1.2 The Simplex Method
n
The simplex method is an iterative procedure
for solving linear programming problems
expressed in standard form.
The simplex algorithm can be used to solve LPs
in which the goal is to maximize the objective
function.
n
n
Step 1 Convert the LP to standard form.
Step 2 Obtain a bfs (if possible) from the standard
form. We will call this bfs the initial bfs. In general,
the most recent bfs will be called the current bfs, so at
the beginning of the problem, the initial bfs is the
current bfs.
n
n
n
Step 3 Determine whether the current bfs is optimal.
(If it is not, find an adjacent bfs that has a larger zvalue.)
Step 4 If the current bfs is not optimal, determine
which nonbasic variable should be come a basic
variable and which basic variable should become a
nonbasic variable to find a bfs with a better objective
function value.
Step 5 Use EROs to find a new bfs with a better
objective function value. Go back to Step 3.
n
Initialization
n
Optimality test
n
If no
if yes
n
iteration
n
stop
1. Initialization
n
“≤” constrains
n
“≥ ” constrains
n
“= ” constrains
“≤” constrains
n
slack variable
max Z =40x1+45x2+24x3
St. 2x1+3x2+x3 ≤100
2 3 1


3 3 2
3x1+3x2+2x3 ≤120
x1 ,x2 ≥0
max Z =40x1+45x2+24x3
S.t.
2x1+3x2+x3 +x4
3x1+3x2+2x3
=100
+x5 =120
x1 ,x2 ,x3 , x4,x5 ≥0
2 3 1 1 0 




 3 3 2 0 1


“≥ ” constrains
n
subtracting the excess variable ei from the i th
constraint and adding the sign restriction ei ≥ 0.
max Z = - 6x1 - 4x2
s.t.
2x1 + x2 ≥2
max Z = - 6x1 - 4x2
s.t.
2x1 + x2 –x3
=2
3x1 +4x2 ≥5
3x1 +4x2
-x4 =5
xj ≥ 0, j=1，2
xj ≥ 0, j=1,2,3,4
“≥ ” constrains
max Z = - 6x1 - 4x2 - M x5- M x6
s.t.
2x1 + x2 –x3
3x1 +4x2
+x5
-x4
=2
+x6 =5
artificial variable
xj ≥ 0 j=1,2,3,4,5,6
M represents a “very large” positive number
（x1，x2， x3，x4， x5，x6）=（0， 0，0，0，2，5）
“≥ ” constrains
max Z = 10x1 + 8x2 + 7x3
s.t.
2x1 + x2 ≥6
x1 +x2 + x3 ≥4
xj ≥ 0 j=1，2 ,3
max Z = 10x1 + 8x2 + 7x3 - Mx6
s.t.
2x1+x2
x1+x2 + x3
- x4
+ x6=6
- x5
xj ≥ 0 j=1,2,3,4,5
=4
“= ” constrains
max Z =5x1 +2x2 +3x3 -x4 +x5
s.t.
x1 +2x2 +2x3 +x4
3x1+ 4x2 +x3
=8
+x5 =7
xj≥0，i=1，2，…5
n
Convert to “≥”and” ≤”
2X1+3X2-4X3=5
2X1+3X2-4X3≤5
2X1+3X2-4X3≥5
1.2.1 Example: Dakota Furniture Company(P134)
n
The Dakota Furniture company manufactures desks,
tables, and chairs.
n The manufacture of each type of furniture
requires lumber and two types of skilled labor:
finishing and carpentry.
n The amount of each resource needed to make
each type of furniture is given in the table below.
Resource
Desk
Table
Chair
Lumber
8 board ft
6 board ft
1 board ft
Finishing hours 4 hours
2 hours
1.5 hours
Carpentry hours 2 hours
1.5 hours
0.5 hours
Example – cont’d
n
n
n
At present, 48 board feet of lumber, 20 finishing
hours, 8 carpentry hours are available. A desk sells
for $60, a table for $30, and a chair for $ 20.
Dakota believes that demand for desks and chairs is
unlimited, but at most 5 tables can be sold.
Since the available resources have already been
purchased, Dakota wants to maximize total revenue.
Example : Solution
n
n
Define:
n
x1 = number of desks produced
n
x2 = number of tables produced
n
x3 = number of chairs produced
The LP is:
max z = 60x1 + 30x2 + 20x3
s.t.
