1. Simplex Method 1
This presentation shows how the tableau method is used to solve a simple linear programming problem in two variables:
Maximising subject to three  constraints.

2. LINEAR PROGRAMMING Example 1 Maximise I = x + 0.8y
subject to x + y  1000
2x + y  1500
3x + 2y  2400
Initial solution:
I = 0
at (0, 0)

3. LINEAR PROGRAMMING
Example 1
Maximise I = x + 0.8y
subject to x + y  1000
2x + y  1500
3x + 2y  2400
Maximise I
where I - x - 0.8y = 0
subject to x + y + s = 1000
2x + y s = 1500
3x + 2y s3 = 2400

4. SIMPLEX TABLEAU Initial solution x y s1 s2 s3 RHS
Basic variable x y s1 s2 s3
RHS 1
1000 2
1500 3
2400 I -1
-0.8
I = 0, x = 0, y = 0, s1 = 1000, s2 = 1500, s3 = 2400

5. Most negative number in objective row
PIVOT 1
Choosing the pivot column
Basic variable x y s1 s2 s3
RHS 1
1000 2
1500 3
2400 I -1
-0.8
Most negative number in objective row

6. Minimum of -values gives 2 as pivot element
Choosing the pivot element
Basic variable x y s1 s2 s3
RHS
-values 1
1000
1000/1 2
1500
1500/2 3
2400
2400/3 I -1
-0.8
Minimum of -values gives 2 as pivot element

7. Divide through the pivot row by the pivot element
Making the pivot
Basic variable x y s1 s2 s3
RHS 1
1000
0.5
750 3 2
2400 I -1
-0.8
Divide through the pivot row by the pivot element

8. Objective row + pivot row
Making the pivot
Basic variable x y s1 s2 s3
RHS 1
1000
0.5
750 3 2
2400 I
-0.3
Objective row + pivot row

9. First constraint row - pivot row
Making the pivot
Basic variable x y s1 s2 s3
RHS
0.5 1
-0.5
250
750 3 2
2400 I
-0.3
First constraint row - pivot row

10. Third constraint row – 3 x pivot row
Making the pivot
Basic variable x y s1 s2 s3
RHS
0.5 1
-0.5
250
750
-1.5
150 I
-0.3
Third constraint row – 3 x pivot row

11. PIVOT 1 New solution x y s1 s2 s3 RHS
Basic variable x y s1 s2 s3
RHS
0.5 1
-0.5
250
750
-1.5
150 I
-0.3
I = 750, x = 750, y = 0, s1 = 250, s2 = 0, s3 = 150

12. LINEAR PROGRAMMING Example Maximise I = x + 0.8y
subject to x + y  1000
2x + y  1500
3x + 2y  2400
Solution after pivot 1:
I = 750
at (750, 0)

13. Most negative number in objective row
PIVOT 2
Choosing the pivot column
Basic variable x y s1 s2 s3
RHS
0.5 1
-0.5
250
750
-1.5
150 I
-0.3
Most negative number in objective row

14. Minimum of -values gives 0.5 as pivot element
Choosing the pivot element
Basic variable x y s1 s2 s3
RHS
-values
0.5 1
-0.5
250
250/0.5
750
750/0.5
-1.5
150
150/0.5 I
-0.3
Minimum of -values gives 0.5 as pivot element

15. Divide through the pivot row by the pivot element
Making the pivot
Basic variable x y s1 s2 s3
RHS
0.5 1
-0.5
250
750 -3 2
300 I
-0.3
Divide through the pivot row by the pivot element

16. Objective row + 0.3 x pivot row
Making the pivot
Basic variable x y s1 s2 s3
RHS
0.5 1
-0.5
250
750 -3 2
300 I
-0.4
0.6
840
Objective row x pivot row

17. First constraint row – 0.5 x pivot row
Making the pivot
Basic variable x y s1 s2 s3
RHS 1 -1
100
0.5
750 -3 2
300 I
-0.4
0.6
840
First constraint row – 0.5 x pivot row

18. Second constraint row – 0.5 x pivot row
Making the pivot
Basic variable x y s1 s2 s3
RHS 1 -1
100 2
600 -3
300 I
-0.4
0.6
840
Second constraint row – 0.5 x pivot row

19. PIVOT 2 New solution x y s1 s2 s3 RHS
Basic variable x y s1 s2 s3
RHS 1 -1
100 2
600 -3
300 I
-0.4
0.6
840
I = 840, x = 600, y = 300, s1 = 100, s2 = 0, s3 = 0

20. LINEAR PROGRAMMING Example Maximise I = x + 0.8y
subject to x + y  1000
2x + y  1500
3x + 2y  2400
Solution after pivot 2:
I = 840
at (600, 300)

21. Most negative number in objective row
PIVOT 3
Choosing the pivot column
Basic variable x y s1 s2 s3
RHS 1 -1
100 2
600 -3
300 I
-0.4
0.6
840
Most negative number in objective row

22. Minimum of -values gives 1 as pivot element
Choosing the pivot element
Basic variable x y s1 s2 s3
RHS
-values 1 -1
100
100/1 2
600
600/2 -3
300 I
-0.4
0.6
840
Minimum of -values gives 1 as pivot element

23. Divide through the pivot row by the pivot element
Making the pivot
Basic variable x y s1 s2 s3
RHS 1 -1
100 2
600 -3
300 I
-0.4
0.6
840
Divide through the pivot row by the pivot element

24. Objective row + 0.4 x pivot row
Making the pivot
Basic variable x y s1 s2 s3
RHS 1 -1
100 2
600 -3
300 I
0.4
0.2
880
Objective row x pivot row

25. Second constraint row – 2 x pivot row
Making the pivot
Basic variable x y s1 s2 s3
RHS 1 -1
100 -2
400 -3 2
300 I
0.4
0.2
880
Second constraint row – 2 x pivot row

26. Third constraint row + 3 x pivot row
Making the pivot
Basic variable x y s1 s2 s3
RHS 1 -1
100 -2
400 3
600 I
0.4
0.2
880
Third constraint row x pivot row

27. PIVOT 3 Optimal solution x y s1 s2 s3 RHS
Basic variable x y s1 s2 s3
RHS 1 -1
100 -2
400 3
600 I
0.4
0.2
880
I = 880, x = 400, y = 600, s1 = 0, s2 = 100, s3 = 0

28. LINEAR PROGRAMMING Example Maximise I = x + 0.8y
subject to x + y  1000
2x + y  1500
3x + 2y  2400
Optimal solution after pivot 3:
I = 880
at (400, 600)