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 PROGRAMMINGExample 1 MaximiseI = x + 0.8y subject tox + y  1000 2x + y  1500 3x + 2y  2400 Initial solution: I = 0 at (0, 0)

3. LINEAR PROGRAMMINGExample 1 MaximiseI = x + 0.8y subject to x + y  1000 2x + y  1500 3x + 2y  2400 MaximiseI whereI - x - 0.8y = 0 subject to x + y + s 1 = 1000 2x + y + s 2 = 1500 3x + 2y + s 3 = 2400

4. Basic variable xys1s1 s2s2 s3s3 RHS s1s1 111001000 s2s2 210101500 s3s3 320012400 I-0.80000 SIMPLEX TABLEAU I = 0, x = 0, y = 0, s 1 = 1000, s 2 = 1500, s 3 = 2400 Initial solution

5. Basic variable xys1s1 s2s2 s3s3 RHS s1s1 111001000 s2s2 210101500 s3s3 320012400 I-0.80000 PIVOT 1Choosing the pivot column Most negative number in objective row

6. Basic variable xys1s1 s2s2 s3s3 RHS  -values s1s1 111001000 1000/1 s2s2 210101500 1500/2 s3s3 320012400 2400/3 I-0.80000 PIVOT 1Choosing the pivot element Minimum of  -values gives 2 as pivot element

7. Basic variable xys1s1 s2s2 s3s3 RHS s1s1 111001000 s2s2 10.50 0750 s3s3 320012400 I-0.80000 PIVOT 1Making the pivot Divide through the pivot row by the pivot element

8. Basic variable xys1s1 s2s2 s3s3 RHS s1s1 111001000 s2s2 10.50 0750 s3s3 320012400 I0-0.300.50750 PIVOT 1Making the pivot Objective row + pivot row

9. Basic variable xys1s1 s2s2 s3s3 RHS s1s1 00.51-0.50250 s2s2 10.50 0750 s3s3 320012400 I0-0.300.50750 PIVOT 1Making the pivot First constraint row - pivot row

10. Basic variable xys1s1 s2s2 s3s3 RHS s1s1 00.51-0.50250 s2s2 10.50 0750 s3s3 00.50-1.51150 I0-0.300.50750 PIVOT 1Making the pivot Third constraint row – 3 x pivot row

11. Basic variable xys1s1 s2s2 s3s3 RHS s1s1 00.51-0.50250 x10.50 0750 s3s3 00.50-1.51150 I0-0.300.50750 PIVOT 1New solution I = 750, x = 750, y = 0, s 1 = 250, s 2 = 0, s 3 = 150

12. LINEAR PROGRAMMINGExample MaximiseI = x + 0.8y subject tox + y  1000 2x + y  1500 3x + 2y  2400 Solution after pivot 1: I = 750 at (750, 0)

13. Basic variable xys1s1 s2s2 s3s3 RHS s1s1 00.51-0.50250 x10.50 0750 s3s3 00.50-1.51150 I0-0.300.50750 PIVOT 2 Most negative number in objective row Choosing the pivot column

14. Basic variable xys1s1 s2s2 s3s3 RHS  -values s1s1 00.51-0.50250 250/0.5 x10.50 0750 750/0.5 s3s3 00.50-1.51150 150/0.5 I0-0.300.50750 PIVOT 2Choosing the pivot element Minimum of  -values gives 0.5 as pivot element

15. Basic variable xys1s1 s2s2 s3s3 RHS s1s1 00.51-0.50250 x10.50 0750 s3s3 010-32300 I0-0.300.50750 PIVOT 2Making the pivot Divide through the pivot row by the pivot element

16. Basic variable xys1s1 s2s2 s3s3 RHS s1s1 00.51-0.50250 x10.50 0750 s3s3 010-32300 I000-0.40.6840 PIVOT 2Making the pivot Objective row + 0.3 x pivot row

17. Basic variable xys1s1 s2s2 s3s3 RHS s1s1 0011100 x10.50 0750 s3s3 010-32300 I000-0.40.6840 PIVOT 2Making the pivot First constraint row – 0.5 x pivot row

18. Basic variable xys1s1 s2s2 s3s3 RHS s1s1 0011100 x1002600 s3s3 010-32300 I000-0.40.6840 PIVOT 2Making the pivot Second constraint row – 0.5 x pivot row

19. Basic variable xys1s1 s2s2 s3s3 RHS s1s1 0011100 x1002600 y010-32300 I000-0.40.6840 PIVOT 2New solution I = 840, x = 600, y = 300, s 1 = 100, s 2 = 0, s 3 = 0

20. LINEAR PROGRAMMINGExample MaximiseI = x + 0.8y subject tox + y  1000 2x + y  1500 3x + 2y  2400 Solution after pivot 2: I = 840 at (600, 300)

21. Basic variable xys1s1 s2s2 s3s3 RHS s1s1 0011100 x1002600 y010-32300 I000-0.40.6840 PIVOT 3Choosing the pivot column Most negative number in objective row

22. Basic variable xys1s1 s2s2 s3s3 RHS  -values s1s1 0011100 100/1 x1002600 600/2 y010-32300 I000-0.40.6840 PIVOT 3Choosing the pivot element Minimum of  -values gives 1 as pivot element

23. Basic variable xys1s1 s2s2 s3s3 RHS s1s1 0011100 x1002600 y010-32300 I000-0.40.6840 PIVOT 3Making the pivot Divide through the pivot row by the pivot element

24. Basic variable xys1s1 s2s2 s3s3 RHS s1s1 0011100 x1002600 y010-32300 I000.400.2880 PIVOT 3Making the pivot Objective row + 0.4 x pivot row

25. Basic variable xys1s1 s2s2 s3s3 RHS s1s1 0011100 x10-201400 y010-32300 I000.400.2880 PIVOT 3Making the pivot Second constraint row – 2 x pivot row

26. Basic variable xys1s1 s2s2 s3s3 RHS s1s1 0011100 x10-201400 y0130600 I000.400.2880 PIVOT 3Making the pivot Third constraint row + 3 x pivot row

27. Basic variable xys1s1 s2s2 s3s3 RHS s2s2 0011100 x10-201400 y0130600 I000.400.2880 PIVOT 3Optimal solution I = 880, x = 400, y = 600, s 1 = 0, s 2 = 100, s 3 = 0

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