This preview shows pages 1–3. Sign up to view the full content.
Question #
Max. marks
Part marks
1
8
2
10
3
14
10
2
2
4
15
3
3
3
3
3
5
8
3
1
1
3
6
5
7
Bonus (20)
(0
6
4
6
4)
8
No Credit
9
No Credit
Total
60
1.
8 marks:
4 marks for auxiliary LP and initial tableau, distributed as:
•
1 for the artiﬁcial variables of auxiliary LP
•
1 for rest of auxiliary LP
•
1 for the
w
row of initial tableau
•
1 for the BFS in initial tableau
2 for pivots and optimal tableau
2 for ”
y
” (as in farkas lemma) and checking infeasibility
Exercise 7.7.2 from the course notes.
Solution
:
We are given the system of equations
Ax
=
b
, where
A
=
1
1

8

1

6
1
1

4
1

8
1
4
,b
=

1
1
2
We begin with Phase 1 by formulating the problem, max
{
c
T
x

Dx
=
d, x
≥
0
}
where
D
=

1

1 8
1 1 0 0

6
1 1

4 0 1 0
1

8 1
4 0 0 1
, d
=
1
1
2
,
and
c
T
=
±
0 0 0 0

1

1

1
²
The initial basis is
B
=
{
5
,
6
,
7
}
and the corresponding basic variables are [
x
5
x
6
x
7
]
T
= [1 1 2]
T
.
The initial tableau is:
w
+ 6
x
1
+ 8
x
2

10
x
3

x
4
=

4

x
1

x
2
+
8
x
3
+
x
4
+
x
5
=
1

6
x
1
+
x
2
+
x
3

4
x
4
+
x
6
=
1
x
1

8
x
2
+
x
3
+ 4
x
4
+
x
7
=
2
.
Iteration 1: We take
x
3
as the entering variable, and calculate
t
= min
{
1
/
8
,
1
/
1
,
2
/
1
}
= 1
/
8. So
x
5
leaves, and our new basis is
{
3
,
6
,
7
}
. The updated tableau is:
1
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Documentw
+ 19
/
4
x
1
+ 27
/
4
x
2
+
1
/
4
x
4
+ 5
/
4
x
5
=

11
/
4

1
/
8
x
1

1
/
8
x
2
+
x
3
+
1
/
8
x
4
+ 1
/
8
x
5
=
1
/
8

47
/
8
x
1
+
9
/
8
x
2

33
/
8
x
4

1
/
8
x
5
+
x
6
=
7
/
8
9
/
8
x
1

63
/
8
x
2
+ 31
/
8
x
4

1
/
8
x
5
+
x
7
=
15
/
8
.
Iteration 2: Since
c
j
≤
0
∀
j
∈
N
, the current solution is optimal. We now proceed to calculate a
certiﬁcate of infeasibility for the original problem. First we calculate the vector
y
as,
D
T
B
y
=
c
B
⇒
y
=
1
/
4

1

1
.
However,
b
1
<
0, so in order to recover a certiﬁcate for the original system, we multiply
y
1
by

1 to
obtain
z
=

1
/
4

1

1
.
We calculate
z
T
A
=
±
5

1
/
4 7

1
/
4 0 1
/
4
²
≥
0, and
z
T
b
= 1
/
4

3
<
0. Thus
z
, together with
Farkas’ Lemma shows that the system
Ax
=
b
is infeasible.
2.
10 marks:
6 marks distributed same as ﬁrst 6 marks in Q1
2 marks for initial tableau of phase 2
2 marks for ﬁnal tableau of phase 2
Exercise 7.7.4 from the course notes, which is reproduced here:
Solve the following LP problem by the twophase simplex method:
(
P
) max
z
=
5
x
1

2
x
2
+
x
3
subject to
x
1
+ 4
x
2
+
x
3
≤
6
2
x
1
+
x
2
+ 3
x
3
≥
2
x
1
,
x
2
≥
0
,
x
3
free
Hint
: Observe that
x
3
is a free variable and the simplex method (the version studied in CO350) applies
only to LPs in standard equality form.
Solution
This is the end of the preview. Sign up
to
access the rest of the document.
 Winter '07
 S.Furino,B.Guenin

Click to edit the document details