CS 601 (Autumn 2009): Quiz 2 solutions
November 3, 2009
Problem 1
Part (a)
For each item
i
we have a 0
/
1 integer variable
x
i
, where
x
i
= 1 indicates that item
i
is picked
while
x
i
= 0 indicates that it is not picked. The integer LP is as follows.
max
n
summationdisplay
i
=1
v
i
x
i
n
summationdisplay
i
=1
w
i
x
i
≤
W
x
i
∈
{
0
,
1
}
for all
i
[5 marks]
Part (b)
The LP relaxation is as follows.
max
n
summationdisplay
i
=1
v
i
x
i
n
summationdisplay
i
=1
w
i
x
i
≤
W
x
i
≤
1 for all
i
x
i
≥
0 for all
i
To construct the dual of the above LP, we take the variable
y
i
for the constraint
x
i
≤
1, for each
i
, and the variable
z
for the constraint
∑
n
i
=1
w
i
x
i
≤
W
. The dual LP is as follows.
min
n
summationdisplay
i
=1
y
i
+
Wz
y
i
+
w
i
z
≥
v
i
for all
i
y
i
≥
0 for all
i
z
≥
0
[5 marks]
Common mistake: Almost everyone has ignored the constraint
x
i
≤
1
in the primal. Note that the
“standard” form of the primal is
max
c
⊤
x
;
Ax
≤
b
,
x
≥
0
. Hence each of the constraints
x
i
≤
1
will contribute a row to
A
and
b
.
1
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Part (c)
If the object
i
is not fully picked in the optimal primal solution then we have
x
i
<
1. By the
property of complementary slackness, since the constraint
x
i
≤
1 is slack, the corresponding
variable
y
i
must be 0 in the optimal dual solution. Thus
z
≥
v
i
/w
i
for each item
i
that is not
fully picked.
[5 marks]
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 Summer '09
 PROF.RANADE
 Algorithms, Trigraph, Yi, WI

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