coonnttiinnuuoouuss lliinneeaarr …tthhee aassssoocciiaatteedd llpp our solution of sclp will...
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CC oo nn tt ii nn uu oo uu ss LL ii nn ee aa rr PP rr oo gg rr aa mm mm ii nn gg ..SS ee pp aa rr aa tt ee dd CC oo nn tt ii nn uu oo uu ss LL ii nn ee aa rr PP rr oo gg rr aa mm mm ii nn gg
Bellman (1953)
max ( ) ( )
( ) ( ) ( , ) ( ) ( )
( ) ,
′
+ ≤
≥ < <
∫
∫
c t u t dt
H t u t G s t u s ds a t
u t t T
T
t
0
0
0 0
CLP
(Dantzig, Tyndall, Grinold, Perold, Anstreicher 60's-80's)
Anderson (1978)
max ( ) ( )
( ) ( ) ( )
( ) ( ) ( )
, , ,
′
+ =
+ =
≥ < <
∫
∫
c t u t dt
Gu s ds x t a t
H u t z t b t
x u z t T
T
t
0
0
0 0
SCLP
(Anderson, Nash, Philpott, Pullan 80's-90's).
Duality, Structure, Solution by Discretization.
In 2000, W. proposed a simplex algorithm to solve:
max ( ) ( )
( ) ( )
( ) ( )
, , ,
T t c u t dt
Gu s ds x t a t
H u t z t b
x u z t T
T
t
− ′
+ = +
+ =
≥ < <
∫
∫
0
0
0 0
α SCLPwith linear data
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PP uu ll ll aa nn ’’ ss DD uu aa ll FF oo rr mm uu ll aa tt ii oo nn ::
Primal Problem:
max ( ) ( )
( ) ( ) ( )
( ) ( ) ( )
, , ,
c T t u t dt
Gu s ds x t a t
H u t z t b t
u x z t T
T
t
− ′
+ =
+ =
≥ < <
∫
∫
0
0
0 0
Pullan s SCLP©
Dual Problem
min ( ) ( ) ( ) ( )
( ) ( ) ( )
( ) , , , ,
a T t d t b T t r t dt
G t H r t q t ct
r q t T
T T− ′ + − ′
′ + ′ − =
= ↑ ≥ < <
∫ ∫π
π
π π
0 0
0 0 0 0
Pullan s SCLP© *
Theorem (Pullan)Assume H u t b t u t( ) ( ), ( )≤ ≥ 0 has bounded feasible region.• For a b c, , piecewise analytic, there is a solution which is piecewise analytic,• For a c, piecewise polynomial of degree n +1,b piecewise polynomial of degree nthere is a solution with u t( ) piecewise polynomial of degree n.• For a c, piecewise linear, b piecewise constant, there is a solution with u t( ) piecewise constant.• Number of breakpoints finite bounded (not polynomial)•• Strong duality holds
Pullan’s Algorithm: Sequence of discretizations, converging (in infinitenumber of steps) to the optimal solution.
Note: Separated is very different from General:in the one x t u s ds( ) ( )= ∫ , direct integrationin the other u t u s ds( ) ( )= +∫ K differential equations
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SS CC LL PP ww ii tt hh SS yy mm mm ee tt rr ii cc PP rr ii mm aa ll // DD uu aa ll
max ( ) ( ) ( ) ( )
( ) ( ) ( ) ( )
( ) ( )
, ,
c T t u t dt d T t y t dt
Gu s ds Fy t x t a t
H u t b t
x u t T
SCLP
T T
t
− ′ + − ′
+ + =
=
≥ < <
∫ ∫
∫
0 0
0
0 0
min ( ) ( ) ( ) ( )
( ) ( ) ( ) ( )
( ) ( )
, ,
*
a T t p t dt b T t r t dt
G p s ds H r t q t c t
F p t d t
p q t T
SCLP
T T
t
− ′ + − ′
′ + ′ − =
′ =
≥ < <
∫ ∫
∫
0 0
0
0 0
Here: u p, are primal and dual controls (≥ 0),x q, are primal and dual states (≥ 0),y r, are primal and dual supplementary (unrestricted)
Dimensions: G K J H I J F K L× × ×, ,
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WW ee aa kk DD uu aa ll ii tt yy aa nn dd CC oo mm pp ll ee mm ee nn tt aa rr yy SS ll aa cc kk nn ee ss ss
Dual Objective = a T t p t dt b T t r t dt
u s G ds y T t F p t u T t H r t dt
u T t G p s ds H r t dt y T
T T
TT t
Tt
( ) ( ) ( ) ( )
{ ( ) ( ) } ( ) ( ) ( )
( ) { ( ) ( )} (
− ′ + − ′
≥ ′ ′ + − ′ ′ + − ′ ′
= − ′ ′ + ′ +
∫ ∫
∫ ∫
∫ ∫
−
0 0
00
00 −− ′ ′
≥ − ′ + − ′ =
∫
∫ ∫
t F p t dt
u T t c t dt y T t d t dt
T
T T
) ( )
( ) ( ) ( ) ( )
0
0 0Primal Objective
Equality will hold if and only if
0 00 0
T Tx T t p t dt u T t q t dt∫ ∫− ′ = − ′ =( ) ( ) , ( ) ( ) .
