Chapter 3 Linear Programming Methods

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Chapter 3 Linear Programming Methods
(II)
Chapter 3 Linear Programming Methods
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Initial BFS
• When the original model contains "greater
than or equal to" inequalities or equations, a
BFS is not immediately available.
• We now show how to find an initial solution
by solving an augmented linear program as
the first phase of a two-phase procedure.
• The second phase involves solving the
original problem using the BFS obtained in
the first phase as the starting point.
Chapter 3 Linear Programming Methods
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Example
Chapter 3 Linear Programming Methods
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(artificial variables:
1 and  2
)
Note: The optimal objective value in phase 1 is w*=0.
Chapter 3 Linear Programming Methods
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Chapter 3 Linear Programming Methods
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Chapter 3 Linear Programming Methods
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7
-3
Chapter 3 Linear Programming Methods
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DUAL SIMPLEX ALGORITHM
• We say that the basis for the tableau is
primal feasible if all elements of the righthand side are nonnegative. Alternatively,
when some of the elements are negative, we
say that the basis is primal infeasible.
• Up to this point, we have always been
concerned with primal feasible bases.
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• For the primal simplex algorithm, some
elements in row 0 will be negative until the
final iteration when the optimality
conditions are satisfied.
• In the event that all elements of row 0 are
nonnegative, we say that the associated
basis is dual feasible.
• Alternatively, if some of the elements of
row 0 are negative, we have a dual
infeasible basis.
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• The primal simplex method works with
primal feasible but dual infeasible
(nonoptimal) bases.
• At the final (optimal) solution, the basis is
both primal and dual feasible.
• Throughout the process, we maintain primal
feasibility and drive toward dual feasibility.
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• For the dual simplex method, until the final
iteration, each basis examined is primal
infeasible (there are some negative values
on the right-hand side) and dual feasible (all
elements in row 0 are nonnegative).
• At the final (optimal) iteration, the solution
is both primal and dual feasible. Throughout
the process, we maintain dual feasibility and
drive toward primal feasibility.
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• The dual simplex algorithm is best suited
for problems in which an initial dual
feasible solution is easily available.
• It is particularly useful for reoptimizing a
problem after a constraint has been added or
some parameters have been changed so that
the previously optimal basis is no longer
feasible.
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Max
s.t.
Example
z  -5x1 - 35x 2 - 20x 3
x1 -x 2 -x 3  -2
-x1 -3x 2
 -3
x1  0, x 2  0, x 3  0
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Chapter 3 Linear Programming Methods
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Leaving variable: the basic variable with most negative value.( X B )
r
Entering variable: min. ratio test min{
cj
arj
: arj  0}
Chapter 3 Linear Programming Methods
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Chapter 3 Linear Programming Methods
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Chapter 3 Linear Programming Methods
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Restarting After Changing the RightHand-Side Constants
Ex:
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Chapter 3 Linear Programming Methods
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• Changing the RHS constants will change
only the entries in the last column of the
tableau. In particular, if we change b 2 from
35 to 20 and b 3 from 20 to 26 in the
original problem statement, the RHS vector
in the tableau shown in Table 3.29 for the
current basis B becomes
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Why?
The inverse matrix B-1 records the operations that have been
done to the system of equations.
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46
Chapter 3 Linear Programming Methods
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Chapter 3 Linear Programming Methods
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Adding a Constraint
• Using the preceding problem, we now add
the constraint x 2  10 . The solution in the
optimal tableau, x 1 = 20 and x 2 = 5, does
not satisfy this constraint, so action must be
taken to incorporate it into the tableau.
• First we subtract a slack variable to get the
equality x 2 - x s4  10 and then multiply it by
-1 to achieve the correct form.
Chapter 3 Linear Programming Methods
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Chapter 3 Linear Programming Methods
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Chapter 3 Linear Programming Methods
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SIMPLEX METHOD USING
MATRIX NOTATION
• Decision variables:
x  (x 1 ,, x n ) T
• Objective coefficients: c  (c1 ,, c n )
• Right-hand-side constants: b  (b1 , ,b m )T
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• Structural coefficients:
 a11 a12

