Principles of Mathematics in Operations Research

228 Solutions

where

A =

' 0

0

1

20

3

L 100

0

0

1

50

0

3

100

1

25

0

0

1 -1

50

1

100

0

0

and the initial condition is W 0 = [100,60,40,30]T.

(b) A = SAS-\ where

S =

0.46791

0.54010

0.64713

0.26563

-0.46791 -0.20890 -0.20890

-0.54010 0.69374 0.69374

0.64713 0.33092 -0.33092

0.26563 -0.60464 0.60464

0.79296 0.23878 0.63090 0.34529

-0.79296 -0.23878 0.63090 0.34529

-0.61736 0.53484 0.27717 -0.67525

-0.61736 0.53484 -0.27717 0.67525

and A —

-0.052845 0.000000

0.000000 0.052845

0.000000 0.000000

0.000000 0.000000

AtC-U

0.000000 0.000000

0.000000 0.000000

-0.010365 0.000000

0.000000 0.010365

The solution is W = SeMS~lW 0

X 1 (t)

Yi(t)

Y 2 (t)

0.46791 -0.46791

0.54010 -0.54010

0.64713 0.64713

0.26563 0.26563

-0.20890 -0.20890

0.69374 0.69374

0.33092 -0.33092

-0.60464 0.60464

-0.052845 t

„0.052845t

-0.010365t

„0.010365«

0.79296 0.23878 0.63090 0.34529

-0.79296 -0.23878 0.63090 0.34529

-0.61736 0.53484 0.27717 -0.67525

-0.61736 0.53484 -0.27717 0.67525

100

60

40

30

Since 5 _1W 0 =

129.220

-58.028

-38.816

-20.475

we have