Download presentation

Presentation is loading. Please wait.

Published byDayton Whedon Modified over 3 years ago

1
Assignment (6) 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: Min. Z= -7T - 5C + 0S 1 + 0S 2 1 st Step, continued

8
Initial Simplex tableau TCS1S2Quantity (Q) S12110100 S24301240 Z-7-5000 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 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

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=80 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 Z-7-500 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 -7-500 0 Pivot column Pivot row Pivot item Pivoting

15
4 th Step: 4 th Step: Pivoting TCS1S2Q T11/2 0 50 S201-21 40 Z0-3/2+7/20 350

16
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.

17
4 th Step: 4 th Step: Pivoting TCS1S2Q T11/2 0 50 S201-21 40 Z0-3/2+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 Z001/23/2 410 All coefficients of objective function are now equal to zero. then the optimal solution is found: T=30, C=40 ; Z=410

Similar presentations

Presentation is loading. Please wait....

OK

Mechanical Engineering Department 1 سورة النحل (78)

Mechanical Engineering Department 1 سورة النحل (78)

© 2018 SlidePlayer.com Inc.

All rights reserved.

Ads by Google

Ppt on job evaluation and job rotation Ppt on section 186 of companies act 2013 Ppt on eddy current suppression Ppt on chromosomes and genes middle school Light coloured backgrounds for ppt on social media Ppt on question tags grammar Ppt on building information modeling salary Ppt on event driven programming visual basic Ppt on ideal gas law formula Ppt on electricity for class 10th book