a 21 a 22

A


 a m1 a m2
a1n 

a 2n 


a mn 
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LP Model
Chapter 3 Linear Programming Methods
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• Suppose we now assume that the n variables are
permuted so that the basic variables are the first m
components of x. Then we can write x = (x B , x N),
where x B and x N refer to the basic and nonbasic
variables, respectively. The matrix A can also be
partitioned similarly into A = (B, N), where B is
the m × m basis matrix and N is m × (n - m). The
equation Ax = b can thus be written as
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• Multiplying through by B-1 yields
=>
=>
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• we introduce the n-dimensional row vector
of dual variables,π, and define it as   c B B -1
, so currently z =πb and xB=B-1b.
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Example
• Max z  2x 1  1.25x 2  3x 3
s.t.
2x 1 
x 2  2x 3  7
3x 1 
x2
6
x 2  6x 3  9
x 1  0, x 2  0, x 3  0
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Chapter 3 Linear Programming Methods
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The dual solution is
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• we see that the objective function value for
a given basis can be written as
z  c B B -1 b  (c N - c B B -1 N)x N
z  c B B -1 b - (c B B -1 N - c N )x N
=>
• The objective value as a function of x k
alone is z  c B B-1b - (c B B-1A k - c k )x k
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• Reduced cost of x k
c k  c B B A k - c k  A K - c K
-1
• Optimality condition: (Q=set of nonbasic
variables)
c k  0 for all k  Q
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• For a particular basis B, we have
x B  B (b - Nx N )
-1
• When we set all the nonbasic variables
equal to zero except xk , this expression
becomes x B  B-1b - B-1A k x k
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• Note
 a 1k 


 a 2k 
-1
Ak  B Ak  




 a mk 


• For the example problem, we start with the
basic solution
 x1 
 2 
 
 
-1
x B   x 3   B b  b   3/2 
x 
 0 
 5
 
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• Allowing x 2 to enter the basis, we compute
1 1/3 
   
-1
-1
A 2  B A 2  B 1  1/6 
1  0 
   
• The minimum ratio is θ= 6 for the first
equation, so , x 1 must leave the basis.
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Iteration
0
The initial and later simplex tableau
BV
Z
Z
1
Original
Variables
-c
XS
0
A
I
Z
Original
Variables
Slack
Variables
Z
1
cB B 1 A  c
cB B 1
XB
0
Iteration BV
Any
Slack
Variables
0
1
B A
B
1
RHS
0
b
RHS
1
cB B b
1
B b
Performing elementary row operations to a system of equations is
equivalent to pre-multiply the system of equations by a certain matrix.
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REVISED SIMPLEX METHOD
This method does not update and store the
entire tableau but only those data elements
needed to construct the current basis inverse
and to reproduce the matrices describing the
original problem.
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• Commercial codes do not store as an m × m
matrix but use an implicit approach such as
LUdecomposition to reconstruct it as
needed. In this approach, the B-matrix is
decomposed into an upper triangular matrix,
U, and a lower triangular matrix, L, such
that B = LU.
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• Components of Revised Simplex Algorithm:
Chapter 3 Linear Programming Methods
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Chapter 3 Linear Programming Methods
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Chapter 3 Linear Programming Methods
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Chapter 3 Linear Programming Methods
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Chapter 3 Linear Programming Methods
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Chapter 3 Linear Programming Methods
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Inverse Update formula
Suppose that xk is the entering variable and the rth basic variable is
the leaving variable.
1
1
The new basis inverse is: Bnew
 EBold
where E is an identity matrix except that its rth column is replaced by
1 
 
 2
   .  where
 
 . 
 m 
 aik
 a

i   rk
 1
 ark
if i  r
if i  r
That is, E  [e1, e2 ,..., er 1,  , er 1,..., em ]
Ex:
0 0
 3
E   1 / 2 1 0 (r  1) - -the previous example


 0
0 1
Chapter 3 Linear Programming Methods

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