This presentation shows how the tableau
method is used to solve a simple linear
programming problem in two variables:
Maximising subject to three  constraints.
LINEAR PROGRAMMING
Example 1
Maximise I = x + 0.8y
subject to x + y  1000
2x + y  1500
3x + 2y  2400
y
960
800
640
480
Initial solution:
320
I = 0
at (0, 0)
160
x
0
0
160
320
480
640
800
960
LINEAR PROGRAMMING
Maximise
subject to
Maximise
where
subject to
I = x+
x+y
2x + y
3x + 2y
Example 1
0.8y
 1000
 1500
 2400
I
I - x - 0.8y
x + y + s1
2x + y
+ s2
3x + 2y
+ s3
=
=
=
=
0
1000
1500
2400
SIMPLEX TABLEAU
Initial solution
Basic
variable
x
y
s1
s2
s3
RHS
s1
1
1
1
0
0
1000
s2
2
1
0
1
0
1500
s3
3
2
0
0
1
2400
I
-1
-0.8
0
0
0
0
I = 0, x = 0, y = 0, s1 = 1000, s2 = 1500, s3 = 2400
PIVOT 1
Choosing the pivot column
Basic
variable
x
y
s1
s2
s3
RHS
s1
1
1
1
0
0
1000
s2
2
1
0
1
0
1500
s3
3
2
0
0
1
2400
I
-1
-0.8
0
0
0
0
Most negative number in objective row
PIVOT 1
Choosing the pivot element
Basic
variable
x
y
s1
s2
s3
RHS
s1
1
1
1
0
0
1000 1000/1
s2
2
1
0
1
0
1500 1500/2
s3
3
2
0
0
1
2400 2400/3
I
-1
-0.8
0
0
0
-values
0
Minimum of -values gives 2 as pivot element
PIVOT 1
Making the pivot
Basic
variable
x
y
s1
s2
s3
RHS
s1
1
1
1
0
0
1000
s2
1
0.5
0
0.5
0
750
s3
3
2
0
0
1
2400
I
-1
-0.8
0
0
0
0
Divide through the pivot row by the pivot element
PIVOT 1
Making the pivot
Basic
variable
x
y
s1
s2
s3
RHS
s1
1
1
1
0
0
1000
s2
1
0.5
0
0.5
0
750
s3
3
2
0
0
1
2400
I
0
-0.3
0
0.5
0
750
Objective row + pivot row
PIVOT 1
Making the pivot
Basic
variable
x
y
s1
s2
s3
RHS
s1
0
0.5
1
-0.5
0
250
s2
1
0.5
0
0.5
0
750
s3
3
2
0
0
1
2400
I
0
-0.3
0
0.5
0
750
First constraint row - pivot row
PIVOT 1
Making the pivot
Basic
variable
x
y
s1
s2
s3
RHS
s1
0
0.5
1
-0.5
0
250
s2
1
0.5
0
0.5
0
750
s3
0
0.5
0
-1.5
1
150
I
0
-0.3
0
0.5
0
750
Third constraint row – 3 x pivot row
PIVOT 1
New solution
Basic
variable
x
y
s1
s2
s3
RHS
s1
0
0.5
1
-0.5
0
250
x
1
0.5
0
0.5
0
750
s3
0
0.5
0
-1.5
1
150
I
0
-0.3
0
0.5
0
750
I = 750, x = 750, y = 0, s1 = 250, s2 = 0, s3 = 150
LINEAR PROGRAMMING
Example
Maximise I = x + 0.8y
subject to x + y  1000
2x + y  1500
3x + 2y  2400
y
960
800
640
480
Solution after
pivot 1:
320
I = 750
160
x
0
0
160
320
480
640
800
960
at (750, 0)
PIVOT 2
Choosing the pivot column
Basic
variable
x
y
s1
s2
s3
RHS
s1
0
0.5
1
-0.5
0
250
x
1
0.5
0
0.5
0
750
s3
0
0.5
0
-1.5
1
150
I
0
-0.3
0
0.5
0
750
Most negative number in objective row
PIVOT 2
Choosing the pivot element
Basic
variable
x
y
s1
s2
s3
RHS
s1
0
0.5
1
-0.5
0
250 250/0.5
x
1
0.5
0
0.5
0
750 750/0.5
s3
0
0.5
0
-1.5
1
150 150/0.5
I
0
-0.3
0
0.5
0
750
