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1
Linear Programming Problem
Optimization Problem
Problems which seek to maximize or minimize an
objective function of a finite number of variables
subject to certain constraints are called optimization
problems
Example
1 1 2 2 n n Maximize z = c x +c x +.............+c x
Subject to;
11 1 12 2 1n n 1
21 1 22 2 2n n 2
m1 1 m2 2 mn n m
a x +a x +.............+a x b
a x +a x +.............+a x b
.
.
.
a x +a x +.............+a x b
jx > 0 j (j = 1, 2,............,n)
Feasible Solution
Any solution of a linear programming problem that
2
satisfies all the constraints of the model is called a
feasible solution.
1 1 2 2 n n Maximize z = c x +c x +.............+c x
Subject to;
11 1 12 2 1n n 1
21 1 22 2 2n n 2
m1 1 m2 2 mn n m
a x +a x +.............+a x b
a x +a x +.............+a x b
.
.
.
a x +a x +.............+a x b
jx > 0 j (j = 1, 2,............,n)
The solution, 1 1 2 2 n n x = s , x = s ,..........,x = s will be a
feasible solution of the given problem if it does not
violate any of the constraints of the given problem.
Programming Problem
Programming problems always deal with determining
optimal allocations of limited resources to meet given
objectives. The constraints or limited resources are
3
given by linear or non-linear inequalities or equations.
The given objective may be to maximize or minimize
certain function of a finite number of variables.
Linear Programming and Linear Programming
Problem
Suppose we have given m linear inequalities or
equations in n unknown variables 1 2 nx , x ,............and,x
and we wish to find non-negative values of these
variables which will satisfy the constraints and
maximize or minimize some linear functions of these
variables (objective functions), then this procedure is
known as linear programming and the problem which is
described is known as linear programming problem.
Mathematically it can be described as, suppose we have
m linear inequalities or equations in n unknown
variables 1 2 nx , x ,............and,x of the form
n
ij j i
j=1
a x { ,=, }b (i= 1, 2,....,m) where for each constraint one
and only one of the signs ,=, holds. Now we wish to
4
find the non-negative values of jx , j = 1, 2,,n.
which will satisfy the constraints and maximize or
minimize a linear function n
j j
j=1
z = c x . Here ija , ib and
jc are known constant.
At the short-cut method mathematically the linear
programming problem can be written as
Optimize (maximize or minimize) n
j j
j=1
z = c x
Subject to
n
ij j i
j=1
a x { ,=, }b (i= 1, 2,....,m)
jx > 0 j (j = 1, 2,............,n)
Application of LPM
(i) Linear programming problem is widely applicable in
business and economic activities
(ii) It is also applicable in government, military and
industrial operations
(iii) It is also extensively used in development and
distribution of planning.
5
Objective Function
In a linear programming problem, a linear function of
the type n
j j
j=1
z = c x of the variables jx , j = 1, 2,,n.
which is to be optimized is called objective function. In
an objective function no constant term will be appeared.
i. e. we cannot write the objective function of the type
n
j j
j=1
z = c x +k
Example of Linear Programming Problem:
Suppose m types of machines 1 2 m A , A ,.........,A are producing
n products namely 1 2 nP , P ,.........P . Let (i) ij a is the hours
required of the ith machine (i = 1, 2,,m) to produce
per unit of the jth product (j=1, 2,..,n) (ii) ib (i= 1,
2,.,m) is the total available hours per week for
machine i, and (iii) jc is the per unit profit on sale of
each of the jth product.
Machines 1P 2P nP Total Available Time
1A
2A
.
.
11a 12 a 1na
21a 22a 2na
.
1b
2b
.
.
6
.
m A m1a m2a mna
.
m b
Unit Profits 1c 2c nc
Construct the LPP
Suppose jx (j = 1, 2, 3,..,n) is the no. of units of
the jth product produced per week. The objective
function is given by;
1 1 2 2 n n z = c x +c x +.............+c x
The constraints are given by;
11 1 12 2 1n n 1
21 1 22 2 2n n 2
m1 1 m2 2 mn n m
a x +a x +.............+a x b
a x +a x +.............+a x b
.
