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Chapra Numerical Analysis 11th Chapter Solution

Chapra Numerical 11th Chapter solution
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2K views

Chapra Numerical Analysis 11th Chapter Solution

Chapra Numerical 11th Chapter solution
Copyright
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1

CHAPTER 11
11.1 First, the decomposition is implemented as
e2 = 0.4/0.8 = 0.5
f2 = 0.8 0.5)(0.4) = 0.6
e3 = 0.4/0.6 = 0.66667
f3 = 0.8 0.66667)(0.4) = 0.53333
Transformed system is
0.4
0.8
0.6
0.5
0
0.66667

0
0.4
0.53333

which is decomposed as
0
1
[L] 0.5
1
0
0.66667

0
0
1

0
0.8 0.4
[U ] 0
0.6
0.4
0
0
0.53333

The right hand side becomes


r1 = 41
r2 = 25 0.5)(41) = 45.5
r3 = 105 0.66667)45.5 = 135.3333
which can be used in conjunction with the [U] matrix to perform back substitution and obtain
the solution
x3 = 135.3333/0.53333 = 253.75
x2 = (45.5 (0.4)253.75)/0.6 = 245
x1 = (41 0.4)245)/0.8 = 173.75
11.2 As in Example 11.1, the LU decomposition is

0.49

1
[L]

0.645
1

0.717 1

2.04 1

1.550 1
[U ]
1.395 1

1.323

To compute the first column of the inverse

1

[ L]{D} 0
0
0

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2
Solving this gives

0.490196
{D}
0.316296
0.226775

Back substitution, [U]{X} = {D}, can then be implemented to give to first column of the
inverse

0.755841

{X } 0.541916
0.349667
0.171406

For the second column

0

[ L]{D} 1
0
0

which leads to

0.541916

{X } 1.105509
0.713322
0.349667

For the third column

0

[ L]{D} 0
1
0

which leads to

0.349667

{X } 0.713322
1.105509
0.541916

For the fourth column

0

[ L]{D} 0
0
1

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which leads to

0.171406

{X } 0.349667
0.541916
0.755841

Therefore, the matrix inverse is

0.755841
[ A] 0.541916
0.349667
0.171406
1

0.541916
1.105509
0.713322
0.349667

0.349667
0.713322
1.105509
0.541916

0.171406
0.349667
0.541916
0.755841

11.3 First, the decomposition is implemented as


e2 = 0.020875/2.01475 = 0.01036
f2 = 2.014534
e3 = 0.01036
f3 = 2.014534
e4 = 0.01036
f4 = 2.014534
Transformed system is

2.01475 0.02875

0.01036 2.014534 0.02875

0.01036 2.014534 0.02875

0.01036 2.014534

which is decomposed as

0.01036

1
[L]

0.01036
1

0.01036 1

2.01475 0.02875

2.014534 0.02875
[U ]
2.014534 0.02875

2.014534

Forward substitution yields


r1 = 4.175
r2 = 0.043258
r3 = 0.000448
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4
r4 = 2.087505
Back substitution
x4 = 1.036222
x3 = 0.01096
x2 = 0.021586
x1 = 2.072441
11.4 We can use MATLAB to verify the results of Example 11.2,
>> L=[2.4495 0 0;6.1237 4.1833 0;22.454 20.916 6.1106]
L =
2.4495
6.1237
22.4540

0
4.1833
20.9160

0
0
6.1106

15.0000
54.9997
224.9995

55.0011
224.9995
979.0006

>> L*L'
ans =
6.0001
15.0000
55.0011

11.5
l11 8 2.828427

l 21

20
7.071068
2.828427

l 22 80 7.071068 2 5.477226

l31

15
5.303301
2.828427

l32

50 7.071068(5.303301)
2.282177
5.477226

l 33 60 5.303301 2 2.282177

5.163978

Thus, the Cholesky decomposition is

2.828427

[L] 7.071068 5.477226


5.303301 2.282177 5.163978

11.6
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5
l11 6 2.44949

l 21

15
6.123724
2.44949

l 22 55 6.123724 2 4.1833

l31

55
22.45366
2.44949

l32

225 6.123724(22.45366)
20.9165
4.1833

l 33 979 22.45366 2 20.9165 2 6.110101

Thus, the Cholesky decomposition is

2.44949

[L] 6.123724 4.1833


22.45366 20.9165 6.110101

The solution can then be generated by first using forward substitution to modify the righthand-side vector,
[ L]{D} {B}

which can be solved for


62 .29869
{D} 48 .78923
11 .36915

Then, we can use back substitution to determine the final solution,


[ L]T { X } {D}

which can be solved for


2.478571
{D} 2.359286
1.860714

11.7 (a) The first iteration can be implemented as

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x1

41 0.4 x2 41 0.4(0)

