1. 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)

2. 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

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

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

5. Ratio test: Min. of 3 ratios gives 2 as pivot element
Choosing the pivot element I x y s1 s2 s3
RHS 1 -1
-0.8
1000
1000/1 2
1500
1500/2 3
2400
2400/3
Ratio test: Min. of 3 ratios gives 2 as pivot element

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

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

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

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

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

11. 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)

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

13. Ratio test: Min. of 3 ratios gives 0.5 as pivot element
Choosing the pivot element I x y s1 s2 s3
RHS 1
-0.3
0.5
750
-0.5
250
250/0.5
750/0.5
-1.5
150
150/0.5
Ratio test: Min. of 3 ratios gives 0.5 as pivot element

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

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

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

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

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

19. 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)

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

21. Ratio test: Min. of 2 ratios gives 1 as pivot element
Choosing the pivot element I x y s1 s2 s3
RHS 1
-0.4
0.6
840 -1
100
100/1 2
600
600/2 -3
300
Ratio test: Min. of 2 ratios gives 1 as pivot element

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

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

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

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

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

27. 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)