Combinatorial Integral Approximation for Mixed-Integer Nonlinear Optimal Control

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1 Combinatorial Integral Approximation for Mixed-Integer Nonlinear Optimal Control Sebastian Sager Junior Research Group Mathematical and Computational Optimization Interdisciplinary Center for Scientific Computing University of Heidelberg October, 2010 Banff Sebastian Sager Combinatorial Integral Approximation 1

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4 Outline MIOC introduction Methods for Mixed-Integer Optimal Control... based on HJB / Dynamic Programming... based on Direct Methods MINLP MS MINTOC Switching Costs and Combinatorial Constraints Sebastian Sager Combinatorial Integral Approximation 4

5 Applications I: Valves and ports I Watergates and on/off pumps in water networks [Burgschweiger, Deuerlein, Gugat, Hante, Leugering, Steinbach,... ] Images: Wikipedia, BASF SE Sebastian Sager Combinatorial Integral Approximation 5

6 Applications I: Valves and ports I Watergates and on/off pumps in water networks [Burgschweiger, Deuerlein, Gugat, Hante, Leugering, Steinbach,... ] I Valves, (de)compressors in gas networks [Leugering, Martin, Schultz, Steinbach, S. Ulbrich,... ] Images: Wikipedia, BASF SE Sebastian Sager Combinatorial Integral Approximation 5

7 Applications I: Valves and ports I Watergates and on/off pumps in water networks [Burgschweiger, Deuerlein, Gugat, Hante, Leugering, Steinbach,... ] I Valves, (de)compressors in gas networks [Leugering, Martin, Schultz, Steinbach, S. Ulbrich,... ] I Valves in separation processes [Biegler, Engell, Findeisen, Grossmann, Kienle, Marquardt, Swartz,... ] I Slop cut recycling in batch distillation [Bock, Diehl, S.,... ] I I Evaporator operation [Sonntag, Engell,... ] Simulated Moving Bed superstructure [Biegler, Engell, Potschka, S., Weismantel, Westerlund,... ] Images: Wikipedia, BASF SE Sebastian Sager Combinatorial Integral Approximation 5

8 Applications II: optimal gear shifts in transport New York subway: energy optimal rides (?), (?) Submarine control (?) Look-ahead control in heavy duty trucks (?), (?), (?) Feedback control, GPS data, 16 gears Elchtest benchmark: automobile testdriving (??), (?), [Borelli,... ] Periodic time-optimal automobile driving, (?) Formula 1 racing, (?) Images: Wikipedia, youtube (Hellström), Pesch, Gerdts in Mathematics Key Technology for the Future Sebastian Sager Combinatorial Integral Approximation 6

9 Applications III: Others Traffic lights in Traffic flow [Aw, Leugering, Rascle,... ], (?), (?) Biology / Medicine: inhibition, treatment [Engelhart, Lebiedz, S.,... ] Economics: yes/no decisions (part time, 2nd job?) [Kübler, Kuhn] Robot swarm movement and communication (?) Optimum experimental design [Bock, Körkel, Hoffmann, Kostina, Lebiedz, Schittkowski, Schlöder, Seidel-Morgenstern,... ] Variables: is measurement done or not, w(x, t) {0, 1}... benchmark library: Sebastian Sager Combinatorial Integral Approximation 7

10 Problem class min f )) x,v,u s.t. ẋ(t) = f (x(t), v(t), u(t)), 0 c(x(t), u(t)), 0 r in (x(t 0 ),..., x(t f )), 0 = r eq (x(t 0 ),..., x(t f )), v(t) Ω := {v 1, v 2,..., v nω }, t [t 0, t f ]. x differential states, u, v control functions All functions sufficiently smooth Generalizations: PDEs, algebraic variables, constant-in-time control values, multi-stage,... Focus of talk: controls v(t) from finite set v i Ω R nv. Sebastian Sager Combinatorial Integral Approximation 8

11 Why is this problem class difficult? min f )) x,v,u s.t. ẋ(t) = f (x(t), v(t), u(t)), 0 c(x(t), u(t)), 0 r in (x(t 0 ),..., x(t f )), 0 = r eq (x(t 0 ),..., x(t f )), v(t) Ω := {v 1, v 2,..., v nω }, t [t 0, t f ]. Nonlinear Differential equations Path and point constraints Integer decisions (in function space) Sebastian Sager Combinatorial Integral Approximation 9

