Principles of Mathematics in Operations Research

Solutions 237

xn = A(2)-%I = 4 -6

-6 12

1.5

1.0

Ab = bi - bu =

xi - xu

-0.5

-0.5

f

-3

=HIA,II = \H

VlO, ||a;/|| = VT=»

\\xi\\

Then, the relative error for this case is is ^2 = 5.0.

2. The maximum error is the condition number

det(sJ - 4(2)) =

s-1

l = (-D(-J)-^o

=>Ai =

6

-v

4-v/l3

13 4 + \/T3

"*2 = ^ •

Therefore, c[4(2)] = % = ^§ = £fff§ = 19.2815 is the upper

bound.

3.

4(2) + AA(2) =

1|

1 I

2 3

+

(^0) 2"
I 1
Ax = xui -xi -
hxin = bi => xiu = bi
Ax\
1.0
0.5
2.0
0.5
=$•
_ y/ 425
\xi + Ax\ ~ y/h25
= 1.84391
||4(2)|| = A 2 = 4 +/^13 = 1.26759
|AA(2)|| is the largest eigenvalue of 1 2
'2 3
, which is 0.9343. Then,
M(2)|

P(2)|

0.9343

1.2676

0.7371

\\xr + Ax

iM2)\

1.84391

0.7371

= 2.5017

ll^(2)||

4. The maximum error is ||4(2)|| ||4(2)_1||, where ||4(2)_1||is the largest
   eigenvalue of 4(2)_1 as calculated below:

det(s/-4(2)-^1 )

s-4 6

6 s-12 = (s-4)(s-12)-36 = 0