IE521 Advanced Optimization
HW4 Solution
1.
(a)
•
Primal Problem:
max
6
x
1
+ 10
x
2
+ 9
x
3
+ 20
x
4
s.t.
4
x
1
+ 9
x
2
+ 7
x
3
+ 10
x
4
≤
600
x
1
+
x
2
+ 3
x
3
+ 40
x
4
≤
400
3
x
1
+ 4
x
2
+ 2
x
3
+
x
4
≤
500
x
1
, x
2
, x
3
, x
4
≥
0
•
Dual Problem:
min
600
p
1
+ 400
p
2
+ 500
p
3
s.t.
4
p
1
+
p
2
+ 3
p
3
≥
6
(1)
9
p
1
+
p
2
+ 4
p
3
≥
10
(2)
7
p
1
+ 3
p
2
+ 2
p
3
≥
9
(3)
10
p
1
+ 40
p
2
+
p
3
≥
20
(4)
p
1
, p
2
, p
3
≥
0
(b) Optimal solution to this LP is
z
= 2800
/
3
, x
1
= 400
/
3
, x
4
= 20
/
3
, x
2
=
x
3
= 0 and
s
1
=
s
2
= 0
, s
3
= 280
/
3.
Since
x
1
>
0 and
x
4
>
0 then the constraints associated with
x
1
, x
4
in the dual problem
(Constraints 1 and 4) must be binding. Therefore:
4
p
1
+
p
2
+ 3
p
3
= 6
(5)
10
p
1
+ 40
p
2
+
p
3
= 20
(6)
Also, since the slack variable for third constraint in primal,
s
3
>
0, then the dual variable
associated with the third constraint,
p
3
must be zero. If we put
p
3
= 0 into equations 5
and 6 and solve the equation, we get
p
1
= 22
/
15,
p
2
= 2
/
15. Therefore the optimal dual
solution is
p
1
= 22
/
15,
p
2
= 2
/
15,
p
3
= 0 and optimal solution value = 2800/3.
2.
•
Primal Problem:
max
5
x
1
+ 3
x
2
+
x
3
s.t.
2
x
1
+
x
2
+
x
3
≤
6
x
1
+ 2
x
2
+
x
3
≤
7
x
1
, x
2
, x
3
≥
0
•
Dual Problem:
min
6
p
1
+ 7
p
2
s.t.
2
p
1
+
p
2
≥
5
(7)
p
1
+ 2
p
2
≥
3
(8)
p
1
+
p
2
≥
1
(9)
p
1
, p
2
≥
0
Dual problem is two-dimensional. Therefore we can draw the constraints 7, 8, and 9 on two
dimension and find the feasible region. When we find the feasible region, we can see that there
are three extreme points which are (0
,
5)
,
(3
,
0)
,
(7
/
3
,
2
/
6). It is a minimization and the optimal
solution is at the point (7
/
3
,
2
/
6) with objective function value of 49/3.
Using the complementary slackness condition, we can say that since
p
1
>
0 and
p
2
>
0, the
constraints associated with
p
1
and
p
2
in primal problem must be binding. Therefore,
2
x
1
+
x
2
+
x
3
= 6
(10)
x
1
+ 2
x
2
+
x
3
= 7
(11)
(12)
1