Unrestricted-in-Sign Variables
In solving LPs with the simplex algorithm, we used the
ratio test to determine the row in which the entering
variable became a basic variable. Recall that the ratio
test depended on the fact that any feasible point
required all variables to be nonnegative. Thus, if some
variables are allowed to be unrestricted in sign (urs),
the ratio test and therefore the simplex algorithm are
no longer valid. In this section, we show how an LP
with unrestricted-in-sign variables can be transformed
into an LP in which all variables are required to be
nonnegative.
• For each urs variable xi, we begin by defining two
new variables xi` and xi``. Then substitute xi `- xi``
for xi in each constraint and in the objective
function. Also add the sign restrictions xi ` and
xi `` . The effect of this substitution is to express xi
as the difference of the two nonnegative
variables xi ‘and xi’’. Because all variables are now
required to be nonnegative, we can proceed with
the simplex.
Example- Baker Problem
• A baker has 30 oz of flour and 5 packages of
yeast. Baking a loaf of bread requires 5 oz of
flour and 1 package of yeast. Each loaf of
bread can be sold for 30¢. The baker may
purchase additional flour at 4¢/oz or sell
leftover flour at the same price. Formulate
and solve an LP to help the baker maximize
profits (revenues costs).
Baker Problem
• Define
x1 = number of loaves of bread baked
x2 = number of ounces by which flour supply
is increased by cash transactions
Therefore, x2 > 0 means that x2 oz of flour were purchased,
and x2 <0 means that x2 ounces of flour were sold (x2 =
0 means no flour was bought or sold). After noting that
x1 ≥ 0 and x2 is urs, the appropriate LP is
• Objective function
z = 30x1-4x2
Subject to
5x1 ≤ 30+x2
x1 ≤ 5
x1 ≥ 0 and x2 is urs
• Put x2=x2`- x2``
Maximize z= 30x1-4x2’+4x2``
S.T
5x1 ≤ 30+x2`-x2``
x1 ≤ 5
x1,x2`,x`` ≥ 0
Maximize z= 30x1-4x2’+4x2``
S.T
5x1-x2`+x2`` ≤ 30
x1 ≤ 5
x1,x2`,x`` ≥ 0
• Maximize z= 30x1-4x2’+4x2``
S.T
5x1-x2`-x2`` +s1=30
x1 +s2
=5
x1,x2`,x`` ≥ 0
• z- 30x1+4x2’-4x2``=0
Simplex Method
z
x1
X2`
X2``
s1
s2
sol
Basic
var
1
-30
4
-4
0
0
0
z=0
0
5
-1
1
1
0
30
S1=30
0
1
0
0
0
1
5
S2=5
ration
Simplex Method
z
x1
X2`
X2``
s1
s2
sol
Basic
var
ration
1
-30
4
-4
0
0
0
z=0
0
5
-1
1
1
0
30
S1=30
6
0
1
0
0
0
1
5
S2=5
5
Simplex Method
z
x1
X2`
X2``
s1
s2
sol
Basic
var
1
-30
4
-4
0
0
0
z=0
0
5
-1
1
1
0
30
S1=30
0
1
0
0
0
1
5
S2=5
ration
Simplex Method
s2 is leaving and x1 in entering
z
x1
X2`
X2``
s1
s2
sol
Basic
var
1
0
4
-4
0
30
150
z=150
0
5
-1
1
1
0
30
S1=30
0
1
0
0
0
1
5
S2=5
ration
Simplex Method
s2 is leaving and x1 is entering
z
x1
X2`
X2``
s1
s2
sol
Basic
var
1
0
4
-4
0
30
150
z=150
0
0
-1
1
1
-5
5
S1=5
0
1
0
0
0
1
5
x1=5
ration
Simplex Method
s1 is leaving and x2`` is entering
z
x1
X2`
X2``
s1
s2
sol
Basic
var
ration
1
0
4
-4
0
30
150
z=150
0
0
-1
1
1
-5
5
S1=5
5
0
1
0
0
0
1
5
x1=5
inf
Simplex Method
s1 is leaving and x2``is entering
z
x1
X2`
X2``
s1
s2
sol
Basic
var
1
0
4
-4
0
30
150
z=150
0
0
-1
1
1
-5
5
S1=5
0
1
0
0
0
1
5
x1=5
ration
Simplex Method
s2 is leaving and x1 is entering
z
x1
X2`
X2``
s1
s2
sol
Basic var
1
0
0
0
4
10
170
z=170
0
0
-1
1
1
-5
5
X2``=5
0
1
0
0
0
1
5
x1=5
ration