Difference between revisions of "Rizky Narendra Putra"

Latest revision as of 00:28, 28 October 2019

Profil: Rizky Narendra Putra / 1606907524 Teknik Mesin Universitas Indonesia

Video Mekanika Fluida Bab 8: https://www.youtube.com/watch?v=42tyCDgc2dA&feature=youtu.be

Video Mekanika Fluida Bab 9: https://www.youtube.com/watch?v=sDhnzQfGJoU&feature=youtu.be

Tugas 1 Kelas Metode Numerik:

Apa yang saya pelajari tentang Python:

Python merupakan bahasa pemograman Python dapat mengerjakan operasi matematika ( integral, penjumlahan, pengurangan, pengalian, pangkat, pembagian, dst ) Jika dalam suatu pembagian hasilnya ada desimal dan kita tidak ingin memperlihatkan desimal tersebut, kita dapat menggunakan rumus misal: 5//2, maka akan keluar hasilnya 2, bukan 2,5 dengan menggunakan // maka desimal dalam pembagian tersebut akan hilang, namun jika kita ingin memperlihatkan desimal tersebut, cukup: 5/2, maka akan keluar hasilnya 2,5 Jika kita ingin menggunakan pangkat, maka dapat dilakukan 2**5 dan hasilnya akan keluar 32, jika kita ingin perkalian saja maka dapat digunakan 2*5 dan akan keluar hasilnya 10

Untuk operasi matematika, python mempunyai prioritas tertinggi sampai yang terendah yaitu: 1. ( .. ) 2. ( * , / , ** ) 3. ( +,- )

Tugas 2 Metode Numerik

Tugas 3

Menggunakan matrix equation untuk eliminasi menggunakan Gaussian Elimination.

6x₁ + 4x₂ = 50
2x₁ + x₃ + 4x₄ = 50
7x₂ + 3x₃ + 4x₄ = 50
4x₁ + 4x₃ = 50

Didapat hasil matrix:

[[6. 4. 0. 0.]
 [2. 0. 1. 4.]
 [0. 7. 3. 4.]
 [4. 0. 4. 0.]]

Mendefinisikan matrix di dalam python:

import numpy as np

A = np.array([[6, 4, 0, 0], [2, 0 ,1, 4], [0, 7, 3, 4], [ 4, 0, 4, 0]], float)
B = np.array([50, 50, 0, 0], float)

n = len(A)

Menggunakan Eliminasi Gaussian

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)
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]

KUIS 1 METNUM :

soal nomer 1 runge kutta: kode:

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.arrange (0,0.03,h) for t in step_size:

   k1 = diff_y (x,y)
   k2 = diff_y ((x+0.5*h),(y+0.5*k1*h))

   y = y+k1*h

   print ('nilai y(0,03) adalah',y)

Hasil:

Soal nomer 2 Gaussian elimination method Kode:

import numpy as np

class GEPP():

   def __init__(self, A, b, doPricing=True):
       #super(GEPP, self).__init__()

       self.A = A                      # input: A is an n x n numpy matrix
       self.b = b                      # b is an n x 1 numpy array
       self.doPricing = doPricing

       self.n = None                   # n is the length of A
       self.x = None                   # x is the solution of Ax=b

       self._validate_input()          # method that validates input
       self._elimination()             # method that conducts elimination
       self._backsub()                 # method that conducts back-substitution

   def _validate_input(self):
       self.n = len(self.A)
       if self.b.size != self.n:
           raise ValueError("Invalid argument: incompatible sizes between" +
                            "A & b.", self.b.size, self.n)

   def _elimination(self):

       # Elimination
       for k in range(self.n - 1):
           if self.doPricing:
               # Pivot
               maxindex = abs(self.A[k:, k]).argmax() + k
               if self.A[maxindex, k] == 0:
                   raise ValueError("Matrix is singular.")
               # Swap
               if maxindex != k:
                   self.Ak, maxindex = self.Amaxindex, k
                   self.bk, maxindex = self.bmaxindex, k
           else:
               if self.A[k, k] == 0:
                   raise ValueError("Pivot element is zero. Try setting doPricing to True.")
           # Eliminate
           for row in range(k + 1, self.n):
               multiplier = self.A[row, k] / self.A[k, k]
               self.A[row, k:] = self.A[row, k:] - multiplier * self.A[k, k:]
               self.b[row] = self.b[row] - multiplier * self.b[k]

   def _backsub(self):
       # Back Substitution

       self.x = np.zeros(self.n)
       for k in range(self.n - 1, -1, -1):
           self.x[k] = (self.b[k] - np.dot(self.A[k, k + 1:], self.x[k + 1:])) / self.A[k, k]

def main():

