1 / 14

LPP :Simplex Method

LPP :Simplex Method. Simplex method. Introduction – S implex method through an iterative process progressively approaches and ultimately reaches to the maximum or minimum values of the objective function.

Download Presentation

LPP :Simplex Method

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. LPP :Simplex Method

  2. Simplex method Introduction – • Simplex method through an iterative process progressively approaches and ultimately reaches to the maximum or minimum values of the objective function. • Helps the decision maker to identify unbounded solution, multiple solution and infeasible problem Definition- “Simplex method is suitable for solving linear programming problems with a larger number of variables.”

  3. Basic terms- 1. Standard form- All constraints are written as equalities. 2. Slack variable- A variable added to the left hand side of a less than or equal to constraint to convert the constraint into an equality. 3. Surplus variable- Variable subtracted from the left hand side of the greater than or equal to constraint, to convert the constraint into an equality. 4. Basic solutions- A solution obtained by setting(n-m) variables equal to zero and solving for remaining m variables. 5. Optimal solution- Any basic feasible solution which optimizes(minimizes or maximizes)the objective function of general L.P problem is called optimal solution.

  4. 6. Zj Row- Represents total contribution of outgoing profit when one unit of non basic variable is introduced into the basis variable place. 7. Cj-Zj(or Net Evaluation or Index)Row- Therow containing net profit(or net loss)that will result from introducing one unit of the variable indicated in that column in the solution. 8. Pivot column- The column with largest positive number(maximization) or largest negative number(minimization)in the net evaluation row. 9. Pivot row- The row corresponding to the variable that will leave the basis in order to make room for the variable entering .smallest positive ratio.

  5. Q- A plant is engaged in production of 2 products which are processed through 3 departments. The number of hours required to finish each are given below- The profit of product is rs.60 per unit of A & Rs.40/- of B. Solve it by simplex method.

  6. Solution- Max Z = 60 A + 40 B Subject to Constraints- 7A + 8B ≤ 1600 8A+ 12B ≤ 1600 15A + 16B ≤ 1600 A,B ≥ 0 Z = 60 A + 40 B + 0S1 + 0S2 + 0S3 7A+ 8B+ S1+ 0S2+ 0S3= 1600 8A+ 12B+ 0S1+ S2+ 0S3= 1600 15A+ 16B+ 0S1 + 0S2+ S3= 1600 A, B, S1, S2, S3 ≥ 0

  7. Working notes- • 1. Zj (for A )= 0*7 + 0*8 + 0*15 = 0 (A*CB) (for B)= 0*8 + 0*12 + 0*16 = 0 (B*CB) (for S1)=0*1+ 0*0 + 0*0 = 0 (S1*CB) same for S2 & S3. • 2. Net evaluation row- Cj- Zj= A= 60-0 =6 B= 40-0=40 S1= 0-0=0 S2= 0-0=0 S3= 0-0=0

  8. Since 60 is highest vertically ,it becomes pivot column, and we divide pivot column to CJ to get minimum ratios. • Here the minimum is 106.67 row wise, so that row gets out.

  9. Now we replace the s3 column 0,0,1 with A in table2. • To make 15 to 1,we have to divide 15 to whole “row”. Cj = 1600/15= i.e 320/3 A =15/15= 1 B=16/15 S1 =0/15=0 S2 =0/15=0 S3 =1/15 Now for S1 row=(consider the first and second table values) 1600-7 *320 = 1600*3 - 7*320 = 4800 – 2240 =2560 3 3 3 • Now for S2 row= 1600 -8*320 = 1600*3 - 8*320 = 4800 -2560 = 2240 3 3 3

  10. For b row wise- (8-7 i.e-B-A from table 1) &(16/15 from table 2) 8 - 7*16 =`15*8 – 7*16 = 120-112 = 8 15 15 15 15 • For S2- 12- 8*16 = 180-128 = 52 15 15 15 • Now, • For s3=(0-7 i.e- s3-A from table 1)&(1/15 we got from slide form slide 11) 0 - 7*1 = 15*0 -7 = -7 15 15 0- 8* 1 = -8 15 15

  11. Now for Zj= table 2 Cb*A = 0*0 + 0*0 + 60*1 = 60 Cb* B = 0* 8/15 + 0*52/15 + 60*16/15 = 64 S1= 0*1 + 0*0 + 60*0 = 0 .same for S2 S3= 0* -7/15 + 0*- 8/15 + 60*1/15 = 4 Cj- zj = A= 60-60=0 B = 40-64 = -24 S1 = 0 - 0 =0 ,same for s2 s3= 0 - 4= -4 ,,since all cj-zj appears in negative and zero,solution is optimal.so, • Max Z = 60 A + 40 B • =60(320/3) + 40 (0) • = 6400

More Related