-values
Minimum of -values gives 0.5 as pivot element
PIVOT 2
Making the pivot
Basic
variable
x
y
s1
s2
s3
RHS
s1
0
0.5
1
-0.5
0
250
x
1
0.5
0
0.5
0
750
s3
0
1
0
-3
2
300
I
0
-0.3
0
0.5
0
750
Divide through the pivot row by the pivot element
PIVOT 2
Making the pivot
Basic
variable
x
y
s1
s2
s3
RHS
s1
0
0.5
1
-0.5
0
250
x
1
0.5
0
0.5
0
750
s3
0
1
0
-3
2
300
I
0
0
0
-0.4
0.6
840
Objective row + 0.3 x pivot row
PIVOT 2
Making the pivot
Basic
variable
x
y
s1
s2
s3
RHS
s1
0
0
1
1
-1
100
x
1
0.5
0
0.5
0
750
s3
0
1
0
-3
2
300
I
0
0
0
-0.4
0.6
840
First constraint row – 0.5 x pivot row
PIVOT 2
Making the pivot
Basic
variable
x
y
s1
s2
s3
RHS
s1
0
0
1
1
-1
100
x
1
0
0
2
-1
600
s3
0
1
0
-3
2
300
I
0
0
0
-0.4
0.6
840
Second constraint row – 0.5 x pivot row
PIVOT 2
New solution
Basic
variable
x
y
s1
s2
s3
RHS
s1
0
0
1
1
-1
100
x
1
0
0
2
-1
600
y
0
1
0
-3
2
300
I
0
0
0
-0.4
0.6
840
I = 840, x = 600, y = 300, s1 = 100, s2 = 0, s3 = 0
LINEAR PROGRAMMING
Example
Maximise I = x + 0.8y
subject to x + y  1000
2x + y  1500
3x + 2y  2400
y
960
800
640
480
Solution after
pivot 2:
320
I = 840
160
x
0
0
160
320
480
640
800
960
at (600, 300)
PIVOT 3
Choosing the pivot column
Basic
variable
x
y
s1
s2
s3
RHS
s1
0
0
1
1
-1
100
x
1
0
0
2
-1
600
y
0
1
0
-3
2
300
I
0
0
0
-0.4
0.6
840
Most negative number in objective row
PIVOT 3
Choosing the pivot element
Basic
variable
x
y
s1
s2
s3
RHS
-values
s1
0
0
1
1
-1
100
100/1
x
1
0
0
2
-1
600
600/2
y
0
1
0
-3
2
300
I
0
0
0
-0.4
0.6
840
Minimum of -values gives 1 as pivot element
PIVOT 3
Making the pivot
Basic
variable
x
y
s1
s2
s3
RHS
s1
0
0
1
1
-1
100
x
1
0
0
2
-1
600
y
0
1
0
-3
2
300
I
0
0
0
-0.4
0.6
840
Divide through the pivot row by the pivot element
PIVOT 3
Making the pivot
Basic
variable
x
y
s1
s2
s3
RHS
s1
0
0
1
1
-1
100
x
1
0
0
2
-1
600
y
0
1
0
-3
2
300
I
0
0
0.4
0
0.2
880
Objective row + 0.4 x pivot row
PIVOT 3
Making the pivot
Basic
variable
x
y
s1
s2
s3
RHS
s1
0
0
1
1
-1
100
x
1
0
-2
0
1
400
y
0
1
0
-3
2
300
I
0
0
0.4
0
0.2
880
Second constraint row – 2 x pivot row
PIVOT 3
Making the pivot
Basic
variable
x
y
s1
s2
s3
RHS
s1
0
0
1
1
-1
100
x
1
0
-2
0
1
400
y
0
1
3
0
-1
600
I
0
0
0.4
0
0.2
880
Third constraint row + 3 x pivot row
PIVOT 3
Optimal solution
Basic
variable
x
y
s1
s2
s3
RHS
s2
0
0
1
1
-1
100
x
1
0
-2
0
1
400
y
0
1
3
0
-1
600
I
0
0
0.4
0
0.2
880
I = 880, x = 400, y = 600, s1 = 0, s2 = 100, s3 = 0
LINEAR PROGRAMMING
Example
Maximise I = x + 0.8y
subject to x + y  1000
2x + y  1500
3x + 2y  2400
y
960
800
640
Optimal solution
after pivot 3:
480
320
I = 880
160
x
0
0
160
320
480
640
800
960
at (400, 600)