lies in identifying the set of always-active constraints of Ax b by solving one linear program generated from the data (A,b). Key Words: Linear Inequalities, Active ...
Identifying the Set of Always-Active Constraints in a System of Linear Inequalities by a Single Linear Program Robert M. Freund Robin Roundy Michael J. Todd Sloan W.P. No.
1674-85 (Rev)
October,
1985
Abstract:
Given a system of m linear
the form Ax < b, a constraint
inequalities in n unknowns of
index i is called always-active if Ax
(P2)
1
We have the following straightforward result: Proposition 1. and finite,
If the system
(*) is
feasible, then (P2)
and for any optimal solution
always-active constraint indices
(x*,y*,a*) to
is the set
ily*
is feasible
(P2), the set of
= 0).
Furthermore,
x*
is an element of the relative interior of (xeRnlAx < b}. the system
(*) is not feasible, then
Another linear program that
(P2)
is
infeasible.
identifies all of
If
[X]
the always-active
constraints is obtained through consideration of the following problem. maximize t x,t subject Obviously,
to:
(P3).
Ax+et < b
(P3) is always feasible,
solution to constraints.
(*),
When t*=O,
it can easily be shown that if x* and X* are
and its dual minimize X
subject
if and only if there is no
> 0 if and only if there are no always-active
and t
optimal values of (P3)
t* < 0
to:
XTb
ATX = 0 eT
=
(D3), namely
(D3),
1
>O
3
then if X i
> 0,
the ith constraint
is always-active, and if Aix*
than the ith-constraint is not always-active. we have a strictly complementary (D3)
(i.e. X i
> 0 or bi -
pair of
for each constraint whether or not it
If we could ensure that
optimal
Aix*> 0 for each i),
< bi,
solutions to
(P3) and
then we could identify
is always-active.
This is
accomplished by solving the following linear program: maximize e x,X,t,e subject to: (1)
Ax + et
< b
(2)
ATX
= 0
(3)
eTX
=
+ t
= 0
(4)
-bTX
(5)
X-Ax-et-ee
(6)
X
(6) represent dual
> -b > 0
feasibility, contraint
duality, and a positive value of strict complementarity of X and
(*) has a solution.
Proposition 2. constraints.
If If
set
(4) represents strong
in constraints (b-Ax) when t=0.
(5) represents Tucker's strict
We have:
infeasible, then
(*) has no always-active
(x*,X*,t*,e*),
we have:
= 0, the set of always-active constraint indices (i*kXi
{xERnAx
(3) and
(P4) is feasible, then it is finite, and for any
optimal solution If t
(P4) is
(2),
ensures that a positive value of e will
complementarity theorem [4]
(i)
(P4)
(1) represent primal feasibility, constraints
Constraints
exist if
1
> 0) and x*
lies
in the relative interior of
< b).
4
is the
III
t* > O,
(ii)
If
(iii)
If t* < 0, Instead of
there are no always-active constraints. the system
(*) has no solution.
the system (*), Ax=b,
[X]
we could have worked with the system
x> O,
(**)
whereby the always-active constraints would correspond to null variables of to
(**),
interest
(**),
i.e.
see Luenberger in the recent
variables xj [2].
such that xj=O
Determining null
in every solution
variables is also of
linear programming algorithm of Karmarkar
5
[1].
References
[1]
N. Karmarkar, "A new polynomial-time algorithm for linear programming," Combinatorica 4 (1984) pp. 373-395.
[2]
D.G. Luenberger, Introduction to linear and nonlinear programming (Addison-Wesley, Reading, Massachusetts, 1973).
[3]
J.B.
[4]
A.W. Tucker, "Dual systems of homogeneous linear relations," in: H.W. Kuhn and A.W. Tucker, eds., Linear Inequalities and Related Systems, Annals of Mathematics Study No. 38 (Princeton University Press, Princeton, New Jersey, 1956) pp. 3-18.
Orlin,
private communication.
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