.
.
a x +a x +.............+a x b
Since the amount of production cannot be negative so,
jx 0 (j = 1, 2, 3, 4) .
The weekly profit is given by 1 1 2 2 n n z = c x +c x +.............+c x .
Now we wish to determine the values of the variables
jx 's for which all the constraints are satisfied and the
objective function will be at maximum. That is
1 1 2 2 n n Maximize z = c x +c x +.............+c x
7
Subject to
11 1 12 2 1n n 1
21 1 22 2 2n n 2
m1 1 m2 2 mn n m
a x +a x +.............+a x b
a x +a x +.............+a x b
.
.
.
a x +a x +.............+a x b
and
jx 0 (j = 1, 2, 3, 4)
Formulation of Linear Programming Problem
(i) Transportation Problem
Suppose given amount of uniform product are available
at each of a no. of origins say warehouse. We wish to
send specified amount of the products to each of a no.
of different destinations say retail stores. We are
interested in determining the minimum cost-routing
from warehouse to the retail stores.
Let use define,
m = no. of warehouses
n = no. of retail stores
8
ijx the amount of product shipped from the ith
warehouse to the jth retail store.
Since negative amounts cannot be shipped so we have
ijx 0 i, j
ia = total no. of units of the products available for
shipment at the ith (i= 1, 2,,m)warehouse.
jb = the no. of units of the product required at the jth
retail store.
Since we cannot supply more than the available amount
of the product from ith warehouse to the different retail
stores, therefore we have
11 11c :x
1a 11x 1b
21x 12 x
2a 22x 2b
.
m Origins m2 x n Destinations
.
m1x 2n x 1nx
ma mn mnc :x nb
1 1
2 2
m n
9
i1 i2 in ix +x +............+x a i= 1, 2,..,m
We must supply at each retail store with the no. of units
desired, therefore. The total amount received at any
retail store is the sum over the amounts received from
each warehouse. That is
1j 2j mj jx +x +.............+x =b ; j = 1, 2,.,n
The needs of the retail stores can be satisfied
m n
i
i=1 j=1
a b j
Let us define ij c is the per unit cost of shifting from ith
warehouse to the jth retail store, then the total cost of
shifting is given by;
m n
ij ij
i=1 j=1
z= c x
We wish to determine ijx s which minimize the cost m n
ij ij
i=1 j=1
z= c x
subject to the constraints
i1 i2 in ix +x +............+x a ;i=1, 2,....,m
10
1j 2j mj jx +x +.............+x =b ; j= 1, 2,........,n
It is a linear programming problem in mn variables with
(m+n) constraints.
(2) The Diet Problem
Suppose we have given the nutrient content of a no. of
different foods. We have also given the minimum daily
requirement for each nutrient and quantities of nutrient
contained in one of each food being considered. Since
we know the cost per ounce of food, the problem is to
determine the diet that satisfy the minimum daily
requirement of nutrient and also the minimum cost diet.
Let us define
m = the no. of nutrients
n = the no. of foods
ija = the quantity (mg) of ith nutrient per (oz) of the jth
food
ib = the minimum quantity of ith nutrient
jc = the cost per (oz) of the jth food
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jx = the quantity of jth food to be purchased
The total amount of ith nutrient contained in all the
purchased foods cannot be less than the minimum daily
requirements
Therefore we have
n
i1 1 i2 2 in n ij j i
j=1
a x +a x +............+a x = a x b
The total cost for all purchased foods is given by;
n
j j
j=1
z = c x
Now our problem is to minimize cost n
j j
j=1
z = c x subject
to the constraints
n
i1 1 i2 2 in n ij j i
j=1
a x +a x +............+a x = a x b and
jx 0
This is called the linear programming problem.