12 Outline MIOC introduction Methods for Mixed-Integer Optimal Control... based on HJB / Dynamic Programming... based on Direct Methods MINLP MS MINTOC Switching Costs and Combinatorial Constraints Sebastian Sager Combinatorial Integral Approximation 10

13 Alternative Methods HJB / Dynamic Programming based on Bellman s Principle of Optimality tabulated enumeration easy to extend to integer controls but: curse of dimensionality Sebastian Sager Combinatorial Integral Approximation 11

14 Alternative Methods HJB / Dynamic Programming based on Bellman s Principle of Optimality tabulated enumeration easy to extend to integer controls but: curse of dimensionality Pontryagin s Maximum Principle use optimality conditions in function space solve resulting boundary value problem u (t) is the pointwise maximum of the Hamiltonian u (t) = arg max H(x(t), u, λ(t)) u U U may also be a discrete set! practice: use switching functions to detect changes difficult for singular and path-constrained arcs analytical work, needs to be well initialized Sebastian Sager Combinatorial Integral Approximation 11

15 Switching time optimization Take given switching order with piecewise fixed controls v( ), optimize interval lengths (after standard time transformation) v(t) t 0 [Gerdts, Kaya, Leineweber, Maurer, Noakes, S.,... ] 1 0 h 0 h 1 h 2 t 1 t 2 h 3 t 3 t 4 h 4 t f Sebastian Sager Combinatorial Integral Approximation 12

16 Switching time optimization Take given switching order with piecewise fixed controls v( ), optimize interval lengths (after standard time transformation) v(t) t 0 [Gerdts, Kaya, Leineweber, Maurer, Noakes, S.,... ] 1 0 h 0 h 1 h 2 t 1 t 2 h 3 t 3 t 4 h 4 t f min φ(x(t f )) x,v,u s.t. ẋ(t) = f (x(t), v(t), u(t)), t [t 0, t f ] Sebastian Sager Combinatorial Integral Approximation 12

17 Switching time optimization Take given switching order with piecewise fixed controls v( ), optimize interval lengths (after standard time transformation) v(t) t 0 [Gerdts, Kaya, Leineweber, Maurer, Noakes, S.,... ] 1 0 h 0 h 1 h 2 t 1 t 2 h 3 t 3 t 4 h 4 t f min φ(x(t f )) x,v,u s.t. ẋ(t) = f (x(t), v(t), u(t)), t [t 0, t f ] min f )) x,h,u s.t. ẋ(t) = f (x(t), v i j, u(t)), t [ t j, t j+1 ]... Sebastian Sager Combinatorial Integral Approximation 12

18 Switching time optimization Take given switching order with piecewise fixed controls v( ), optimize interval lengths (after standard time transformation) v(t) t 0 [Gerdts, Kaya, Leineweber, Maurer, Noakes, S.,... ] 1 0 h 0 h 1 h 2 t 1 t 2 h 3 t 3 t 4 h 4 t f min φ(x(t f )) x,v,u s.t. ẋ(t) = f (x(t), v(t), u(t)), t [t 0, t f ] min f )) x,h,u s.t. ẋ(t) = f (x(t), v i j, u(t)), t [ t j, t j+1 ]... Main issue: initialization Sebastian Sager Combinatorial Integral Approximation 12

19 Linear approximations MILP way faster than MINLP solvers Idea: approximate nonlinearities and obtain MILP A) Use linearizations in NMPC context [Jones, Morari, Borrelli,... ] B) Use piecewise linear approximation (?) Main issues: problem size, adaptivity, restarts Image: Wikipedia Sebastian Sager Combinatorial Integral Approximation 13

20 Direct Single Shooting [Hicks, Ray 1971; Sargent, Sullivan 1977] Discretize controls u(t) on fixed grid 0 = t 0 < t 1 <... < t N = t f. x 0 states x(t; q) discretized controls u(t; q) q 0 q N 1 0 q 1 t Regard states x(t) on [t 0, t f ] as dependent variables. Use numerical integration to obtain state as function x(t; q, x 0 ) of finitely many control parameters q = (q 0, q 1,..., q N 1 ) and the initial value x 0. t f Sebastian Sager Combinatorial Integral Approximation 14

21 Direct Multiple Shooting [Bock and Plitt, 1981, 1984,... ] x i (s i, q i, p, h) s x i+1 s i s i+1 s 0 s 1 q i s N 1 s N q 0 Idea: Decouple intervals, add extra continuity constraints. Denote each interval s variables by w i := (s x i, s z i, q i ). Summarize all in large vector w := (w 0,..., w N ). Sebastian Sager Combinatorial Integral Approximation 15

