[2022 Fall] Assignment6-5¶

Course: AP3021

Q6-5¶

In [1]:

import numpy as np

import numpy as np

In [2]:

a = np.array([
    [ 0.8, -0.4,  0.0],
    [-0.4,  0.8, -0.4],
    [ 0.0, -0.4,  0.8]]
)
b = np.array([
    [ 41],
    [ 25],
    [105]]
)

a = np.array([ [ 0.8, -0.4, 0.0], [-0.4, 0.8, -0.4], [ 0.0, -0.4, 0.8]] ) b = np.array([ [ 41], [ 25], [105]] )

In [3]:

x1 = 0
x2 = 1
x3 = 0
count = 0
c = 1.2

x1 = 0 x2 = 1 x3 = 0 count = 0 c = 1.2

In [4]:

while True:
    xold1 = x1
    xold2 = x2
    xold3 = x3
    es = 0.5 * 10 ** (-2)

    count += 1
    x1 = ((b[0][0]) - (a[0][1]) * x2 - (a[0][2]) * x3) / a[0][0]
    x1 = c * x1 + (1 - c) * xold1
    x2 = ((b[1][0]) - (a[1][0]) * x1 - (a[1][2]) * x3) / a[1][1]
    x2 = c * x2 + (1 - c) * xold2
    x3 = ((b[2][0]) - (a[2][0]) * x1 - (a[2][1]) * x2) / a[2][2]
    x3 = c * x3 + (1 - c) * xold3
    if ((x1 - xold1) / x1 < es) and \
           ((x2-xold2)/x2 < es) and \
           ((x3-xold3)/x3 < es) :
        break

print('x1 =', x1)
print('x2 =', x2)
print('x3 =', x3)
print()
print('Total used', count, "times.")

while True: xold1 = x1 xold2 = x2 xold3 = x3 es = 0.5 * 10 ** (-2) count += 1 x1 = ((b[0][0]) - (a[0][1]) * x2 - (a[0][2]) * x3) / a[0][0] x1 = c * x1 + (1 - c) * xold1 x2 = ((b[1][0]) - (a[1][0]) * x1 - (a[1][2]) * x3) / a[1][1] x2 = c * x2 + (1 - c) * xold2 x3 = ((b[2][0]) - (a[2][0]) * x1 - (a[2][1]) * x2) / a[2][2] x3 = c * x3 + (1 - c) * xold3 if ((x1 - xold1) / x1 < es) and \ ((x2-xold2)/x2 < es) and \ ((x3-xold3)/x3 < es) : break print('x1 =', x1) print('x2 =', x2) print('x3 =', x3) print() print('Total used', count, "times.")

x1 = 173.79293976576002
x2 = 245.05792502579206
x3 = 253.76707266232324

Total used 6 times.

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Last update: 2024-04-27