Objectives
Linear Programming
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To introduce linear programming problems (LPP)
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To discuss the standard and canonical form of LPP
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To discuss elementary operation for linear set of
Preliminaries
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D Nagesh Kumar, IISc
equations
LP_1: Intro
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Introduction
D Nagesh Kumar, IISc
LP_1: Intro
Example
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Linear Programming (LP) is the most useful optimization
technique
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Objective function and constraints are the ‘linear’ functions of
‘nonnegative’ decision variables
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Thus, the conditions of LP problems are
Maximize
Z = 6x + 5 y
≺ Objective Function
subject to
2x − 3 y ≤ 5
≺ 1st Constraint
x + 3 y ≤ 11
≺ 2nd Constraint
4 x + y ≤ 15
≺ 3rd Constraint
x, y ≥ 0
≺ Nonnegativity Condition
1. Objective function must be a linear function of decision
variables
This is in “general” form
2. Constraints should be linear function of decision variables
3. All the decision variables must be nonnegative
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LP_1: Intro
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Standard form of LP problems
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D Nagesh Kumar, IISc
LP_1: Intro
General form Vs Standard form
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Standard form of LP problems must have following
three characteristics:
1. Objective function should be of maximization type
General form
Minimize
Z = −3 x1 − 5 x 2
subject to
2 x1 − 3 x 2 ≤ 15
x1 + x 2 ≤ 3
2. All the constraints should be of equality type
4 x1 + x 2 ≥ 2
3. All the decision variables should be nonnegative
x1 ≥ 0
x 2 unrestrict ed
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Violating points for
standard form of LPP:
1. Objective function is of
minimization type
2. Constraints are of inequality
type
3. Decision variable, x2, is
unrestricted, thus, may take
negative values also.
How to transform a general form of a LPP to the standard form ?
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D Nagesh Kumar, IISc
LP_1: Intro
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D Nagesh Kumar, IISc
LP_1: Intro
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Transformation
General form
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General form
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1. Objective function
Minimize
Z = −3 x1 − 5 x 2
2. First constraint
Standard form
General form
Standard form
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1. Objective function
Maximize
3. Second constraint
x1 + x 2 ≤ 3
3. Second constraint
Variables
Variable
Standard form
Standard form
4. Third constraint
4x1 + x2 − x5 = 2
4 x1 + x 2 ≥ 2
x5 is known as surplus variable
5. Constraints for decision
variables, x1 and x2
5.
x1 ≥ 0
Constraints for decision
variables, x1 and x2
x1 ≥ 0
x 2 = x 2′ − x ′2′
and x2′ , x′2′ ≥ 0
x2 unrestricted
x1 + x2 + x4 = 3
x3 and x 4 are known as slack variables
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LP_1: Intro
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Canonical form of LP Problems
z
z
Third constraint
2. First constraint
2 x1 − 3 x 2 + x3 = 15
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General form
4.
Z ′ = − Z = 3 x1 + 5 x 2
2 x 1 − 3 x 2 ≤ 15
Transformation
D Nagesh Kumar, IISc
LP_1: Intro
Canonical form of a set of linear equations
The ‘objective function’ and all the ‘equality constraints’
Let us consider the following example of a set of linear equations
(standard form of LP problems) can be expressed in canonical
z
form.
3 x + 2 y + z = 10
Canonical form of LP problems is essential for simplex method
x − 2 y + 3z = 6
(B0)
2x + y − z = 1
(C0)
(will be discussed later)
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Canonical form of a set of linear equations will be discussed
The system of equation will be transformed through ‘Elementary
Operations’.
next.
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D Nagesh Kumar, IISc
LP_1: Intro
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LP_1: Intro
Set of equation (A0, B0 and C0) is transformed through elementary
operations (shown inside bracket in the right side)
The following operations are known as elementary operations:
1. Any equation Er can be replaced by kEr, where k is a nonzero
constant.
2. Any equation Er can be replaced by Er + kEs, where Es is
another equation of the system and k is as defined above.
Note: Transformed set of equations through elementary operations is
equivalent to the original set of equations. Thus, solution of
transformed set of equations is the solution of original set of
equations too.
D Nagesh Kumar, IISc
D Nagesh Kumar, IISc
Transformation to Canonical form:
An Example
Elementary Operations
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(A0)
LP_1: Intro
1 ⎞
⎛
⎜ A1 = A 0 ⎟
3 ⎠
⎝
3 x + 2 y + z = 10
x+
2
1
10
y+ z=
3
3
3
x − 2 y + 3z = 6
0−
8
8
8
y+ z =
3
3
3
(B1 = B0 − A1 )
2x + y − z = 1
0−
1
5
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y− z =−
3
3
3
(C1 = C0 − 2A1 )
Note that variable x is eliminated from B0 and C0 equations to
obtain B1 and C1. Equation A0 is known as pivotal equation.
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LP_1: Intro
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Transformation to Canonical form:
Example contd.
Transformation to Canonical form:
Example contd.