51.25
0.8
0.8

x2

25 0.4 x1 0.4 x3 25 0.4(51.25) 0.4(0)

56.875
0.8
0.8

x3

105 0.4 x 2 105 0.4(56.875)

159.6875
0.8
0.8

Second iteration:

x1

41 0.4(56.875)
79.6875
0.8

x2

25 0.4(79.6875) 0.4(159.6875)
150.9375
0.8

x3

105 0.4(150.9375)
206.7188
0.8

The error estimates can be computed as

a,1

79 .6875 51 .25
100 % 35 .69 %
79 .6875

a,2

150 .9375 56 .875


100 % 62 .32 %
150 .9375

a ,3

206 .7188 159 .6875


100 % 22 .75 %
206 .7188

The remainder of the calculation proceeds until all the errors fall below the stopping criterion
of 5%. The entire computation can be summarized as
iteration
1

unknown
x1
x2
x3
x1
x2
x3
x1
x2
x3
x1
x2

value
51.25
56.875
159.6875
79.6875
150.9375
206.7188
126.7188
197.9688
230.2344
150.2344
221.4844

a
100.00%
100.00%
100.00%
35.69%
62.32%
22.75%
37.11%
23.76%
10.21%
15.65%
10.62%

maximum a

100.00%

62.32%

37.11%

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x3
x1
x2
x3
x1
x2
x3

241.9922
161.9922
233.2422
247.8711
167.8711
239.1211
250.8105

4.86%
7.26%
5.04%
2.37%
3.50%
2.46%
1.17%

15.65%

7.26%

3.50%

Thus, after 6 iterations, the maximum error is 3.5% and we arrive at the result: x1 = 167.8711,
x2 = 239.1211 and x3 = 250.8105.
(b) The same computation can be developed with relaxation where = 1.2.
First iteration:

x1

41 0.4 x 2 41 0.4(0)

51.25
0.8
0.8

Relaxation yields: x1 1.2(51 .25) 0.2(0) 61 .5

x2

25 0.4 x1 0.4 x3 25 0.4(61.5) 0.4(0)

62
0.8
0.8

Relaxation yields: x 2 1.2(62 ) 0.2(0) 74 .4

x3

105 0.4 x 2 105 0.4(74.4)

168.45
0.8
0.8

Relaxation yields: x3 1.2(168 .45) 0.2(0) 202 .14


Second iteration:

x1

41 0.4(74.4)
88.45
0.8

Relaxation yields: x1 1.2(88 .45) 0.2(61 .5) 93 .84

x2

25 0.4(93.84) 0.4(202.14)
179.24
0.8

Relaxation yields: x 2 1.2(179 .24 ) 0.2(74 .4) 200 .208

x3

105 0.4(200.208)
231.354
0.8

Relaxation yields: x3 1.2(231 .354 ) 0.2(202 .14 ) 237 .1968


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The error estimates can be computed as

a ,1

93 .84 61 .5
100 % 34 .46 %
93 .84

a,2

200 .208 74 .4
100 % 62 .84 %
200 .208

a ,3

237 .1968 202 .14


100 % 14 .78 %
237 .1968

The remainder of the calculation proceeds until all the errors fall below the stopping criterion
of 5%. The entire computation can be summarized as
iteration
1

unknown
x1
x2
x3
x1
x2
x3
x1
x2
x3
x1
x2
x3

value
51.25
62
168.45
88.45
179.24
231.354
151.354
231.2768
249.99528
169.99528
243.23898
253.44433

relaxation
61.5
74.4
202.14
93.84
200.208
237.1968
162.8568
237.49056
252.55498
171.42298
244.38866
253.6222

a
100.00%
100.00%
100.00%
34.46%
62.84%
14.78%
42.38%
15.70%
6.08%
5.00%
2.82%
0.42%

maximum a

100.000%

62.839%

42.379%

4.997%

Thus, relaxation speeds up convergence. After 6 iterations, the maximum error is 4.997% and
we arrive at the result: x1 = 171.423, x2 = 244.389 and x3 = 253.622.
11.8 The first iteration can be implemented as

c1

3800 3c 2 c3 3800 3(0) 0

253.3333
15
15

c2

1200 3c1 6c3 1200 3(253.3333) 6(0)

