Transcript Document

The Simplex Method
Maximize
Subject to
and
Z  3x1  5x 2 ,
4
2 x2  12
3x1  2 x2  18
x1  0, x2  0
x1
From a geometric viewpoint
x2 x1  0
8
: CPF solutions
(Corner-Point Feasible)
: Corner-point infeasible
solutions
3x1  2 x2  18
( 4,6)
2 x2  12
6
x1  4
4
Feasible
2 region
0
2
4
x2  0
6
8
10
x1
Optimality test:
There is at least one optimal solution.
If a CPF solution has no adjacent CPF solutions
that are better (as measured by Z) than itself,
then it must be an optimal solution.
Initialization
Optimal
Solution?
No
Iteration
Yes
Stop
x2
Z  30
( 2,6)
(0,6) 1
2
Z  36
(4,3)
Feasible
region
0
(0,0) Z  0
Z  27
Z  12
( 4,0)
x1
The Key Solution Concepts
Solution concept 1:
The simplex method focuses solely on CPF
solutions.
For any problem with at least one optimal
solution, finding one requires only finding a best
CPF solution.
Solution concept 2:
The simplex method is an iterative
algorithm ( a systematic solution procedure
that keeps repeating a fixed series of steps,
called an iteration).
Solution concept 3:
The initialization of the simplex method
chooses the origin to be the initial CPF
solution.
Solution concept 4:
Given a CPF solution, it is much quicker
computationally to gather information about its
adjacent CPF solutions than other CPF solutions.
Therefore, each time the simplex method
performs an iteration to move from the current
CPF solution to a better one, it always chooses a
CPF solution that is adjacent to the current one.
Solution concept 5:
After the current CPF solution is identified, the
simplex method identifies the rate of improvement
in Z that would be obtained by moving along edge.
Solution concept 6:
The optimality test consists simply of checking
whether any of the edges give a positive rate of
improvement in Z. If no improvement is identified,
then the current CPF solution is optimal.
Simplex Method
To convert the functional inequality
constraints to equivalent equality constraints,
we need to incorporate slack variables.
Original Form
of Model
Augmented Form
of the Model
Slack variables
Max Z  3x1  5x 2 ,
s.t.
x1
4
Max Z  3x1  5x 2 ,
s.t. x
x
2x2  x4
2 x2  12
3x1  2 x2  18
and
x1  0, x2  0
3
1
3x1  2 x2
and
4
 12
 x5  18
x j  0, for j  1,2,3,4,5.
A basic solution is an augmented corner-point
solution.
A Basic Feasible (BF) solution is an
augmented CPF solution.
Properties of BF Solution
1. Each variable is designated as either a nonbasic
variable or a basic variable.
2. # of basic variables = # of functional
constraints.
3. The nonbasic variables are set equal to zero.
4. The values of the basic variables are obtained
from the simultaneous equations.
5. If the basic variables satisfy the
nonnegativity constraints, the basic solution
is a BF solution.
Simplex in Tabular Form
(a) Algebraic Form
(0) Z  3x1  5 x2
x1
 x3
(1)
2x 2  x4
(2)
(3)
 x5
3x1  2 x2
(b) Tabular Form
Basic
Variable Eq. Z
(0) 1
Z
x3
(1) 0
x4
(2) 0
x5
(3) 0
0
4
 12
 18
Coefficient of:
x1 x2 x3
x4
x5
-3 -5 0
1 0 1
0 2 0
3 2 0
0
0
1
0
0
0
0
1
Right
Side
0
4
12
18
The most negative coefficient
minimum
(b) Tabular Form
Coefficient of:
BV Eq. Z x1 x2 x3 x4
Z (0) 1 -3 -5 0 0
x3 (1) 0 1 0 1 0
x4 (2) 0 0 2 0 1
x5 (3) 0 3 2 0 0
Right
Side
0
12
4
6
2
12
18 18  9
x5
0
0
0
1
2
12
18
12
6
2
18
9
2
minimum
The most negative coefficient
(b) Tabular Form
Coefficient of:
Iteration BV Eq.
Z (0)
x3 (1)
0
x4 (2)
x5 (3)
Z
1
0
0
0
x1 x2 x3
x4
x5
-3 -5 0
1 0 1
0 2 0
3 2 0
0
0
1
0
0
0
0
1
Right
Side
0
4
12
18
4
6
4
4
1
6
2
3
minimum
The most negative coefficient
(b) Tabular Form
Coefficient of:
Iteration BV Eq.
Z (0)
x3 (1)
1
x2 (2)
x5 (3)
Z
1
0
0
0
x1 x2 x3
x4
x5
-3
1
0
3
5
0
0
0
1
0
0
1
0
0
1
0
0
2
0
1
2
-1
Right
Side
30
4
