192 Introduction to Linear Optimization and Extensions with MATLAB
R
Go to Step 1.
Iteration 4
Step 1: f
B
=
s
1
f
1
2
f
1
3
f
2
2
=
0
(c
1
)
T
v
1
2
(c
1
)
T
v
1
3
(c
2
)
T
v
2
2
=
0
−6
−4
−9
, then solving B
T
π = f
B
gives
π =
π
1
π
2
1
π
2
2
=
π
1
1
π
1
2
π
2
1
π
2
2
=
0
−1
−4
−7
.
Step 2:
The objective functions of SP
1
and SP
2
change since π
1
=
0
−1
.
Now the objective function of SP
1
is σ
1
= (
c
1
T
−(π
1
)
T
L
1
)x
1
= −2x
1
−
2x
2
, so SP
1
is
minimize −2x
1
−2x
2
subject to 2x
1
+ x
2
≤ 4
x
1
+ x
2
≤ 2
x
1
≥ 0 x
2
≥ 0.
The optimal solution is x
1
=
x
1
x
2
=
0.0224
1.9776
with objective function
value σ
∗
1
= −4, and so r
1
∗
= σ
∗
1
− π
2
1
= −4 − (−4) = 0.
Now the objective function of SP
2
is σ
2
= (
c
2
T
−(π
1
)
T
L
2
)x
2
= −4x
3
−
3x
4
, so SP
2
is
minimize −4x
3
−3x
4
subject to x
3
+ x
4
≤ 2
3x
3
+ 2x
4
≤ 5
x
3
≥ 0 x
4
≥ 0.
The optimal solution is x
2
=
x
3
x
4