8x1 + 6x2 +
x3 ≤ 48 (lumber constraint)
4x1 + 2x2 + 1.5x3 ≤ 20 (finishing constraint)
2x1 + 1.5x2 + 0.5x3 ≤ 8
(carpentry constraint)
x2
≤ 5 (table demand constraint)
x1, x2, x3 ≥ 0
Convert the LP to standard form
n
n
n
n
Begin the simplex algorithm by converting the
constraints of the LP to the standard form.
To put the constraints in standard form, simply add
slack variables s1, s2, s3, s4, respectively to the four
constraints.
Label the constraints row 1, row 2, row 3, row 4, and
add the sign restrictions si ≥ 0
Then convert the LP’s objective function to the row 0
format.
Example : Solution cont’d
Basic
Variable
Canonical Form 0
Row 0
z – 60x1 – 30x2 – 20x3
Row 1
8x1 +
6x2 +
Row 2
4x1 +
2x2 + 1.5x 3
Row 3
2x1 + 1.5x 2 + 0.5x 3
Row 4
x2
x 3 + s1
+ s2
+ s3
+ s4
= 0
z= 0
= 48
s1 = 48
= 20
s2 = 20
= 8
s3 = 8
= 5
s4 = 5
Example: Solution cont’d
n
initial canonical form has:
n
n
n
BV = {z, s1, s2, s3, s4}
n
NBV = {x1, x2, x3}
For this initial bfs, z=0, s1=48, s2= 20, s3=8, s4=5,
x1=x2=x3=0
A slack variable can be used as a basic variable if the
rhs of the constraint is nonnegative.
Is the current Basic Feasible Solution Optimal?
n
if there is any way z can be increased by increasing
some nonbasic variable from its current value of zero
while holding all other nonbasic variables at their
current values of zero.
n
Solving for z in row 0 yields: z = 60x1 + 30x2 + 20x3
n
Observe x1 has the most negative coefficient in row 0
n
entering variable
Determine the Entering Variable
n
To keep all the basic variables nonnegative, the
largest we can make x1 is min {(48/8)= 6, (20/4) =5,
(8/2) =4} = 4, without forcing one of the constraints
to be broken.
Example : Solution cont’d
n
Rule for determining how large an entering variable
can be.
n When entering a variable into the basis, compute
the ratio
min{
Right − hand side of row
Coefficient of entering variable in row
}
for every constraint in which the entering variable
has a positive coefficient.
n The constraint with the smallest ratio is called
the winner of the ratio test.
n The smallest ratio is the largest value of the
entering variable that will keep all the current
basic variables nonnegative.
Pivot in the Entering Variable
n
To make x1 a basic variable in row 3, we use EROs to
make x1 have a coefficient of 1 in row 3 and a
coefficient of 0 in all other rows.
n
n
This procedure is called pivoting on row 3; and
row 3 is called the pivot row.
The final result is that x1 replaces s3 as the basic
variable for row 3. The term in the pivot row that
involves the entering basic variable is called the
pivot term.
Example : Solution cont’d
The result is:
n
Canonical Form 1
Row 0
z
+
15x2 -
Row 1
Row 2
Row 3
Row 4
Basic
Variable
5x 3
-
-
x 3 + s1
x2 + 0.5 x3
x1 + 0.75x2 + 0.25x3
x2
+ 30s3
-
= 240
= 16
s1 = 16
2 s3
= 4
s2 = 4
+ 0.5s3
= 4
x1 = 4
= 5
s4 = 5
+ s2 -
4s3
z = 240
+ s4
The procedure used to go from one bfs to a better
adjacent bfs is called an iteration of the simplex
algorithm.
Example: Solution cont’d
n
n
Attempt to determine if the current bfs is optimal.
Rearranging row 0 from Canonical Form 1, and
solving for z yields: z = 240 – 15x2 +5x3 -30s3
Example : Solution cont’d
n
n
n
Recall the rule for determining the entering variable is
the row 0 coefficient with the greatest negative value.
Since x3 is the only variable with a negative coefficient,
x3 should be entered into the basis.
Performing the ratio test using x3 as the entering
variable yields the following results (holding other NBVs
to zero):
n
From row 1, s1 ≥ 0 for all values of x3 since s1 = 16 + x3
n
From row 2, s2 ≥ 0 if x3 > 4 / 0.5 = 8
n
From row 3, x1 ≥ 0 if x3 >4 / 0.25 = 16
n
From row 4, s4 ≥ 0 for all values of x3 since s4 = 5
Example : Solution cont’d
So, row 2 becomes the pivot row.
