Principles of Mathematics in Operations Research

248 Solutions

xB = B~lb =

[10-1]

01 0

10 0

[10"

8

2

=

[ 6"

10

8

We are on point D.

z = cTBxB = [0,2,2]

6

10 = 36.

cl-clB^N =[1,0,0] -[0,2,2]

Thus, D is the optimal point.

2. simplex tableau:

"10 -1"

0 1 0

10 0

"20 1"

010

000

= [-3,-2,-2].

«1

Xz

x 2

z

X\ X 2 X 3 Si S 2

2 0 0 10

S3

1

0 0 10 1 0

0 1 0 0 0-1

-10 0 0 2 -2

RHS~

6

10

2

24

S3

2^3

x 2

z

Xi X 2 X 3 Si S 2 S 3

2 0 0 10 1

0 0 10 10

2 10 0 0 0

3 0 0 2 2 0

RHS'

6

10

8

36.

3. revised simplex with product form of the inverse:

Let xB = (s 1 ,x 3 ,a;2)T, xN = (a;i,s 2 ,s 3 )T- Then, B~l

10-1

0 1 0

00 1

w = c

T

BB~

l = [0,2,2].

rXl =cXl ~wNXl =1-[0,2,2]

cS2 - wNS2 = 0 - [0,2,2]

uNS3 =0-[0,2,2]

0

1

0

0

0

-1

S3 is the entering variable and Si leaves. Ex l =

xB = £ 1 -^1 6= (6,8,10)T.

= 1>0.

= -2<0.

= 2>0.

f00

-flO

-^01