Simplex Method
MSci331—Week 3~4
1
Simplex Algorithm
• Consider the following LP, solve using Simplex:
MAX Z  3x1  2 x2
s.t.
2 x1  x2  100
x1  x2  80
x1  x2  40
x1 , x2  0
2
Step 1: Preparing the LP
LP Model
•Is the LP
model in
normal form?
LP in a standard
form
•All
constraints
are “=“
•All RHS >0
•All
variables>0
If there are = or
>
•Add an
artificial
variables to
these
constraints
Write Row 0
•Move all
variables in
the objective
function
equation to
LHS. Keep all
constants in
the RHS
•A min
problem can
be treated as
a –MAX
problem
Obtain an initial
BFS
•If the original
LP is not in
normal form
apply the Big
M method to
obtain the
initial BFS.
3
Step 2: Express the LP in a tableau form
Row 0
Row 1
Row 2
Z
1
-
X1
-3
2
1
X2
-2
1
1
S1
0
1
0
S2
0
0
1
S3
0
0
0
RHS
0
100
80
Row 3
-
1
0
0
0
1
40
Ratio
--
4
Step 3: Obtain the initial basic feasible
solution (if available)
a) Set n-m variables equal to 0
These n-m variables the NBV
b) Check if the remaining m variables satisfy the condition of BV
= If yes, the initial feasible basic solution (bfs) is readily a available
= else, carry on some ERO to obtain the initial bfs
Row 0
Row 1
Row 2
Row 3
Z
1
-
X1
-3
2
1
1
X2
-2
1
1
0
S1
0
1
0
0
S2
0
0
1
0
S3
0
0
0
1
RHS
0
100
80
40
Ratio
--
5
Step 4: Apply the Simplex Algorithm
a) Is the initial bfs optimal? (Will bringing a NBV improve the value of Z?)
b) If yes, which variable from the set of NBV to bring into the set of BV?
- The entering NBV defines the pivot column
c) Which variable from the set of BV has to become NBV?
- The exiting BV defines the pivot row
Row 0
Row 1
Row 2
Row 3
Pivot cell
Z
1
X1
-3
X2
-2
S1
0
S2
0
S3
0
RHS
0
Ratio
--
-
2
1
1
1
1
0
1
0
0
0
1
0
0
0
1
100
80
40
100/2
80/1
40/1
6
Summary of Simplex Algorithm for Papa Louis
Set: n-m=0
m≠0
BFS
(intial)
BFS (1)
BFS (2)
BFS (3)
1
The optimal solution is x1=20, x2=60 The optimal value is Z=180
The BFS at optimality x1=20, x2=60, s3=20
7
Geometric Interpretation of Simplex Algorithm
8
Class activity
• Consider the following LP:
This is a maximizing LP, in normal form. So an initial BFS exists.
9
Class activity
10
Class activity
Z
1
RHS
100 -----
x1
-3
x2
-3
s1
0
s2
0
s3
0
s4
0
1
1
2
1
2
-3
1
0
0
0
1
0
0
0
1
0
0
0
4
6
2
-1
2
0
0
0
1
4
11
Class activity
Make this coefficient equal 1 and pivot
all other rows relative to it
Z
1
x1
-3
x2
-3
s1
0
s2
0
s3
0
s4
0
1
1
2
1
2
-3
1
0
0
0
1
0
0
0
1
0
0
0
-1
2
0
0
0
1
RHS
100 ----4/1
4
6
2
6/1
4
---
2/2*
12
Class activity
Z
1
RHS
103
100
-----
x1
0
-3
x2
-7.5
-3
s1
00
s2
00
s3
3/2
0
s4
00
10
10
1
2.5
1
00
11
0
-1/2
0
1/2
00
00
0
34
-3/2
11
00
0
0
-1
1/2
2
00
00
1/2
0
11
54
3.5
2
-1/2
0
56
1
13
Class activity
Make this coefficient equal 1 and pivot
all other rows relative to it
Z
1
x1
0
-3
x2
-7.5
-3
s1
00
s2
00
s3
3/2
0
s4
00
10
10
1
2.5
1
00
11
0
-1/2
0
-3/2
11
00
0
1/2
00
00
0
0
-1
1/2
2
00
00
1/2
0
11
3.5
2
-1/2
0
RHS
103
100
----3/2.5*
34
56
5/3.5
1
---
54
5/0.5
14
Class activity
Z
1
x1
x2
s1
s2
s3
s4
RHS
0
-7.5
0
03
0
3/2
0
0
103
112
0
1
2/5
0
-1/5
0
6/5
0
3.5
0
-1.4
0
11
-1/2
1/5
00
0.8
5
1
-3/2
0
0
3/5
0
1/2
0.8
00
1
2.8
0
0
1/2
-1/5
0
0
0.6
1/2
11
4.4
5
-----
15
Example: LP model with Minimization Objective
•
•
Solve the following LP model:
Min z  2 x1  3 x2
MAX  z  2 x1  3x2
s.t.
s.t.
