−2(0)−(0)
B = = 0,5
6
−5−4(0)−2(0)−(0)
C = = -0,5555
9
2−(0)+3(0)−(−)
D = = 0,2857
7
0,8−0
𝜀A = | | × 100% = 100%
0,8
0,5−0
𝜀B = | | × 100% = 100%
0,5
−0,5555−0
𝜀C = | | × 100% = 100%
−0,5555
0,2857−0
𝜀D = | | × 100% = 100%
0,2857
Iterasi 2
A = 0,8
B = 0,5
C = -0,5555
D = 0,2857
4−(0,5)−3(−0,5555)
A = = 1,0333
5
3+(0,8)−2(−0,5555)−(0,2857)
B = = 0,7709
6
−5−4(0,8)−2(0,5)−(0,2857)
C = = -1,0540
9
2−(0,8)+3(0,5)−(−0,5555)
D = = 0,4651
7
1,0333−0,8
𝜀A = | | × 100% = 22,5806%
1,0333
0,7709−0,5
𝜀B = | | × 100% = 35,1407%
0,7709
−1,0540−(−0,5555)
𝜀C = | | × 100% = 47,2892%
−1,0540
0,4651−0,2857
𝜀D = | | × 100% = 38,5666%
0,4651
Iterasi 3
A = 1,0333
B = 0,7709
C = -1,0540
D = 0,4651
4−(0,7709)−3(1,0540)
A = = 1,2782
5
3+(1,0333)−2(1,0540)−(0,4651)
B = = 0,9460
6
−5−4(1,0333)−2(0,7709)−(0,4651)
C = = -1,2378
9
2−(1,0333)+3(0,7709)−(1,0540)
D = = 0,6190
7
1,2782−1,0333
𝜀A = | | × 100% = 19,5172%
1,2782
0,9460−0,77095
𝜀B = | | × 100% = 18,5123%
0,9460
−1,2378−(−1,0540)
𝜀C = | | × 100% = 14,8516%
−1,2378
−0,6190−0,4651
𝜀D = | | × 100% = 24,8718%
0,6190
Iterasi 4
A = 1,2782
B = 0,9460
C = -1,2378
D = 0,6190
4−(0,9460)−3−(−1,2378)
A = = 1,3535
5
3+(1,2782)−2(−1,2378)−(0,6190)
B = = 1,0225
6
−5−4(1,2782)−2(0,9460)−(0,6190)
C = = -1,4027
9
2−(1,2782)+3(0,9460)−(−1,2378)
D = = 0,6854
7
1,3535−1,2782
𝜀A = | | × 100% = 5,5615%
1,3535
1,0225−0,9460
𝜀B = | | × 100% = 7,4749%
1,0225
−1,4027−(−1,2378)
𝜀C = | | × 100%= 11,7531%
−1,4027
0,6854−0,6190
𝜀D = | | × 100% = 9,6789%
0,6854