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A helpful presentation to understand Big-M method by animations by Nitesh Singh Patel, IMT, Nagpur
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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.
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