108.8889
18
18

c3

2350 4c1 c 2 2350 4(253.3333) 108.8889

289.3519
12
12

Second iteration:

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c1

3800 3c 2 c3 3800 3(108.8889) 289.3519

294.4012
15
15

c2

1200 3c1 6c3 1200 3(294.4012) 6(289.3519)

212.1842
18
18

c3

2350 4c1 c2 2350 4(294.4012) 212.1842

311.6491
12
12

The error estimates can be computed as

a ,1

294 .4012 253 .3333


100 % 13 .95 %
294 .4012

a,2

212 .1842 108 .8889


100 % 48 .68 %
212 .1842

a ,3

311 .6491 289 .3519


100 % 7.15 %
311 .6491

The remainder of the calculation proceeds until all the errors fall below the stopping criterion
of 5%. The entire computation can be summarized as
iteration
1

unknown
c1
c2
c3
c1
c2
c3
c1
c2
c3
c1
c2
c3

value
253.3333
108.8889
289.3519
294.4012
212.1842
311.6491
316.5468
223.3075
319.9579
319.3254
226.5402
321.1535

a
100.00%
100.00%
100.00%
13.95%
48.68%
7.15%
7.00%
4.98%
2.60%
0.87%
1.43%
0.37%

maximum a

100.00%

48.68%

7.00%

1.43%

Thus, after 4 iterations, the maximum error is 1.43% and we arrive at the result: c1 =
319.3254, c2 = 226.5402 and c3 = 321.1535.
11.9 The first iteration can be implemented as

c1

3800 3c 2 c3 3800 3(0) 0

253.3333
15
15

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10

c2

1200 3c1 6c3 1200 3(0) 6(0)

66.6667
18
18

c3

2350 4c1 c 2 2350 4(0) 0

195.8333
12
12

Second iteration:

c1

3800 3c2 c3 3800 3(66.6667) 195.8333

279.7222
15
15

c2

1200 3c1 6c3 1200 3(253.3333) 6(195.8333)

174.1667
18
18

c3

2350 4c1 c 2 2350 4(253.3333) 66.6667

285.8333
12
12

The error estimates can be computed as

a ,1

279 .7222 253 .3333


100 % 9.43 %
279 .7222

a,2

174 .1667 66 .6667


100 % 61 .72 %
174 .1667

a ,3

285 .8333 195 .8333


100 % 31 .49 %
285 .8333

The remainder of the calculation proceeds until all the errors fall below the stopping criterion
of 5%. The entire computation can be summarized as
iteration
1

unknown
c1
c2
c3
c1
c2
c3
c1
c2
c3
c1
c2
c3

value
253.3333
66.66667
195.8333
279.7222
174.1667
285.8333
307.2222
208.5648
303.588
315.2855
219.0664
315.6211

a
100.00%
100.00%
100.00%
9.43%
61.72%
31.49%
8.95%
16.49%
5.85%
2.56%
4.79%
3.81%

maximum a

100.00%

61.72%

16.49%

4.79%

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11
Thus, after 4 iterations, the maximum error is 4.79% and we arrive at the result: c1 =
315.5402, c2 = 219.0664 and c3 = 315.6211.
11.10 The first iteration can be implemented as

x1

27 2 x 2 x3 27 2(0) 0

2.7
10
10

x2

61.5 3x1 2 x3 61.5 3(2.7) 2(0)

8.9
6
6

x3

21.5 x1 x2 21.5 (2.7) 8.9

6.62
5
5

Second iteration:

x1

27 2(8.9) 6.62
0.258
10

x2

61.5 3(0.258) 2(6.62)


7.914333
6

x3

21.5 (0.258) 7.914333


5.934467
5

The error estimates can be computed as

a,1

0.258 2.7
100 % 947 %
0.258

a,2

7.914333 8.9
100 % 12 .45 %
7.914333

a ,3

5.934467 (6.62 )
100 % 11 .55 %
5.934467

The remainder of the calculation proceeds until all the errors fall below the stopping criterion
of 5%. The entire computation can be summarized as
iteration
1

unknown
x1
x2
x3
x1
x2
x3

value
2.7
8.9
-6.62
0.258
7.914333
-5.93447

a
100.00%
100.00%
100.00%
946.51%
12.45%
11.55%

maximum a

100%

946%

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12

x1
x2
x3
x1
x2
x3
x1
x2
x3

0.523687
8.010001
-6.00674
0.497326
7.999091
-5.99928
0.500253
8.000112
-6.00007

50.73%
1.19%
1.20%
5.30%
0.14%
0.12%
0.59%
0.01%
0.01%

50.73%

5.30%

0.59%

Thus, after 5 iterations, the maximum error is 0.59% and we arrive at the result: x1 =
0.500253, x2 = 8.000112 and x3 = 6.00007.
11.11 The equations should first be rearranged so that they are diagonally dominant,

