Big M Method

Submitted by: Kanika Sood Jatinder Kaur Yashu Janartha

Big M MethodWe need to assign, in the objective function, coefficients to the artificial variables that are either very small (maximization problem) or very large(minimization problem); whatever this value, let us call it Big M. In fact, this notion is an old trick in optimization in general; we simply associate a penalty value with variables that we do not want to be part of an ultimate solution.

Such variables will never be part of any feasible solution. This method removes artificial variables from the basis. Here, we assign a large undesirable (unacceptable penalty) coefficients to artificial variables from the objective function point of view. If the objective function (Z) is to be minimized, then a very large positive price (penalty, M) is assigned to each artificial variable and if Z is to be minimized, then a very large negative price is to be assigned. The penalty will be designated by +M for minimization problem and by M for a maximization problem and also M>0.

Big M Method Minimize Z=12x1+20x2 subject to 6x1+8x2>100 7x1+12x2>120 x1, x2 >0

Step 1: Express the problem in standard form Minimize Z= 12x1+20x2+0S1+0S2 Subject to: 6x1+8x2-S1-0S2= 100 7x1+12x2-0S1-S2= 120 x1, x2, S1, S2 > 0 Initial basic feasible solution Putting x1=0, x2=0 We get S1= -100 and S2= -120 Which is not feasible, as it does not fulfill the non-negativity constraint To make it positive introduce artificial variables in the constraints 6x1+8x2-S1+A1=100 7x1+12x2-S2+A2=120 Since they are artificial variables A1 and A2 should not occur in the final solution. Hence, they are assigned a large unit penalty (a large +M) in the objective function which can be written as

Since they are artificial variables A1 and A2 should not occur in the final solution. Hence, they are assigned a large unit penalty (a large +M) in the objective function which can be written as Minimize Z= 12x1+20x2+0S1+0S2+MA1+MA2 Now there are 6 variables and 2 constraints and hence, we have to assume 4 variables as zero to get the initial basic feasible solution. Assume x1=x2=S1=S2=0 subject to: 6x1+8x2-S1-0S2+A1+0A2=100 7x1+12x2-0S1-S2+0A1+A2=120

Now putting the value of x1,x2,S1,S2=0 We get A1=100 and A2=120 Now Step 2: Prepare a simplex table

Simplex TableCj Cb M M Bv A1 A2 Zj 12 x1 6 7 13M 20 X2 8 (12) 20M 20-20M 0 S1 -1 0 -M M 0 S2 0 -1 -M M M A1 1 0 M 0 M A2 0 1 M 0 Bi 100 120 12.5 10

Cj-Zj 12-13M

Next Iteration TableCj Cb R1(R2*8) M 20 Bv A1 x2 Zj Cj-Zj 12 x1 (4/3) 7/12 20 x2 0 1 0 S1 -1 0 -M M 0 S2 2/3 -1/12 2/3M-5/3 M A1 Bi 1 0 M 20 15 10 120/ 7

4/3M+35/ 20 3 4/3M+1/3 0

0 2/3M+5/3

Next Iteration TableCj Cb 12 R2(R1*7/12) 20 Bv x1 x2 Zj 12 x1 1 0 12 20 x2 0 1 20 0 0 S1 -3/4 7/16 -1/4 1/4 0 S2 1/2 -3/8 -3/2 3/2 Bi 15 5/4

Cj-Zj 0

Optimal SolutionSince it is a minimization problem and there is no negative value found so it is the optimal solution. Hence, we get the value of x1=15 and x2=5/4. Putting these values in the objective function: Min Z= 12x1+20x2 =12*15+20*5/4 =180+25 =205

Maximize Z= 3x1-x2 Subject to: 2x1+x23 x20

Step 1: Standardization Max Z= x1+2x2+3x3-x4-MA1-MA2-MA3 Subject to : x1+2x2+3x3+0x4+A1+0A2+0A4=15 2x1+x2+5x3+0x4+0A1+A2+0A3=20 x1+2x2+x3+x4+0A1+0A2+A3=10 Step 2: Prepare a simplex table

Simplex TableCj Cb -M -M -M Bv A1 A2 A3 Zj 1 x1 1 2 1 -5M 2 x2 2 1 2 -5M 3 x3 3 (5) 1 -9M -1 x4 0 0 1 -M -M A1 1 0 0 -M -M A2 0 1 0 -M 0 -M A3 0 0 1 Bi 15 20 10 5 4 10

-M -45M 0

Cj-Zj 1+5M

2+5M 3+9M

-1+M 0

Next Iteration TableCj Cb R1-M (R2*3) 3 -M Bv A1 x3 A3 Zj Cj-Zj 1 x1 -1/5 2/5 3/5 6/5-2/5M 1/5+2/5M 2 x2 (7/5) 1/5 9/5 3 -1 M M A1 A 3 1 0 0 M 0 0 1 M 0 Bi 3 4 6 12-9M 15/7 20 10/3 x3 X4 0 1 0 0 0 1 -M

3/5-16/5M 3 7/5+16/5 M 0

0 1+M

Next Iteration TableCj Cb 2 R2(R1*1/5) R3(R1*9/5) 3 -M Bv x2 x3 A3 Zj Cj-Zj 1 x1 -1/7 3/7 6/7 16/7M 6/7M 2 3 -1 -M A3 0 0 1 -M Bi 15/7 25/7 15/7 105/7-15/7M 15/7 x2 x3 x4 1 0 0 2 0 0 1 0 3 0 0 0 (1) -M

0 1+M

Next Iteration TableCj Cb 2 3 -1 Bv x2 x3 x4 Zj Cj-Zj 1 x1 -1/7 3/7 (6/7) 1/7 6/7 2 x2 1 0 0 2 0 3 x3 0 1 0 3 0 -1 x4 0 0 1 -1 0 Bi 15/7 25/7 15/7 90/7 -15 25/3 5/2

Next Iteration TableCj Cb R1-(R3*1/7) R2(R3*3/7) 2 3 1 Bv x2 x3 x1 Zj Cj-Zj 1 x1 0 0 1 1 0 2 x2 1 0 0 2 0 3 x3 0 1 0 3 0 -1 x4 1/6 -1/2 7/6 0 -1 Bi 5/2 5/2 5/2 15

Optimal Solution

THANK YOU!!!