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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.

andrew-chang
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LPP :Simplex Method

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  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

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