6 x1 x 2 x3 3
6 x1 9 x 2 x3 40
3x1 x 2 12 x3 50
Each can be solved for the unknown on the diagonal as
x1

3 x 2 x3
6

x2

40 6 x1 x3
9

x3

50 3x1 x 2
12

(a) The first iteration can be implemented as


x1

300
0.5
6

x2

40 6(0.5) 0
4.11111
9

x3

50 3(0.5) 4.11111
3.949074
12

Second iteration:

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13

x1

3 4.11111 3.949074
1.843364
6

x2

40 6(1.843364 ) 3.949074
2.776749
9

x3

50 3(1.843364 ) 2.776749
4.396112
12

The error estimates can be computed as

a ,1

1.843364 0.5
100 % 72 .88 %
1.843364

a,2

2.776749 4.11111
100 % 48 .05 %
2.776749

a ,3

4.396112 3.949074
100 % 10 .17 %
4.396112

The remainder of the calculation proceeds until all the errors fall below the stopping criterion
of 5%. The entire computation can be summarized as
iteration
1

unknown
x1
x2
x3
x1
x2
x3
x1
x2
x3
x1
x2
x3

value
0.5
4.111111
3.949074
1.843364
2.776749
4.396112
1.695477
2.82567
4.355063
1.696789
2.829356
4.355084

a
100.00%
100.00%
100.00%
72.88%
48.05%
10.17%
8.72%
1.73%
0.94%
0.08%
0.13%
0.00%

maximum a

100.00%

72.88%

8.72%

0.13%

Thus, after 4 iterations, the maximum error is 0.13% and we arrive at the result: x1 =
1.696789, x2 = 2.829356 and x3 = 4.355084.
(b) First iteration: To start, assume x1 = x2 = x3 = 0

x1new

300
0.5
6

Apply relaxation

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14
x1 0.95(0.5) (1 0.95 )0 0.475

40 6(0.475) 0
4.12778
9

x2new

x 2 0.95 (4.12778 ) (1 0.95 )0 3.92139

50 3(0.475) 3.92139
3.95863
12

x3new

x3 0.95(3.95863 ) (1 0.95)0 3.76070

Note that error estimates are not made on the first iteration, because all errors will be 100%.
Second iteration:

x1new

3 3.92139 3.76070
1.78035
6

x1 0.95 (1.78035 ) (1 0.95 )(0.475 ) 1.71508

At this point, an error estimate can be made

a ,1

1.71508 0.475
100 % 72 .3%
1.71508

Because this error exceeds the stopping criterion, it will not be necessary to compute error
estimates for the remainder of this iteration.

x2new

40 6(1.71508) 3.76070
2.88320
9

x 2 0.95 (2.88320 ) (1 0.95 )3.92139 2.93511

x3new

50 3(1.71508) 2.93511
4.35084
12

x3 0.95(4.35084 ) (1 0.95)3.76070 4.32134

The computations can be continued for one more iteration. The entire calculation is
summarized in the following table.
iteration
1
2
3

x1
0.50000
1.78035
1.70941

x1r
0.47500
1.71508
1.70969

a1
100.0%
72.3%
0.3%

x2
4.12778
2.88320
2.82450

x2r
3.92139
2.93511
2.83003

a2
100.0%
33.6%
3.7%

x3
3.95863
4.35084
4.35825

x3r
3.76070
4.32134
4.35641

a3
100.0%
13.0%
0.8%

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15

After 3 iterations, the approximate errors fall below the stopping criterion with the final
result: x1 = 1.70969, x2 = 2.82450 and x3 = 4.35641. Note that the exact solution is x1 =
1.69737, x2 = 2.82895 and x3 = 4.35526
11.12 The equations must first be rearranged so that they are diagonally dominant
8 x1 x 2 2 x3 20
2 x1 6 x 2 x3 38
3x1 x 2 7 x3 34