22 NLP in Direct Multiple Shooting [Bock and Plitt, 1981, 1984,... ] min w N φ i (w i ) s.t. i=0 si+1 x x i(w i ) = 0 (continuity) g i (w i ) = 0 (algebraic consistency) c i (w i ) 0 (path constraints) N i=0 r i(w i ) 0 (multipoint inequality constraints) N i=0 r e(w i ) 0 (multipoint equality constraints) Sebastian Sager Combinatorial Integral Approximation 16

23 Intermediate summary Large problem class with many applications Different approaches to solve control problems Dynamic Programming Maximum Principle Direct Methods: discretize controls, solve NLP Direct Methods: different ways to parameterize states Sebastian Sager Combinatorial Integral Approximation 17

24 MINLP approach to solve MIOCPs min f )) x,v,u s.t. ẋ(t) = f (t, x(t), v(t), u(t)), t [t 0, t f ], 0 c(t, x(t), v(t), u(t)), 0 r i (x(t 0 ),..., x(t f )), 0 = r e (x(t 0 ),..., x(t f )), v(t) Ω. Consider the infinite dimensional problem Discretize Sebastian Sager Combinatorial Integral Approximation 18

25 MINLP approach to solve MIOCPs min f )) x,v,u s.t. ẋ(t) = f (t, x(t), v(t), u(t)), t [t 0, t f ], 0 c(t, x(t), v(t), u(t)), 0 r i (x(t 0 ),..., x(t f )), 0 = r e (x(t 0 ),..., x(t f )), v(t) Ω. Consider the infinite dimensional problem Discretize Let variables inherit integer constraint min γ,β F(γ, β) s.t. 0 = G(γ, β) 0 H(γ, β) β i Ω, i = 1..N Sebastian Sager Combinatorial Integral Approximation 18

26 Mixed Integer Nonlinear Programming Generic algorithms: Nonlinear Branch & Bound, Outer Approximation, Disjunctive Programming,... Active research area [Bonami, Wächter,... ] (Bonmin) [Leyffer, Linderoth,... ] (FilMint) [Biegler, Floudas, Grossmann, Lee, Marquardt, Oldenburg, Sahinidis, Weismantel,... ]... But: extremly expensive for control problems Sebastian Sager Combinatorial Integral Approximation 19

27 New Approach min f )) x,v,u s.t. ẋ(t) = f (x(t), v(t), u(t)), 0 c(x(t), u(t)), 0 r in (x(t 0 ),..., x(t f )), 0 = r eq (x(t 0 ),..., x(t f )), v(t) Ω := {v 1, v 2,..., v nω }, t [t 0, t f ]. Important: lower bounds Relaxation v Ω v conv Ω is too weak! Is there a better relaxation? Sebastian Sager Combinatorial Integral Approximation 20

28 (Partial) Outer convexification [Sager, 2005] Easy to show: v( ) Ω := {v 1, v 2,..., v nω } ẋ(t) = f (x(t), v(t), u(t)) Sebastian Sager Combinatorial Integral Approximation 21

29 (Partial) Outer convexification [Sager, 2005] Easy to show: v( ) Ω := {v 1, v 2,..., v nω } ẋ(t) = f (x(t), v(t), u(t)) n ω i=1 ω( ) {0, 1} nω ẋ(t) = n ω i=1 ω i (t) = 1, t [t 0, t f ] f (x(t), v i, u(t)) ω i (t) Sebastian Sager Combinatorial Integral Approximation 21

30 Problem class after Outer Convexification min φ(x(t f )) x,u,ω s.t. ẋ(t) = n ω i=1 f (x(t), v i, u(t)) ω i (t), 0 c(x(t), u(t)), 0 r i (x(t 0 ),..., x(t f )), 0 = r e (x(t 0 ),..., x(t f )), 1 = n ω i=1 ω i (t), ω(t) {0, 1} nω, t [t 0, t f ]. Important: lower bounds Is relaxation ω {0, 1} nω α [0, 1] nω good? Sebastian Sager Combinatorial Integral Approximation 22

31 Approximating the state [S., Bock, Diehl, Mathematical Programming, to appear] LEMMA. Let x( ) and y( ) be solutions of ẋ(t) = A(t, x(t)) α(t), x(0) = x 0, ẏ(t) = A(t, y(t)) ω(t), y(0) = y 0, with t [0, t f ], for given measurable functions α, ω : [0, t f ] [0, 1] nω and A : R nx +1 R nx nω. Sebastian Sager Combinatorial Integral Approximation 23

