dy/dx - = 1 - y,y(0) = 0 is an example of

[zaasmi]

@zaasmi said in MTH603 Grand Quiz Solution and Discussion:

8. Solve the system of equations by Jacobi’s iteration method.
   20x + y – 2z = 17
   3x + 20y – z = -18
   2x – 3y + 20z = 25

a) x = 1, y = -1, z = 1
b) x = 2, y = 1, z = 0
c) x = 2, y = 1, z = 0
d) x = 1, y = 2, z = 1

Answer: a
Explanation: We write the equations in the form
x = 120 (17 – y +2z)
y = 120 (-18 -3x + z)
z = 120 (25 -2x +3y)
We start from an approximation x = y = z = 0.
Substituting these in the right sides of the equations (i), (ii), (iii), we get
First iteration:
x = 0.85, y = -0.9, z = 1.25
Putting these values again in equations (i), (ii), (iii), we obtain,
x = [17 – (-0.9) + 2(1.25)] = 1.02
y = [-18 -3(0.85) + 1.25] = -0.965
z = [25 – 2(0.85) + 3(-0.9)] = 1.03
Substituting these values again in equations (i), (ii), (iii), we obtain,
Second iteration:
x = 1.00125, y = -1.0015, z = 1.00325
Proceeding in this way, we get,
Third iteration:
x = 1.0004, y = -1.000025, z = 0.9965
Fourth iteration
x = 0.999966, y = -1.000078, z = 0.999956
Fifth iteration
x = 1.0000, y = -0.999997, z = 0.999992
The values in the last iterations being practically the same, we can stop.
Hence the solution is
x = 1, y = -1, z = 1.

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9. Solve the system of equations by Jacobi’s iteration method.

10x = y – x = 11.19
x + 10y + z = 28.08
-x + y + 10z = 35.61

correct to two decimal places.
a) x = 1.00, y = 2.95, z = 3.85
b) x = 1.96, y = 2.63, z = 3.99
c) x = 1.58, y = 2.70, z = 3.00
d) x = 1.23, y = 2.34, z = 3.45

[zaasmi]

@zaasmi said in MTH603 Grand Quiz Solution and Discussion:

9. Solve the system of equations by Jacobi’s iteration method.

10x = y – x = 11.19
x + 10y + z = 28.08
-x + y + 10z = 35.61

correct to two decimal places.
a) x = 1.00, y = 2.95, z = 3.85
b) x = 1.96, y = 2.63, z = 3.99
c) x = 1.58, y = 2.70, z = 3.00
d) x = 1.23, y = 2.34, z = 3.45

Answer: d
Explanation: Rewriting the equations as,
x = 110 (11.19 – y + z)
y = 110 (28.08 – x – z)
z = 110 (35.61 + x – y)
We start from an approximation, x = y = z = 0.
First iteration, x = 1.119, y = 2.808, z = 3.561
Second iteration,
x = 110 (11.19 – 2.808 + 3.651) = 1.19
y = 110 (28.08 – 1.119 – 3.561) = 2.34
z = 110 (35.61 + 1.119 – 2.808) = 3.39
Third Iteration:
x = 1.22, y = 2.35, z = 3.45
Fourth iteration:
x = 1.23, y = 2.34, z = 3.45
Fifth iteration:
x = 1.23, y = 2.34, z = 3.45
Hence, x = 1.23, y = 2.34, z = 3.45.

[zaasmi]

10. Solve the system of equations by Jacobi’s iteration method.

10a - 2b - c - d = 3

• 2a + 10b - c - d = 15
• a - b + 10c - 2d = 27
• a - b - 2c + 10d = -9
  a) a = 1, b = 2, c = 3, d = 0
  b) a = 2, b = 1, c = 9, d = 5
  c) a = 2, b = 2, c = 9, d = 0
  d) a = 1, b = 1, c = 3, d = 5

[zaasmi]

@zaasmi said in MTH603 Grand Quiz Solution and Discussion:

10. Solve the system of equations by Jacobi’s iteration method.

10a - 2b - c - d = 3

• 2a + 10b - c - d = 15
• a - b + 10c - 2d = 27
• a - b - 2c + 10d = -9
  a) a = 1, b = 2, c = 3, d = 0
  b) a = 2, b = 1, c = 9, d = 5
  c) a = 2, b = 2, c = 9, d = 0
  d) a = 1, b = 1, c = 3, d = 5

Answer: a
Explanation: Rewriting the given equations as
a = 110(3 + 2b + c + d)
b = 110(15 + 2z + c + d)
c = 110(27 + a + b + 2d)
d = 110(-9 + a + b + 2d)
We start from an approximation a = b = c = d = 0.
First iteration: a = 0.3, b = 1.5, c = 2.7, d = -0.9
Second iteration:
a = 110[3 + 2(1.5) + 2.7 + (-0.9)] = 0.78
b = 110[15 + 2(0.3) + 2.7 + (-0.9)] = 1.74
c = 110[27 + 0.3 + 1.5 + 2(-0.9)] = 2.7
d = 110[-9 + 0.3 + 1.5 + 2(-0.9)] = -0.18
Proceeding in this way we get,
Third iteration, a = 0.9, b = 1.908, c = 2.916, d = -0.108
Fourth iteration, a = 0.9624, b = 1.9608, c = 2.9592, d = -0.036
Fifth iteration, a = 0.9845, b = 1.9848, c = 2.9851, d = -0.0158
Sixth iteration, a = 0.9939, b = 1.9938, c = 2.9938, d = -0.006
Seventh iteration, a = 0.9939, b = 1.9975, c = 2.9976, d = -0.0025
Eighth iteration, a = 0.999, b = 1.999, c = 2.999, d = -0.001
Ninth iteration, a = 0.9996, b = 1.9996, c = 2.9996, d = -0.004
Tenth iteration, a = 0.9998, b = 1.9998, c = 2.9998, d = -0.0001
Hence, a = 1, b = 2, c = 3, d = 0.

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More MCQs for Numerical Analysis click here

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A series 16+8+4+2+1 is replaced by the series 16+8+4+2, then it is called

[zaasmi]

@zaasmi said in MTH603 Grand Quiz Solution and Discussion:

A series 16+8+4+2+1 is replaced by the series 16+8+4+2, then it is called

Each number in the sequence is half the value of the number receding it. So the common difference in the series is dividing by two.

16÷2=8

8÷2=4

4÷2=2

2÷2=1

1÷2=½

The answer is ½ or 0.5

When you keep dividing by two, you will notice an interesting pattern: the denominator continues to increase by two, while the numerator value remains the same. That’s fascinating because in natural, whole numbers the numbers in the series would decrease by two.

1/4 , 1/8 , 1/16 etc.