ORIE 3300/5300
ASSIGNMENT 13 SOLUTION
Fall
2010
Problem 1
After introducing a slack variable and solving the linear programming relaxation
using the simplex method, we arrive at the following optimal tableau:
z
+
1
5
x
1
+
3
5
x
3
=
27
5
2
5
x
1
+
x
2
+
1
5
x
3
=
9
5
.
We can see the optimal solution is (
x
1
, x
2
) = (0
,
9
/
5), with optimal value
z
= 27
/
5.
Now we will construct the branchandbound tree.
Since
x
2
has a fractional
value now, we branch on
x
2
and consider two subproblems with extra con
straint
x
2
≥
2 or
x
2
≤
1.
It is easy to see the subproblem with constraint
x
2
≥
2 is infeasible since 2
x
1
+ 5
x
2
≥
0 + 5
×
2 = 10
>
9, but in a large instance
we will have to determine infeasibility as we reoptimize using the dual simplex
method. We will show how to do that below.
For the subproblem with constraint
x
2
≥
2, after introducing an additional
surplus variable for this constraint, and adding the resulting equation to the
optimal tableau above, we obtain:
z
+
1
5
x
1
+
3
5
x
3
=
27
5
2
5
x
1
+
x
2
+
1
5
x
3
=
9
5
x
2

x
4
=
2
.
x
4
will be a basic variable, but we want to eliminate the basic variable
x
2
from
this last row, so we multiply the last equation by 1 and add the second equation
to it to obtain the dualfeasible tableau:
z
+
1
5
x
1
+
3
5
x
3
=
27
5
2
5
x
1
+
x
2
+
1
5
x
3
=
9
5
2
5
x
1
+
1
5
x
3
+
x
4
=

1
5
.
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 Fall '08
 TODD
 Linear Programming, Optimization, dual simplex method

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