Upload
potnuru-vinay
View
215
Download
0
Embed Size (px)
Citation preview
7/29/2019 05 Simplex 3
1/26
Artificial VariableTechniques
Big M-method
7/29/2019 05 Simplex 3
2/26
Lecture 6
Abstract If in a starting simplex tableau,
we dont have an identity submatrix (i.e. an
obvious starting BFS), then we introduce
artificial variables to have a starting BFS.
This is known as artificial variable
technique. There are two methods to find
the starting BFS and solve the problem
the Big M method and two-phase method.
In this lecture, we discuss the Big M
method.
7/29/2019 05 Simplex 3
3/26
Suppose a constraint equation i does not
have a slack variable. i.e. there is no ith
unit vector column in the LHS of theconstraint equations. (This happens for
example when the ith constraint in the
original LPP is either or = .) Then weaugment the equation with an artificial
variable Ri to form the ith unit vector
column. However as the artificialvariable is extraneous to the given LPP,
7/29/2019 05 Simplex 3
4/26
we use a feedback mechanism in which the
optimization process automatically attempts
to force these variables to zero level. This isachieved by giving a large penalty to the
coefficient of the artificial variable in the
objective function as follows:
Artificial variable objective coefficient
= - M in a maximization problem,= M in a minimization problem
where M is a very large positive number.
7/29/2019 05 Simplex 3
5/26
Consider the LPP:
Minimize 212 xxz
Subject to the constraints
0,
6
93
21
21
21
xx
xx
xx
7/29/2019 05 Simplex 3
6/26
Putting this in the standard form, the LPP is:
212 xxz
Subject to the constraints
1 2 1
1 2 2
1 2 1 2
3 96
, , , 0
x x s
x x s
x x s s
Heres1,s2 are surplus variables.
Minimize
Note that we do NOT have a 2x2 identity
submatrix in the LHS.
7/29/2019 05 Simplex 3
7/26
Introducing the artificial variables, R1,
R2, the LPP is modified as follows:
Minimize
Subject to the constraints
1 2 1 1
1 2 2 2
1 2 1 2 1 2
3 9
6
, , , , , 0
x x s R
x x s R
x x s s R R
Note that we now have a 2x2 identity
submatrix in the coefficient matrix of the
constraint equations.
1 2 1 22z x x MR MR
7/29/2019 05 Simplex 3
8/26
Now we solve the above LPP by theSimplex method.
7/29/2019 05 Simplex 3
9/26
R2 0 1 1 0 -1 0 1 6
-2+4M -1+2M -M -M 0 0 15M
Basic z x1 x2 s1 s2 R1 R2 Sol.
z 1 -2 -1 0 0 -M -M 0R1 0 3 1 -1 0 1 0 9
R2 0 0 2/3 1/3 -1 -1/3 1 3
z 1 0 -1/3+ -2/3+ -M 2/3- 0 6+3M2M/3 M/3 4M/3
x1 0 1 1/3 -1/3 0 1/3 0 3
x1 0 1 0 -1/2 1/2 1/2 -1/2 3/2
z 1 0 0 -1/2 -1/2 1/2M 1/2-M 15/2
x2 0 0 1 1/2 -3/2 -1/2 3/2 9/2
7/29/2019 05 Simplex 3
10/26
Note that we have got the optimal solution
to the given problem as
1 2
3 9,
2 2x x
15
2
z
It is illuminating to look at the graphical
solution also.
Optimal z = Minimum
7/29/2019 05 Simplex 3
11/26
(3,0) (6,0)
(0.9)
(0,6)3 9
( , )2 2
(0,0)
SF
z minimum at3 9
( , )2 2
7/29/2019 05 Simplex 3
12/26
Problem 5(a) Problem Set 3.4A Page 97
Maximize 1 2 32 3 5z x x x
Subject tothe constraints
1 2 3
1 2 3
1 2 3
7
2 5 10
, , 0
x x x
x x x
x x x
7/29/2019 05 Simplex 3
13/26
Introducing surplus and artificial variables,
s2, R1and R2, the LPP is modified as follows:
Maximize
Subject to the constraints
1 2 3 1 22 3 5z x x x MR MR
1 2 3 1
1 2 3 2 2
1 2 3 2 1 2
7
2 5 10, , , , , 0
x x x R
x x x s R
x x x s R R
Now we solve the above LPP by the Simplex
method.
7/29/2019 05 Simplex 3
14/26
R2 0 2 -5 1 -1 0 1 10
-2-3M -3+4M 5-2M M 0 0 -17M
Basic z x1 x2 x3 s2 R1 R2 Sol.
z 1 -2 -3 5 0 M M 0R1 0 1 1 1 0 1 0 7
x1 0 1 -5/2 1/2 -1/2 0 1/2 5
z 1 0 -8 - 6 - -1 - 0 1 + 10 -7M/2 M/2 M/2 3M/2 2M
R1 0 0 7/2 1/2 1/2 1 -1/2 2
x2 0 0 1 1/7 1/7 2/7 -1/7 4/7
z 1 0 0 50/7 1/7 16/7 + -1/7 102/7
M +M
x1 0 1 0 6/7 -1/7 5/7 1/7 45/7
7/29/2019 05 Simplex 3
15/26
The optimum (Maximum) value of
z = 102/7
and it occurs at
x1 = 45/7,x2 = 4/7,x3 = 0
7/29/2019 05 Simplex 3
16/26
Remarks
If in any iteration, there is a tie for entering
variable between an artificial variable and
other variable (decision, surplus or slack),
we must prefer the non-artificial variable to
enter the basis.