Feasible Solution
Any set of values of the variables jx which satisfies
the constraints
n
ij j i
j=1
a x { , , b , where ija and ib
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are constant is called a solution to the linear
programming problem and any solution which satisfies
the non-negative restrictions i. e. jx 0 is called a
feasible solution.
Optimal Feasible Solution
In a linear programming problem there is an infinite no.
of feasible solutions and out of all these solutions we
must find one feasible solution which optimize the
objective function
n
j j
j=1
z = c x is called optimal feasible
solution
In other words, any feasible solution which satisfies the
following conditions;
(i)
n
ij j i
j=1
a x { , , b
(ii) jx 0
(iii) optimize objective function
n
j j
j=1
z = c x , is called a
optimal feasible solution.
13
Slack and Surplus Variables
In LP problems, generally the constraints are not all
equations. Since equations are easy to handle as
compared to inequalities, a simple conversion is needed
to make the inequalities into equality. Let us consider
first, the constraints having less than or equal signs ().
Any constraint of this category can be written as
h1 1 h2 2 hn n ha x +a x +..............+a x b (1)
Let us introduce a new variable n+h x which satisfies
that n+h x 0 where n
n+h h hj j
j=1
x b a x 0 , to convert the inequalities to the
equality such that
h1 1 h2 2 hn n n+h ha x +a x +..............+a x +x b (2)
The new variable n+h x is the difference between the
amount available of resource and the amount actually
used and it is called the slack variables.
Next we consider the constraints having signs greater
than or equal (). A typical inequality in this set can
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be written as;
k1 1 k2 2 kn n ka x +a x +..............+a x b (3)
Introducing a new variable n+k x 0 , the inequality can
be written as equality which is given by;
k1 1 k2 2 kn n n+k ka x +a x +..............+a x -x b (4)
Here the variable n+k x is called the surplus variable,
because it is the difference between resources used and
the minimum amount to be produced is called the
surplus.
Therefore using algebraic method for solving a linear
programming problem, the linear programming problem
with original constraints can be transformed into a LP
problem with constraints of simultaneously linear
equation form by using slack and surplus variable
Example: Considering the LP problem
1 2Min: -x -3x
St
1 2x -2x 4
1 2-x +x 3
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1 2x , x > 0
Now introducing two new variables 3 4x and x , the
problem can be written as;
1 2 3 4Min: -x -3x +0.x +0.x
St:
1 2 3
1 2 4
1 2 3 4
x -2x x =4
-x +x x = 3
x , x , x , x > 0
Here 3x is the slack variable and 4x is the surplus
variable.
Effect of Introducing Slack and Surplus Variables
Suppose we have a linear programming problem 1P
such that
Optimize
1 1 2 2 n nZ= c x +c x +..............+c x (1)
Subject to the condition
h1 1 h2 2 hn n ha x +a x +..............+a x { , , }b (2)
Where one and only one of the signs in the bracket hold
for each constraint
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The problem is converted to another linear
programming problem 2P such that
1 1 2 2 n n n+1 mZ= c x +c x +..............+c x +0.x +............+0.x (3)
Subject to the condition
h1 1 h2 2 hn n hn+1 n+1 hm m hAx = a x +a x +..............+a x a x ........... a x b (4)
Where ij n mA= a and ja ( j = 1, 2,.,m) is the jth
column of A.
We claim that optimizing (3) subject to (4) with
jx 0 is completely equivalent to optimizing (1)
subject to (2) with jx 0
To prove this, we first note that if we have any feasible
solution to the original constraints, then our method of
introducing slack or surplus variables will yield a set of
non-negative slack or surplus variables such that
equation (4) is satisfied with all variables non-negative
consequently if we have a feasible solution to (4) with
all variables non-negative, then its first n components
will yield a feasible solution to (2) .Thus there exist one
to-one correspondence between the feasible solutions
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to the original set of constraints and the feasible
solution to the set of simultaneous linear equations.