The result of using EROs, to make x3 a basic variable in row 2.
n
n
Canonical Form 2
Row 0
Row 1
Row 2
Row 3
Row 4
z
Basic
Variable
+
5x 2
+ 10s2
+ 10s3
= 280
-
2x 2
+ s1 + 2s2
- 8s3
= 24
s1 = 24
- 4s3
= 8
x3 = 8
+ 1.5s3
= 2
x1 = 2
= 5
s4 = 5
-
2x 2 +
x 1 + 1.25x 2
x2
x3
+ 2s2
- 0.5 s2
+ s4
z = 280
Example : Solution cont’d
n
n
n
In Canonical Form 2:
n BV = {z, s1, x3, x1, s4}
n NBV = {s3, s2, x2 }
n Yielding the bfs z = 280, s1=24, x3=8, x1=2, s4=5,
s2=s3=x2=0
Solving for z in row 0 yields:
z = 280 - 5x2 - 10s2 -10s3
We can see that increasing x2, s2, or s3 (while holding
the other NBVs to zero) will cause the value of z to
decrease.
Example : Solution cont’d
n
The solution at the end of iteration 2 is therefore
optimal. The following rule can be applied to
determine whether a canonical form’s bfs is optimal.
n
A canonical form is optimal (for a max problem) if
each nonbasic variable has a nonnegative
coefficient in the canonical form’s row 0.
2. rule applied to determine
whether the current bfs is optimal
max z = c1 x1 + c2 x2 + ⋅ ⋅ ⋅ + cn xn
a11 x1 + a12 x2 + ⋅ ⋅ ⋅ + a1n xn = b1
a x + a x + ⋅ ⋅ ⋅ + a x = b
2n n
2
 21 1 22 2
⋅ ⋅ ⋅ ⋅ ⋅ ⋅
a x + a x + ⋅ ⋅ ⋅ + a x = b
mn n
m
 m1 1 m 2 2
 x1 , x2 ,⋅ ⋅ ⋅, xn ≥ 0
n
n
maxz=
n
s.t
∑a x
j =1
n
j
i =1
n
n
∑c x
xj≥0
ij
j
j
= bi ,
, j=1,2,…,n
i=1,2,…,m
n
maxz=
n
∑c x
j
i =1
n
n
n
s.t
∑P x
j =1
n
xj≥0
j
j
j
=b
, j=1,2,…,n
n
x1
x2
n
n
n
n
n
n
n
n
+ a1,m xm + a1,m+1 xm+1 + … + a1,n xn = b1
+ a2,m xm + a2,m+1 xm+1 + … + a2,n xn = b2
……
……
xm+ am,m+1 xm+1 + a2,m+1 xm+1 + … + am,n xn = bm
x1 ，x2 ，… ，xn ≥ 0
x1 = b1 - a1,m+1 xm+1 - … - a1n xn
x2 = b2 -a2 ,m+1 xm+1 - … - a2n xn
……
……
xm = bm - am ,m+1 xm+1 - … - amn xn
xi = bi −
n
∑ a x (i = 1,2,..., m)
j = m +1
ij
j
xi = b i −
n
'
a
∑ ij x j (i = 1,2,..., m)
'
n
j = m +1
m
z = ∑ c j x j = ∑ ci xi +
j =1
i =1
m
= ∑ ci b 'i +
i =1
n
∑
j = m +1
n
∑c x
j = m +1
j
j
= ...
m
(c j − ∑ ci a 'ij ) x j
i =1
m
z j = ∑ ci a 'ij ( j = m + 1,..., n)
i =1
m
z0 = ∑ ci b 'i
i =1
n
m
σ j = c j − ∑ ci a 'ij = c j − z j ( j = m + 1,..., n)
i =1
Theorem 1 P24
z = z0 +
n
∑σ
j = m +1
j
xj
 xk' = 1 > 0 (m + 1 ≤ k ≤ n)
 '
 x j = 0 (m + 1 ≤ j ≤ n, 且j ≠ k )
 '
'
'
=
−
1
x
b
a
i
ik (i = 1,2,..., m)
 i
m
z = z 0 + ( ck − z k ) = z 0 + σ k
z k = ∑ ci a 'ik
i =1
n
Theorem 2 (P24)
 xk' = λ > 0 (m + 1 ≤ k ≤ n)
 '
 x j = 0 (m + 1 ≤ j ≤ n, 且j ≠ k )
 '
'
'
x
=
b
−
λ
a
i
ik (i = 1,2,..., m )
 i
z = z0 + λσ k
n
Theorem 3 (P24) An unbounded LP occurs in a
max problem if there is a nonbasic variable xj with
cj-zj≥0 and there is no constraint that limits how
large we can make this NBV.
