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Question #
Max. marks
Part marks
1
5
1
1
1
2
2
5
2
3
3
7
3
4
4
43
18
13
12
5
No Credit
6
No Credit
7
No Credit
Total
60
1.
5 marks = 1+1+1+2
The following is a tableau of a maximization problem, where the entries
α
,
β
,
γ
,
δ
and
ε
in the tableau
are unknown parameters.
z
+
δx
4
+
γx
5
=
0
x
2
+
αx
4
=
3
x
3

2
x
4
+
εx
5
=
2
x
1
+
2
x
5
=
β
For each of the following statements, ﬁnd conditions on the values of
α
,
β
,
γ
,
δ
and
ε
such that the
statement is true.
(a) The basic solution is not feasible.
(b) The basic solution is feasible, but the tableau is not optimal.
(c) The basic solution is feasible and the next pivot, under the smallest subscript rule (see Section 6.6),
indicates that the problem is unbounded.
(d) The basic solution is feasible,
x
5
is a candidate for entering the basis, and when
x
5
enters the basis,
then
x
3
leaves the basis.
Solution
:
(a)
β <
0.
The basic solution
x
*
deﬁned by
x
*
i
=
b
i
for
i
∈
B
and
x
*
i
= 0 for
i /
∈
B
is feasible if and only if
x
*
≥
0.
(b)
β
≥
0 and either
δ <
0 or
γ <
0.
A tableau is optimal if and only if it is feasible and
c
j
≤
0 for all
j
.
(c)
β
≥
0,
δ <
0 and
α
≤
0.
To satisfy the simplex “test” for unboundedness, the entering column
k
must have
a
ik
≤
0 for all
i
∈
B
. Thus we must have
k
= 4.
(d)
β
≥
0,
γ <
0,
± >
0 and
β±
≥
4.
Since
x
5
enters, we must have
c
5
=

γ >
0. Since
x
3
leaves, we must have
a
35
=
± >
0, and by the
ratio test
b
3
/
a
35
= min
{
b
i
/
a
i
5
:
i
∈
B,
a
i
5
>
0
}
.
2.
5 marks = 2+3
(Based on Exercises 6.9.5 and 6.9.6 in the course notes)
1
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View Full Document (a) Give an example to show that the entering variable on one iteration of the simplex method can
become the leaving variable on the next iteration.
(b) Is it possible for the leaving variable on one iteration of the simplex method to become an entering
variable on the next iteration? Justify your answer.
Hint
: Suppose that
x
k
enters and
x
r
leaves on an iteration of the simplex method. Derive a formula
for the new value of
c
r
(the reduced cost of
x
r
in the new tableau) in terms of the coeﬃcients of the
old tableau.
Solution
:
(a) The following tableau has basis
B
=
{
2
}
.
z

x
1

2
x
3
= 0
x
1
+
x
2
+
x
3
= 1
We choose
x
1
as the entering variable. Then
x
2
is the leaving variable, and we get a new tableau
with basis
B
=
{
1
}
:
z
+
x
2

x
3
= 1
x
1
+
x
2
+
x
3
= 1
Now, the entering variable is
x
3
, and the leaving variable is
x
1
.
This shows that the entering variable on one iteration (
x
1
in this example) can be the leaving
variable on the next iteration.
(b) No.
The new tableau necessarily has a negative reduced cost for the leaving variable of the previous
iteration, so that variable cannot be chosen as an entering variable.
To see this in detail, write the formula for the old tableau, denoting the old basis by
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This note was uploaded on 02/05/2010 for the course CO 350 taught by Professor S.furino,b.guenin during the Spring '07 term at Waterloo.
 Spring '07
 S.Furino,B.Guenin

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