full symmetric duality in continuous linear programming evgeny shindingideon weiss
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Full symmetric duality in continuous linear programming
Evgeny Shindin Gideon Weiss
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Introduction
Bellman CLP (1953), economic models
0
0
max ( ) ( )
( ) ( ) ( , ) ( ) ( )
( ) 0, 0
T
t
c s u s ds
H t u t G s t u s ds a t
u t t T
Anderson SCLP (1975), job-shop scheduling
0
0
max ( ) ( )
s.t. ( ) ( )
( )
( ) 0, [0, ].
T
u
t
c t u t dt
Gu s ds a t
Hu t b
u t t T
, 0 0max ( ) ( ) ( ) ( )
s.t. ( ) ( ) ( )
( ) 0, ( ) 0, ( ) non-decreasing [0, ].
T T
q a t d t b t q t dt
G t Hq t c t
q t T t t T
Pullan (1993) non-symmetric dual problem, strong duality, convergent (not finite) algorithm
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Introduction
Most past approaches (including Pullan) used discrete approximation. In discrete time CLP and SCLP become LP
(1) LP is large (2) the solution is only approximate (3) error bound don’t always known before problem is solved. (4) problem structure is lost
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Introduction
Weiss SCLP, symmetric dual, finite algorithm, exact solution
0
0
max ( ) ( ) ( )
s.t. ( ) ( )
( )
( ), ( ) 0, 0 .
T
t
T t c u t d x t dt
Gu s ds Fx t at
Hu t b
x t u t t T
0
0
min ( ) ( ) ( )
s.t. ( ) ( )
( )
( ), ( ) 0, 0 .
T
t
T t a p t b q t dt
G p s ds H q t ct
F p t d
p t q t t T
There exist problems which are primal and dual feasible and bounded, but bounded measurable controls do not achieve optimum and strong duality.
Infeasible
Solution in the space of measures
Solution in the
space of densities
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0
0
min ( )
s
is a non-decreasing,
.t.
0
T
t
T t b dP t
A dP s q t ct
P q t
0
0
ma
is a n
x ( )
s.t.
on-decreasing, 0
T
t
T t c dU t
AdU s x t bt
U x t
0
0
max ( )
s
is a non-decreasin
.t
g
.
T
t
T t c dU t
AdU s bt
U
0
0
is a non
min ( )
s.t.
-decreasing
T
t
T t b dP t
A dP s ct
P
M-CLP: Generalization of SCLP
M-CLP problem and symmetric dual (dual runs in reversed time).
We look for solution (U,P) in space of measures. The integrals are Lebeq-Steltjes integrals.
Weiss’ SCLP can be formulated as M-CLP. Dual of Pullan (which isn’t SCLP) can be formulated as M-CLP. Weak duality holds between M-CLP and its dual. Feasibility condition for M-CLP – ordinary LP problem.
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I-CLP problem
I-CLP problem.
T 0
0
0
0
0
0
0
max ( )
s.t. ( ) ,
( ) ,
, ( ), 0, 0,
T
t
N
N
N
T
Tc T t c u t dt
A Au s ds A bT
A Au s ds bt
u t t T
u u
u u
u
u u
Controls have impulses at 0,T and densities over (0,T). In SCLP we haven’t impulses. In dual of Pullan we haven’t impulses at 0.
I-CLP(T) feasible M-CLP(T) feasible.Optimal solution of I-CLP are also optimal for M-CLP. Strong duality asymptotically holds between I-CLP and it’s dual under Strong duality asymptotically holds between I-CLP and it’s dual under
Slater type condition.Slater type condition.
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0 0 0 00 0 1 1
0 0 0 00 0 1 1
: 0, 0; : 0, 0 : 0, 0; : 0, 0
: 0, 0; : 0, 0 : 0, 0; : 0, 0
N N N NN N
N N N NN N
j q j q j q j q
k x j x k x j x
u u u u
p p p p
J J J J
K K K K
1
1
N
N
J
K
0
0
J
K
I-CLP: Structure of optimal solution
Controls are piecewise constant, states are piecewise linear and continuous on [0,T) with possible jump at T.
max min
. . . .
: ; : : ; :
: ; :: ; :
j j k k
j jk k
c u b p
s t Au x b s t A p q b
j u j u k p k p
j q j qk x k x
P P
U PU P
J J K K
J JK K
Controls and slopes of states determined by Rates-LP. Impulses, boundary values and interval lengths determined by set of equations.
Time partition0t 1t 2t 1Nt Nt
1 1,J K 2 2,J K ,N NJ K
1 2 N
,i iJ KBases
Interval length
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I-CLP: Structure of optimal solution
Given an optimal base sequence optimal solution can be found by solving set of equations. The equations determine boundary values as well as time interval lengths.
Optimal base sequence is valid for all time horizons T1 ≤ T ≤ T2
Given an optimal time partition optimal solution can be found by solving discrete time LP. But this solution is valid only for specified time horizon T, because time partition is a function of time horizon.
Under non-degeneracy assumption optimal solution of I-CLP is unique. This solution is also unique solution for M-CLP with same time horizon.
Our final goal to find algorithm that can solve I-CLP exactly (in progress).
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Example
5 2 8 3 5 1
3 4 10 1 6 2A b c
5
Solution for 012
T
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Example
5Solution for 1
12T
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Example
Solution for 1T
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Example