Now if * * * *
1 2 mX = (x x ,........,x ) 0 is a feasible optimal
solution to linear programming 2P then the first n
components of *X that is * * *
1 2 n(x x ,........,x ) is an optimal
solution by annexing the slack and surplus variables
to any optimal solution to 1P we obtain an optimal
solution to 2P
Therefore, wet may conclude that if slack and surplus
variables having a zero cost are introduced to convert
the original set of constraint into a set of simultaneous
linear equations, so the resulting problem is equivalent
to the original problem.
Example: Consider the following inequalities
1 2Maximize: z= 3x +2x
Subject to constraints
1 2
2
1 2
x +x 6
x 3
x , x 0
Find basic solutions, and optimal feasible solution.
18
Solution. By introducing slack variables 3 4x and x , the
problem is put into the following standard format
1 2 3
2 4
1 2 3 4
x +x x =6
x x =3
x , x ,x ,x 0
So, the constraint matrix A is given by;
1 1 1 0A =
0 1 0 1
= 1 2 3 4(a , a , a , a ) , 6
b=3
Rank(A) = 2
Therefore, the basic solutions will be corresponding to
finding a 2 2 basis B. Following are the possible ways
of extracting B out of A
(i) 1 21 1
B=(a , a ) =0 1
, -1 1 -1B =0 1
, -1B1 -1 6 3
x =B b= =0 1 3 3
, 3n4
x 0x = =
x 0
(ii) 1 31 1
B=(a , a )= 0 0
, Since |B|=0, it is not possible to find
-1B and hence Bx
(iii) 1 41 0
B=(a , a )=0 1
; -1 1 0B =0 1
21 -1
B n
34
xx 1 0 6 6 0x = =B b= = x = =
xx 0 1 3 3 0
(iv) 2 31 1
B=(a , a )=1 0
-10 1
B =1 1
2 1-1
B n
3 4
x x0 1 6 3 0x = =B b= = x = =
x x1 1 3 3 0
(v) 2 41 0
B=(a , a )=1 1
; -1 1 0B =-1 1
; 12 -1B n34
xx 1 0 6 6 0x = =B b= = x = =
xx -1 1 3 -3 0
(vi) 3 41 0
B=(a , a )=0 1
; -1 1 0B =0 1
; 3 1-1B n4 2
x x1 0 6 6 0x = =B b= = x = =
x x0 1 3 3 0
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Hence we have the following five basic solutions
1
3
3x =
0
0
; 2
6
0x =
0
3
; 3
0
3x =
3
0
; 4
0
6x =
0
-3
; 5
0
0x =
6
3
Of which except 4x are BFS because it violates
non-negativity restrictions. The BFS belong to a four
dimensional space. These basic feasible solutions are
projected in the 1 2(x , x ) space gives rise to the following
four points.
3 6 0 0 0, , ,
3 0 3 6 0
From the graphical representation the extreme points
are (0, 0), (0, 3), (3, 3) and (6,0) which are the same as
the BFSs. Therefore the extreme points are precisely the
BFS. The no. of BFS is 4 less than 6.
The optimal BFS is 60
Corner Point Feasible Solution
A feasible solution which does not lie on the line
segment, connection any other two feasible solution is
called a corner point feasible solution.
20
The corner point feasible solution is 33
Properties of Corner Point Feasible Solution
(i) If there is an exactly one optimal solution of the
linear programming problem, then it is a corner point
feasible solution.
(ii) If there are more than two optimal solutions of the
given problem, then at least two of them are adjacent
corner points.
(iii) In a linear programming problem there are a finite
number of corner points
(iv) If a corner point feasible solution is better than its
adjacent corner point solution, then it is better than all
other feasible solutions.
Methods for Solving Linear Programming Problems
(1) Graphical Method
(2) Algebraic Method
(3) Simplex Method
Graphical Method
The graphical method to solve a linear programming
21
problem involves two basic steps
(1) At the first step we have to determine the feasible
solution space.
