Difference between revisions of "Ramandika Garindra Putra"
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PT : Teknik Mesin Universitas Indonesia (2016 - sekarang) | PT : Teknik Mesin Universitas Indonesia (2016 - sekarang) | ||
+ | |||
+ | == KUIS (14 Oktober 2019) == | ||
+ | |||
+ | ==='''1. Problem Set 2.1 No. 6 hal. 55'''=== | ||
+ | |||
+ | Solve the equations '''Ax = b''' by Gauss elimination, where | ||
+ | |||
+ | A =[[0, 0, 2, 1, 2], | ||
+ | [0, 1, 0, 2, -1], | ||
+ | [1, 2, 0, -2, 0], | ||
+ | [0, 0, 0, -1, 1], | ||
+ | [0, 1, -1, 1, -1]] | ||
+ | |||
+ | B =[1, 1, -4, -2, -1] | ||
+ | |||
+ | Matrix dikonfigurasi ulang | ||
+ | |||
+ | A = [[1, 2, 0, -2, 0], [0, 1, 0, 2, -1],[0, 1, -1, 1, -1], [0, 0, 0, -1, 1], [0, 0, 2, 1, 2]] | ||
+ | |||
+ | B = [-4, 1, -1, -2, 1] | ||
+ | |||
+ | Hasil operasi matrix | ||
+ | |||
+ | X1 = 2 | ||
+ | |||
+ | X2 = -2 | ||
+ | |||
+ | X3 = 1 | ||
+ | |||
+ | X4 = 1 | ||
+ | |||
+ | X5 = -1 | ||
+ | |||
+ | '''Python''' | ||
+ | |||
+ | import numpy as np | ||
+ | |||
+ | #1. define matrix | ||
+ | |||
+ | A = np.array ([[0, 0, 2, 1, 2], [0, 1, 0, 2, -1],[1, 2, 0, -2, 0], [0, 0, 0, -1, 1], [0, 1, -1, 1, -1]], float) | ||
+ | B = np.array ([[1], [1],[-4], [-2], [-1]], float) | ||
+ | |||
+ | n=len(A) | ||
+ | |||
+ | #2. eliminasi gauss | ||
+ | |||
+ | for k in range (0,n-1): | ||
+ | for i in range (k+1, n): | ||
+ | if A[i,k]! = 0 : | ||
+ | lam = A [i, k] / A [k, k] | ||
+ | A [i, k : n] = A [i, k : n] - ( A [k, k : n] * lam) | ||
+ | B[i] = B[i] - (B[k]*lam) | ||
+ | |||
+ | print ('matrix A:', '\n', A) | ||
+ | print ('matrix B:', '\n', B) | ||
+ | |||
+ | #3. back subtitution | ||
+ | |||
+ | x = np.zeros(n,float) | ||
+ | for m in range (n-1, -1, -1): | ||
+ | x[m]=(B[m]-np.dot(A[m, m+1:n], x[m+1:n]))/A[m,m] | ||
+ | print ('nilai X', m+1, '=', x[m]) | ||
+ | |||
+ | |||
+ | |||
+ | |||
+ | [[File:gauss.jpg]] | ||
+ | |||
+ | [[File:coding.jpg]] | ||
+ | |||
+ | ==='''2. Problem Set 7.1 No. 1 hal. 263'''=== | ||
+ | |||
+ | import numpy as np | ||
+ | |||
+ | def diff_y (x,y): | ||
+ | fungsi = x**2 - 4*y | ||
+ | return (fungsi) | ||
+ | x = 0 | ||
+ | y = 1 | ||
+ | h = 0.01 | ||
+ | step_size = np.range (0,0.03,h) | ||
+ | |||
+ | for t in step_size: | ||
+ | k1 = diff_y (x,y) | ||
+ | k2 = diff_y ((x+0.5h),(y+0.05*k1*h) | ||
+ | |||
+ | y = y + k1*h | ||
+ | |||
+ | print ('maka y(0.03) adalah', y) | ||
+ | |||
+ | |||
+ | == UTS == | ||
+ | === Soal A === | ||
+ | === Soal B === | ||
+ | |||
+ | import numpy as np | ||
+ | |||
+ | a = 10 | ||
+ | |||
+ | b = 2 | ||
+ | |||
+ | c = 2 | ||
+ | |||
+ | def diff_y (t,v): | ||
+ | fungsi = a - b - c | ||
+ | return (fungsi) | ||
+ | |||
+ | v = 0 | ||
+ | |||
+ | h = 1 | ||
+ | step_size = np.arange (0,10,h) | ||
+ | |||
+ | for t in step_size: | ||
+ | k1 = diff_y (t,v) | ||
+ | k2 = diff_y ((t+0.5*h), (v+0.5*k1*h)) | ||
+ | k3 = diff_y ((t+0.5*h), (v+0.5*k2*h)) | ||
+ | k4 = diff_y ((t+h), (v+k1*h)) | ||
+ | |||
+ | v = v + 1/6*(k1+2*k2+2*k3+k4)*h | ||
+ | |||
+ | print ('maka v ketika 10 detik adalah', v) |
Latest revision as of 15:12, 21 October 2019
Contents
Biodata
Nama : Ramandika Garindra Putra
NPM : 1606907625
Departemen : Teknik Mesin
Program Studi : Teknik Mesin
Biografi
Nama saya Ramandika Garindra Putra, orang - orang biasa memanggil Rama. Saya lahir di Jakarta 11 Agustus 1998. Saya berasal dari keluarga bersuku Sunda dan Jawa. Ayah saya keturunan Sunda yang lahir di Jakarta, dan Ibu saya keturunan Jawa yang lahir di Ponorogo. Saya merupakan anak pertama dari dua bersaudara, adik saya perempuan lahir di Jakarta tahun 2003.
