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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 must solve Phase 1, max
{
c
T
x

Ax
=
b, x
≥
0
}
where
c
T
= [0 0 0 0

1

1

1]
A
=

1

1 8
1 1 0 0

6
1 1

4 0 1 0
1

8 1
4 0 0 1
, b
=
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:
w
+ 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
.
1
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View Full DocumentIteration 2: Since
c
j
≤
0
∀
j
∈
N
, the current solution is optimal.
A
T
B
y
=
c
B
⇒
y
=
1
/
4

1

1
Since
c
j
≤
0
∀
j
∈
N
, the current solution is optimal. The vector
y
satisﬁes the conditions of Farkas’
Lemma (compute for yourself) and hence acts as a certiﬁcate of infeasibility.
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
:
We need to put the LP into standard equality form (using
v
1

v
2
to replace the free variable
x
3
):
max
z
= 5
x
1

2
x
2
+
v
1

v
2
s.t.
x
1
+ 4
x
2
+
v
1

v
2
+
s
1
= 6
2
x
1
+
x
2
+ 3
v
1

3
v
2

s
2
= 2
x
i
, v
i
, s
i
≥
0
Now to formulate the auxiliary problem we use
s
1
as the auxiliary variable for the ﬁrst constraint and
add auxiliary variable
u
2
for the second constraint. The auxiliary problem is:
max
w
=

s
1

u
2
s.t.
x
1
+ 4
x
2
+
v
1

v
2
+
s
1
+
= 6
2
x
1
+
x
2
+ 3
v
1

3
v
2

s
2
+
u
2
= 2
x
i
, v
i
, s
i
, u
i
≥
0
The ﬁrst tableau, using the basis
{
s
1
,u
2
}
:
w

3
x
1

5
x
2

4
v
1
+ 4
v
2
+
s
2
=

8
x
1
+ 4
x
2
+
v
1

v
2
+
s
1
=
6
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 Spring '07
 S.Furino,B.Guenin

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