3. Simplex Method in Tableau Form
cj
cB
XB
c1
x1
M
M
M
cm
M
xm
cj −zj
n
c1 L L c m c m +1 L L c n
b
b1
M
M
bm
x1 L L
1 L L
M
M
0 L L
0L L
x m x m +1 L L x n
0 a1,m+1 L L a1n
M
M
M
M
M
M
1 am,m+1 L L amn
0 σ j = c j − ∑ c i a ij
θi
θ1
M
M
θm
cj is the coefficients of variables in objective function,
CB is the coefficients of the initial basic variables in
objective function.
Example
max z = 60x1 + 30x2 + 20x3
s.t.
8x1 + 6x2 +
x3 ≤ 48
4x1 + 2x2 + 1.5x3 ≤ 20
2x1+ 1.5x2 + 0.5x3 ≤ 8
x2
≤ 5
x1, x2, x3 ≥ 0
n
(1) we can convert the LP to the following form to get
the initial basic feasible solution X (0) = (0, 0, 0, 48, 20,8,5)T
max z = 60x1 + 30x2 + 20x3
s.t.
8x1 + 6x2 +
x3 +
4x1 + 2x2 + 1.5x3 +
2x1+ 1.5x2 + 0.5x3 +
x2
+
x1, x2, x3 , x4, x5,
n
n
x4 = 48
x5 = 20
x6 = 8
x7= 5
x6 ≥ 0
x4 , x5 , x6 , x7 are the basic variables.
Then the initial tableau can be obtained as Tableau 1.
First tableau for the problem
cj
60 30
0
0
0
0
x3
x4
x5
x6
x7
CB
XB
b
x1
0
0
0
0
x4
x5
x6
x7
48
20
8
5
8
6
4
2
[2] 3/2
0
1
1
3/2
1/2
0
1
0
0
0
0
1
0
0
0
0
1
0
0
0
0
1
60
20
0
0
0
0
cj-zj
x2
20
30
θi
6
5
4
-
(2) Because all cj-zj≥0, the the initial basic feasible
is not the optimal solution, go to the next step;
n (3) max(σ 1 , σ 2 , σ 3 ) = max(60,30, 20) = 60 , so x1 is the
entering variable, and
n
bi
θ = min( | ai1 > 0) = min(48 / 8, 20 / 4,8 / 2, −) = 4
i
ai1
x6 is the leaving Basic Variable and is replaced by x1. In
the tableau, the [2] is called pivot element.
n
(4) Performing a pivot operation on the pivot
element, converting P1 to (0 0 1 0)T , x6 is replaced
by x1 in XB, and obtain Tableau 2.
cj
60
30
20
0
x2
x3
x4 x5
x6 x7
θi
1
0
0
0
-4
-2
1/2
0
8
16
-
CB
XB
b
x1
0
0
60
0
x4
x5
x1
x7
16
4
4
5
0
0
1
0
0
-1
3/4
1
-1
[1/2]
1/4
0
0
-15
5
cj-zj
0
0
0
1
0
0
0
0
-30
0
0
0
0
1
0
n
From the tableau 2, the new basic feasible solution is
X (1) = ( 4,0,0,16,4,0,5)
the value of objective function z=240.
n
(5)Checking all the cj-zj of the Tableau 2, we get
c3-z3=5, it means x3 should be entering variable.
Doing step(2)~(4), and we can get Tableau 3.
Tableau 3 (Optimal Tableau for the problem)
cj
60
30
20
0
0
0
0
x6
x7
CB
XB
b
x1
x2
x3
x4
x5
0
20
60
0
x4
x3
x1
x7
24
8
2
5
0
0
1
0
-2
-2
5/4
1
0
1
0
0
1
0
0
0
2 -8
2 -4
1/2 3/2
0
0
0
-5
0
0
-5
cj-zj
0
0
0
1
θi
-10 0
(6)In tableau 3, all cj-zj are negative. we can conclude
that the larger value of objective function do not exist
X * = ( 2,0,8,24,0,0,5)
n the optimal solution is
n
z * = 280
1.2.3 Using the Simplex Algorithm to solve
Minimization Problems
n
n
There are two different ways the simplex method can
be used to solve minimization problems.
