t dt t SCLP Bellman (1953) CLP (Dantzig, Tyndall, Grinold, Perold, Anstreicher 60's-80's) Anderson (1978) SCLP

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1 Coninuous Linear Programming. Separaed Coninuous Linear Programming Bellman (1953) max c () u() d H () u () + Gsusds (,) () a () u (), < < CLP (Danzig, yndall, Grinold, Perold, Ansreicher 6's-8's) Anderson (1978) max c () u() d Gu() s ds + x() = a() Hu () + z () = b () xuz,,, < < (Anderson, Nash, Philpo, Pullan 8's-9's). SCLP Dualiy, Srucure, Soluion by Discreizaion. In 2, W. proposed a simplex algorihm o solve: max ( ) c u( ) d Gu() s ds + x() = a+ α Hu () + z () = b xuz,,, < < SCLP wih linear daa MS&E324, Sanford Universiy, Spring 22 6B-1 Gideon Weiss manufacuring & conrol

2 Pullan s Dual Formulaion: Primal Problem: max c ( ) u ( ) d Gu() s ds+ x() = a() Hu () + z () = b () uxz,,, < < Pullan s SCLP Dual Problem min a ( ) dπ( ) + b ( ) rd ( ) G π() + H r() q() = c π( ) =, π, rq,, < < * Pullan s SCLP heorem (Pullan) Assume Hu () b (), u () has bounded feasible region. For abc,, piecewise analyic, here is a soluion which is piecewise analyic, For ac, piecewise polynomial of degree n +1, b piecewise polynomial of degree n here is a soluion wih u () piecewise polynomial of degree n. For ac, piecewise linear, b piecewise consan, here is a soluion wih u () piecewise consan. Number of breakpoins finie bounded (no polynomial) Srong dualiy holds Pullan s Algorihm: Sequence of discreizaions, converging (in infinie number of seps) o he opimal soluion. Noe: Separaed is very differen from General: in he one x () = usds (), direc inegraion in he oher u () = usds () + K differenial equaions MS&E324, Sanford Universiy, Spring 22 6B-2 Gideon Weiss manufacuring & conrol

3 SCLP wih Symmeric Primal / Dual max c ( ) u ( ) d+ d ( ) y ( ) d Gu() s ds + Fy() + x() = a() Hu () = b () xu,, < < SCLP min a ( ) p ( ) d+ b ( ) r ( ) d Gpsds () + Hr () q () = c () F p() = d() pq,, < < SCLP * Here: up, are primal and dual conrols ( ), xq, are primal and dual saes ( ), yr, are primal and dual supplemenary (unresriced) Dimensions: G K J, H I J, F K L MS&E324, Sanford Universiy, Spring 22 6B-3 Gideon Weiss manufacuring & conrol

4 Weak Dualiy and Complemenary Slackness Dual Objecive = a ( ) p ( ) d+ b ( ) r ( ) d { u () s G ds+ y( ) F } p() + u( ) H r() d = u( ) { G p( s) ds + H r( )} d + y( ) F p( ) d u ( ) c ( ) d+ y ( ) d ( ) d= Primal Objecive Equaliy will hold if and only if x ( ) pd ( ) =, u ( ) qd ( ) =. Complemenary Slackness: almos everywhere x () > p ( ) = u () > q ( ) = Corollary: he following are equivalen (a) uxy,, pqr,, are complemenary slack feasible primal and dual soluions (b) hey are opimal and have he same objecive value (no dualiy gap) (Srong dualiy no dualiy gap - exiss in LP bu no in oher problems, unless addiional condiions are imposed). MS&E324, Sanford Universiy, Spring 22 6B-4 Gideon Weiss manufacuring & conrol

5 Boundary Values For infiniesimal values of we solve: max c ( ) u ( ) + d ( ) y ( ) Fy() + x() = a() Hu () = b () xu,, < < min a ( ) p ( ) + b ( ) r ( ) Hr () q () = c () F p() = d() pq,, < < hese separae ino 2 ses of primal and dual problems: max d ( ) y( ) Fy( ) + x( ) = a( ) boundary LP x( ), min b ( ) r( ) Hr ( ) q( ) = c( ) q( ), boundary LP * Assumpion (Feasibiliy and Boundedness): he boundary problems are feasible and bounded. SCLP is feasible and bounded for some range of MS&E324, Sanford Universiy, Spring 22 6B-5 Gideon Weiss manufacuring & conrol