(a) The first iteration can be implemented as

x1

20 x2 2 x3 20 0 2(0)

2.5
8
8

x2

38 2 x1 x3 38 2(2.5) 0

7.166667
6
6

x3

34 3x1 x 2 34 3(2.5) 7.166667

2.761905
7
7

Second iteration:

x1

20 7.166667 2(2.761905)
4.08631
8

x2

38 2 x1 x3 38 2(4.08631) (2.761905)

8.155754
6
6

x3

34 3x1 x2 34 3(4.08631) 8.155754

1.94076
7
7

The error estimates can be computed as

a ,1

4.08631 2.5
100 % 38 .82 %
4.08631

a,2

8.155754 7.166667
100 % 12 .13 %
8.155754

a ,3

1.94076 (2.761905 )
100 % 42 .31 %
1.94076

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16
The remainder of the calculation proceeds until all the errors fall below the stopping criterion
of 5%. The entire computation can be summarized as
iteration
0

unknown
x1
x2
x3
x1
x2
x3
x1
x2
x3
x1
x2
x3

value

0
0
0
2.5
7.166667
-2.7619
4.08631
8.155754
-1.94076
4.004659
7.99168
-1.99919

100.00%
100.00%
100.00%
38.82%
12.13%
42.31%
2.04%
2.05%
2.92%

maximum a

100.00%

42.31%

2.92%

Thus, after 3 iterations, the maximum error is 2.92% and we arrive at the result: x1 =
4.004659, x2 = 7.99168 and x3 = 1.99919.
(b) The same computation can be developed with relaxation where = 1.2.
First iteration:

x1

20 x2 2 x3 20 0 2(0)

2.5
8
8

Relaxation yields: x1 1.2(2.5) 0.2(0) 3

x2

38 2 x1 x3 38 2(3) 0

7.333333
6
6

Relaxation yields: x 2 1.2(7.333333 ) 0.2(0) 8.8

x3

34 3x1 x2 34 3(3) 8.8

2.3142857
7
7

Relaxation yields: x3 1.2(2.3142857 ) 0.2(0) 2.7771429


Second iteration:

x1

20 x 2 2 x3 20 8.8 2(2.7771429)

4.2942857
8
8

Relaxation yields: x1 1.2(4.2942857 ) 0.2(3) 4.5531429

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17

x2

38 2 x1 x3 38 2(4.5531429) 2.7771429

8.3139048
6
6

Relaxation yields: x 2 1.2(8.3139048 ) 0.2(8.8) 8.2166857

x3

34 3x1 x 2 34 3(4.5531429) 8.2166857

1.7319837
7
7

Relaxation yields: x3 1.2(1.7319837 ) 0.2(2.7771429 ) 1.5229518


The error estimates can be computed as

a ,1

4.5531429 3
100 % 34 .11 %
4.5531429

a,2

8.2166857 8.8
100 % 7.1%
8.2166857

a ,3

1.5229518 (2.7771429 )
100 % 82 .35 %
1.5229518

The remainder of the calculation proceeds until all the errors fall below the stopping criterion
of 5%. The entire computation can be summarized as
iteration
1

unknown
x1
x2
x3
x1
x2
x3
x1
x2
x3
x1
x2
x3
x1
x2
x3
x1
x2
x3

value
2.5
7.3333333
-2.314286
4.2942857
8.3139048
-1.731984
3.9078237
7.8467453
-2.12728
4.0336312
8.0695595
-1.945323
3.9873047
7.9700747
-2.022594
4.0048286
8.0124354
-1.990866

relaxation
3
8.8
-2.777143
4.5531429
8.2166857
-1.522952
3.7787598
7.7727572
-2.248146
4.0846055
8.12892
-1.884759
3.9678445
7.9383056
-2.050162
4.0122254
8.0272613
-1.979007

a
100.00%
100.00%
100.00%
34.11%
7.10%
82.35%
20.49%
5.71%
32.26%
7.49%
4.38%
19.28%
2.94%
2.40%
8.07%
1.11%
1.11%
3.60%

maximum a

100.000%

82.353%

32.257%

19.280%

8.068%

3.595%

Thus, relaxation actually seems to retard convergence. After 6 iterations, the maximum error
is 3.595% and we arrive at the result: x1 = 4.0122254, x2 = 8.0272613 and x3 = 1.979007.
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18

