Principles of Mathematics in Operations Research

222 Solutions

{AIM -i AT A^1 =

107

90

_iZ

45

47

45

83

90

147 10

2 58 11

J 7 1_ 2

10 15 30 5

4 13 J 3_

5 30 15 10

x = (AUsr

1

A

T

b 3 Xi

X2

_9 7 L 2

10 15 30 5

4 13 J _3_

5 30 15 10

[21

5

6

LsJ

34 "

30

53

30

The A3 — QR decomposition is given below:

Q =

-0.07762 -0.83305 -0.39205 -0.38249

-0.31046 -0.45124 0.23763 0.80220

-0.54331 -0.06942 0.70087 -0.45693

-0.77615 0.31239 -0.54646 0.03722

R =

-12.8840 -14.5920

0.0000 -1.0413

0.0000 0.0000

0.0000 0.0000

The equivalent system Rx — QTb$ is solved below:

QTh =

-0.07762 -0.31046 -0.54331 -0.77615

-0.83305 -0.45124 -0.06942 0.31239

-0.39205 0.23763 0.70087 -0.54646

-0.38249 0.80220 -0.45693 0.03722

(9)

-11.1770

-1.8397

0.2376

0.8022

Rx =

*2 = E$S = 1-7667

12.8840 -

0.0000

0.0000

0.0000

-14.5920

-1.0413

0.0000

0.0000

Xi

X2

-11.1770

-1.8397

0.2376

0.8022

Xi -11.177-1.7667(-14.592) -12.884 = -1.1333

(0)

The two solutions, (9) and (), are equivalent.

A 3 x =

" 1 2"

4 5

7 8

.10 11.

-1.1333

1.7667

=

2.4201

4.3503

6.2805

8.2107

^

"2"

5

6

.8.

= 63