Following similar procedure, y is eliminated from equation A1 and C1
considering B1 as pivotal equation:
Finally, z is eliminated form equation A2 and B2 considering C2 as
pivotal equation :
2 ⎞
⎛
⎜ A 2 = A1 − B 2 ⎟
3 ⎠
⎝
x+0+ z = 4
0 + y − z = −1
3 ⎞
⎛
⎜ B 2 = − B1 ⎟
8 ⎠
⎝
0 + 0 − 2 z = −6
1 ⎞
⎛
⎜ C 2 = C1 + B 2 ⎟
3 ⎠
⎝
x + 0 + 0 =1
0+ y+0 = 2
0+0+ z = 3
(A 3 = A 2 − C3 )
(B3 = B2 + C3 )
1 ⎞
⎛
⎜ C3 = − C 2 ⎟
2 ⎠
⎝
Note: Pivotal equation is transformed first and using the transformed pivotal equation
other equations in the system are transformed.
The set of equations (A3, B3 and C3) is said to be in Canonical form which
is equivalent to the original set of equations (A0, B0 and C0)
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D Nagesh Kumar, IISc
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LP_1: Intro
Consider the following system of n equations with n variables
Operation at each step to eliminate one variable at a time,
from all equations except one, is known as pivotal
operation.
a11 x1 + a12 x 2 +
+ a1n x n = b1
( E1 )
Number of pivotal operations are same as the number of
variables in the set of equations.
a 21 x1 + a 22 x 2 +
+ a 2 n x n = b2
(E2 )
Three pivotal operations were carried out to obtain the
canonical form of set of equations in last example having
three variables.
a n1 x1 + a n 2 x 2 +
+ a nn x n = bn
(En )
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LP_1: Intro
Transformation to Canonical form:
Generalized procedure
General procedure for one pivotal operation consists of
following two steps,
1. Divide j th equation by a ji . Let us designate it as ( E′j ) , i.e., E ′j =
2. Subtract a ki times of ( E′j ) equation from
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j − 1, j + 1,
LP_1: Intro
After repeating above steps for all the variables in the system
of equations, the canonical form will be obtained as
follows:
Variable x i (i = 1 n ) is eliminated from all equations
except j th equation for which a ji is nonzero.
k th equation (k = 1, 2,
D Nagesh Kumar, IISc
Transformation to Canonical form:
Generalized procedure
Canonical form of above system of equations can be
obtained by performing n pivotal operations
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LP_1: Intro
Transformation to Canonical form:
Generalized procedure
Pivotal Operation
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D Nagesh Kumar, IISc
+ 0 x n = b1′′
( E1c )
0 x1 + 1x 2 +
+ 0 x n = b2′′
( E 2c )
0 x1 + 0 x 2 +
+ 1x n = bn′′
( E nc )
Ej
a ji
, n ) , i.e., Ek − a ki E ′j
LP_1: Intro
1x1 + 0 x 2 +
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It is obvious that solution of above set of equation such as xi = bi′′
is the solution of original set of equations also.
D Nagesh Kumar, IISc
LP_1: Intro
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Transformation to Canonical form:
More general case
Transformation to Canonical form:
More general case
Consider more general case for which the system of equations
has m equation with n variables (n ≥ m )
By performing n pivotal operations for any m variables (say, x1 , x 2 ,
called pivotal variables) the system of equations is reduced to
canonical form as follows
a11 x1 + a12 x 2 +
+ a1n x n = b1
( E1 )
1x1 + 0 x 2 +
+ 0 x m + a1′′, m +1 x m +1 +
+ a1′′n x n = b1′′
( E1c )
a 21 x1 + a 22 x 2 +
+ a 2 n x n = b2
(E2 )
0 x1 + 1x 2 +
+ 0 x m + a ′2′,m +1 x m +1 +
+ a 2′′n x n = b2′′
( E 2c )
a m1 x1 + a m 2 x 2 +
+ a mn x n = bm
(Em )
0 x1 + 0 x 2 +
+ 1x m + a m′′ ,m +1 x m +1 +
′′ x n = bm′′
+ a mn
( E mc )
It is possible to transform the set of equations to an equivalent
canonical form from which at least one solution can be easily
deduced
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D Nagesh Kumar, IISc
LP_1: Intro
x m,
Variables, x m +1 , , x n , of above set of equations is known as
nonpivotal variables or independent variables.
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LP_1: Intro
Basic variable, Nonbasic variable,
Basic solution, Basic feasible solution
One solution that can be obtained from the above set of equations is
xi = bi′′
xi = 0
for i = 1,
,m
for i = (m + 1),
,n
This solution is known as basic solution.
Pivotal variables, x1 , x 2 ,
x m, are also known as basic variables.
Nonpivotal variables, x m +1 ,
, x n , are known as nonbasic variables.
Basic solution is also known as basic feasible solution because it
satisfies all the constraints as well as non-negativity criterion for all the
variables
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D Nagesh Kumar, IISc
LP_1: Intro
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