11.13 As shown below, for slopes of 1 and 1 the Gauss-Seidel technique will neither converge
nor diverge but will oscillate interminably.

x2
v

x1
u
11.14 As ordered, none of the sets will converge. However, if Set 1 and 2 are reordered so that
they are diagonally dominant, they will converge on the solution of (1, 1, 1).
Set 1:

9x + 3y + z = 13
2x + 5y z = 6
6x
+ 8z = 2
Set 2:

4x + 2y z = 4
x + 5y z = 5
x + y + 6z = 8

At face value, because it is not strictly diagonally dominant, Set 2 would seem to be
divergent. However, since it is very close to being diagonally dominant, a solution can be
obtained.
The third set is not diagonally dominant and will diverge for most orderings. However, the
following arrangement will converge albeit at a very slow rate:
Set 3:

3x + 4y + 5z = 6
2y z = 1
2x + 2y 3z = 3

11.15 Using MATLAB:


(a) The results for the first system will come out as expected.
>> A=[1 4 9;4 9 16;9 16 25]
>> B=[14 29 50]'
>> x=A\B
x =
1.0000
1.0000
1.0000
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19

>> inv(A)
ans =
3.8750
-5.5000
2.1250

-5.5000
7.0000
-2.5000

2.1250
-2.5000
0.8750

>> cond(A,inf)
ans =
750.0000

(b) However, for the 44 system, the ill-conditioned nature of the matrix yields poor results:
>> A=[1 4 9 16;4 9 16 25;9 16 25 36;16 25 36 49];
>> B=[30 54 86 126]';
>> x=A\B
Warning: Matrix is close to singular or badly scaled.
Results may be inaccurate. RCOND = 3.037487e-019.
x =
0.5496
2.3513
-0.3513
1.4504
>> cond(A,inf)
Warning: Matrix is close to singular or badly scaled.
Results may be inaccurate. RCOND = 3.037487e-019.
> In cond at 48
ans =
3.2922e+018

Note that using other software such as Excel yields similar results. For example, the condition
number computed with Excel is 51017.
11.16 (a) As shown, there are 4 roots, one in each quadrant.
8

(0.618,3.236)
4

(1, 2)

0
-4

-2

0
-4

2
(1.618, 1.236)

(2, 4)
-8

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20
(b) It might be expected that if an initial guess was within a quadrant, the result would be the
root in the quadrant. However a sample of initial guesses spanning the range yield the
following roots:
6
3
0
-3
-6

(-2, -4)
(-0.618,3.236) (-0.618,3.236)
(1,2)
(-0.618,3.236) (-0.618,3.236) (-0.618,3.236)
(1,2)
(1,2)
(1.618, -1.236) (1.618, -1.236) (1.618, -1.236)
(-2, -4)
(-2, -4)
(1.618, -1.236) (1.618, -1.236)
(-2, -4)
(-2, -4)
(-2, -4)
(1.618, -1.236)
-6
-3
0
3

(-0.618,3.236)
(-0.618,3.236)
(1.618, -1.236)
(1.618, -1.236)
(-2, -4)
6

We have highlighted the guesses that converge to the roots in their quadrants. Although some
follow the pattern, others jump to roots that are far away. For example, the guess of (6, 0)
jumps to the root in the first quadrant.
This underscores the notion that root location techniques are highly sensitive to initial guesses
and that open methods like the Solver can locate roots that are not in the vicinity of the initial
guesses.
11.17 Define the quantity of transistors, resistors, and computer chips as x1, x2 and x3. The system
equations can then be defined as

4 x1 3x 2 2 x3 960
x1 3x 2 x3 510
2 x1 x 2 3x3 610
The solution can be implemented in Excel as shown below:

The following view shows the formulas that are employed to determine the inverse in cells
A7:C9 and the solution in cells D7:D9.

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21
Here is the same solution generated in MATLAB:
>> A=[4 3 2;1 3 1;2 1 3];
>> B=[960 510 610]';
>> x=A\B
x =
120
100
90

In both cases, the answer is x1 = 120, x2 = 100, and x3 = 90


11.18 The spectral condition number can be evaluated as
>> A = hilb(10);
>> N = cond(A)
N =
1.6025e+013

The digits of precision that could be lost due to ill-conditioning can be calculated as
>> c = log10(N)
c =
13.2048

