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Faculty of Engineering and Natural Sciences
Industrial Engineering
Solution Midterm exam MS 301
May 7th 2011, 11.00  13.00, FENS G029, G035 and FMAN L014, L018
Remark: keep your answers as short as possible but of course complete. Also read very
carefully the questions. The total number of points of this exam is 100. At the beginning
of each part of the exercise the total number of points are listed. Please write your student
number at each page and write clearly. To avoid noise during the whole exam it is only
allowed during the ﬁrst hour of the exam to ask questions. Within two weeks the results of
the exam will be known.
Exercise 1 (30 points).
Consider the linear programming problem
max 5
x
1
+4
x
2
+3
x
3
2
x
1
+3
x
2
+
x
3
≤
5
4
x
1
+
x
2
+2
x
3
≤
11
3
x
1
+4
x
2
+2
x
3
≤
8
x
1
x
2
x
3
≥
0
1. (
15 points)
It is claimed that the optimal solution is given by
x
∗
1
= 2
, x
∗
2
= 0
, x
∗
3
= 1
.
However,you do not trust this claim and want to check whether it is correct. Without solving
the above problem by the simplex method give a procedure to check this claim and show in
detail your calculations.
2. (
10 points)
How are the relations you try to verify called in the literature?
3. (5
points)
Identify without any calculations for the above problem the optimal basis
B
.
Explain shortly your answer!
Solution.
1
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View Full Document 1. It follows that
2
×
2 + 0 + 1
= 5
4
×
2 + 0 + 2
= 10
<
11
and hence
y
∗
2
= 0
3
×
2 + 0 + 2
= 8
Hence the solution is primal feasible and by the complementary slackness relations for the
primal problem it follows that
y
∗
2
= 0
.
The dual problem is given by (use principle of
Lagrange to construct it!)
min 5
y
1
+11
y
2
+8
y
3
2
y
1
+4
y
2
+3
y
3
≥
5
3
y
1
+
y
2
+4
y
3
≥
4
y
1
+2
y
2
+2
y
3
≥
3
y
1
y
2
y
3
≥
0
Since
x
∗
1
>
0
and
x
∗
3
>
0
, (observe that
x
∗
j
is the dual variable associated with the inequality
j
in the dual problem) we obtain by the complementary slackness relations for both the
primal and dual problem that
2
y
∗
1
+4
y
∗
2
+3
y
∗
3
= 5
y
∗
1
+2
y
∗
2
+2
y
∗
3
= 3
y
∗
2
= 0
This system has solution
y
∗
1
=
y
∗
3
= 1
, y
∗
2
= 0
and this solution is dual feasible. Hence
by the complementary slackness relations this is indeed an optimal dual solution. To check
whether we did not make any mistake we observe that the objective value of both the primal
and dual problem equals
13
(for
x
⊤
= (2
,
0
,
1)
en
y
⊤
= (1
,
0
,
1))
. Hence by the weak
duality theorem the given vector is indeed an optimal solution.
2. Complementary slackness relations or conditions.
3. Since any nonbasic variable in the optimal solution must be zero and we have in total
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This note was uploaded on 03/17/2012 for the course MS 301 taught by Professor H during the Spring '12 term at Sabancı University.
 Spring '12
 H

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