We represent the values of the variable 1x to the X
axis and the their corresponding values of the variable
2x to the Y axis. Any point lying in the first quadrant
satisfies 1x > 0 and 2x 0 . The easiest way of
accounting for the remaining constraints for
optimization objective function is to replace inequalities
with equations and then plot the resulting straight lines
Next we consider the effect of the inequality. All the
inequality does is to divide the 1 2(x , x ) -plane into two
spaces that occur on both sides of the plotted line: one
side satisfies the inequality and the other one does not.
Any point lying on or below the line satisfies the
inequality. A procedure to determine the feasible side is
to use the origin (0, 0) as a reference point.
Step 2: At the second step we have to determine the
optimal solution.
22
Problem: Find the non-negative value of the
variables 1 2x and x which satisfies the constraints
1 23x +5x 15
1 25x +2x 10
And which maximize the objective function
1 2z = 5x +3x
Solution: We introduce an 1 2x x co-ordinate system.
Any point lying in the first quadrant has 1 2x ,x 0 .
Now we show the straight lines 1 23x +5x =15 and
1 25x +2x =10 on the graph. Any point lying below
the line 1 23x +5x =15 satisfies the 1 23x +5x 15 .
Similarly any point lying below the line 1 25x +2x =10
satisfies the constraint
2x
C(0,5)
1 25x +2x =10
A(3.0)
F(1.053, 2.3684)
23
1 2z = 5x +3x 1 2z = 5x +3x
So, the region FDOA containing the set of points
satisfying both the constraints and the non negative
restriction. So, the points in this region are the feasible
solution. Now we wish to find the line with the largest
value of 1 2z = 5x +3x which has at least one point in
common with the region of feasible solution. The line is
drawn in the graph above. It shows that the value of 1x
and 2x at the point A are the required solution.
Here 1x =1.053 and 2x 2.368 approximate.
Now from the objective function we get the maximum
value of z which is given by z = 5 1.053+3 2.368=12.37
24
Existence of Extreme Basic Feasible Solution:
Reduction of any feasible solution to a basic feasible
solution
Let us consider a linear programming problem with m
linear equations in n unknowns such that
AX = b
X 0
Which has at least one basic feasible solution without
loss of generality suppose that Rank(A) = m and let
1 2 nX=(x , x ,......,x )be as feasible solution. Further suppose
that 1 2 px , x ,......,x >0 and that p+1 P+2 nx , x ,......,x =0 . And let
1 2 pa , a ,......,a be the respective columns of A
corresponding to the variables 1 2 px , x ,......,x . If
1 2 pa , a ,......,a are linearly independent then X is a basic
feasible solution. in such case p m . If p=m from the
theory of system of linear equation, the solution is
non-degeneracy basic feasible solution.
If p
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zero.
If 1 2 pa , a ,......,a are dependent then there exist scalars
1 2, ,......, p with at least one positive j such that
1
0p
j j
j
a
Considering the following point X with
j 0
j
x ; 1,2,....,x =
0; 1, 2,.....
j j p
j p p n
where j k
oj=1,2,....,p
k
x x=Minimum ; 0 = >0j
j
If j 0 , then jx >0 , since both jx and 0 are positive.
If 0j , then by the definition of 0 we have
j
o j 0
xx j
j
. Thus jx >0
Furthermore
kk k 0 k
k
xx = x - =x - =0k k
. Hence x has at most (p-1)
positive components.
Also,
26
n
j j
j=1
n
j j 0
j=1
n n
j j 0 j
j=1 j=1
Ax = a x
= a (x )
= a x a
= b
j
j
Thus we have a constructed feasible solution x since
Ax =b , x 0 with at most (p-1) positive
components. If the columns of A corresponding to these
positive components are linearly independent then x
is basic feasible solution. Otherwise the process is
repeated. Eventually a basic feasible solution (BFS) will
be obtained.