If in any iteration, there is a tie for leaving
variable between an artificial variable andother variable (decision, surplus or slack),
we must prefer the artificial variable to
leave the basis.
7/29/2019 05 Simplex 3
17/26
If in the final optimal tableau, an
artificial variable is present in thebasis at a non-zero level, this means
our original problem has no feasible
solution.
7/29/2019 05 Simplex 3
18/26
Problem 4(a) Problem Set 3.4A Page 97
Maximize 1 25 6z x x
Subject tothe constraints
1 2
1 2
1 2
1 2
2 3 32 5
6 7 3, 0
x x
x x
x x
x x
7/29/2019 05 Simplex 3
19/26
Introducing slack and artificial variables, s2,
s3
, and R1
, the LPP is modified as follows:
Maximize
Subject to the constraints
1 2 1
1 2 2
1 2 3
1 2 1 2 3
2 3 3
2 56 7 3
, , , , 0
x x R
x x s
x x s
x x R s s
1 2 15 6z x x MR
i 1 2 1 2 3 S l
7/29/2019 05 Simplex 3
20/26
s2 0 1 2 0 1 0 5
-5+2M -6-3M 0 0 0 -3M
Basic z x1 x2 R1 s2 s3 Sol
z 1 -5 -6 M 0 0 0R1 0 -2 3 1 0 0 3
s3 0 6 7 0 0 1 3
s2 0 -12/7 0 0 1 -2/7 29/7
z 1 1/7+ 0 0 0 6/7+ 18/7-
32M/7 3M/7 12M/7
R1 0 -32/7 0 1 0 -3/7 12/7
x2 0 6/7 1 0 0 1/7 3/7
This is the optimal tableau. As R1 is not zero, there is NO feasible
7/29/2019 05 Simplex 3
21/26
Problem 4(d) Problem Set 3.4A Page 97
Minimize 21 64xxz
Subject tothe constraints
0,584
1054332
21
21
21
21
xx
xx
xx
xx
7/29/2019 05 Simplex 3
22/26
Introducing the surplus and artificial variables,
R1
, R2
, the LPP is modified as follows:
Minimize
Subject to the constraints1 2 1
1 2 2 2
1 2 3 3
1 2 2 3 1 2 3
2 3 3
4 5 10
4 8 5
, , , , , , 0
x x R
x x s R
x x s R
x x s s R R R
1 2 1 2 34 6z x x M R M R M R
B i 1 2 2 3 R1 R2 R3 S l
7/29/2019 05 Simplex 3
23/26
R2 0 4 5 -1 0 0 1 0 10
-4+6M -6+16M -M -M 0 0 0 18M
Basic z x1 x2 s2 s3 R1 R2 R3 Sol.
z 1 -4 -6 0 0 -M -M -M 0R1 0 -2 3 0 0 1 0 0 3
R1 0 -7/2 0 0 3/8 1 0 -3/8 9/8
z 0 -1-2M 0 -M -3/4 0 0 3/4 15/4
+M -2M +8M
R2 0 3/2 0 -1 5/8 0 1 -5/8 55/8
x2 0 1/2 1 0 -1/8 0 0 1/8 5/8
R3 0 4 8 0 -1 0 0 1 5
B i 1 2 2 3 R1 R2 R3 S l
7/29/2019 05 Simplex 3
24/26
R2 0 3/2 0 -1 5/8 0 1 -5/8 55/8
Basic z x1 x2 s2 s3 R1 R2 R3 Sol.
z 1 -1-2M 0 -M -3/4 0 0 3/4 15/4
+M -2M +8MR1 0 -7/2 0 0 3/8 1 0 -3/8 9/8
s3 0 -28/3 0 0 1 8/3 0 -1 3
z 1 -8 + 0 -M 0 2 0 -M 6
22M/3 -8M/3 +5M
R2 0 22/3 0 -1 0 -5/3 1 0 5
x2 0 -2/3 1 0 0 1/3 0 0 1
x2 0 1/2 1 0 -1/8 0 0 1/8 5/8
B i 1 2 2 3 R1 R2 R3 S l
7/29/2019 05 Simplex 3
25/26
Basic z x1 x2 s2 s3 R1 R2 R3 Sol.
s3 0 -28/3 0 0 1 8/3 0 -1 3
z 1 -8 + 0 -M 0 2 0 -M 6
22M/3 -8M/3 +5M
R2 0 22/3 0 -1 0 -5/3 1 0 5
x2 0 -2/3 1 0 0 1/3 0 0 1
s3 0 0 0 -14/11 1 6/11 14/11 -1
z 1 0 0 -6/11 0 2/11 12/11 -M
- M -M
x1 0 1 0 -3/22 0 -5/22 3/22 0
x2 0 0 1 -1/11 0 2/11 1/11 0
10311
126
11
15
22
16
11
7/29/2019 05 Simplex 3
26/26
This is the optimal tableau.
The Optimal solution is:
1 215 16,22 11
x x
And Optimal z = Min z =126
11