   A = np.array([[1., 2., 0., -2., 0.],
                 [0., 1., 0., 2., -1.],
                 [0., 0., 2., 1., 2.],
                 [0., 0., 0., -1., 1.],
                 [0., 1., -1., 1., -1.]])
   b = np.array([[-4.],
                 [1.],
                 [1.],
                 [-2.],
                 [-1.]])
   GaussElimPiv = GEPP(np.copy(A), np.copy(b), doPricing=False)
   print(GaussElimPiv.x)
   print(GaussElimPiv.A)
   print(GaussElimPiv.b)
   GaussElimPiv = GEPP(A, b)
   print(GaussElimPiv.x)

if __name__ == "__main__":

   main()

Hasil:

UTS METODE NUMERIK 2019

Video Muasabah Diri: https://www.youtube.com/watch?v=jkp6N7AG6_g

Soal B:

import numpy as np def diff_y (x,y):

  fungsi = x**2 - 2*y
  return (fungsi)

x=0 y=10 h=10 a = 2 j = 1 step_size = -np.arange (0,2,h) for t in step_size:

  k1 = diff_y (x,y) - (a + j)
  k2 = diff_y ((x+0.5*h),(y+0.05*k1*h)) - (a + j)

w1 = y + 1/3*(k1+2*k2) print ('maka x(2) adalah', w1)

a = hambatan angin j = gesekan jalan

Video Penjelasan: https://www.youtube.com/watch?v=TAMz5momeSc

Soal A:

print ('Penghitungan Soal UTS A') import numpy as np class GEPP():

   def __init__(self, A, b, doPricing=True):
       #super(GEPP, self).__init__()
       self.A = A                      # input: A is an n x n numpy matrix
       self.b = b                      # b is an n x 1 numpy array
       self.doPricing = doPricing

       self.n = None                   # n is the length of A
       self.x = None                   # x is the solution of Ax=b

       self._validate_input()          # method that validates input
       self._elimination()             # method that conducts elimination
       self._backsub()                 # method that conducts back-substitution

   def _validate_input(self):
       self.n = len(self.A)
       if self.b.size != self.n:
           raise ValueError("Invalid argument: incompatible sizes between" +
                            "A & b.", self.b.size, self.n)

   def _elimination(self):
       """
       k represents the current pivot row. Since GE traverses the matrix in the
       upper right triangle, we also use k for indicating the k-th diagonal
       column index.
       :return
       """
       # Elimination
       for k in range(self.n - 1):
           if self.doPricing:
               # Pivot
               maxindex = abs(self.A[k:, k]).argmax() + k
               if self.A[maxindex, k] == 0:
                   raise ValueError("Matrix is singular.")
               # Swap
               if maxindex != k:
                   self.Ak, maxindex = self.Amaxindex, k
                   self.bk, maxindex = self.bmaxindex, k
           else:
               if self.A[k, k] == 0:
                   raise ValueError("Pivot element is zero. Try setting doPricing to True.")
           # Eliminate
           for row in range(k + 1, self.n):
               multiplier = self.A[row, k] / self.A[k, k]
               self.A[row, k:] = self.A[row, k:] - multiplier * self.A[k, k:]
               self.b[row] = self.b[row] - multiplier * self.b[k]

   def _backsub(self):
       # Back Substitution
       self.x = np.zeros(self.n)
       for k in range(self.n - 1, -1, -1):
           self.x[k] = (self.b[k] - np.dot(self.A[k, k + 1:], self.x[k + 1:])) / self.A[k, k]

def main():

   A = np.array([[1., 0., 0., 0.],
                 [-1., 1., 0., 0.],
                 [0., 0., -1., 1.],
                 [0., 0., 0., 1.]])
   b = np.array([[50.],
                 [20.],
                 [5.],
                 [10.]])
   GaussElimPiv = GEPP(np.copy(A), np.copy(b), doPricing=False)
   print(GaussElimPiv.x)
   print(GaussElimPiv.A)
   print(GaussElimPiv.b)
   GaussElimPiv = GEPP(A, b)
   print(GaussElimPiv.x)

if __name__ == "__main__":

   main()

Video Penjelasan: https://www.youtube.com/watch?v=6SqbgmkvWQc

Difference between revisions of "Rizky Narendra Putra"