Method 1 - Consider this LP:
min z = 2x1 – 3x2
s.t.
x1 + x2 ≤ 4
x1 – x2 ≤ 6
x1, x2 ≥ 0
The optimal solution is the point (x1,x2) that makes
z = 2x1 – 3x2 the smallest.
n Equivalently, this point makes max z’=-z = - 2x1 +
3x2 the largest.
n
n
Method 2 - A simple modification of the simplex
algorithm can be used to solve min problems directly.
n
Modify step 3 of the simplex algorithm.
n
n
n
If cj-zj≥0 for all nonbasic variables (NBV), the current
bfs is optimal.
If any nonbasic variable has cj-zj≤0 , choose the variable
with the “most negative” cj-zj as the entering variable.
This modification of the simplex algorithm works
because increasing a nonbasic variable (NBV) with cjzj≤0 will decrease z.
1.2.4 Alternative Optimal Solutions
n
If there is a nonbasic variable (NBV) with cj-zj=0
of the optimal tableau, the LP has alternative
optimal solutions.
max z = 40 x1 + 45 x2 + 25 x3
s.t. 2 x1 + 3x2 + x3 ≤ 100
3 x1 + 3x2 + 2 x3 ≤ 120
x j ≥ 0, j = 1,2,3
Optimal Tableau for the problem
c j→
40 45 25
0
0
x4
x5
CB
XB
b
x1
x2 x3
45
40
x2
x1
20
20
0
1
1
0
40
45
25
0
0
0 -5 -10
zj
cj-zj
-1/3 1
1 -1
5
0
1
10
θi
Another Optimal Tableau for the problem
c j→
40 45 25
0
0
x4
x5
CB
XB
b
x1
45
x2
26.67
1/3
1
0
2/3 -1/3
25
x3
20
1
0
1
-1
1
5
10
-5
-10
zj
40
cj-zj
0
x2 x3
45 25
0
0
θi
Convex combinations
n
n
If we have two (or more) optimal solutions called two
bfs X1 and X2,Then use α X1 + (1- α)X2 to represent
the set of optimal solutions.
N.B. You must state that 0≤α≤1.
1.2.5 Unbounded LPs
n
an unbounded LP for a max problem occurs when a
nonbasic variable cj with cj-zj≥0 has a non positive
coefficient in each constraint.
c j→
1
1
0
0
0
CB
XB
b
x1
x2
x3 x4 x5
0
1
x3
x1
8
2
0
1
-1
-1
1
0
2
1
0
0
0
x5
9
0
-2
0
3
1
zj
1
-1
0
1
0
cj-zj
0
2
0 -1
0
θi
1.2.6 The Big M Method
n
n
If an LP have ≥ or = constraints, however, a
starting bfs may not be readily apparent. In
such a case, the Big M method may be used to
solve the problem.
Example 6
min z = −3 x1 + x2 + x3
 x1 − 2 x2 + x3 ≤ 11
−4 x + x + 2 x ≥ 3

1
2
3

2 x1 − x3 = −1
 x1 , x2 , x3 ≥ 0
n
First converting it to the standard form as follows:
max z = 3 x1 − x2 − x3
 x1 − 2 x2 + x3 + x4 = 11
 −4 x + x + 2 x − x = 3

1
2
3
5

 −2 x1 + x3 = 1
 x1 , x2 , x3 , x4 , x5 ≥ 0
n
（6.1）
（6.2）
（6.3）
we add artificial variables x6 and x7 to (6.2) and (6.3)
−4 x1 + x2 + 2 x3 − x5 + x6 = 3
−2 x1 + x3 + x7 = 1
x1 , x2 , x3 , x4 , x5 , x6 , x7 ≥ 0
x7 can be written by follows:
x6 function
the objective
max z = 3x1 − x2 − x3 − Mx6 − Mx7
where M is a very large positive number
Now, we used simplex method in tableau form to
solve it. The initial simplex tableau is the Tableau 6-1.
