BIG-M METHOD

A VARIANT OF SIMPLEX METHOD

PRESENTED BY:NISHIDH VILAS LAD-2013176

NITESH BERIWAL-2013177NITESH SINGH PATEL-2013178

NITIN BORATWAR-2013179NITIN KUMAR SHUKLA-2013180NOOPUR MANDHYAN-2013181

INTRODUCTION

WHAT IS BIG-M METHOD? The Big M method is a method of solving linear programming

problems. It is a variation of the simplex method designed for solving problems

typically encompassing "greater-than" constraints as well as "less-than" constraints - where the zero vector is not a feasible solution.

The "Big M" refers to a large number associated with the artificial variables, represented by the letter M.

DRAWBACKS

If optimal solution has any artificial variable with non-zero value, original problem is infeasible

Four drawbacks of BIG-M method: How large should M be? If M is too large, serious numerical difficulties in a computer Big-M method is inferior than 2 phase method Here feasibility is not known until optimality

Steps In The Big-M Method

Add artificial variables in the model to obtain a feasible solution. Added only to the ‘>’ type or the ‘=‘ constraints A value M is assigned to each artificial variable The transformed problem is then solved using simplex eliminating the

artificial variables

Important Points To Remember

Solve the modified LPP by simplex method, untilany one of the three cases may arise. If no artificial variable appears in the basis and the optimality

conditions are satisfied If at least one artificial variable in the basis at zero level and the

optimality condition is satisfied If at least one artificial variable appears in the basis at positive level

and the optimality condition is satisfied, then the original problem has no feasible solution.

Big M Method: Example 1

Minimize Z = 40x1 + 24x2

Subject to 20x1 + 50x2 >= 4800 80x1 + 50x2 >= 7200 x1 , x2 >= 0

Introducing Surplus Variable and Artificial Variable to obtain an Initial Solution

Minimize Z = 40x1 + 24x2+ 0s1 + 0s2 + MA1 + MA1

Subject to 20x1 + 50x2 –S1 + A1 = 4800

80x1 + 50x2 –S2 + A2 = 7200 x1 , x2 >= 0

S1 and S2 Surplus VariableA1 and A2 Artificial Variable

Basic Variabl

es

Cb Xb

X1 X2 S1 S2 A1 A2

Cj -40 -24 0 0 -M -M

A1-M 4800

A2-M 7200

20 50 -1 0 1 0

80 50 0 -1 0 1

Z∑ (Cb-Xb) = 12000M

-100M+4

0

-100M+2

4M M 0 0 ∆j= ∑CbXj-Cj

Min Ratio =

Xb/Xk4800/50

7200/50

Maximize Z = -40x1 – 24x2 – 0S1 – 0S2 - MA1 – MA2

Basic Variabl

esCb Xb

X1 X2 S1 S2 A1 A2

Cj -40 -24 0 0 -M -M

X2-24

96A2

-M 2400

60 0 1 -1 -1 1

Z∑ (Cb-Xb) = -2400M - 2304

152/5-60M 0 12/25

-M M -12/25+2M 0 ∆j= ∑CbXj-Cj

Min Ratio = Xb/Xk480/2

2400/602/5 1 -1/50 0 1/50 0

BV Cb Xb X1 X2 S1 S2 A1 A2

A1 -M 4800 20 50 -1 0 1 0

A2 -M 7200 80 50 0 -1 0 1

Basic Variabl

esCb Xb

X1 X2 S1 S2 A1 A2

Cj -40 -24 0 0 -M -M

X2-24

80X1

-40 40

1 0 1/60 -1/60 -1/60 1/60

Z∑ (Cb-Xb) = -3520 0 0 -2/75 78/150 2/75+M 38/75

+M ∆j= ∑CbXj-Cj

Min Ratio = Xb/Xk-6000/2

2400/10 1 -2/75 1/150 2/75 -1/150

BV Cb Xb X1 X2 S1 S2 A1 A2

X2 -24 96 2/5 1 -1/50 0 1/50 0

A2 -M 24 60 0 1 -1 -1 1

Basic Variabl

es

Cb Xb

X1 X2 S1 S2 A1 A2

Cj -40 -24 0 0 -M -M

X2-24 144

S10 2400

60 0 1 -1 -1 1

Z∑ (Cb-Xb) = 3456 8/5 0 0 12/25 M M-

12/25 ∆j= ∑CbXj-Cj

Min Ratio = Xb/Xk_

_8/5 1 0 -1/50 0 1/50

THE VALUE OF OBJECTIVE FUNCTION IS 3456

BV Cb Xb X1 X2 S1 S2 A1 A2

X2 -24 80 0 1 -2/5 1/150 2/75 -1/150

X1 -40 40 1 0 1/60 -1/60 -1/60 1/60

Hence Optimal Solution is ACHIEVED.