6 he Associaed LP Our soluion of SCLP will evolve hrough soluion of he following LP / LP*, under varying sign resricions. max c «( ) u ( ) + d ( ) y «( ) Gu() + Fy«( ) + ξ() = a«( ) Hu () = b () u (), y «( )" U", < < LP min a «( ) p ( ) + b ()«( r ) G p( ) + H r«( ) + θ( ) = c«( ) F p( ) = d( ) p, r«" U", < < LP * Noe: he ime indices on he variables are only labels, we label he variables o indicae heir role: ξ = x«, y«, θ = q«, r«assumpion (Non-Degeneracy): he r.h.s. of LP / LP* is (a.e. ) in general posiion o G F I column spaces of H, G H I F SCLP is non-degenerae: Soluion is unique and a an exreme poin in he funcion space of he soluions. MS&E324, Sanford Universiy, Spring 22 6B-6 Gideon Weiss manufacuring & conrol

7 he Srucure heorem Consider any funcions u ( ), x ( ), y ( ), p ( ), q ( ), r ( ), < < which saisfy up,, xyqr,,, are absoluely coninuous. Hence xyqr,,, have derivaives a.e. x( ), y( ) solve he boundary-lp, q( ), r( ) solve he boundary-lp*. For xy, hese are values a he lef boundary, qr, run in reversed ime so his is a he righ boundary. A all imes, he rae funcions (derivaives) u(), ξ(),«( y ), p( ), θ( ), r«( ), < < are complemenary slack feasible soluions for LP / LP*. hus hey are opimal soluions for LP / LP* under some sign configuraion of ξ, θ. By non-degeneracy hey are basic and unique soluions. A all imes, if xk() >, ξ k() is basic, if q ( ) >, θ ( ) is basic. j x () q (), < <. j heorem: A soluion o SCLP / SCLP* is opimal if (and only if for linear daa) i has hese properies. MS&E324, Sanford Universiy, Spring 22 6B-7 Gideon Weiss manufacuring & conrol

8 Resricion o he Linear Daa Case Problem: max ( γ + ( ) c ) u() d+ d y() d Gu() s ds + Fy() + x() = α + a Hu () xu,, < < min ( α + ( ) a ) p() d+ b r() d Gpsds () + Hr () q () = γ + c F p() = d pq,, < < Boundary: max dy min N br Fy + x = α = b SCLP ( linear daa) SCLP Hr q = γ * ( linear daa) x, q, Associaed LP max cu + d y«min ap + br «Gu+ Fy«+ ξ = a Gp + Hr «θ = c Hu = b Fp = d u, y«" U", p, r«" U" N N N MS&E324, Sanford Universiy, Spring 22 6B-8 Gideon Weiss manufacuring & conrol

9 he Algorihm he algorihm consrucs a soluion for SCLP / SCLP* for he whole range of ime horizons, < <. Corollary 1: he soluion for any ime horizon wih he 1 excepion of a finie lis: = ( ) < ( ) L < ( R) <, is characerized by a sequence of adjacen bases of LP, B 1, K, B N. For a paricular here are ime breakpoins = < 1L < N = such ha for n < < n n n n n n u, ξ,«y, p, θ,«r are basic B n. 1 n, In he single pivo Bn Bn +1 le v leave he basis if v = ξ k hen we have equaion xk( n)=. if v= hen we have equaion q ( ) =. u j Corollary 2: he soluion of he equaions xk( n)=, qj( n) =, ogeher wih: τ1+ L+ τ N = ( τ n = n n 1) deermines he soluion for a whole range of ime horizons, ( r 1) ( r) < < <. j n MS&E324, Sanford Universiy, Spring 22 6B-9 Gideon Weiss manufacuring & conrol

10 Algorihmic Srucure of Soluion: For ime horizon, Break poins: = < 1 < L < N = τ n = n n 1 N 1 Equaions: Adjacen Bn Bn+1 : = vn leaves: n n k k n k k m v = ξ x ( ) = x + ξ τ =, m= 1 n j j n j N N m j v = u q ( ) = q + θ τ =. and equaion N: τ1 + L+ τ N = m m= n+ 1 Slack Inequaliies, a local minima of x m k, q : k n k n n + 1 k n k k m ξ <, ξ > or n= N x ( ) = x + ξ τ, n j n+ 1 j n= or θ >, θ < q ( ) = q + θ τ. j m= 1 m j n j N N m j m= n+ 1 m MS&E324, Sanford Universiy, Spring 22 6B-1 Gideon Weiss manufacuring & conrol

11 Algorihm: ( ) ( 1) Sep 1: Soluion for < < is given by he soluion of boundary-lp / boundary-lp*. Sep r: Exend he soluion o nex ime horizon range. ( r) 1L 1 A d + τ = σ B I g Sep R+1 Sop when nex can be, or when soluion infeasible. MS&E324, Sanford Universiy, Spring 22 6B-11 Gideon Weiss manufacuring & conrol

