In order to obtain an initial basic feasible solution, it is necessary to convert the given LPP into its standard form; in order to obtain the standard form; a non-negative variable is added to the left side of each of the equation that lacks the much needed starting basic variables. So added variable is called an artificial variable and it places the role in providing initial basic feasible solution.
Big-M Method [Introduction]
In order to obtain an initial basic feasible
solution, it is necessary to convert the given LPP into its standard form; in
order to obtain the standard form; a non-negative variable is added to the left
side of each of the equation that lacks the much needed starting basic
variables. So added variable is called an artificial variable and it places the
role in providing initial basic feasible solution.
The starting point of the big-M method is a basic
solution that is feasible to the artificial problem. This solution allows us to
start the Simplex
algorithm expeditiously, but it is not a feasible solution to the original
problem. Our goal is to iterate toward solutions that are inside the original
feasible set, assuming that it is not empty.
The artificial variables
have no physical meaning from the view of the original problem; the method will
be valid only if it is possible to force these variables at zero level when the
optimum solution is attained.
To accomplish this goal, we designed an artificial
objective function that is an aggregation of two functional parts, of which one
is a copy of the original objective function and the other is a “penalty
function” associated with the artificial variables. The magnitude of the penalty, M, needs to be chosen
to ensure that the contribution to this aggregate objective function from the
second part, whenever it is positive, always outweighs that from the first
part, for any solution. Artificial
Variable
If there is a constraint which is of > = or
equality type then it is needed to write the constraint in standard form
because it is not possible to get unit column vector. The additional variable,
which is added in order to get unit column vector,
is called an artificial variable. Two
methods are generally employed for the solution of linear programming problems
having artificial variables: 1.
Two-Phase
Method. 2. Big-M
Method (or) Method of Penalties. We will have discussion only on Big-M Method
here. The Big M Method Procedure
If an LP has any > or = constraints, the Big M
method or the two-phase simplex method may be used to solve the problem. The Big M method is a version of the Simplex
Algorithm that first finds a best feasible solution by adding “artificial”
variables to the problem. The objective function of the original LP must, of
course, be modified to ensure that the artificial variables are all equal to 0
at the conclusion of the simplex algorithm. The iterative procedure of the
algorithm is given below. Step-1 Modify the constraints so that the RHS of each
constraint is non-negative (This requires that each constraint with a negative
RHS be multiplied by -1. Remember that if any negative number multiplies an
inequality, the direction of the inequality is reversed). After modification,
identify each constraint as a <, >, or = constraint. Step-2 Convert each inequality constraint to standard form
(If a constraint is a ≤ constraint, then add a slack variable Xi; and if any
constraint is a ≥ constraint, then subtract an excess variable Xi , known as
surplus variable). Step-3 Add an artificial variable a1 to the constraints
identified as ‘≥’ or with ‘=’ constraints at the end of Step2. Also add the
sign restriction ai ≥ 0. Step-4 Let M denote a very large positive number. If the
LP is a minimization problem, add (for each artificial variable) Mai to the objective
function. If the LP is a maximization problem, add (for each artificial
variable) -Mai to the objective function. Step-5 Since each artificial variable will be in the
starting basis, all artificial variables must be eliminated from row 0 before beginning
the simplex. Now we
solve the transformed problem by the simplex (In choosing the entering
variable, remember that M is a very large positive number). If all artificial
variables are equal to zero in the optimal solution, we have found the optimal
solution to the original problem. If any artificial variables are positive in
the optimal solution, the original problem is infeasible!!! 1. If at
least one artificial variable is present in the basis with zero value, in such
a case the current optimum basic feasible solution is a degenerate solution. 2. If at
least one artificial variable is present in the basis with a positive value,
then in such a case, the given LPP does not have an optimum basic feasible
solution. The given problem is said to have a pseudo-optimum basic feasible
solution. Example-1 Solve the following LPP by using
Big -M Method Maximize Z = 6X1+4X2 Subject to constraints: 2X1+3X2<=30 3X1+2X2<=24 X1+X2>=3 Solution Introducing slack variables S1>=0, S2>=0 to
the first and second equations in order to convert <= type to equality and
add surplus variable to the third equation S3 >=0 to convert >= type to
equality. Then the standard form of LPP is MAX
Z=6X1+4X2+0S1+0S2+0S3 Subject to constraints 2X1+3X2+S1=30 3X1+2X2+S2=24 X1+X2-S3=3 Clearly there is no initial basic feasible
solution. So an artificial variable A1>=0 is added in the third equation.
Now the standard form will be MAX Z=6X1+4X2+0S1+0S2+0S3+A1 Subject to constraints 2X1+3X2+S1=30 3X1+2X2+S2=24 X1+X2-S3+A1=3 Now the
first simplex table is placed below with following additional information. It clearly shows the net evaluations for each
column variable; from these calculations, it is clear that X1 is the entering
variable and A1, the artificial variable leaves the basis. Introduce Y1 and
drop Y6
Then, in
the usual row operations, we modify this table and the new table is arrived. Table-2
Introduce
Y5 and drop Y4 from the basis; the new modified table-3 is given below;
Since all
Zj –Cj >=0, an optimum solution has reached. Thus an optimum basic feasible
solution to the LPP is X1=8 X2=0 MAX Z =48 Example-2 Use big M
method to solve a given LPP. Minimize Z = 5X1-6X2-7X3 Subject to constraints X1+5X2-3X3>=15 5X1-6X2+10X3<=20 X1+X2+X3=5 X1, X2
& X3 >= 0 Solution Introducing slack variables X4>=0 to the first
equations in order to convert <= type to equality and add surplus variable
X5>=0 to the second equation in order to convert >= type to equality. Then the standard form of LPP is MIN
Z=5X1-6X2-7X3+X4-X5 Subject to constraints 5X1-6X2+10X3+X4=20 X1+5X2-3X3-X5=15 X1+X2+X3=5 Clearly
there is no initial basic feasible solution. So two artificial variables
A1>=0 and A2>=0 are added in the second and third equation. Now the
standard form will be MIN
Z=5X1-6X2-7X3+X4-X5+A1+A2 Subject to constraints 5X1-6X2+10X3+X4=20 X1+5X2-3X3-X5+A1=15 X1+X2+X3+A2=5
From
the calculations related to entering column and leaving variable, which is
summarized in the above table, it is clear that introduces X2 and drop X6 from
the basis. The new simplex table is given below;
The new modified table- 3 is
given below -1/5 +32/5*1/5*5/8
Since all Zj –Cj <=0, an optimum solution has
reached. Thus an optimum basic feasible solution to the LPP is X3=5/4 X2=15/4 MIN Z=
(5*0) + (-6*15/4) + (-7* 5/4) =
(-90/4) +
(-35/4) =
-125/4
Tags : Operations Management - Introduction to Operations Research
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