9.5
THE SIMPLEX METHOD: MIXED CONSTRAINTS
In Sections 9.3 and 9.4, you looked at linear programming problems that occurred in
stan
dard form.
The constraints for the maximization problems all involved
inequalities, and
the constraints for the minimization problems all involved
inequalities.
Linear programming problems for which the constraints involve
both
types of inequali
ties are called
mixedconstraint
problems. For instance, consider the following linear
programming problem.
MixedConstraint Problem:
Find the maximum value of
Objective function
subject to the constraints
where
and
Because this is a maximization problem, you would
expect each of the inequalities in the set of constraints to involve
Moreover, because the
first inequality does involve
you can add a slack variable to form the following equation.
For the other two inequalities, a new type of variable, called a
surplus variable,
is intro
duced as follows.
Notice that surplus variables are
subtracted from
(not added to) their inequalities. The vari
ables
and
are called surplus variables because they represent the amount that the left
side of the inequality exceeds the right side. Surplus variables must be nonnegative.
Now, to solve the linear programming problem, form an initial simplex tableau as follows.
Basic
x
1
x
2
x
3
s
1
s
2
s
3
b
Variables
2
1
1
1
0
0
50
s
1
2
1
0
0
0
36
s
2
1
0
1
0
0
10
s
3
←
Departing
0
0
0
0
↑
Entering
You will soon discover that solving mixedconstraint problems can be difficult. One
reason for this is that there is no convenient feasible solution to begin the simplex method.
Note that the solution represented by the initial tableau above,
x
1
,
x
2
,
x
3
,
s
1
,
s
2
,
s
3
0, 0, 0, 50,
36,
10 ,
2
1
1
1
1
s
3
s
2
x
1
x
3
s
3
10
2
x
1
x
2
s
2
36
2
x
1
x
2
x
3
s
1
50
≤
,
≤
.
x
3
≥
0.
x
2
≥
0,
x
1
≥
0,
x
1
x
3
≥
10
2
x
1
x
2
≥
36
2
x
1
x
2
x
3
≤
50
z
x
1
x
2
2
x
3
≥
≤
SECTION 9.5
THE SIMPLEX METHOD: MIXED CONSTRAINTS
557
Constraints
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is not a feasible solution because the values of the two surplus variables are negative. In fact,
the values
do not even satisfy the constraint equations. In order to elim
inate the surplus variables from the current solution, “trial and error” is used. That is, in an
effort to find a feasible solution, arbitrarily choose new entering variables. For instance, in
this tableau, it seems reasonable to select
as the entering variable. After pivoting, the new
simplex tableau becomes the following.
Basic
x
1
x
2
x
3
s
1
s
2
s
3
b
Variables
1
1
0
1
0
1
40
s
1
2
1
0
0
0
36
s
2
←
Departing
1
0
1
0
0
10
x
3
1
0
0
0
20
↑
Entering
The current solution
is still not feasible, so
choose
as the entering variable and pivot to obtain the following simplex tableau.
Basic
x
1
x
2
x
3
s
1
s
2
s
3
b
Variables
0
0
1
1
1
4
s
1
←
Departing
2
1
0
0
0
36
x
2
1
0
1
0
0
10
x
3
3
0
0
0
56
↑
Entering
At this point, the following feasible solution is finally obtained.
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 Spring '08
 Staff
 Linear Algebra, Algebra, Linear Programming, Optimization, objective function, basic variables

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