The Simplex Method
Maximize
Subject to
and
Z  3x1  5x 2 ,
x1
4
2x2  12
3x1  2 x2  18
x1  0, x2  0
Software Operation
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(1) Select “Linear Programming” and “OK”
(2) Select “File” and Click “New”
(3) Specify Number of Decision Variables
(4) Specify Number of Constraints
(5) Specify Objective Type and “OK”
(6) Put “Coefficients”
(7) Solve
Example
Embassy Motorcycle (EM) manufactures two lightweight
motorcycles designed for easy handling and safety. The EZRider model has a new engine and a low profile that make it
easy to balance. The Lady-Sport model is slightly larger,
uses a more traditional engine, and is specifically designed
to appeal to women riders. Embassy produces the engines
for both models at its Des Moines, Iowa, plant. Each EZRider engine requires 6 hours of manufacturing time and
each Lady-Sport engine requires 3 hours of manufacturing
time. The Des Moines plant has 2100 hours of engine
manufacturing time available for the next production period.
Embassy’s motorcycle frame supplier can supply as many
EZ-Rider frames as needed.
However, the Lady-Sport frame is more complex and the
supplier can provide only up to 280 Lady-Sport frames
for the next production period. Final assembly and
testing requires 2 hours for each EZ-Rider model and 2.5
hours for each Lady-Sport model. A maximum of 1000
hours of assembly and testing time are available for the
next production period. The company’s accounting
department projects a profit contribution of $2400 for
each EZ-Rider produced and $1800 for each Lady-Sport
produced. Formulate a linear programming model that
can be used to determine the number of units of each
model that should be produced in order to maximize the
total contribution to profit. Find the optimal solution
using the graphical solution procedure. Which
constraints are binding.
Example
(a)
Let E = number of units of the EZ-Rider produced
L = number of units of the Lady-Sport produced
Max
2400E
+
1800L
6E
+
3L
s.t.
L
2E
+
2.5L
 2100
Engine time

280
Lady-Sport maximum
 1000
Assembly and testing
E, L  0
L
Example
(b)
700
Number of EZ-Rider Produced
600
Engine
Manufacturing Time
500
400
Frames for Lady-Sport
300
Optimal Solution E = 250, L = 200
Profit = $960,000
200
100
Assembly and Testing
0
E
100
200
300
400
Number of Lady-Sport Produced
500
Example
(c)
The binding constraints are the manufacturing time and the
assembly and testing time.
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
2x2  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 nonbasic 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  5x2
x1
 x3
(1)
2x 2  x4
(2)
 x5
(3)
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 1 3
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 Solution occur if a non
BF solution has zero or at its coefficient
at row 0.
Algebraic Form
(0)
(1)
(2)
(3)
Z  3x1  2x2
0
4
 x3
2x2
 x4  12
 x5  18
3x1  2 x2
x1
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
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
x4
0
2
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  13 13
Extra x
3 (2) 0 0 0
1 1 3  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
2x2  12
3x1  2 x2  18
x1
and
x1  0, x2  0
Artificial Problem
Max Z  3x1  5 x2  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
(M: a large positive
number.)
x5 : Artificial Variable
Max:
s.t.
3x1  5 x2  Mx5
3x1  2 x2  x5  18
(0)
(3)
Eq (3) can be changed to
x5  18  3x1  2 x2
(4)
Put (4) into (0), then
Max: 3x1  5x2  M (18  3x1  2 x2 )
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
x1
 x3
2x2
3x1  2 x2
4
 x4  12
 x5  18
Coefficient of:
x2 x3 x4
x1
BV Eq. Z
Z (0) 1 -3M-3 -2M-5 0 0
x3 (1) 0 1
0
1 0
0
x4 (2) 0
3
2
0 1
x5 (3) 0 3
2
0 0
(1)
(2)
(3)
Right
x5 Side
0 -18M
0
4
0
12
1
18
Coefficient of:
1
x2 x3 x4
x1
BV Eq. Z
Z (0) 1 -2M-5 3M+3 0 0
x1 (1) 0 1
0
1 0
x4 (2) 0
0
2
0 1
x5 (3) 0 3
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
0  13 1 3
4
0
0
1
1
0
1
0
1
Coefficient of:
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  5x2
4
2x2  12
3x1  2 x2  18
x1 : URS , x2  0
Min
s.t.

1

1
3x  3x  5 x2

1

1
x x
4
2x2  12

1

1
3x  3x  2 x2  18

1

1
x  0, x ,0 x2  0
Homework
(1) P. 78 : Prob. 2-31
(2) P. 79 : Prob. 2-38
(3) P. 132 : Prob. 3-7
(4) P. 134 : Prob. 3-10
Due day: September 8 (M)