Difference between revisions of "Metode Numerik soal 21 - Jaringan pipa"

From ccitonlinewiki
Jump to: navigation, search
(Replaced content with "Pada tugas hiburan ini File:S__18382858.jpg <comments voting="Plus" />")
 
Line 2: Line 2:
  
 
[[File:S__18382858.jpg]]
 
[[File:S__18382858.jpg]]
 
dikerjakan menggunakan hukum kontinuitas massa dimana masa yang masuk akan sama dengan yang dikeluarkan sehingga didapati rumus Q*''p''=Q*''p'' sehingga akan mendapatakan 4 persamaan dengan 4 variabel
 
 
 
6C1 - 4C2 = 50
 
 
-2C1 - 1C3 + 4C4 = 50
 
   
 
7C2 - 3C3 - 4C4 = 0
 
 
-4C1 + 4C3 = 0
 
 
 
 
 
6C1 - 4C2 + 0C3 + 0C4 = 50
 
 
-2C1 + 0C2 - 1C3 + 4C4 = 50
 
 
0C1 + 7C2 - 3C3 - 4C4 = 0
 
 
-4C1 + 0C2 + 4C3 + 0C4 = 0
 
 
yang kemudian dijadikan dalam modue python lalu didapatkan nilai C1 = 275/9 , C2 = 100/3 , C3 = 275/9, C4 = 425/12
 
 
 
 
<div border-style: inset;">
 
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.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([[6., -4., 0., 0.],
 
                  [-4., 0., 4., 0.],
 
                  [-2., 0., -1., 4.],
 
                  [0., 7., -3., -4.]])
 
    b = np.array([[50.],
 
                  [0.],
 
                  [50.],
 
                  [0.]])
 
    print("ini matriks awal nya")
 
    print(A)
 
    print("ini hasil yang awal")
 
    print(b)
 
    GaussElimPiv = GEPP(np.copy(A), np.copy(b), doPricing=False)
 
    print("ini hasil akhirnya")
 
    print(GaussElimPiv.x)
 
    print(GaussElimPiv.A)
 
    print(GaussElimPiv.b)
 
    GaussElimPiv = GEPP(A, b)
 
    print(GaussElimPiv.x)
 
if __name__ == "__main__":
 
    main()
 
 
</div>
 
  
 
<comments voting="Plus" />
 
<comments voting="Plus" />

Latest revision as of 14:24, 30 September 2019

Pada tugas hiburan ini

S 18382858.jpg


Edosyafei

63 months ago
Score 0+
Ingin sedikit berkomenta mengenai penyelesaian tugas mengenai kestimbagan 4 tangki konsentrasi. Untuk usaha mencari cara menyelesaikan penyelesaian persamaan aljabar di internet sudah baik. Namun, alangkah lebih baiknya jika saudara menyelesaikannya dengan mengimplentasikan langsung konsep penyelesaian aljabar dengan membuat sendiri algoritma dan kode nya.Diharapkan hal tersebut dapat menambah kemampuan saudara dalam memahami metode numerik dengan menggunakan bantuan komputer

IlliyyinLafi

63 months ago
Score 0+

Berikut kodingan yang saya kerjakan

http://air.e...ran_02-_LAFI

Bagus fadhlurrohman

63 months ago
Score 0+

di bawah ini merupakan hasil tugas saya

http://air.e...an#Hiburan_3

IlliyyinLafi

63 months ago
Score 0+

Anggitoz

62 months ago
Score 0+

Tugas M. Anggito Zhafranny/1706027761

Muhammad_Anggito_Zhafranny#4th_Session

Renoandib

62 months ago
Score 0+

Tugas Ananda Reno Andi Bahar/1706024860

Ananda_Reno_Andi_Bahar#HIBURAN_III
Add your comment
ccitonlinewiki welcomes all comments. If you do not want to be anonymous, register or log in. It is free.