BIG-M Method : An Example

Maximize Z = 2x1 + 4x2

Subject to 2x1 + x2 <= 18 3x1 + 2x2 >= 30 x1 + 2x2 = 26 x1 , x2 >= 0

Introduce Surplus , Slack and Artificial variables

Maximize Z = 2x1 + 4x2 + 0S1 + 0S2 – MA1 – MA2

Subject to 2x1 + x2 + S1 = 18 3x1 + 2x2 – S2 + A1 = 30 x1 + 2x2 + A2 = 26 x1 , x2 , S1 , S2 , A1 , A2 >= 0S1 -> Slack VariableS2 -> Surplus VariableA1 , A2 -> Artificial VariableM -> Large value

Max: Z= 2x1 + 4x2 + 0S1 + 0S2 – MA1 – MA2

2x1 + x2 + S1 = 18 3x1 + 2x2 – S2 + A1 = 30 x1 + 2x2 + A2 = 26 x1 , x2 , S1 , S2 , A1 , A2 >= 0

Z∑ (Cb*Xb) = -56M

Basic Variables

Cb Xb

X1 X2 S1 S2 A1 A2

S1 0 18

A1 -M 30

A2 -M 26

2 1 1 0 0 0

3 2 0 -1 1 0

1 2 0 0 0 1

Min Ratio = Xb/Xk18/1

30/2

26/2

Cj 2 4 0 0 -M -M

Calculation:∑(Cb*Xb) = 0*18 + (-M*30) + (-M*26) = -56M∆j= ∑CbXj-Cj = (0*2)+(-M*3)+(-M*1)-2= -4M-2∆j= ∑CbXj-Cj = (0*1)+(-M*2)+(-M*2)-4 = -4M-4∆j= ∑CbXj-Cj = (0*1)+(-M*0)+(-M*0)-0 = 0∆j= ∑CbXj-Cj = (0*0)+(-M*(-1))+(-M*0)-0 = M∆j= ∑CbXj-Cj = (0*0)+(-M*1)+(-M*0)-(-M) = 0∆j= ∑CbXj-Cj = (0*0)+(-M*0)+(-M*1)-(-M) = 0

∆j= ∑CbXj-Cj

0M0-4M-4-4M-2 0

BV Cb Xb X1 X2 S1 S2 A1 A2

S1 0 18 2 1 1 0 0 0A1 -M 30 3 2 0 -1 1 0A2 -M 26 1 2 0 0 0 1

Cj 2 4 0 0 -M -MBasic

VariablesCb Xb

X1 X2 S1 S2 A1 A2

S1

A1

X2

0-M 4

Replace A2 by X2.Divide the key row X2 by key element 2. Now operate row X2 & S1. i.e. S1- X2

13 ½ 1 0 0 0 ½

5 3/2 0 1 0 0 1

Now operate A1 & X2i.e. A1-2 X2

4 2 0 0 -1 1 -1

Z∑ (Cb*Xb) = 52-4M

Calculation:∑(Cb*Xb) = 0*5 + (-M*4) + 4*13 = 52-4M∆j= ∑CbXj-Cj = (0*3/2)+(-M*2)+(4*1/2)-2= -2M∆j= ∑CbXj-Cj = (0*0)+(-M*0)+(4*1)-4 = 0∆j= ∑CbXj-Cj = (0*1)+(-M*0)+(4*0)-0 = 0∆j= ∑CbXj-Cj = (0*0)+(-M*(-1))+(4*0)-0 = M∆j= ∑CbXj-Cj = (0*0)+(-M*1)+(4*0)-(-M) = 0∆j= ∑CbXj-Cj = (0*1)+(-M*(-1))+(4*1/2)-(-M) = 2+2M

-2M 0 0 M 0 2+2M

Select the least negative element i.e. -2M, this column will be taken as Xk .Now select the min. ratio which is 2, corresponding key element will be 2, & the key row would be the A1.

Min. Ratio=

Xb/Xk10/32

26

BV Cb Xb X1 X2 S1 S2 A1 A2

S1 0 5 3/2 0 1 0 0 -.5A1 -M 4 2 0 0 -1 1 -1A2 4 13 1/2 1 0 0 0 .5

Cj 2 4 0 0 -M -MBasic

VariablesCb Xb

X1 X2 S1 S2 A1 A2

S1

X1

X2

024

Replace A1 by X1.Divide the key row X1 by key element 2Now operate row X1 & S1. i.e. S1- 1.5 X1

2 0 0 1 ¾ -3/4 ¼

Now operate X1 & X2i.e. X2-0.5 X1

Z∑ (Cb*Xb) = 52

Calculation:∑(Cb*Xb) = 0*2 + (2*2) + 4*12 = 52∆j= ∑CbXj-Cj = (0*0)+(2*1)+(4*0)-2= 0∆j= ∑CbXj-Cj = (0*0)+(2*0)+(4*1)-4 = 0∆j= ∑CbXj-Cj = (0*1)+(2*0)+(4*0)-0 = 0∆j= ∑CbXj-Cj = (0*3/4)+(2*(-1/2))+(4*1/4)-0 = 0∆j= ∑CbXj-Cj = (0*-3/4)+(2*1/2)+(4*(-1/4))-(-M) = M∆j= ∑CbXj-Cj = (0*1/4)+(2*(-1/2))+(4*3/4)-(-M) = 2+M

0 0 0 0 M 2+M

2 1 0 0 -1/2 ½ -1/212 0 1 0 ¼ -1/4 3/4

The optimum solution to the problem is X1=2, X2=12, S1=2 & other variable is 0. The objective function value is 52.