Complementary Slackness: almost everywherex t p T t
u t q T t
( ) ( )
( ) ( )
> ⇒ − =
> ⇒ − =
0 0
0 0Corollary: The following are equivalent(a) u x y p q r, , , , are complementary slack feasible primaland dual solutions(b) they are optimal and have the same objective value (noduality gap)
(Strong duality no duality gap - exists in LP but not in other
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problems, unless additional conditions are imposed).
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BB oo uu nn dd aa rr yy VV aa ll uu ee ss
For infinitesimal values of T we solve:max ( ) ( ) ( ) ( )
( ) ( ) ( )
( ) ( )
, ,
c T t u t d T t y t
Fy t x t a t
H u t b t
x u t T
− ′ + − ′
+ =
=
≥ < <0 0
min ( ) ( ) ( ) ( )
( ) ( ) ( )
( ) ( )
, ,
a T t p t b T t r t
H r t q t c t
F p t d t
p q t T
− ′ + − ′
′ − =
′ =
≥ < <0 0
These separate into 2 sets of primal and dual problems:max ( ) ( )
( ) ( ) ( )
( ) ,
d T y
Fy x a
x
boundary LP
′
+ =
≥
0
0 0 0
0 0
min ( ) ( )
( ) ( ) ( )
( ) ,
*b T r
H r q c
q
boundary LP
′
′ − =
≥
0
0 0 0
0 0
Assumption (Feasibility and Boundedness):The boundary problems are feasible and bounded.
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⇒ SCLP is feasible and bounded for some range of T
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TT hh ee AA ss ss oo cc ii aa tt ee dd LL PP
Our solution of SCLP will evolve through solution of thefollowing LP / LP*, under varying sign restrictions.
max «( ) ( ) ( ) «( )
( ) «( ) ( ) «( )
( ) ( )
( ) , «( ) " ",
c T t u t d T t y t
G u t Fy t t a t
H u t b t
u t y t U t T
LP
− ′ + − ′
+ + =
=
≥ < <
ξ
0 0
min «( ) ( ) ( ) «( )
( ) «( ) ( ) «( )
( ) ( )
, «" ",
*
a t p T t b t r T t
G p T t H r T t T t c T t
F p T t d T t
p r U t T
LP
′ − + ′ −
′ − + ′ − + − = −
′ − = −
≥ < <
θ
0 0
Note: the time indices on the variables are only labels,we label the variables to indicate their role:
ξ θ= =«, «, «, «x y q r
Assumption (Non-Degeneracy):The r.h.s. of LP / LP* is (a.e. t) in general position to
column spaces ofG F I
H 0 0
,
′ ′
′
G H I
F 0 0
⇒ SCLP is non-degenerate: Solution is unique and at an
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extreme point in the function space of the solutions.
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TT hh ee SS tt rr uu cc tt uu rr ee TT hh ee oo rr ee mm
Consider any functionsu t x t y t p t q t r t t T( ), ( ), ( ), ( ), ( ), ( ), 0 < < which satisfy
• u p x y q r, , , , ,≥ 0 are absolutely continuous.
Hence x y q r, , , have derivatives a.e.
• x y( ), ( )0 0 solve the boundary-LP,q r( ), ( )0 0 solve the boundary-LP*.
For x y, these are values at the left boundary,q r, run in reversed time so this is at the right boundary.
• At all times t , the rate functions (derivatives)u t t y t p T t T t r T t t T( ), ( ), «( ), ( ), ( ), «( ),ξ θ− − − < <0are complementary slack feasible solutions for LP / LP*.
Thus they are optimal solutions for LP / LP* under somesign configuration of ξ θ, . By non-degeneracy they are basicand unique solutions.
• At all times t , if x t tk k( ) , ( )> 0 ξ is basic, ifq T t T tj j( ) , ( )− > −0 θ is basic.
• x t q t t T( ) ( ) ,≥ ≥ < <0 0 0 .