Latest revision as of 00:28, 28 October 2019

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@@ Line 1: / Line 1: @@
+Profil:
+Rizky Narendra Putra / 1606907524
+Teknik Mesin Universitas Indonesia
+[[File:Muka.jpg]]
+----
 [[File:Studi Kasus Bab 8 Rizky.jpg]]
 [[File:Studi Kasus Bab 9 Rizky.jpg]]
@@ Line 30: / Line 36: @@
 ----
-[[Media:program.py]]
+Tugas 2 Metode Numerik
+[[File:program python limit.png]]
+[[File:hasil program python limit.png]]
+----
+Tugas 3
+Menggunakan matrix equation untuk eliminasi menggunakan Gaussian Elimination.
+<div border-style: inset;">
+x<sub>1</sub> + 4x<sub>2</sub> = 50
+x<sub>1</sub> + x<sub>3</sub> + 4x<sub>4</sub> = 50
+x<sub>2</sub> + 3x<sub>3</sub> + 4x<sub>4</sub> = 50
+x<sub>1</sub> + 4x<sub>3</sub> = 50
+Didapat hasil matrix:
+ [[6. 4. 0. 0.]
+  [2. 0. 1. 4.]
+  [0. 7. 3. 4.]
+  [4. 0. 4. 0.]]
+</div>
+Mendefinisikan matrix di dalam python:
+<div border-style: inset;">
+ import numpy as np
+ <br>A = np.array([[6, 4, 0, 0], [2, 0 ,1, 4], [0, 7, 3, 4], [ 4, 0, 4, 0]], float)
+ B = np.array([50, 50, 0, 0], float)
+ <br>n = len(A)
+</div>
+Menggunakan Eliminasi Gaussian
+<div border-style: inset;">
+ 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)
+ 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]
+</div>
+----
+KUIS 1 METNUM :
+soal nomer 1 runge kutta:
+kode:
+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.arrange (0,0.03,h)
+for t in step_size:
+    k1 = diff_y (x,y)
+    k2 = diff_y ((x+0.5*h),(y+0.5*k1*h))
+    y = y+k1*h
+    print ('nilai y(0,03) adalah',y)
+Hasil:
+Soal nomer 2 Gaussian elimination method
+Kode:
+import numpy as np
+class GEPP():
+    def __init__(self, A, b, doPricing=True):
+        #super(GEPP, self).__init__()
+        self.A = A                      # input: A is an n x n numpy matrix
+        self.b = b                      # b is an n x 1 numpy array
+        self.doPricing = doPricing
+        self.n = None                   # n is the length of A
+        self.x = None                   # x is the solution of Ax=b
+        self._validate_input()          # method that validates input
+        self._elimination()             # method that conducts elimination
+        self._backsub()                 # method that conducts back-substitution
+    def _validate_input(self):
+        self.n = len(self.A)
+        if self.b.size != self.n:
+            raise ValueError("Invalid argument: incompatible sizes between" +
+                             "A & b.", self.b.size, self.n)
+    def _elimination(self):
+        # Elimination
+        for k in range(self.n - 1):
+            if self.doPricing:
+                # Pivot
+                maxindex = abs(self.A[k:, k]).argmax() + k
+                if self.A[maxindex, k] == 0:
+                    raise ValueError("Matrix is singular.")
+                # Swap
+                if maxindex != k:
+                    self.A[[k, maxindex]] = self.A[[maxindex, k]]
+                    self.b[[k, maxindex]] = self.b[[maxindex, k]]
+            else:
+                if self.A[k, k] == 0:
+                    raise ValueError("Pivot element is zero. Try setting doPricing to True.")
+            # Eliminate
+            for row in range(k + 1, self.n):
+                multiplier = self.A[row, k] / self.A[k, k]
+                self.A[row, k:] = self.A[row, k:] - multiplier * self.A[k, k:]
+                self.b[row] = self.b[row] - multiplier * self.b[k]
+    def _backsub(self):
+        # Back Substitution
+        self.x = np.zeros(self.n)
+        for k in range(self.n - 1, -1, -1):
+            self.x[k] = (self.b[k] - np.dot(self.A[k, k + 1:], self.x[k + 1:])) / self.A[k, k]
+def main():
+    A = np.array([[1., 2., 0., -2., 0.],
+                  [0., 1., 0., 2., -1.],
+                  [0., 0., 2., 1., 2.],
+                  [0., 0., 0., -1., 1.],
+                  [0., 1., -1., 1., -1.]])