6
6
The optimal solution
x1  2, x2  6
None of the coefficient is negative.
(b) Tabular Form
Coefficient of:
Iteration BV Eq. Z
Z (0) 1
x3 (1) 0
2
x2 (2) 0
x1 (3) 0
x1 x2 x3 x4
0
0
0
0
0
1
0
1
1
0
3
2
Right
x5 Side
1 36
1 1
3
3
0 12 0
0  13 13
2
6
2
(a) Optimality Test
(b) Minimum Ratio Test:
(a) Optimality Test:
The current BF solution is optimal
if and only if every coefficient in row 0 is
nonnegative ( 0) .
Pivot Column:
A column with the most negative coefficient
(b) Minimum Ratio Test:
1. Pick out each coefficient in the pivot column
that is strictly positive (>0).
2. Divide each of these coefficients into
the right side entry for the same row.
3. Identify the row that has the smallest of
these ratios.
4. The basic variable for that row is the leaving
basic variable, so replace that variable by the
entering basic variable in the basic variable
column of the next simplex tableau.
Breaking in Simplex Method
(a) Tie for Entering Basic Variable
Several nonbasic variable have largest and
same negative coefficients.
(b) Degeneracy
Multiple Optimal Solutions occur if a
nonbasic variable has zero as its coefficient
at row 0 in the final tableau.
Algebraic Form
(0)
(1)
(2)
(3)
Z  3x1  2 x2
x1
 x3
2x2
3x1  2 x2
 x4
 x5
0
4
 12
 18
0
0
Right Solution
x5 Side Optimal?
0
0
1
0
0
4
0
1
0
12
x5 (3) 0 3 2 0 0
1
18
Coefficient of:
0
BV Eq. Z x1 x2
Z (0) 1 -3 -2
x3 (1) 0 1 0
x4 (2) 0 0 2
x3 x4
No
3
0
Right Solution
x5 Side Optimal?
0 12
1
0
0
4
0
1
0
12
x5 (3) 0 0 2 -3 0
1
6
Coefficient of:
1
BV Eq. Z x1 x2
Z (0) 1 0 -2
x1 (1) 0 1 0
x4 (2) 0 0 2
x3 x4
No
0
Right Solution
x5 Side Optimal?
1 18
0
0
4
1
-1
6
0
1
3
Coefficient of:
2
BV Eq. Z x1 x2 x3
Z (0) 1 0 0 0
x1 (1) 0 1 0 1
x4 (2) 0 0 0 3
x2 (3) 0 0 1  3
2
x4
2
Yes
Coefficient of:
Right Solution
x5 Side Optimal?
1 18
BV Eq. Z x1 x2 x3 x4
Z (0) 1 0 0 0 0
x1 (1) 0 1 0 0  1 3 1 3
Extra x
3 (2) 0 0 0
1 13  13
x2 (3) 0 0 1 0 1 0
2
2
2
6
Yes
(c) Unbounded Solution
If An entering variable has zero in these coefficients
in its pivoting column, then its solution can be
increased indefinitely.
Basic
Coefficient of:
Right
x2 x3 side Ratio
Variable Eq. Z x1
(0) 1 -3
-5
0
0
Z
x3
(1) 0
1
0
1
4
None
Other Model Forms
(a) Big M Method
(b) Variables - Allowed to be Negative
(a) Big M Method
Original Problem
Max Z  3x1  5x2
s.t.
4
2 x2  12
3x1  2 x2  18
x1
and
x1  0, x2  0
(M: a large positive
number.)
Artificial Problem
Max Z  3x1  5x2  Mx5
s.t.
x1
 x3
2x2  x4
3x1  2 x2
4
 12
 x5  18
and
x j  0, for j  1,2,3,4,5.
x3 , x4: Slack Variables
x5 : Artificial Variable
Max:
s.t.
3x1  5x2  Mx5
3x1  2 x2  x5  18
(0)
(3)
Eq (3) can be changed to
x5  18  3x1  2x2
(4)
Put (4) into (0), then
Max: 3x1  5x2  M (18  3x1  2x2 )
or Max: (3M  3) x1  (2M  5) x2 18M
or Max: Z  (3M  3) x1  (2M  5) x2  18M
Max
s.t.
Z  (3M  3) x1  (2M  5) x2  18M
 4 (1)
x
x
1
3
2x2  x4  12 (2)
3x1  2 x2
 x5  18 (3)
Coefficient of:
x2
x3
x1
BV Eq. Z
Z (0) 1 -3M-3 -2M-5 0
x3 (1) 0 1
0
1
0
x4 (2) 0
3
2
0
x5 (3) 0
3
2
0
0
Right
x5 Side
0 -18M
0
0
4
1
0
12
0
1
18
x4
1
BV Eq. Z
Z (0) 1
x1 (1) 0
x4 (2) 0
x5 (3) 0
x1
Coefficient of:
x2
x3
x4
0 -2M-5 3M+3 0
0
1
0
1
2
0
1
0
0
2
0
0
Right
x5 Side
0 -6M+12
0
4
0
12
1
18
2
BV Eq. Z
Z (0) 1
x1 (1) 0
x4 (2) 0
x2 (3) 0
Right
x3 x4 x5
Side
M 5
9
0
2 27
2
0
1
0
4
Coefficient of:
x1
x2
0
0
1
0
0
0
3
1
-1
6
1
3
0
1
3
0
2
2
BV Eq. Z
Z (0) 1
x1 (1) 0
Extra
x3 (2) 0
x2 (3) 0
x1
x2
0
0
1
0
Right
x3 x4
x5 Side
0 3 2 M  1 27
1
1

0
4
3
3
0
0
1
1
0
1
Coefficient of:
0
1
3
2
1
0
3
6
3
(b) Variables with a Negative Value

j

j

j

j
x j  x  x , x  0, x  0
Min
s.t.
x1
3x1  5 x2
4
2 x2  12
3x1  2 x2  18
x1 : URS, x2  0
Min
s.t.

1

1
3x  3x  5x2

1

1
x x
4
2 x2  12

1

1
3x  3x  2x2  18

1

1
x  0, x ,0 x2  0