The Simplex Method
Standard Linear Programming
Problem
Standard Maximization Problem
1. All variables are nonnegative.
2. All the constraints (the conditions) can be
expressed as inequalities of the form:
ax + by ≤ c, where c is a positive constant
Illustrating Example (1)
Maximize the objective function:
P(x,y) = 5x + 4y
Subject to:
x + y ≤ 20
2x + y ≤ 35
-3x + y ≤ 12
x≥0
y≥0
Solution
First :
We rewrite the given conditions
and formula
We had :
x  y  20 ,
2 x  y  35 ,
 3 x  y  12 and p  5 x  4 y
Re write the inequaliti es , by in troducing
" slack va riables "
u , v and w , as follows : as follows :
x  y  u  20 ,
2 x  y  v  35 ,
 3 x  y  w  12
and rewrite the formula of the objective
 5x  4 y  p  0
Second :
We construct
the following
table :
x
y
u
v
w
p
const an ts
 1

2

 3


  5
1
1
0
0
0
1
0
1
0
0
1
0
0
1
0





4
0
0
0
1
20 

35

12 


0 
function as follows :
x
y
u
v
w
p
const an ts
 1

2

 3


 5

1
1
0
0
0
1
0
1
0
0
1
0
0
1
0





4
0
0
0
1
20 

35

12 


0 
Third :
We locate the pivot element ( entry ), as follows :
a.
We locate the pivot column :
which is the column containing
( whch is here
b.
the most negative entry in the last row ( left to the line ) :
 5 ). Thus the pivot column is the x  column .
We divide the consta nt in each row by the correspond ing positive entry in the pivot column
to obtain a ( positive ) quotient . If that entry is negative then we do not have a quotient
The entry correspond ing to the smallest
quotient is the pivot element ( and the row where it is
located is the pivot row )
Here , the quotients are : 20 / 1  20 and 35 / 2 . The quotient 35 / 2 is the smallest , and so the
correspond ing entry , which is the 2 is the pivot element ( entry )
Fourth
We perform the necessary
elementary
row operations
to transform the pivot element to 1 and
the other entries in the pivot column to 0 ' s ; thus transform ing that column to a unit column .
x
y
u
v
w
p
const an ts
 1

2

 3


  5
1
1
0
0
0
1
0
1
0
0
1
0
0
1
0





4
0
0
0
1
20 

35

12 


0 
x
y
u
v
w
p
const an ts
 1

1

 3


  5
1
20 
35 
2 
12 


0 
1
2
R2

1
0
0
0
1
2
0
1
2
0
0
1
0
0
1
0





4
0
0
0
1
( R ) R
(3 R ) R
(5 R ) R
1
4
2
  2 

3   2 

x
0

1

0


0

,
y
u
v
1
2
1

1
2
5
2


3
2
,
w
1
2
p
0
const an ts
0
0
1
2
0
0
0
3
2
1
0




0
5
2
0
1



129

2


175 
2 
5
2
35
2
Fifth :
If after that , the last row still contains negativ entries , we repeat step three and four .
3
Here , we have a negative entry in the last row , which is : 
2
x
0

1

0


0

y
u
v
1
2
1

1
2
0
5
2


3
2
w
1
2
p
const an ts
0
0
1
2
0
0
0
3
2
1
0




0
5
2
0
1

35 
2 
129

2


175 
2 
5
2
R epeat steps ( 3 ) and ( 4 )
Step ( 3 )
a.
We locate the pivot column :
which is the column containing

( whch is here
3
the most negative entry in the last row ( left to the line ) :
). Thus the pivot column is the y  column .
2
b.
We divide the consta nt in each row by the correspond ing positive entry in the pivot column
to obtain a ( positive ) quotient . If that entry is negative then we do not have a quotient
The entry correspond ing to the smallest
quotient is the pivot element ( and the row where it is
located is the pivot row )
Here , the quotients
are :
5
2
The quotient
5
/
1
2
5,
35
2
/
1
.  35 and
2
129
2
/
5
2

129
5
 25
4
5
is the smallest , and so the correspond ing entry , which is the
2
in the first row is the pivot element ( entry )
1
2
Step ( 4 )
We perform the necessary
elementary
row operations
to transform the
pivot element to 1 and the other entries in the pivot column to 0 ' s
; thus transform ing that column to a unit column .
x
0

