Presentation on theme: "Assignment (6) Simplex Method for solving LP problems with two variables."— Presentation transcript:
Assignment (6) Simplex Method for solving LP problems with two variables
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.
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
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)
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
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
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: Min. Z= -7T - 5C + 0S 1 + 0S 2 1 st Step, continued
Initial Simplex tableau TCS1S2Quantity (Q) S12110100 S24301240 Z-7-5000 We start by a basic solution Z=0. Non-Basic variables Basic variables
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 negative coefficient in the objective function. Here the entering variable will be T and the corresponding column is called pivot column. TCS1S2Quantity (Q) S12110100 S24301240 Z-7-5000 Pivot column
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=80 Then the leaving variable will be S1 and the corresponding row is called pivot row.
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
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.
2 nd Step: 2 nd Step: Entering variable The new largest negative 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.