+    b = np.array([[-4.],
+                  [1.],
+                  [1.],
+                  [-2.],
+                  [-1.]])
+    GaussElimPiv = GEPP(np.copy(A), np.copy(b), doPricing=False)
+    print(GaussElimPiv.x)
+    print(GaussElimPiv.A)
+    print(GaussElimPiv.b)
+    GaussElimPiv = GEPP(A, b)
+    print(GaussElimPiv.x)
+if __name__ == "__main__":
+    main()
+Hasil: [[File:hasil ss.png]]
+----
+UTS METODE NUMERIK 2019
+Video Muasabah Diri: https://www.youtube.com/watch?v=jkp6N7AG6_g
+Soal B:
+import numpy as np
+def diff_y (x,y):
+   fungsi = x**2 - 2*y
+   return (fungsi)
+x=0
+y=10
+h=10
+a = 2
+j = 1
+step_size = -np.arange (0,2,h)
+for t in step_size:
+   k1 = diff_y (x,y) - (a + j)
+   k2 = diff_y ((x+0.5*h),(y+0.05*k1*h)) - (a + j)
+w1 = y + 1/3*(k1+2*k2)
+print ('maka x(2) adalah', w1)
+a = hambatan angin
+j = gesekan jalan
+[[File:Hasil UTS 2019 ss.jpg]]
+Video Penjelasan: https://www.youtube.com/watch?v=TAMz5momeSc
+Soal A:
+print ('Penghitungan Soal UTS A')
+import numpy as np
+class GEPP():
+    def __init__(self, A, b, doPricing=True):
+        #super(GEPP, self).__init__()
+        self.A = A                      # input: A is an n x n numpy matrix
+        self.b = b                      # b is an n x 1 numpy array
+        self.doPricing = doPricing
+        self.n = None                   # n is the length of A
+        self.x = None                   # x is the solution of Ax=b
+        self._validate_input()          # method that validates input
+        self._elimination()             # method that conducts elimination
+        self._backsub()                 # method that conducts back-substitution
+    def _validate_input(self):
+        self.n = len(self.A)
+        if self.b.size != self.n:
+            raise ValueError("Invalid argument: incompatible sizes between" +
+                             "A & b.", self.b.size, self.n)
+    def _elimination(self):
+        """
+        k represents the current pivot row. Since GE traverses the matrix in the
+        upper right triangle, we also use k for indicating the k-th diagonal
+        column index.
+        :return
+        """
+        # Elimination
+        for k in range(self.n - 1):
+            if self.doPricing:
+                # Pivot
+                maxindex = abs(self.A[k:, k]).argmax() + k
+                if self.A[maxindex, k] == 0:
+                    raise ValueError("Matrix is singular.")
+                # Swap
+                if maxindex != k:
+                    self.Ak, maxindex = self.Amaxindex, k
+                    self.bk, maxindex = self.bmaxindex, k
+            else:
+                if self.A[k, k] == 0:
+                    raise ValueError("Pivot element is zero. Try setting doPricing to True.")
+            # Eliminate
+            for row in range(k + 1, self.n):
+                multiplier = self.A[row, k] / self.A[k, k]
+                self.A[row, k:] = self.A[row, k:] - multiplier * self.A[k, k:]
+                self.b[row] = self.b[row] - multiplier * self.b[k]
+    def _backsub(self):
+        # Back Substitution
+        self.x = np.zeros(self.n)
+        for k in range(self.n - 1, -1, -1):
+            self.x[k] = (self.b[k] - np.dot(self.A[k, k + 1:], self.x[k + 1:])) / self.A[k, k]
+def main():
+    A = np.array([[1., 0., 0., 0.],
+                  [-1., 1., 0., 0.],
+                  [0., 0., -1., 1.],
+                  [0., 0., 0., 1.]])
+    b = np.array([[50.],
+                  [20.],
+                  [5.],
+                  [10.]])
+    GaussElimPiv = GEPP(np.copy(A), np.copy(b), doPricing=False)
+    print(GaussElimPiv.x)
+    print(GaussElimPiv.A)
+    print(GaussElimPiv.b)
+    GaussElimPiv = GEPP(A, b)
+    print(GaussElimPiv.x)
+if __name__ == "__main__":
+    main()
+[[File:utsmetnum.jpeg]]
+Video Penjelasan: https://www.youtube.com/watch?v=6SqbgmkvWQc