1

0


0

y
u
v
1
2
1

1
2
0
5
2


3
2
w
1
2
p
const an ts
0
1
2
0
0
0
3
2
1
0




0
5
2
0
1
( R ) R
( 5 R ) R

35 
2 
129

2


175 
2 
5
2
0
(3 R ) R
2
3
1
 1   1 
 1  

4 
2R
,
,
x
y
u
v
0

1

0


 0
1
1
0
,
w
p
1
0
0
1
1
0
0
0
5
4
1
0





0
3
1
0
1
const an ts
5 

15

52 


95 
Sixth :
Now all etries in the last row to the left of the line are positive ,
which means that the optimal solution has been reached .
The correspond ing values of the v ariables to this optial solutin are :
zero for all nonbasic
( the ones not assiciated
v ariables
with ( heading ) a unit column )
& the correspond ing con sta nt lying in the row containing
( the ones assiciated
the 1 for basic v ariables
with a unit column ).
Thus :
x  15 , y  5 , u  0  v , w  52 and p  95
Therfore the max value of the objective
function
p is 95
This max value occurs when x  15 , y  5 , u  0 , v  0 and w  52
What about when all of the
constraints (the inequalities) are of
the type “≤ positive constant”
But we want to minimize the
objective function instead of
maximizing.
Minimization with “≤” constraints
Illustrating Example (2)
Minimize the objective function:
p(x,y) = -2x - 3y
Subject to:
5x + 4y ≤ 32
x + 2y ≤ 10
x≥0
y≥0
Solution
Let
q(x) = - p(x) = - ( -2x -3y) = 2x + 3y
To minimize p is to maximize q. Thus, we solve the
following standard maximization linear programming
problem:
Maximize the objective function:
q(x) = 2x + 3y
Subject to:
5x + 4y ≤ 32
x + 2y ≤ 10
x≥0
y≥0
Rewriting the inequalities as equations, by
introducing the “slack” variables u and v
and the formula of the objective function
as done in example (1).
5x + 4y ≤ 32 , x + 2y ≤ 10 and q = 2x +3y
Are transformed to:
5x + 4y + u = 32
x + 2y + v = 10
- 2x - 3y + q = 0
We have ,
5x  4y  u  32, x  2y  v  10, - 2x  3y  q  0
We construct
the simplex
tableaus :
x
y
u v
q
const ants
 5

1

 .

 2
4
1
0
0
2
0
1
0
.
.
.
.
3
0
0
1
32 

10 
. 

0 
Applying
the method as e xplained
x
u
1

0

.

0
y
0
1

1
3

1
6
v
q
2
3
0
5
6
0
.
.
.
.
0
1
6
7
6
1
Thus : x  4 ,
y3
in example (1), we arrive at :
const ants
4 

3 
. 

17 
and
q  17

q attains the max value , which is 17 at ( x , y )  ( 4 , 3 )

Since p   q , then :
p attains the min value , which is  17 at ( x , y )  ( 4 , 3 )
Standard Linear Programming
Problem
Standard Minimization Problem
1. All variables are nonnegative.
2. All the constraints (the conditions) can be
expressed as inequalities of the form:
ax + by ≥ c, where c is a positive constant
Solving
The Standard Minimization
Problem
We use the fundamental theorem of
Duality
Illustrating Example (3)
Minimize the objective function:
p(x,y) = 6x + 8y
Subject to:
40x + 10y ≥ 2400
10x + 15y ≥ 2100
5x + 15y ≥ 1500
x≥0
y≥0
Minimize the objective function: p(x,y) = 6x + 8y
Subject to:
40x + 10y ≥ 2400, 10x + 15y ≥ 2100 , 5x + 15y ≥ 1500, x ≥ 0 and y ≥ 0
We will refer to the above given problem by the primal (original) problem
First: We construct the following table, which we will refer to by the “primal” table:
x y
constant
--------------------------------40 10
2400
10 15
2100
5 15
1500
--------------------------------6 8
Second: We construct a dual (twin) table from interchanging the rows and columns in the primal table:
x'
y'
z'
constant
----------------------------------------------------------40
10
5
6
10
15
15
8
--------------------------------------------------------2400
2100
1500
Third: We interpret the “dual table” as a standard maximization problem, which will refer to as the
“dual problem” or “twin problem” of the “primal problem” or the “original problem”
Miaximoze the objective function: q( x ' , y ' , z ' ) = 2400x' + 2100y' + 1500z'
Subject to:
40x' + 10y' + 5z' ≤ 6, 10x' + 15y' + 15z' ≤ 8 , x' ≥ 0 and y' ≥ 0, z' ≥ 0
Fourth: We apply the simplex method explained in example (1) to solve this problem
Maximize the objective function: q(x,y,z) = 2400x' + 2100y' + 1500z'
Subject to:
40x' + 10y' + 5z' ≤ 6, 10x' + 15y' + 15z' ≤ 8 , x' ≥ 0 and y' ≥ 0, z' ≥ 0
4.a.Rewriting the inequalities and the formula of the objective function, with the slack
variables being the same x and y (in that order) of the original (minimization) problem
:
40x' + 10y' + 5z' + x = 6
10x' + 15y' + 15z' + y = 8
- 2400x' - 2100y' - 1500z‘ + q = 0
4.b. We construct the simplex table for this problem
40x'  10y'  5z'  x  6
10x'  15y'  15z'  y  8
- 2400x' - 2100y' - 1500 z   q  0