12 Algorihm Case iii Buffer empies a x 1 x 2 < (l) x 2 x 1 ξ 1 leaves = (l) x 1 x 2 u 2 eners > (l) MS&E324, Sanford Universiy, Spring 22 6B-12 Gideon Weiss manufacuring & conrol

13 Algorihm Case ii Inerval shrinks a n B' C ξ 1 leaves x 1 q C B'' u 3 leaves 3 B' C B'' < (l) B' B'' ξ 1 u 3 leave x 1 q 3 B' B'' = (l) x 1 q 3 D B'' ξ 1 leaves B' D u 3 leaves B' D B'' > (l) MS&E324, Sanford Universiy, Spring 22 6B-13 Gideon Weiss manufacuring & conrol

14 Algorihm Case ii, Subproblem Inerval shrinks a n : subproblem for addiional inervals B' C ξ 1 leaves x 1 q C B'' u 3 leaves 3 B' C B'' < (l) B' B'' ξ 1 u 3 leave x 1 q 3 B' B'' = (l) subproblem E 1 D E 2 B' B'' ξ 1 u 3 leave x 1 q 3 B' B'' = (l) E 1 D 3u leaves x 1 q E B'' 1ξ leaves 3 2 B' E 1 D E 2 B'' > (l) Subproblem solved by recursive call o he algorihm, wih smaller, modified problem. MS&E324, Sanford Universiy, Spring 22 6B-14 Gideon Weiss manufacuring & conrol

15 Example 1: Pullan solved he following ASSE DISPOSAL PROBLEM in his 1993 paper: G= 1 H, = [ 1 2], 1 a1() = 4+, a2() = 3+ 2, b1() = 1, c1() = c2() = 2, < < 2. 2 max ( 2 u ) ( ) + ( 2 u ) ( d ) 1 2 u () s ds + x () = u2() s ds + x2() = 3+ 2 u1() + u2() + z1() = 1 uxz,,, < < Our formulaion, for arbirary ime horizon is: max ( ) u1( ) + ( ) u2( ) d u1() s ds + x1() = 4 + u2() s ds + x2() = 3+ 2 u1() + u2() + z1() = 1 uxz,,, < < MS&E324, Sanford Universiy, Spring 22 6B-15 Gideon Weiss manufacuring & conrol

16 Example 1: he opimal soluion is: 1 2 I is solved in 3 seps: For ime horizons < < 4 9 : Empy buffer 1 only, u () = 1, x () = 4 9, x () = For ime horizons 4 < <2 : keep buffer 1 empy, and empy buffer 2: u1() = 1, x1() =, u2() =, x2() = For ime horizons 2 < < : keep buffers 1 and 2 empy: u () = 1, x () =, u () = 2, x () = NOW WACH PULLAN SOLVE I: MS&E324, Sanford Universiy, Spring 22 6B-16 Gideon Weiss manufacuring & conrol

17 MS&E324, Sanford Universiy, Spring 22 6B-17 Gideon Weiss manufacuring & conrol

18 Example 2: A 2 machine 3 buffer example, m2 > m1+ m3: m m1 2 m 2 m > m + m Variaion 2 1 (Soluion of his example is join wih Florin Foram, 1992) MS&E324, Sanford Universiy, Spring 22 6B-18 Gideon Weiss manufacuring & conrol

19 S o l u i o n () 1, Buffer 3 drains B = { ξ, ξ, ξ, u, z } () 1 ( 2), Buffer 2 drains Opimal basis B = { ξ, ξ, u, u, z } Subproblem B = { ξ, ξ, u, ξ, u } ( 2) ( 3), Flow ou of buffer 1 unil inerval 3 shrinks B = { ξ, u, u, u, z } ( 3) ( 4), Basis B 3 repalced by B = { ξ, u, u, ξ, u } MS&E324, Sanford Universiy, Spring 22 6B-19 Gideon Weiss manufacuring & conrol ( 4), empy B = { u, u, u, z, z }

20 Example 3: 5 machine 2 buffer re-enran example I II III IV V MS&E324, Sanford Universiy, Spring 22 6B-2 Gideon Weiss manufacuring & conrol

21 S o l u i o n : MS&E324, Sanford Universiy, Spring 22 6B-21 Gideon Weiss manufacuring & conrol

22 Example 4: A coninuous ime Leonief Sysem X MS&E324, Sanford Universiy, Spring B-22 3 Gideon Weiss 4 manufacuring & conrol 5

23 Evoluion of he soluion: x 4 q 6 r2 q 6 x 3 x 4 x 3 r 2 r 2 x 3 x 4 q 12 x 2 MS&E324, Sanford Universiy, Spring 22 6B-23 Gideon Weiss manufacuring & conrol