Example: Consider the following inequalities
1 2Maximize: z= 3x +2x
Subject to constraints
1 2
2
1 2
x +x 6
x 3
x , x 0
Find basic solution, BFS and extreme points.
Solution. By introducing slack variables 3 4x and x , the
problem is put into the following standard format
27
1 2 3
2 4
1 2 3 4
x +x x =6
x x =3
x , x ,x ,x 0
So, the constraint matrix A is given by;
1 1 1 0A =
0 1 0 1
= 1 2 3 4(a , a , a , a ) , 6
b=3
Rank(A) = 2
Therefore, the basic solutions corresponding to finding
a 2 2 basis B. Following are the possible ways of
extracting B out of A
(i) 1 21 1
B=(a , a ) =0 1
, -1 1 -1B =0 1
, -1B1 -1 6 3
x =B b= =0 1 3 3
, 3n4
x 0x = =
x 0
(ii) 1 31 1
B=(a , a )= 0 0
, Since |B|=0, it is not possible to find
-1B and hence Bx
(iii) 1 41 0
B=(a , a )=0 1
; -1 1 0B =0 1
21 -1
B n
34
xx 1 0 6 6 0x = =B b= = x = =
xx 0 1 3 3 0
(iv) 2 31 1
B=(a , a )=1 0
-10 1
B =1 1
2 1-1
B n
3 4
x x0 1 6 3 0x = =B b= = x = =
x x1 1 3 3 0
(v) 2 41 0
B=(a , a )=1 1
; -1 1 0B =-1 1
; 12 -1B n34
xx 1 0 6 6 0x = =B b= = x = =
xx -1 1 3 -3 0
(vi) 3 41 0
B=(a , a )=0 1
; -1 1 0B =0 1
; 3 1-1B n4 2
x x1 0 6 6 0x = =B b= = x = =
x x0 1 3 3 0
Hence we have the following five basic solutions
28
1
3
3x =
0
0
; 2
6
0x =
0
3
; 3
0
3x =
3
0
; 4
0
6x =
0
-3
; 5
0
0x =
6
3
Of which except 4x are BFS because it violates
non-negativity restrictions. The BFS belong to a four
dimensional space. These basic feasible solutions are
projected in the 1 2(x , x ) space gives rise to the following
four points.
3 6 0 0 0, , ,
3 0 3 6 0
From the graphical representation the extreme points
are (0, 0), (0, 3), (3, 3) and (6,0) which are the same as
the BFSs. Therefore the extreme points are precisely the
BFS. The no. of BFS is 4 less than 6.
General Mathematical Formulation for Linear
Programming
Let us define the objective function which to be
optimized
1 1 2 2 n n z = c x +c x +...................+c x
We have to find the values of the decision variables
1 2 nx , x ,.........,x on the basis of the following m
29
constraints;
11 1 12 2 1n n 1
21 1 22 2 2n n 2
m1 1 m2 2 mn n m
a x +a x +.........+a x ( ,=, )b
a x +a x +.........+a x ( ,=, )b
a x +a x +.........+a x ( ,=, )b
and
jx 0; j = 1, 2,.......,n
The above formulation can be written as the following
compact form by using the summation sign;
Optimize (maximize or minimize)
n
j j
j=1
z = c x
Subject to the conditions;
n
ij j i
j=1
a x ( ,=, )b ;i=1, 2,.......,m
and
jx 0; j = 1, 2,.......,n
The constants jc ; j =1, 2,......,n are called the cost
coefficients; the constants ib ; i =1, 2,.......,m are called
stipulations and the constants ij a ; i =1, 2,.....,m; j=1,2,.....,n
are called structural coefficients. In matrix notation the
30
above equations can be written as;
Optimize z = CX
Subject to the conditions
AX( ,=, )B
where
1 2 n 1 nC= c c ... ... c ;
1
2
n n 1
x
x
.
X= .
.
.
x
;
11 12 1n
21 22 2n
m1 m2 mn m n
a a ...... a
a a ...... a
A= . . . .
