Homework 4 Solution
February 17, 2009
Solution to Exercise 3, page 111
By the twophase method, we need to maximize
z
= 2
x
1
+ 3
x
2

5
x
3
subject to
x
1
+
x
2
+
x
3
= 7, 2
x
1

5
x
2
+
x
3
≥
10, and
x
i
≥
0 for all
i
.
We first transform the problem to standard form:
maximize
z
= 2
x
1
+ 3
x
2

5
x
3
subject to
x
1
+
x
2
+
x
3
= 7
2
x
1

5
x
2
+
x
3

s
1
= 10
x
1
, x
2
, x
3
, s
1
≥
0
.
In each equality , we introduce an artificial variable. Thus, the problem in standard
form and with artificial variables is given by:
maximize
z
= 2
x
1
+ 3
x
2

5
x
3
subject to
x
1
+
x
2
+
x
3
+
R
1
= 7
2
x
1

5
x
2
+
x
3

s
1
+
R
2
= 10
x
1
, x
2
, x
3
, s
1
, R
1
, R
2
≥
0
.
In Phase I, we always solve the feasibility problem, i.e., we minimize the sum of
the artificial variables to determine a basic feasible solution. Thus, in this case we
solve (by the standard simplex method) the following problem:
mimimize
v
=
R
1
+
R
2
subject to
x
1
+
x
2
+
x
3
+
R
1
= 7
2
x
1

5
x
2
+
x
3

s
1
+
R
2
= 10
x
1
, x
2
, x
3
, s
1
, R
1
, R
2
≥
0
.
We start the initial simplex table for this problem by choosing
R
1
and
R
2
as basic
variables. The initial table is:
Basic
x
1
x
2
x
3
s
1
R
1
R
2
Solution
v
0
0
0
0

1

1
0
R
1
1
1
1
0
1
0
7
R
2
2

5
1

1
0
1
10
1