Last updated: October 7, 2014
Page 9 of 18
Ch 2: The Simplex Method
MATH2230/MATH3205/Mark Lau
1. (Initialization) We choose (
x
1
, x
2
) = (0
,
0) to be the initial CFP.
2. (Optimality test) This CFP is not optimal since
z
increases when either
x
1
or
x
2
increases, in order to satisfy the first constraint.
3. (Initial BF solution) Set up the initial tabular form
z
(
x
1
)
(
x
2
)
x
3
x
4
x
5
x
6
RHS
1
3
2
0
0
0
0
0
0
1
2
1
0
0
0
6
0
2
1
0
1
0
0
8
0
1
1
0
0
1
0
1
0
0
1
0
0
0
1
2
and see that (0
,
0
,
6
,
8
,
1
,
2) is our first BF solution.
4. (Iteration 1, step 1) We choose
x
1
as the entering variable, since
z
increases faster in
the direction of
x
1
.
5. (Iteration 1, step 2) We choose
x
4
as the leaving variable, so that
x
1
yields the greatest
increase without leaving the feasible set. Note that
x
5
cannot be the leaving variable
since

x
1
+
x
5
= 1, and hence
x
1
would be negative if
x
5
= 0.
6. (Iteration 1, step 3) We perform row operations on the simplex tableau in 3, in order
to obtain a BF solution:
z
x
1
(
x
2
)
x
3
(
x
4
)
x
5
x
6
RHS
1
0
1/2
0
3/2
0
0
12
0
1
1/2
0
1/2
0
0
4
0
0
3/2
1
1/2
0
0
2
0
0
3/2
0
1/2
1
0
5
0
0
1
0
0
0
1
2
The BF solution found in the current iteration is (4
,
0
,
2
,
0
,
5
,
2). More iterations are
need since the objective function value can be improved further by moving to one of
the adjacent BF solutions.
7. (Iteration 2, step 1) Assign
x
2
as the entering variable.
Last updated: October 7, 2014
Page 10 of 18
Ch 2: The Simplex Method
MATH2230/MATH3205/Mark Lau
8. (Iteration 2, step 2) Assign
x
3
the leaving variable.
9. (Iteration 2, step 3) Perform row operations on the tabular form in Iteration 1 to get
z
x
1
x
2
(
x
3
)
(
x
4
)
x
5
x
6
RHS
1
0
0
1/3
3/2
0
0
38/3
0
1
0
1/3
2/3
0
0
10/3
0
0
1
2/3
1/3
0
0
4/3
0
0
0
1
1
1
0
3
0
0
0
2/3
1/3
0
1
2/3
The BF solution found in the current iteration is (10
/
3
,
4
/
3
,
0
,
0
,
3
,
2
/
3). The optimal
solution is found with objective function value 38/3. Keep in mind that the algorithm
stops when an optimal solution has been obtained. In problems having multiple optimal
solutions, one can perform more iterations to obtain other optimal solutions.
2.1
Solving Minimization Problems
The following minimization
min
c
1
x
1
+
c
2
x
2
+
· · ·
+
c
n
x
n
subject to a set of constraints can be converted to a maximization,
max

c
1
x
1

c
2
x
2
 · · · 
c
n
x
n
,
over the same set of constraints. For example, the following minimization problem
min
x
1
,x
2
∈
R
0
.
4
x
1
+ 0
.
5
x
2
s.t.
0
.
3
x
1
+ 0
.
1
x
2
≤
2
.
7
,
0
.
5
x
1
+ 0
.
5
x
2
= 6
,
0
.
6
x
1
+ 0
.
4
x
2
≥
6
,
x
1
, x
2
≥
0
Last updated: October 7, 2014
Page 11 of 18
Ch 2: The Simplex Method