LP Standard Form

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LP Standard Form

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LP Standard Form

• A linear program is in standard form if the
  following are all true
• Objective expressed as a maximization.
• All constraints expressed as equalities.
• Constraint right-hand side constants are
  non-negative.
• All variables are restricted to be non-negative.

not equality
not equality
-ve rhs
x3,x4 may be negative
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Converting Inequalities into Equalities
s1 is called a slack variable, which measures
the amount of unused resource. Note that s1
5 - x1 - 2x2 - x3 x4.
To convert a constraint to an equality, add a
non-negative slack variable to the lhs.
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Converting -ve rhs constraints

• Consider the inequality -2x1 - 4x2 x3 x4 -1

Consider the inequality -2x1 - 4x2 x3 x4
-1 Step 1. Convert the negative RHS constant
2x1 4x2 - x3 - x4 1
Consider the inequality -2x1 - 4x2 x3 x4
-1 Step 1. Convert the negative RHS constant
2x1 4x2 - x3 - x4 1 Step 2.
Convert inequality to an equality
2x1 4x2 - x3 - x4 s2 1
s2 0
Consider the inequality -2x1 - 4x2 x3 x4
-1 Step 1. Convert the negative RHS constant
2x1 4x2 - x3 - x4 1 Step 2.
Convert inequality to an equality
2x1 4x2 - x3 - x4 s2 1
s2 0 The new variable s2 is called
a surplus variable
To convert a constraint to an equality,
subtract a non-negative surplus variable from the
lhs.
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More Transformations
Has the same optimum solution(s) as
Maximize -3W 2P Subject to constraints
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The Last Transformations (for now)
Transforming variables that can take on negative
values maximize 3x1 4x2 5x3 subject
to 2x1 - 5x2 2x3 7
... other constraints x1 0, x2 is
unrestricted in sign, x3 0

• Transforming x1 replace x1 by y1 -x1.

maximize -3y1 4x2 5x3 subject to
-2y1 - 5x2 2x3 7 ...
other modified constraints
y1 0, x2 is unrestricted in sign, x3 0
One can recover x1 from y1.
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Transforming variables that can take on negative
values.
maximize -3y1 4x2 5x3
-2y1 - 5x2 2x3 7 y1 0,
x2 is unrestricted in sign, x3 0
Note y1 1, x2 -1, x3 2 is feasible.
Transforming x2 replace x2 by x2 x2 - x2-
x2 0, x2- 0
maximize -3y1 4x2 - 4x2- 5x3
-2y1 - 5x2 5x2- 2x3 7
y1 0, x2 0, x2- 0, x3 0
One can recover x2 from x2and x2-.
Note y1 1, x2 0, x2- 1, x3 2 is
feasible.
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For Practice

• Exercise transform the following into standard
  form
• Minimize x1 3x2
• Subject to 2x1 5x2 12
• x1 x2 1
• x1 0

Maximize -x1 - 3(x2 - x2- ) Subject to
2x1 5(x2 - x2- ) s1 12
x1 x2 - x2- - s2 1
x1 0, x2 0, x2- 0, s1 0, s2 0

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LP Standard Form

• Suppose an LP with m constraints and n variables1
  has been converted into standard form. The form
  of such an LP is

1includes the original decision variables other
variables added to bring problem to standard form
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LP Standard Form
If we define the (cost or profit) coefficient
vector
cT (c1, c2, ... , cn )
and the constraint coefficient matrix A and
requirement vector b
then we can (compactly) write our standard form
LP as
maximize z cTx subject to Ax b
x 0
where b 0.
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Solution of Linear Equations

• If the m equations Axb in the n variables x are
  independent then mn and
• mn ? solution is unique (Is it feasible?)
• mltn ? n-m degrees of freedom and an infinity of
  solutions to Axb
• ?
• assign any values to any (n-m) variables and
  solve Axb for other m variables
• Basic solutions solutions obtained by setting
  n-m variables (called non-basic) to zero and
  solving Axb for the remaining m variables
  (called basic)

There are nCm n!/m!(n-m)! basic solutions to
Axb. For n20 and m12 there are 126,000 basic
solutions!
Basic solutions that are feasible are called
basic feasible solutions
Basic feasible solutions are in 1-to-1
correspondence with the extreme points of the
feasible region
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The Simplex Method

• The Simplex Method examines the basic feasible
  solutions in its pursuit for the optimal
  solution.
• On each iteration it moves from one BFS to an
  adjacent BFS with a non-inferior objective value
• Two BFSs (alternatively, two extreme points) are
  adjacent if they share all but one basic
  variables
• ?
• Each Simplex iteration results in exchanging the
  roles of one basic variable and one non-basic
  variable