MATRIX FORMULATION OF THE LPps

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MATRIX FORMULATION OF THE LPps
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In this lecture we shall look at the matrix
formulation of the LPPs. We see that the Basic
feasible solutions are got by solving the matrix
equation
where B is a m?m nonsingular submatrix of the
contraint matrix of the LPP.
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Look at the LPP (in standard form) Maximize
Subject to the constraints
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Using the notations
,
,
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We can write the LPP in matrix form as Maximize
Subject to
Here
denotes the row vector
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Basic Feasible Solutions We assume that the rank
of the matrix A is m. (This means that all the
constraints are LI. We also assume m ? n.) After
possibly rearranging the columns of A, let A
B, N where B is a mxm invertible submatrix of A
and N is the mx(n-m) submatrix formed by the
remaining columns of A. The solution
to the equations
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where
and
is called a basic
then
is
solution of the system. If
called a Basic feasible solution (BFS). We give
an example illustrating these.
Consider the LPP Maximize
Subject to
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Adding slack variables x3, x4 the LPP becomes
Maximize
Subject to
In matrix form this can be written as
Maximize
Subject to
9
,
,
where
There are 6 basic solutions of which 4 are
feasible. We give them below.
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B B-1 B-1b XT
Feasible? z
Y
8
N
-
12
Y
Maximum
Y
5
N
-
Y
0
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Problem 2 Problem Set 7.1 C Page 296
Consider the following LPP
Maximize
Subject to
In matrix form this can be written as
Maximize
Subject to
12
where
There are only 5
basic solutions of
which 3 are feasible. We give them below.
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B B-1 B-1b XT Feasible?
z
Y
56
Maximum
44
Y
Y
5
Does Not exist
-
-
N
-
N
-