CO 227  Homework assignment 6
Fall ‘10
Page 1
CO 227  Fall ‘10
Homework assignment #6:
(Due at 5pm, on Monday, Dec 6th)
Instructions:
•
Please show all your work and justify your answers. Answers without proper justification
will not be considered.
•
Please list who you collaborated with.
•
Please staple your homework and write your name and student id in pen.
•
Total value = 90 points
Question 1
 Branchandbound
(30 points)
Consider the following integer program:
max
z
=
x
1
+5
x
2
s.t.
−
4
x
1
+3
x
2
≤
6
3
x
1
+2
x
2
≤
18
x
1
, x
2
≥
0 and integer
(IP)
The picture below shows the branchandbound tree in the middle of solving this problem (only the
nodes that were already explored are shown).
Subp. 1
z
*
= 28
.
94
x
*
1
= 2
.
47
x
*
2
= 5
.
29
Subp. 2
z
*
= 25
.
5
x
*
1
= 3
x
*
2
= 4
.
5
Subp. 4
INFEASIBLE
Subp. 5
z
*
= 23
.
33
x
*
1
= 3
.
33
x
*
2
= 4
Subp. 6
z
*
= 19
x
*
1
= 4
x
*
2
= 3
Subp. 7
z
*
= 23
x
*
1
= 3
x
*
2
= 4
Subp. 3
z
*
= 25
.
33
x
*
1
= 2
x
*
2
= 4
.
67
x
1
≥
3
x
1
≤
2
x
2
≥
5
x
2
≤
4
x
1
≥
4
x
1
≤
3
(a) (3 points) Suppose that we know
z
2
which is the value of the optimal value to the subproblem 2
and
z
3
which is the optimal value to the subproblem 3. How would you express the optimal value
to (IP) in terms of
z
2
and
z
3
?
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CO 227  Homework assignment 6
Fall ‘10
Page 2
[CAREFUL: the optimal solution to the subproblems is NOT the optimal solution to the LP relax
ations of the subproblems. It does take into account the integrality constraints]
Solution:
The original problem (IP) is divided into two subproblems 2 and 3.
Therefore the optimal value for (IP) is given by:
max
{
z
2
, z
3
}
(b) (4 points) Solve the following IP: (show all your work)
max
z
=
x
1
+5
x
2
s.t.
−
4
x
1
+3
x
2
≤
6
3
x
1
+2
x
2
≤
18
x
1
≥
3
x
1
, x
2
≥
0 and integer
Solution:
Notice that this problem is exactly the subproblem 2 in the branchandbound tree.
From the tree, we notice that all the nodes that correspond to solving subproblem 2 have been
explored and there are no more nodes to branch on. Hence, the optimal solution to subproblem
2 is the best solution found in that subtree, which is:
x
= (3
,
4)
T
with objective value 23.
(c) (5 points) Suppose you stopped branching now, and just used the best integral solution you have
found so far.
Give a bound on how far from the optimal value for (IP) that your best integral
solution’s objective value could be. Show all your work.
Solution:
The best solution found in the given branchandbound tree has value 23.
Now suppose that
z
IP
is the optimal value for (IP). We know that:
23
≤
z
IP
= max
{
z
1
, z
2
}
But, from what we’ve seen in the previous question,
z
1
= 23. Moreover, since the LP relaxation
of
IP
2
has value 25.33, we know that
z
2
≤
25
.
33.
In particular, since
z
2
must be an integer number, we know that
z
2
≤
25.
Hence
23
≤
z
IP
= max
{
z
1
, z
2
} ≤
max
{
23
,
25
}
= 25
So our best integral solution’s objective value is at most 2 units away from the optimal value.
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 Linear Programming, Optimization, Yi, perfect matching

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