Tableau 6-1
cj
3
-1
-1
0
0
-M -M
x5
x6
x7
0
-1
0
0
1
0
0
0
1
-M
0
0
CB
XB
b
x1
x2
x3
x4
0
-M
-M
x4
x6
x7
11
3
1
1
-4
-2
-2
1
0
1
2
1
1
0
0
cj-zj
3-6M M-1 3M-1 0
θi
11
3/2
1
Tableau 6-2
cj
3
-1
-1
0
0
-M -M
x5
x6
x7
-1
-2
1
CB
XB
b
x1
x2
x3
x4
0
-M
-1
x4
x6
x3
10
1
1
3
0
-2
-2
1
0
0
0
1
1
0
0
0
-1
0
0
1
0
1
M-1
0
0
-M
0 1-3M
cj-zj
n
Finally, we get Tableau 6-3 which is optimal
θi
1
-
Tableau 6-3
cj
3
-1
-1
0
0
-M -M
x4
x5
x6
CB
XB
b
x1
x2
x3
3
-1
-1
x1
x2
x3
4
1
9
1
0
0
0
1
0
0
0
1
0
0
0 -1/3 -2/3 -1/3 2/3-M
cj-zj
n
n
θi
x7
1/3 -2/3 2/3 -5/3
0
-1
1 -2
2/3 -4/3 4/3 -7/3
1
-
If all artificial variables in the optimal solution equal
zero, the solution is optimal.
If any artificial variables are positive in the optimal
solution, the problem is infeasible.
1.2.7 The two-Phase Method
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This is another approach to handle the artificial
variables whenever they are added.
Phase1. the objective function is to minimize the sum of
all artificial variables.. If the minimum value of the
artificial problem is zero, we have a basic feasible
solution to the original problem. We then go to Phase2.
If at least one of the artificial variables is positive, it
means that the original problem without the artificial
variables is infeasible, and we terminate.
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Phase2. the basic feasible solution found at the end
of Phase1 is optimized with respect to the original
objective function. Then the simplex method is once
again applied to determine the optimal solution.
Let us consider Example 6-1 again.
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Example 7-1
min w = x6 + x7
 x1 − 2 x2 + x3 + x4 = 11
− 4 x + x + 2 x − x + x = 3

1
2
3
5
6

− 2 x1 + x3 + x7 = 1
 x1 ,..., x7 ≥ 0
The initial basic feasible solution for the Phase 1 problem is
given below:
Tableau 7-1 (PHASE 1)
cj
0
0
0
0
0
1
1
x5
x6
x7
CB
XB
b
x1
x2
x3
x4
0
1
1
x4
x6
x7
11
3
1
1
-4
-2
-2
1
0
1
2
1
1
0
0
0
-1
0
0
1
0
0
0
1
6
-1
-3
0
1
0
0
cj-zj
θi
11
3/2
1
The simplex method is used until we get the final Tableau 7-2.
Tableau 7-2 (PHASE 1)
cj
0
0
0
0
0
1
1
CB
XB
b
x1
x2
x3
x4
x5
x6
x7
0
0
0
x4
x2
x3
4
1
9
3
0
-2
0
1
0
0
0
1
1
0
0
-2
-1
0
2
1
0
-5
-2
1
0
0
0
0
0
1
1
cj-zj
θi
We now have an optimal solution to the Phase1 LP, and the
artificial variables x6 = 0, x7 = 0 and W=0. We now begin Phase 2 of
the simplex method. The initial tableau for Phase 2 is constructed
by deleting the column corresponding to the artificial variable, and
computing the new last row coefficient with respect to the original
objective function, max z = 3x1 − x2 − x3 , as shown below;
Tableau 7-3 (PHASE 2)
cj
3
-1
-1
0
0
CB
XB
b
x1
x2
x3
x4
x5
0
-1
-1
x4
x2
x3
4
1
9
3
0
-2
0
1
0
0
0
1
1
0
0
-2
-1
0
cj-zj
1
0
0
0
-1
θi
4
-
Tableau 7-3 is not optimal, the nonbasic variable x1 replaces the
basic variable x4 to increase the value of Z further.
Tableau 7-4 (PHASE 2)
CB
cj
XB
3
-1
-1
x1
x2
x3
cj-zj
b
4
1
9
3
x1
-1
x2
-1
x3
0
x4
0
x5
1
0
0
0
1
0
0
0
1
1/3 -2/3
0
-1
2/3 -4/3
0
0
0
-1/3 -1/3
θi
Z=2
An optimal solution has been reached, and it is given by x1 = 4,
x2 = 1, x3 = 9, x4 = 0, x5 = 0, and maximum Z=2.
1.2.8 Degeneracy of the Simplex Algorithm
n
n
n
n
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An LP is degenerate if it has at least one bfs in
which a basic variable is equal to zero.
Any bfs that has at least one basic variable equal to
zero is a degenerate bfs.
unique optimal solution
alternative or multiple optimal solution
cycling