Thus, about 13 digits could be suspect. A right-hand side vector can be developed
corresponding to a solution of ones:
>> b=[sum(A(1,:)); sum(A(2,:)); sum(A(3,:)); sum(A(4,:)); sum(A(5,:));
sum(A(6,:)); sum(A(7,:)); sum(A(8,:)); sum(A(9,:)); sum(A(10,:))]
b =
2.9290
2.0199
1.6032
1.3468
1.1682
1.0349
0.9307
0.8467
0.7773
0.7188

The solution can then be generated by left division


>> x = A\b
x =
1.0000
1.0000
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22
1.0000
1.0000
0.9999
1.0003
0.9995
1.0005
0.9997
1.0001

The maximum and mean errors can be computed as


>> e=max(abs(x-1))
e =
5.3822e-004
>> e=mean(abs(x-1))
e =
1.8662e-004

Thus, some of the results are accurate to only about 3 to 4 significant digits. Because
MATLAB represents numbers to 15 significant digits, this means that about 11 to 12 digits
are suspect.
11.19 First, the Vandermonde matrix can be set up
>> x1 = 4;x2=2;x3=7;x4=10;x5=3;x6=5;
>> A = [x1^5 x1^4 x1^3 x1^2 x1 1;x2^5 x2^4 x2^3 x2^2 x2 1;x3^5 x3^4
x3^3 x3^2 x3 1;x4^5 x4^4 x4^3 x4^2 x4 1;x5^5 x5^4 x5^3 x5^2 x5 1;x6^5
x6^4 x6^3 x6^2 x6 1]
A =
1024
32
16807
100000
243
3125

256
16
2401
10000
81
625

64
8
343
1000
27
125

16
4
49
100
9
25

4
2
7
10
3
5

1
1
1
1
1
1

The spectral condition number can be evaluated as


>> N = cond(A)
N =
1.4492e+007

The digits of precision that could be lost due to ill-conditioning can be calculated as
>> c = log10(N)
c =
7.1611

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23
Thus, about 7 digits might be suspect. A right-hand side vector can be developed
corresponding to a solution of ones:
>> b=[sum(A(1,:));sum(A(2,:));sum(A(3,:));sum(A(4,:));sum(A(5,:));
sum(A(6,:))]
b =
1365
63
19608
111111
364
3906

The solution can then be generated by left division


>> format long
>> x=A\b
x =
1.00000000000000
0.99999999999991
1.00000000000075
0.99999999999703
1.00000000000542
0.99999999999630

The maximum and mean errors can be computed as


>> e = max(abs(x-1))
e =
5.420774940034789e-012
>> e = mean(abs(x-1))
e =
2.154110223528960e-012

Some of the results are accurate to about 12 significant digits. Because MATLAB represents
numbers to about 15 significant digits, this means that about 3 digits are suspect. Thus, for
this case, the condition number tends to exaggerate the impact of ill-conditioning.
11.20 The flop counts for the tridiagonal algorithm in Fig. 11.2 can be determined as
Sub Decomp(e, f, g, n)
Dim k As Integer
For k = 2 To n
e(k) = e(k) / f(k - 1)
f(k) = f(k) - e(k) * g(k - 1)
Next k
End Sub

mult/div

add/subt

'(n 1)
'(n 1)

(n 1)

Sub Substitute(e, f, g, r, n, x)
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24
Dim k As Integer
For k = 2 To n
r(k) = r(k) - e(k) * r(k - 1)
Next k
x(n) = r(n) / f(n)
For k = n - 1 To 1 Step -1
x(k) = (r(k) - g(k) * x(k + 1)) / f(k)
Next k
End Sub
Sum =

'(n 1)
'

(n 1)

'2(n 1)

(n 1)

5(n-1) + 1

(3n 3)

The multiply/divides and add/subtracts can be summed to yield 8n 7 as opposed to n3/3 for
naive Gauss elimination. Therefore, a tridiagonal solver is well worth using.
1000000
100000

Tridiagonal
Naive Gauss

10000
1000
100
10
1
1

10

100

11.21 Here is a VBA macro to obtain a solution for a tridiagonal system using the Thomas
algorithm. It is set up to duplicate the results of Example 11.1.
Option Explicit
Sub TriDiag()
Dim i As Integer, n As Integer
Dim e(10) As Double, f(10) As Double, g(10) As Double
Dim r(10) As Double, x(10) As Double
n = 4
e(2) = -1: e(3) = -1: e(4) = -1
f(1) = 2.04: f(2) = 2.04: f(3) = 2.04: f(4) = 2.04
g(1) = -1: g(2) = -1: g(3) = -1
r(1) = 40.8: r(2) = 0.8: r(3) = 0.8: r(4) = 200.8
Call Thomas(e, f, g, r, n, x)
For i = 1 To n
MsgBox x(i)
Next i
End Sub
Sub Thomas(e, f, g, r, n, x)
Call Decomp(e, f, g, n)
Call Substitute(e, f, g, r, n, x)
End Sub
Sub Decomp(e, f, g, n)
Dim k As Integer
For k = 2 To n
e(k) = e(k) / f(k - 1)