x1  x2  4
x1  x2  4
x1  x2  6
x1  x2  6
x1 , x2  0
x1 , x2  0
Initial Tableau
Row
0
1
2
Basic
Variable
s1
s2
Z
x1
x2
s1
s2
RHS
-1
0
0
2
1
1
-3
1
-1
0
1
0
0
0
1
0
4
6
Ratio
test
16
Example: LP model with Minimization Objective
• Iteration 0
Row
Basic
Variable
0
1
2
s1
s2
Z
x1
x2
s1
s2
RHS
-1
0
0
2
1
1
-3
1
-1
0
1
0
0
0
1
0
4
6
Z
x1
x2
s1
s2
RHS
-1
0
0
5
1
2
0
1
0
3
1
1
0
0
1
12
4
10
Ratio
test
4
-
• Iteration 1
Row
Basic
Variable
0
1
2
• Optimality test:
x2
s2
Ratio
test
z  12  5x1  3s1
17
> Constraint
x2
Max
40
Z  2 x1  3 x2
s.t.
 x1  x2
35
5
30
x1  3 x2  35
25
x1 
Constraint 1
3.5 x1  5 x2  35
20
Constraint 3
15
20
 20
xx11,,xx22 00
Z
10
Constraint 2
5
5
Constraint 4
10
15
20
25
30
New feasible region
35
40
x1
18
Equality Constraint
x2
Max
s.t.
40
35
30
25
Constraint 1
Z  2 x11  3 x22
 x1  x22
x1  3 x22
5
 35
x1 
20
 20
20
Constraint 3
15
xx11,,xx22 00
Z
10
Constraint 2
5
5
10
15
20
25
30
New feasible region
35
40
x1
19
The Problem of Finding an Initial Feasible BV
An LP Model
Max Z  2 x1  3x2  4 x3
s.t.
x1  x2  x3  30
2 x1  x2  3x3  60
x1  x2  2 x3  20
x1, x2 , x3  0
Standard
Form
Max Z  2 x1  3x2  4 x3  0s1  0e2
s.t.
x1  x2  x3  s1  30
2 x1  x2  3x3  e2  60
x1  x2  2 x3
 20
x1, x2 , x3, s1, e2  0
Cannot find an initial basic variable
that is feasible.
20
Example: Solve Using the Big M Method
MAX 3 x1  x2  y1  z
s.t.
x1  x2  y1  5 z  25
3 x1  3 x2  4 y1  2 z  12
x1  x2  y1  z  0
x1 , x2 , y1 , z  0
MAX
s.t.
3x1  x2
x1  x2
3x1 3x2
x1  x2
x1 ,
x2 ,
Write in standard form
 y1  z
 y1 5 z  s1
4 y1 2 z
 y1  z
e3
y1 ,
z,
s1 , e3
 25
 12
0
0
21
Example: Solve Using the Big M Method
MAX
s.t.
3x1  x2
x1  x2
3x1 3x2
x1  x2
x1 ,
x2 ,
 y1  z
 y1 5 z  s1
4 y1 2 z
 y1  z
e3
y1 ,
z,
s1 , e3
 25
 12
0
0
Adding artificial variables
MAX
s.t.
3x1  x2
x1  x2
3x1 3x2
x1  x2
x1 ,
x2 ,
 y1  z 0s1 0e3
 y1 5 z  s1
4 y1 2 z
 y1  z
e3
y1 ,
z,
s1 ,
e3 ,
 Ma2  Ma3
 a2
a2 ,
 25
 12
 a3  0
a3  0
22
Example: Solve Using the Big M Method
3x1  x2
x1  x2
3x1 3x2
x1  x2
x1 ,
x2 ,
MAX
s.t.
 y1  z 0s1 0e3
 y1 5 z  s1
4 y1 2 z
 y1  z
e3
y1 ,
z,
s1 ,
e3 ,
 Ma2  Ma3
 25
 12
 a3  0
a3  0
 a2
a2 ,
Put in tableau form
W
x1
x2
Coefficient of
y1
s1
z
0
1
-3
1
1
1
0
M
0
M
0
s1
1
0
1
-1
1
5
1
0
0
0
25
a2
2
0
3
-3
-4
2
0
1
0
0
12
a3
3
0
1
-1
-1
-1
0
0
-1
1
0
Basic
Row/Eq. no.
RHS
a2
e3
a3
MRT
23
Example: Solve Using the Big M Method
Eliminating a2 from row 0 by operations: new Row 0 = old Row 0 -M*old Row 2
W
x1
x2
Coefficient of
y1
s1
z
0
1
-3
1
1
1
0
M
0
M
0
s1
1
0
1
-1
1
5
1
0
0
0
25
a2
2
0
3
-3
-4
2
0
1
0
0
12
a3
3
0
1
-1
-1
-1
0
0
-1
1
0
Basic
Basic
Row/Eq. no.
Row/Eq. no.
RHS
a2
e3
a3
Coefficient of
y1
z
RHS
s1
a2
e3
a3
M
0
-12M
25
W
x1
x2
0
1
-3-3M
1+3M
1+4M
1-2M
0
0
s1
1
0
1
-1
1
5
1
0
0
0
a2
2
0
3
-3
-4
2
0
1
0
0
12
a3
3
0
1
-1
-1
-1
0
0
-1
1
0
MRT
MRT
24
Example: Solve Using the Big M Method
Eliminating a3 from the new row 0 by operations: new Row=old Row-M*old Row 3
Basic
Row/Eq. no.
