228 Linear Programming II: Additional Topics and Extensions
Step 1We choose the initial feasible point asX(^1 )=
1 3 1 3 1 3
and setk=1.
Step 2Since|f (X(^1 )) = ||^23 | > 0. 0 5, we go to step 3.
Step 3Since [a]= { 0 , 1 , − 1 },c= { 2 , 1 ,− 1 }T, ||c|| =√
( 2 )^2 +( 1 )^2 +(− 1 )^2 =
√
6,
we find that[D(X(^1 ))]=
1
3 0 0
0 13 0(^0013)
[a][D(X(^1 )) ]={ 0 13 −^13 }[P]=
[
[a][D(X(^1 ))]
1 1 1]
=
[
0 13 −^13
1 1 1
]
([P][P]T)−^1 =
[ 2
9 0
0 3
]− 1
=
[ 9
2 0
(^013)
]
[D(X(^1 ))]c=
1
3 0 0
0 13 0(^0013)
2
1
− 1
=
2
3
1
3
−^13
([I]−[P]T( [[P]P]T)−^1 [ P])[D(X(^1 ))]c=
1 0 0
0 1 0
0 0 1
−
0 1
1
3 1
−^131
[ 9
2 0
(^013)
] [
0 13 −^13
1 1 1
]
2
3
1
3
−^13
=
2
3 −1
3 −1
3
−^131616
−^131616
2
3
1
3
−^13
=
4
9
−^29
−^29
Usingα=^14 , Eq. (4.68) givesY(^2 )=
1 3 1 3 1 3
−^14
4
9
−^29−^29
1
√
3 ( 2 )
√
6
=
34
108
37
108
37
108