x
y
z
x
y q
40
10
5
1
0
0
10
15
15
0
1
0
.
.
.
.
.
.
2400
 2100
 1500
0
0
1
4 .c . We perform
all the elementary
const ant
6 

8 
. 

0 
row operations
necessiate d
by the simplex method which will lead to the following
x y
1

0

.

0
0

z
x
3
20
3
100
1
11
10
.
.
0
450

1
50
y

1
50
q
0
2
25
0
.
.
.
30
120
1
const ant


13
25


.

1140 
1
50
table :
Fifth: We read the solution from the table
x
 1

1

 .

 1
y
x
y
0
1
1
0
1
0
1
0
.
.
.
0
2
1
.
.
0
q
const ant
3 

2 
. 

4 
The solution to the primal ( original ) min imization
appears in the last row under the " slack " v ariables
problem
x and y :
Thus : x  0 and y  2
The min value of p  the max value of q  4
We can also get this answer
from the formula of p :
p (0,2 )  0  2 ( 2 )  0  4  4
Checking :
q is max when : x   0 , y   2
q ( 0 , 2 ))  0  2 ( 2 )  4
Illustrating Example (4)
Minimize the objective function:
p(x,y) = x + 2y
Subject to:
-2x + y ≥ 1
-x+y≥2
x≥0
y≥0
Minimize the objective function: p(x,y) = x + 2y
Subject to:
-2x + y ≥ 1, - x + y ≥ 2 We will refer to the above given problem by the primal (original) problem
First: We construct the following table, which we will refer to by the “primal” table:
x
y
constant
---------------------------------2 1
1
-1 1
2
--------------------------------1 2
Second: We construct a dual (twin) table from interchanging the rows and columns in the primal table:
x'
y'
constant
-------------------------------------------2
-1
1
1
1
2
----------------------------------------1
2
Third: We interpret the “dual table” as a standard maximization problem, which will refer to as the
“dual problem” or “twin problem” of the “primal problem” or the “original problem”
Maximize the objective function: q( x ' , y ‘ ) = x' + 2y'
Subject to:
-2x' - y' ≤ 1, x' + y' ≤ 2 , x' ≥ 0 and y' ≥ 0
Fourth: We apply the simplex method explained in example (1) to solve this problem
Maximize the objective function: q( x ' , y ‘ ) = x' + 2y'
Subject to:
- 2x' - y' ≤ 1, x' + y' ≤ 2 , x' ≥ 0 and y' ≥ 0
4.a.Rewriting the inequalities and the formula of the objective function, with the slack
variables being the same x and y (in that order) of the original (minimization) problem
:
- 2x' - y' ' + x = 1
x' + y' + y = 2
- x' - 2y' + q = 0
4.b. We construct the simplex table for this problem
- 2x' - y' '  x  1
x'  y'  y  2
- x' - 2y'  q  0
x
y
 2

1

 .

1
1
1
0
0
1
0
1
0
.
.
.
0
0
1
x
.
.
2
4 .c . We perform
y
q
const ant
1 

2 
. 

0 
all the elementary
row operations
necessiate d
by the simplex method which will lead to the following
x
 1

1

 .

 1
y
x
y
0
1
1
0
1
0
1
0
.
.
.
0
2
1
.
0
.
q
const ant
3 

2 
. 

4 
table :
Homework
1. Using the simplex method, maximize: p = x + (6/5)y subject to:
2x + y ≤ 180 , x + 3y ≤ 300 , x ≥ 0 , y ≥ 0
Solution: p(48,84) = 148.8
2. Minimize: p(x,y) = - 5x - 4y
Subject to: x + y ≤ 20 , 2x + y ≤ 35 , -3x + y ≤ 12 , x ≥ 0
y≥0
Solution: p(15,5) = - 95
3. Using the dual theorem, minimize: p = 3x + 2y subject to:
8x + y ≥ 80 , 8x + 5y ≥ 240 , x + 5y ≥ 100, x ≥ 0 , y ≥ 0
Solution: p(20,16) = 92
Maximize the objective function: