1. Simplex Method for solving LP problems with two variables

2. 1- Introduction  Graphical method presented in last chapter is fine for 2 variables. But most LP problems are too complex for simple graphical procedures.  The Simplex Method: ois appropriate for problems with more than 2 variables; ouses algebra rules, to find optimal solutions; ois an algorithm and is a series of steps that will accomplish a certain task.  In this chapter will introduce the simplex method for problems with 2 variables.

3. 2- Simplex algorithm through an example: Assume one company producing flair furniture: Tables (T) and Chairs (C). The following table provides the information available: Department ProductionAvailable Working Hours TablesChairs Carpentry43240 Painting & varnishing 21100 Profit7SR5SR

4. Formulation: Decision variables: T = Number of tables C = Number of chairs Objective function: Maximize Z = 7T + 5C Constraints: 2T + 1C  100 (Painting & varnishing) 4T + 3C  240 (Carpentry) T, C  0 (non-negativity constraints)

5. Graphical method solution Number of Chairs 100 80 60 40 20 020406080100 T C Number of Tables Feasible Region  4T + 3C  2T + 1C   Optimal Solution: T=30, C=40 Z= 410 SR

6. 1 st Step: 1 st Step: Built initial Simplex tableau  Less-than-or-equal-to constraints (≤) are converted to equations by adding a variable called “slack variable”. oSlack variables represent unused resources.  For the flair furniture problem, define the slacks as: oS 1 = unused hours in the painting department oS 2 = unused hours in the carpentry department  The constraints are now written as: o2T + 1C + S 1 = 100 o4T + 3C + S 2 = 240

7.  Slack variables not appearing in an equation are added with a coefficient of 0.This allows all the variables to be monitored at all times.  The final Simplex equations appear as: o2T + 1C + 1S 1 + 0S 2 = 100 o4T + 3C + 0S 1 + 1S 2 = 240 oT, C, S 1, S 2  0  The slacks are added to the objective coefficient with 0 profit coefficients. The objective function, then, is: Max. Z= 7T + 5C + 0S 1 + 0S 2 1 st Step, continued

8. Initial Simplex tableau TCS1S2Quantity (Q) S12110100 S24301240 Z75000 We start by a basic solution Z=0. Non-Basic variables Basic variables

9. 2 nd Step: 2 nd Step: Entering variable Choose one entering variable from non-basic variables (T or C) for which we have the largest positive coefficient in the objective function. Here the entering variable will be T and the corresponding column is called pivot column. TCS1S2Quantity (Q) S12110100 S24301240 Z75000 Pivot column

10. 3 rd Step: 3 rd Step: Leaving variable Choose one leaving variable from the basic variables (S1 or S2) for which we have the smallest value of quantities (Q) divided by items of pivot column: for S1 we have 100/2=50 for S2 we have 240/4=60 Then the leaving variable will be S1 and the corresponding row is called pivot row.

11. 3 rd Step: 3 rd Step: Leaving variable TCS1S2(Q) S12110 100 S24301 240 Z7500 0 Pivot column Pivot row Pivot item

12. 4 th Step: 4 th Step: Pivoting The pivoting is the changing of simplex tableau values as follow:  The entering variable (T) takes the place of leaving variable (S1).  All items of pivot column are =0 except pivot item =1.  All items of pivot row are divided by pivot item.  Other items of the tableau are calculated as follow: AB CD Pivot column Pivot row New A = A – B*C/D

13. 4 th Step: 4 th Step: Pivoting The New value of the objective function is calculated as follow: Capacity of pivot row Z’ = Z + Largest coefficient * Pivot item If all coefficients of objective function are negatives or equal to zero the optimal solution is found. Otherwise go to step2.

14. 4 th Step: 4 th Step: Pivoting TCS1S2Q S12110 100 S24301 240 Z 7500 0 Pivot column Pivot row Pivot item Pivoting

15. 4 th Step: 4 th Step: Pivoting All coefficients of objective function aren’t negatives or equal to zero, the optimal solution is not found. Then go to step2. TCS1S2Q T11/2 0 50 S201-21 40 Z03/2-7/20 350

16. 2 nd Step: 2 nd Step: Entering variable The new largest positive coefficient in the objective function is 3/2 then the entering variable will be C. 3 rd Step: 3 rd Step: Leaving variable The new smallest value of quantities (Q) divided by items of pivot column is 40/1=40 then the leaving variable will be S2.

17. 4 th Step: 4 th Step: Pivoting TCS1S2Q T11/2 0 50 S201-21 40 Z03/2-7/20 350 Pivot column Pivot row Pivot item

18. 4 th Step: 4 th Step: Pivoting TCS1S2Q T103/2-1/2 30 C01-21 40 Z00-1/2-3/2 410 All coefficients of objective function are now equal to zero. then the optimal solution is found: T=30, C=40 ; Z=410

19. Using Simplex method, solve the LP problems modeled in chapter1. 3- Applications

20. The mathematical model of problem is: Max Z= 3x 1 + 5x 2 S.t. x 1  4 (Plant 1 constraint) 2x 2  12 (Plant 2 constraint) 3x 1 +2x 2  18 (Plant 3 constraint) x 1, x 2  0 3.1- Solving of problem1

21. 1 st Step: 1 st Step: Initial Simplex tableau We start by a basic solution Z=0. Non-Basic variables Basic variables X1X2 S1S2S3Quantity (Q) S1 101004 S2 0201012 S3 3200118 Z350000

22. 2 nd Step: 2 nd Step: Entering variable The largest positive coefficient in the objective function is 5 then the entering variable will be X2. 3 rd Step: 3 rd Step: Leaving variable The smallest value of quantities (Q) divided by items of pivot column is 12/2=6 then the leaving variable will be S2.

23. 4 th Step: 4 th Step: Pivoting Pivot column Pivot row Pivot item X1X2 S1S2S3 (Q) S1 101004 S2 0201012 S3 3200118 Z350000

24. 4 th Step: 4 th Step: Pivoting X1X2 S1S2S3 (Q) S1 101004 X2 0101/206 S3 30016 Z300-5/2030 All coefficients of objective function aren’t negatives or equal to zero, the optimal solution is not found. Then go to step2.

25. 2 nd Step: 2 nd Step: Entering variable The new largest positive coefficient in the objective function is 3 then the entering variable will be X1. 3 rd Step: 3 rd Step: Leaving variable The smallest value of quantities (Q) divided by items of pivot column is 6/3=2 then the leaving variable will be S3.

26. 4 th Step: 4 th Step: Pivoting Pivot column Pivot row Pivot item X1X2 S1S2S3 (Q) S1 101004 X2 0101/206 S3 30016 Z300-5/2030

27. 4 th Step: 4 th Step: Pivoting All coefficients of objective function are now equal to zero. then the optimal solution is found. X1=2, X2=6, Z=36. X1X2 S1S2S3 (Q) S1 0011/3-1/32 X2 0101/206 X1 100-1/31/32 Z0007/236

28. The mathematical model of problem is: Max Z= 10x 1 + 15x 2 S.t. 2x1 + 4x2  100 ( aluminum constraint) 3x1 + 2x2  80 ( steel constraint) x1, x2  0 3.2- Solving of problem2

29. 1 st Step: 1 st Step: Initial Simplex tableau X1X2S1S2Quantity (Q) S12410100 S2320180 Z1015000 We start by a basic solution Z=0. Non-Basic variables Basic variables

30. 2 nd Step: 2 nd Step: Entering variable The largest positive coefficient in the objective function is 15 then the entering variable will be X2. 3 rd Step: 3 rd Step: Leaving variable The smallest value of quantities (Q) divided by items of pivot column is 100/4=25 then the leaving variable will be S1.

31. 4 th Step: 4 th Step: Pivoting Pivot column Pivot row Pivot item X1X2S1S2Q S12410100 S2320180 Z1015000

32. 4 th Step: 4 th Step: Pivoting X1X2S1S2Q X21/211/4025 S220-1/2130 Z5/2015/40375 All coefficients of objective function aren’t negatives or equal to zero, the optimal solution is not found. Then go to step2.

33. 2 nd Step: 2 nd Step: Entering variable The new largest positive coefficient in the objective function is 5/2 then the entering variable will be X1. 3 rd Step: 3 rd Step: Leaving variable The smallest value of quantities (Q) divided by items of pivot column is 30/2=15 then the leaving variable will be S2.

34. 4 th Step: 4 th Step: Pivoting Pivot column Pivot row Pivot item X1X2S1S2Q X21/211/4025 S220-1/2130 Z5/2015/40375

35. 4 th Step: 4 th Step: Pivoting All coefficients of objective function are now equal to zero, the optimal solution is found: X1=15, X2=17.5, Z= 412.5. X1X2S1S2Q X2013/8-1/417.5 X110-1/41/215 Z0030/8-5/4412.5