Theorem: A solution to SCLP / SCLP* is optimal if (and
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only if for linear data) it has these properties.
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RR ee ss tt rr ii cc tt ii oo nn tt oo tt hh ee LL ii nn ee aa rr DD aa tt aa CC aa ss eeProblem:max ( ( ) ) ( ) ( )
( ) ( ) ( )
( )
, ,
( )
′ + − ′ + ′
+ + = +
=
≥ < <
∫ ∫
∫
γ
α
T t c u t dt d y t dt
Gu s ds Fy t x t at
H u t b
x u t T
SCLP
linear data
T T
t
0 0
0
0 0
min ( ( ) ) ( ) ( )
( ) ( ) ( )
( )
, ,
( )
*
α
γ
+ − ′ + ′
′ + ′ − = +
′ =
≥ < <
∫ ∫
∫
T t a p t dt b r t dt
G p s ds H r t q t ct
F p t d
p q t T
SCLP
linear data
T T
t
0 0
0
0 0Boundary:
max
,
min
,
′
+ =
≥
′
′ − =
≥
d y
Fy x
x
b r
H r q
q
N
N N
N
0
0 0
0 0 0
α γ
Associated LP
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max «
«
, « " ",
min «
«
, «" "
′ + ′
+ + =
=
≥
′ + ′
′ + ′ − =
′ =
≥
c u d y
G u Fy at
H u b
u y U
a p b r
G p H r c
F p d
p r U
ξ θ
0 0
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TT hh ee AA ll gg oo rr ii tt hh mm
The algorithm constructs a solution for SCLP / SCLP* forthe whole range of time horizons, 0 < < ∞T .
Corollary 1: The solution for any time horizon T with theexception of a finite list: 0 0 1= < < < ∞T T T R( ) ( ) ( )L , is characterized by a sequence of adjacent bases of LP, B BN1, ,K .
For a particular T there are time breakpoints 0 0 1= < < =t t t TNL such that for t t tn n− < <1 ,
u y p rn n n n n n, , « , , , «ξ θ are basic Bn.
In the single pivot B Bn n→ +1 let v leave the basisif v k= ξ then we have equation x tk n( ) = 0.if v uj= then we have equation q T tj n( )− = 0.
Corollary 2: The solution of the equations x tk n( ) = 0,q T tj n( )− = 0, together with:
τ τ τ1 1+ + = = − −L N n n nT t t( )determines the solution for a whole range of time horizons,T T Tr r( ) ( )− < < < ∞1 .
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AA ll gg oo rr ii tt hh mm ii cc SS tt rr uu cc tt uu rr ee oo ff SS oo ll uu tt ii oo nn ::
For time horizon T , Break points: 0 0 1= < < < =t t t TNLτn n nt t= − −1
N −1 Equations:
Adjacent B B vn n n→ =+1 : leaves:
v x t x
v u q T t q
n k k n k km
m
n
m
n j j n jN
jm
m n
N
m
= ⇒ = + =
= ⇒ − = + =
=
= +
∑
∑
ξ ξ τ
θ τ
( ) ,
( ) .
0
1
1
0
0
and equation N : τ τ1 + + =L N T
Slack Inequalities, at local minima of x qk j, :
ξ ξ ξ τ
θ θ θ τ
kn
kn
k n k km
m
n
m
jn
jn
j n jN
jm
m n
N
m
n N x t x
n q T t q
< > = ⇒ = + ≥
= > < ⇒ − = + ≥
+
=
+
= +
∑
∑
0 0 0
0 0 0 0
1 0
1
1
1
, ( ) ,
, ( ) .
or
or
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Algorithm:
Step 1: Solution for T T T( ) ( )0 1< < is given by the solution of boundary-LP / boundary-LP*.
Step r: Extend the solution to next time horizon range.
1 1 0
0
L
A
B I
T
d
g
r
=
+
τ
σ
( ) ∆
Step R+1 Stop when next ∆ can be ∞,or when solution infeasible.
