2
(a) Introduce slack variables
x
4
,
x
5
,
x
6
≥
0 and transform the given problem to an LP problem
[5]
in standard equality form.
Solution.
maximize
10
x
1
+ 20
x
2
+ 30
x
3
subject to
4
x
1

x
2
+
3
x
3
+
x
4
=
5
6
x
1
+
5
x
2
+
x
3
+
x
5
= 10
2
x
1
+
8
x
2
+
8
x
3
+
x
6
=
2
x
1
,
x
2
,
x
3
, x
4
, x
5
, x
6
≥
0
.
(b) Using the simplex method with
the smallest subscript rule
solve the LP problem you
[15]
obtained in part (a). Show all your work. Remember to give, after each simplex iteration,
the new basis and the new basic feasible solution.
Solution.
z

10
x
1

20
x
2

30
x
3
=
0
4
x
1

x
2
+
3
x
3
+
x
4
=
5
6
x
1
+
5
x
2
+
x
3
+
x
5
= 10
2
x
1
+
8
x
2
+
8
x
3
+
x
6
=
2
x
1
enters, min
{
5
4
,
10
6
,
2
2
}
= 1,
x
6
leaves the basis. We pivot on (6
,
1) and obtain the
tableau:
z
+ 20
x
2
+ 10
x
3
+
5
x
6
= 10

17
x
2

13
x
3
+
x
4

2
x
6
=
1

19
x
2

23
x
3
+
x
5

3
x
6
=
4
x
1
+
4
x
2
+
4
x
3
+
1
2
x
6
=
1
which is optimal. The new basis is
B
:=
{
1
,
4
,
5
}
, the corresponding basic feasible solution
is ¯
x
:= [1
,
0
,
0
,
1
,
4
,
0]
T
with the objective value 10. Thus, an optimal solution of the
original LP is
x
*
:= [1
,
0
,
0]
T
, with objective value
z
*
= 10. (Actually, it is the only
optimal solution of the LP.)
3. [10 points]
Let
F
be the set of
x
satisfying
x
1

4849845
x
3
+
x
4
= 3
2
x
2
+ 1048576
x
3
= 2
x
1
,
x
2
,
x
3
, x
4
≥
0
(a) Find two extreme points of
F
. Justify your answer.
[8]
Solution.
Note that
F
=
{
x
:
Ax
=
b, x
≥
0
}
where
A
=
±
1 0

4849845 1
0 2
1048576 0
²
,
b
= [3
,
2]
T
.
Let
x
0
:= [3
,
1
,
0
,
0]
T
and
x
00
:= [0
,
1
,
0
,
3]
T
. Clearly,
x
0
,x
00
∈
F
.