32 Approximating the state [S., Bock, Diehl, Mathematical Programming, to appear] LEMMA. Let x( ) and y( ) be solutions of ẋ(t) = A(t, x(t)) α(t), x(0) = x 0, ẏ(t) = A(t, y(t)) ω(t), y(0) = y 0, with t [0, t f ], for given measurable functions α, ω : [0, t f ] [0, 1] nω and A : R nx +1 R nx nω. If C, L, M R + exist such that d A(t, x(t)) dt C, A(t, y(t)) A(t, x(t)) L y(t) x(t), for t [0, t f ] almost everywhere and A(, x( )) essentially bounded by M on [0, t f ], Sebastian Sager Combinatorial Integral Approximation 23

33 Approximating the state [S., Bock, Diehl, Mathematical Programming, to appear] LEMMA. Let x( ) and y( ) be solutions of ẋ(t) = A(t, x(t)) α(t), x(0) = x 0, ẏ(t) = A(t, y(t)) ω(t), y(0) = y 0, with t [0, t f ], for given measurable functions α, ω : [0, t f ] [0, 1] nω and A : R nx +1 R nx nω. If C, L, M R + exist such that d A(t, x(t)) dt C, A(t, y(t)) A(t, x(t)) L y(t) x(t), for t [0, t f ] almost everywhere and A(, x( )) essentially bounded by M on [0, t f ], and it exists ɛ R + such that t α(τ) ω(τ) dτ ɛ for all t [0, t f ], 0 Sebastian Sager Combinatorial Integral Approximation 23

34 Approximating the state [S., Bock, Diehl, Mathematical Programming, to appear] LEMMA. Let x( ) and y( ) be solutions of ẋ(t) = A(t, x(t)) α(t), x(0) = x 0, ẏ(t) = A(t, y(t)) ω(t), y(0) = y 0, with t [0, t f ], for given measurable functions α, ω : [0, t f ] [0, 1] nω and A : R nx +1 R nx nω. If C, L, M R + exist such that d A(t, x(t)) dt C, A(t, y(t)) A(t, x(t)) L y(t) x(t), for t [0, t f ] almost everywhere and A(, x( )) essentially bounded by M on [0, t f ], and it exists ɛ R + such that t α(τ) ω(τ) dτ ɛ for all t [0, t f ], then for all t [0, t f ] it holds 0 y(t) x(t) ( x 0 y 0 + (M + C)tɛ) e Lt Sebastian Sager Combinatorial Integral Approximation 23

35 Approximating the state [S., Bock, Diehl, Mathematical Programming, to appear] LEMMA. Let x( ) and y( ) be solutions of ẋ(t) = A(t, x(t)) α(t), x(0) = x 0, ẏ(t) = A(t, y(t)) ω(t), y(0) = y 0, with t [0, t f ], for given measurable functions α, ω : [0, t f ] [0, 1] nω and A : R nx +1 R nx nω. If C, L, M R + exist such that d A(t, x(t)) dt C, A(t, y(t)) A(t, x(t)) L y(t) x(t), for t [0, t f ] almost everywhere and A(, x( )) essentially bounded by M on [0, t f ], and it exists ɛ R + such that t α(τ) ω(τ) dτ ɛ for all t [0, t f ], then for all t [0, t f ] it holds 0 y(t) x(t) ( x 0 y 0 + (M + C)tɛ) e Lt Related results: [Gonzalez, Häckl, Pietrus, Tidball, Veliov] Sebastian Sager Combinatorial Integral Approximation 23

36 Approximating the state [S., Bock, Diehl, Mathematical Programming, to appear] In other words: y(t) x(t) will be small for all t, if t α(τ) ω(τ) dτ 0 is small for all times t [0, t f ]. How to get binary variables ω(τ) from continuous variables α( )? Sebastian Sager Combinatorial Integral Approximation 24

37 Sum Up Rounding [Sager, 2005] Assume continuous solution q j,i [0, 1]. Goal: p j,i {0, 1}. For t i = t i+1 t i and all j = 1... n ω, i = 1... n int set p j,i = { 1 if i k=0 q j,k t k i 1 k=0 p j,k t k 0.5 t i 0 else. Sebastian Sager Combinatorial Integral Approximation 25