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AA ll gg oo rr ii tt hh mm CC aa ss ee ii ii ii
Buffer empties at T
T<T (l)
T=T (l)
T>T (l)
ξ leaves
u enters
x 1
x 2
1
2
x 1
x 2
x 2
x 1
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AA ll gg oo rr ii tt hh mm CC aa ss ee ii ii
Interval shrinks at tn
x 1 q 3
B'→C ξ leaves
C→B'' u leaves
1
3
T<T (l)
x 1 q 3
B'→B'' ξ u leave1 3
T=T (l)
x 1 q 3
B'→D u leaves
D→B'' ξ leaves
3
1
T>T (l)
B'
B'
B'
B''
B''
B''
C
D
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AA ll gg oo rr ii tt hh mm CC aa ss ee ii ii ,, SS uu bb pp rr oo bb ll ee mm
Interval shrinks at tn : subproblem for additional intervals
x 1 q 3
B'→C ξ leaves
C→B'' u leaves
1
3
T<T (l)
x 1 q 3
B'→B'' ξ u leave1 3
T=T (l)
B'
B'
B''
B''
C
x 1 q 3
→D u leaves
→B'' ξ leaves
3
1
T>T (l)
B' B''D
x 1 q 3
B'→B'' ξ u leave1 3
T=T (l)
B' B''
D EE1 2
E1 E2
E1
E2
subproblem
Subproblem solved by recursive call to the algorithm, with smaller,modified problem.
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EE xx aa mm pp ll ee 11 ::Pullan solved the following ASSET DISPOSAL PROBLEM in his 1993 paper:
G H
a t t a t t
b t c t c t t t
=
= [ ]
= + = +
= = = − < <
1 0
0 11 2
4 3 2
10 2 0 21 2
1 1 2
, ,
( ) , ( ) ,
( ) , ( ) ( ) , .
max ( ) ( ) ( ) ( )
( ) ( )
( ) ( )
( ) ( ) ( )
, , ,
2 2
4
3 2
10
0 0
1 202
1 10
2 20
1 2 1
− + −
+ = +
+ = +
+ + =
≥ < <
∫
∫
∫
t u t t u t dt
u s ds x t t
u s ds x t t
u t u t z t
u x z t T
t
t
Our formulation, for arbitrary time horizon is:
max ( ) ( ) ( ) ( )
( ) ( )
( ) ( )
( ) ( ) ( )
, , ,
T t u t T t u t dt
u s ds x t t
u s ds x t t
u t u t z t
u x z t T
T
t
t
− + −
+ = +
+ = +
+ + =
≥ < <
∫
∫
∫
1 20
1 10
2 20
1 2 1
4
3 2
10
0 0
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EE xx aa mm pp ll ee 11 ::
The optimal solution is:
12
It is solved in 3 steps:
For time horizons 0 49
< <T : Empty buffer 1 only,
u t x t t x t t1 1 210 4 9 3 2( ) , ( ) , ( ) .= = − = +
For time horizons 49
2< <T : keep buffer 1 empty, and empty buffer 2:
u t x t u t x t t1 1 2 21 0 592
52
( ) , ( ) , ( ) , ( )= = = = −
For time horizons 2 < < ∞T : keep buffers 1 and 2 empty:
u t x t u t x t1 1 2 21 0 2 0( ) , ( ) , ( ) , ( )= = = =
NOW WATCH PULLAN SOLVE IT:
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EE xx aa mm pp ll ee 22 ::
A 2 machine 3 buffer example, m m m2 1 3> + :
m1
1m2
m3
2
3
m > m + 2 1 3
Variation
1
3
2
m
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(Solution of this example is joint with Florin Foram, 1992)
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SS oo ll uu tt ii oo nn
0 1
1 1 2 3 3 2
≤ ≤
=
T T
B u z
( ),
{ , , , , }
Buffer 3 drains
ξ ξ ξ
T T T
B u u z
B u u
( ) ( ),
{ , , , , }
{ , , , , }
1 2
3 1 2 2 3 1
2 1 2 2 3 3
≤ ≤
=
=
Buffer 2 drains
Optimal basis
Subproblem
ξ ξ
ξ ξ ξ
T T T
B u u u z
( ) ( ),
{ , , , , }
2 3
4 1 1 2 3 1
≤ ≤
=
Flow out of buffer 1
until interval 3 shrinks
ξ
T T T
B
B u u u
( ) ( ),
© { , , , , }
3 4
3
3 1 1 2 3 3
≤ ≤
=
Basis repalced by
ξ ξ
MS&E324, Stanford University, Spring 2002 6B-19 Gideon Weiss© manufacturing & control
T T
B u u u z z
( ) ,
{ , , , , }
4
5 1 2 3 1 2
≤ ≤ ∞
=
empty
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MS&E324, Stanford University, Spring 2002 6B-20 Gideon Weiss© manufacturing & control
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MS&E324, Stanford University, Spring 2002 6B-21 Gideon Weiss© manufacturing & control
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MS&E324, Stanford University, Spring 2002 6B-22 Gideon Weiss© manufacturing & control1 2 3 4 5
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MS&E324, Stanford University, Spring 2002 6B-23 Gideon Weiss© manufacturing & control
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