. . . .
a a ...... a
;
1
2
m m n
b
b
.
B= .
.
.
b
Where, A is called the coefficient matrix, X is called the
decision vector, B is called the requirement vector and
C is called the cost vector of linear programming
problem
The Standard Form of LP Problem
The use of basic solutions to solve the general LP
models requires putting the problem in standard form.
The followings are the characteristics of the standard
form
(i) All the constraints are expressed in the form of
31
equations except the non-negative restrictions on the
decision variables which remain inequalities
(ii) The right hand side of each constraint equation is
non-negative
(iii) All the decision variables are non-negative
(iv) The objective function may be of the maximization
or the minimization type
Conversion of Inequalities into Equations:
The inequality constraint of the type ,( ) can be
converted to an equation by adding or subtracting a
variable from the left-hand sides of such constraints.
These new variables are called the slack variables or
simply slacks. They are added if the constraints are of
the types and subtracted if the constraints are of the
types. Since in the cases of type the subtracted
variables represent the surplus of the left-hand side over
right-hand side, it is commonly known as the surplus
variables and is in fact a negative slack.
For example
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1 2 1x +x b
Is equivalent to
1 2 1 1x +x s = b
If 1 2 2x +x b
Is equivalent to
1 2 1 2x +x =bs
The general LP problem that discussed above can be
expressed as the following standard form;
n
j j
j=1
z = c x
Subject to the conditions
n
ij j i i
j=1
a x s =b ;i=1, 2,.......,m
jx 0; j = 1, 2,.......,n
and
is 0; i = 1, 2,.....,m
In the matrix notation, the general LP problem can be
written as the following standard form;
Optimize z = CX
Subject to the conditions
AX S = B
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X 0
S 0
Example: Express the following LP problem in a
standard form;
Maximize 1 2z= 3x +2x
Subject to the conditions;
1 2
1 2
1 2
2x +x 2
3x +4x 12
x ,x 0
Solution: Introducing slack and surplus variables, the
problem can be expressed as the standard form and is
given below;
Maximize 1 2z= 3x +2x
Subject to the conditions;
1 2 1
1 2 2
1 2 1 2
2x +x =2
3x +4x =12
x ,x , , 0
s
s
s s
Conversion of Unrestricted Variable into
Non-negative Variables
An unrestricted variable j x can be expressed in terms
of two non-negative variables by using substitution
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such that + -
j j jx = x -x ; + -
j jx ,x 0
For example, if jx = -10 , then + -
j jx =0, and x = 10 . If jx = 10 ,
then + -
j jx =10, and x = 0 .
The substitution is effected in all constraints and in the
objective function. After solving the problem in terms
of +
jx and -
jx , the value of the original variable jx is
then determined through back substitution.
Example: Express the following linear programming
problem in the standard form;
Maximize, 1 2 3z= 3x +2x +5x
Subject to
1 2
1 2 3
1 3
2x -3x 3
x +2x 3x 5
3x +2x 2
1 2x , x 0
Solution: Here 1x and 2x are restricted to be
non-negative while 3x is unrestricted. Let us express
as,+ -
3 3 3x = x -x where, + -
3 3x 0 and x 0 . Now introducing
slack and surplus variable the problem can be written as
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the standard form which is given by;
Maximize, + -
1 2 3 3z= 3x +2x +5(x -x )
Subject to the conditions;
1 2 1
+ -
1 2 3 3 2
+ -
1 3 3 3
+ -
1 2 3 3 1 2 3
2x -3x 3
x +2x 3x -3x =5
3x +2x -2x =2
x ,x ,x , x , , , 0
s
s
s
s s s
Conversion of Maximization to Minimization:
The maximization of a function 1 2 nf(x , x ,.....,x ) is
equivalent to the minimization of 1 2 n-f(x , x ,.....,x ) in the
sense that both problems yield the same optimal values
of 1 2x ,x ,......, and nx