206 Linear Programming II: Additional Topics and Extensions
X( 12 )= ointp S={
0
0
}
X( 22 )= ointp T={
1000
2000
}
X( 32 )= ointp U={
1000
0
}
Thus any point in the convex feasible sets defined by Eqs. (E 6 ) nd (Ea 7 ) anc
be represented, respectively, asX 1 =μ 11{
0
0
}
+μ 12{
0
600
}
+μ 13{
400
200
}
=
{
400 μ 13
600 μ 12 + 002 μ 13}
with
μ 11 +μ 12 +μ 13 = 1 , 0 ≤μ 1 i≤ 1 , i= 1 , 2 , 3
(E 8 )
andX 2 =μ 21{
0
0
}
+μ 22{
1000
2000
}
+μ 23{
1000
0
}
=
{
1000 μ 22 + 0001 μ 23
2000 μ 22}
with
μ 21 +μ 22 +μ 23 = 1 ; 0 ≤μ 2 i≤ 1 , i= 1 , 2 , 3
(E 9 )
Step 2By substituting the relations of (E 8 ) nd (Ea 9 ) the problem stated in Eqs. (E, 5 )
can be rewritten asMaximizef (μ 11 , μ 12 ,... , μ 23 )=( 12 ){
400 μ 13
600 μ 12 + 002 μ 13}
+( 23 )
{
1000 μ 22 + 0001 μ 23
2000 μ 22}
= 800 μ 13 + 2001 μ 12 + 0008 μ 22 + 0002 μ 23subject to
[
1 1
1 0] {
400 μ 13
600 μ 12 + 002 μ 13}
+
[
11
1 0
] {
1000 μ 22 + 0001 μ 23
2000 μ 22}
≤
{
1000
500
}
that is,600 μ 12 + 006 μ 13 + 0003 μ 22 + 0001 μ 23 ≤ 0001400 μ 13 + 0001 μ 22 + 0001 μ 23 ≤ 005