Coefficient of
y1
z
RHS
s1
a2
e3
a3
M
0
-12M
25
W
x1
x2
0
1
-3-3M
1+3M
1+4M
1-2M
0
0
s1
1
0
1
-1
1
5
1
0
0
0
a2
2
0
3
-3
-4
2
0
1
0
0
12
a3
3
0
1
-1
-1
-1
0
0
-1
1
0
Basic
Row/Eq. no.
Coefficient of
y1
z
RHS
s1
a2
e3
a3
0
0
-12M
25
W
x1
x2
0
1
-3-4M
1+4M
1+5M
1-M
0
0
s1
1
0
1
-1
1
5
1
0
M
0
a2
2
0
3
-3
-4
2
0
1
0
0
12
a3
3
0
1
-1
-1
-1
0
0
-1
1
0
MRT
MRT
25
Example: Solve Using the Big M Method
The initial basic variables are s1=25, a2=12, and a3=0. Now ready to proceed for the
simplex algorithm.
Basic
Row/Eq. no.
Coefficient of
y1
z
RHS
s1
a2
e3
a3
0
0
-12M
25
W
x1
x2
0
1
-3-4M
1+4M
1+5M
1-M
0
0
s1
1
0
1
-1
1
5
1
0
M
0
a2
2
0
3
-3
-4
2
0
1
0
0
12
a3
3
0
1
-1
-1
-1
0
0
-1
1
0
MRT
The initial Tableau
Basic
Row/Eq. no.
Coefficient of
y1
z
RHS
MRT
s1
a2
e3
a3
0
0
-12M
25
25
W
x1
x2
0
1
-3-4M
1+4M
1+5M
1-M
0
0
s1
1
0
1
-1
1
5
1
0
M
0
a2
2
0
3
-3
-4
2
0
1
0
0
12
4
a3
3
0
1
-1
-1
-1
0
0
-1
1
0
0
26
Example: Solve Using the Big M Method
Using EROs change the column of x1 into a unity vector.
Basic
Row/Eq. no.
Coefficient of
y1
z
RHS
MRT
s1
a2
e3
a3
0
0
-12M
25
25
W
x1
x2
0
1
-3-4M
1+4M
1+5M
1-M
0
0
s1
1
0
1
-1
1
5
1
0
M
0
a2
2
0
3
-3
-4
2
0
1
0
0
12
4
a3
3
0
1
-1
-1
-1
0
0
-1
1
0
0
Iteration 1
Basic
Row/Eq. no.
Coefficient of
s1
z
W
x1
x2
y1
0
1
0
-2
-2+M
-2-5M
s1
1
0
0
0
2
a2
2
0
0
0
x1
3
0
1
-1
RHS
MRT
a2
e3
a3
0
0
-3-3M
-3+4M
-12M
7
1
0
1
-1
25
3.57
-1
5
0
1
3
-3
12
2.4
-1
-1
0
0
-1
1
0
-27
Example: Solve Using the Big M Method
Using EROs change the column of z into a unity vector.
Basic
Row/Eq. no.
Coefficient of
s1
z
W
x1
x2
y1
0
1
0
-2
-2+M
-2-5M
s1
1
0
0
0
2
a2
2
0
0
0
x1
3
0
1
-1
RHS
MRT
a2
e3
a3
0
0
-3-3M
-3+4M
-12M
7
1
0
1
-1
25
3.57
-1
5
0
1
3
-3
12
2.4
-1
-1
0
0
-1
1
0
--
Iteration 2
Basic
s1
z
x1
Row/Eq. no.
Coefficient of
s1
a2
z
W
x1
x2
y1
0
1
0
-2
-12/5
0
0
1
2
3
0
0
0
0
0
1
0
0
-1
17/5
-1/5
-6/5
0
1
0
1
0
0
RHS
e3
a3
(2+5M)/5
-9/5
(-21/5)+M
4.8
-7/5
1/5
1/5
-16/5
3/5
--2/5
16/5
-3/5
2/5
8.2
2.4
2.4
MRT
2.35
---
Students to try more iterations. The solution is infeasible. See the attached solution.
28
Special case 1: Alternative Optima
Max z  x1  0.5 x2
s.t.
2 x1  x2  4
x1  2 x2  3
x1 , x2  0
.
See Notes on this slide (below) for more information
29
Special case 1: Alternative Optima
Max z  x1  0.5 x2
s.t.
2 x1  x2  4
x1  2 x2  3
x1 , x2  0
30
Special case 2: Unbounded LPs
MAX z  2 x1  x2
s.t.
 x1  x2  1
x1  2 x2  2
x1, x 2  0
s1
x1
0
1
See Notes on this slide (below) for more information
31
Special Case 3: Degeneracy
32
Special Case 5: Degeneracy
Iteration 0
Iteration 1
Iteration 2
33
Special Case 5: Degeneracy
Degeneracy reveals from practical standpoint that the model has at least one
redundant constraint.
34