Riwayat Pendidikan
KB : Al Hanif Kemang Pratama (2001 - 2002)
TK : Al Azhar 24 Jatikramat (2002 - 2004)
SD : Al Azhar 23 Jatikramat (2004 - 2010)
SMP : Global Islamic School Condet (2010 - 2013)
SMA : Al Azhar 4 Kemang Pratama (2013 - 2016)
PT : Teknik Mesin Universitas Indonesia (2016 - sekarang)
KUIS (14 Oktober 2019)
1. Problem Set 2.1 No. 6 hal. 55
Solve the equations Ax = b by Gauss elimination, where
A =[[0, 0, 2, 1, 2], [0, 1, 0, 2, -1], [1, 2, 0, -2, 0], [0, 0, 0, -1, 1], [0, 1, -1, 1, -1]]
B =[1, 1, -4, -2, -1]
Matrix dikonfigurasi ulang
A = [[1, 2, 0, -2, 0], [0, 1, 0, 2, -1],[0, 1, -1, 1, -1], [0, 0, 0, -1, 1], [0, 0, 2, 1, 2]]
B = [-4, 1, -1, -2, 1]
Hasil operasi matrix
X1 = 2
X2 = -2
X3 = 1
X4 = 1
X5 = -1
Python
import numpy as np
- 1. define matrix
A = np.array ([[0, 0, 2, 1, 2], [0, 1, 0, 2, -1],[1, 2, 0, -2, 0], [0, 0, 0, -1, 1], [0, 1, -1, 1, -1]], float) B = np.array ([[1], [1],[-4], [-2], [-1]], float)
n=len(A)
- 2. eliminasi gauss
for k in range (0,n-1):
for i in range (k+1, n): if A[i,k]! = 0 : lam = A [i, k] / A [k, k] A [i, k : n] = A [i, k : n] - ( A [k, k : n] * lam) B[i] = B[i] - (B[k]*lam)
print ('matrix A:', '\n', A) print ('matrix B:', '\n', B)
- 3. back subtitution
x = np.zeros(n,float) for m in range (n-1, -1, -1):
x[m]=(B[m]-np.dot(A[m, m+1:n], x[m+1:n]))/A[m,m] print ('nilai X', m+1, '=', x[m])
2. Problem Set 7.1 No. 1 hal. 263
import numpy as np
def diff_y (x,y):
fungsi = x**2 - 4*y return (fungsi)
x = 0 y = 1 h = 0.01 step_size = np.range (0,0.03,h)
for t in step_size:
k1 = diff_y (x,y) k2 = diff_y ((x+0.5h),(y+0.05*k1*h)
y = y + k1*h
print ('maka y(0.03) adalah', y)
UTS
Soal A
Soal B
import numpy as np
a = 10
b = 2
c = 2
def diff_y (t,v):
fungsi = a - b - c return (fungsi)
v = 0
h = 1 step_size = np.arange (0,10,h)
for t in step_size:
k1 = diff_y (t,v) k2 = diff_y ((t+0.5*h), (v+0.5*k1*h)) k3 = diff_y ((t+0.5*h), (v+0.5*k2*h)) k4 = diff_y ((t+h), (v+k1*h))
v = v + 1/6*(k1+2*k2+2*k3+k4)*h
print ('maka v ketika 10 detik adalah', v)