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25
f(k) = f(k) - e(k) * g(k - 1)
Next k
End Sub
Sub Substitute(e, f, g, r, n, x)
Dim k As Integer
For k = 2 To n
r(k) = r(k) - e(k) * r(k - 1)
Next k
x(n) = r(n) / f(n)
For k = n - 1 To 1 Step -1
x(k) = (r(k) - g(k) * x(k + 1)) / f(k)
Next k
End Sub

11.22 Here is a VBA macro to obtain a solution of a symmetric system with Cholesky
decomposition. It is set up to duplicate the results of Example 11.2.
Option Explicit
Sub TestChol()
Dim i As Integer, j As Integer
Dim n As Integer
Dim a(10, 10) As Double
n = 3
a(1, 1) = 6: a(1, 2) = 15: a(1, 3) = 55
a(2, 1) = 15: a(2, 2) = 55: a(2, 3) = 225
a(3, 1) = 55: a(3, 2) = 225: a(3, 3) = 979
Call Cholesky(a, n)
'output results to worksheet
Sheets("Sheet1").Select
Range("a3").Select
For i = 1 To n
For j = 1 To n
ActiveCell.Value = a(i, j)
ActiveCell.Offset(0, 1).Select
Next j
ActiveCell.Offset(1, -n).Select
Next i
Range("a3").Select
End Sub
Sub Cholesky(a, n)
Dim i As Integer, j As Integer, k As Integer
Dim sum As Double
For k = 1 To n
For i = 1 To k - 1
sum = 0
For j = 1 To i - 1
sum = sum + a(i, j) * a(k, j)
Next j
a(k, i) = (a(k, i) - sum) / a(i, i)
Next i
sum = 0
For j = 1 To k - 1
sum = sum + a(k, j) ^ 2
Next j
a(k, k) = Sqr(a(k, k) - sum)
Next k
End Sub

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26

11.23 Here is a VBA macro to obtain a solution of a linear diagonally-dominant system with the
Gauss-Seidel method. It is set up to duplicate the results of Example 11.3.
Option Explicit
Sub Gausseid()
Dim n As Integer, imax As Integer, i As Integer
Dim a(3, 3) As Double, b(3) As Double, x(3) As Double
Dim es As Double, lambda As Double
n = 3
a(1, 1) = 3: a(1, 2) = -0.1: a(1, 3) = -0.2
a(2, 1) = 0.1: a(2, 2) = 7: a(2, 3) = -0.3
a(3, 1) = 0.3: a(3, 2) = -0.2: a(3, 3) = 10
b(1) = 7.85: b(2) = -19.3: b(3) = 71.4
es = 0.1
imax = 20
lambda = 1#
Call Gseid(a, b, n, x, imax, es, lambda)
For i = 1 To n
MsgBox x(i)
Next i
End Sub
Sub Gseid(a, b, n, x, imax, es, lambda)
Dim i As Integer, j As Integer, iter As Integer, sentinel As Integer
Dim dummy As Double, sum As Double, ea As Double, old As Double
For i = 1 To n
dummy = a(i, i)
For j = 1 To n
a(i, j) = a(i, j) / dummy
Next j
b(i) = b(i) / dummy
Next i
For i = 1 To n
sum = b(i)
For j = 1 To n
If i <> j Then sum = sum - a(i, j) * x(j)
Next j
x(i) = sum
Next i
iter = 1
Do
sentinel = 1
For i = 1 To n
old = x(i)
sum = b(i)
For j = 1 To n
If i <> j Then sum = sum - a(i, j) * x(j)
Next j
x(i) = lambda * sum + (1# - lambda) * old
If sentinel = 1 And x(i) <> 0 Then
ea = Abs((x(i) - old) / x(i)) * 100
If ea > es Then sentinel = 0
End If
Next i
iter = iter + 1
If sentinel = 1 Or iter >= imax Then Exit Do
Loop
End Sub

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