38 Sum Up Rounding [Sager, 2005] Assume continuous solution q j,i [0, 1]. Goal: p j,i {0, 1}. For t i = t i+1 t i and all j = 1... n ω, i = 1... n int set p j,i = { 1 if i k=0 q j,k t k i 1 k=0 p j,k t k 0.5 t i 0 else. q(t) t Sebastian Sager Combinatorial Integral Approximation 25

39 Sum Up Rounding [Sager, 2005] Assume continuous solution q j,i [0, 1]. Goal: p j,i {0, 1}. For t i = t i+1 t i and all j = 1... n ω, i = 1... n int set p j,i = { 1 if i k=0 q j,k t k i 1 k=0 p j,k t k 0.5 t i 0 else. q(t) 1 p(t) t t Sebastian Sager Combinatorial Integral Approximation 25

40 Sum Up Rounding [Sager, 2005] Assume continuous solution q j,i [0, 1]. Goal: p j,i {0, 1}. For t i = t i+1 t i and all j = 1... n ω, i = 1... n int set p j,i = { 1 if i k=0 q j,k t k i 1 k=0 p j,k t k 0.5 t i 0 else. q(t) 1 p(t) t t Sebastian Sager Combinatorial Integral Approximation 25

41 Sum Up Rounding [Sager, 2005] Assume continuous solution q j,i [0, 1]. Goal: p j,i {0, 1}. For t i = t i+1 t i and all j = 1... n ω, i = 1... n int set p j,i = { 1 if i k=0 q j,k t k i 1 k=0 p j,k t k 0.5 t i 0 else. q(t) 1 p(t) t t Sebastian Sager Combinatorial Integral Approximation 25

42 Sum Up Rounding [Sager, 2005] Assume continuous solution q j,i [0, 1]. Goal: p j,i {0, 1}. For t i = t i+1 t i and all j = 1... n ω, i = 1... n int set p j,i = { 1 if i k=0 q j,k t k i 1 k=0 p j,k t k 0.5 t i 0 else. q(t) 1 p(t) t t Sebastian Sager Combinatorial Integral Approximation 25

43 Sum Up Rounding [Sager, 2005] Assume continuous solution q j,i [0, 1]. Goal: p j,i {0, 1}. For t i = t i+1 t i and all j = 1... n ω, i = 1... n int set p j,i = { 1 if i k=0 q j,k t k i 1 k=0 p j,k t k 0.5 t i 0 else. q(t) 1 p(t) t t Sebastian Sager Combinatorial Integral Approximation 25

44 Sum Up Rounding [Sager, 2005] Assume continuous solution q j,i [0, 1]. Goal: p j,i {0, 1}. For t i = t i+1 t i and all j = 1... n ω, i = 1... n int set p j,i = { 1 if i k=0 q j,k t k i 1 k=0 p j,k t k 0.5 t i 0 else. q(t) 1 p(t) t t Sebastian Sager Combinatorial Integral Approximation 25

45 Sum Up Rounding [Sager, 2005] Assume continuous solution q j,i [0, 1]. Goal: p j,i {0, 1}. For t i = t i+1 t i and all j = 1... n ω, i = 1... n int set p j,i = { 1 if i k=0 q j,k t k i 1 k=0 p j,k t k 0.5 t i 0 else. q(t) 1 p(t) t t Sebastian Sager Combinatorial Integral Approximation 25

46 Sum Up Rounding [Sager, 2005] Assume continuous solution q j,i [0, 1]. Goal: p j,i {0, 1}. For t i = t i+1 t i and all j = 1... n ω, i = 1... n int set p j,i = { 1 if i k=0 q j,k t k i 1 k=0 p j,k t k 0.5 t i 0 else. q(t) 1 p(t) t t Sebastian Sager Combinatorial Integral Approximation 25

47 Sum Up Rounding [Sager, 2005] Assume continuous solution q j,i [0, 1]. Goal: p j,i {0, 1}. For t i = t i+1 t i and all j = 1... n ω, i = 1... n int set p j,i = { 1 if i k=0 q j,k t k i 1 k=0 p j,k t k 0.5 t i 0 else. q(t) 1 p(t) t t Sebastian Sager Combinatorial Integral Approximation 25

48 Sum Up Rounding [Sager, 2005] Assume continuous solution q j,i [0, 1]. Goal: p j,i {0, 1}. For t i = t i+1 t i and all j = 1... n ω, i = 1... n int set p j,i = { 1 if i k=0 q j,k t k i 1 k=0 p j,k t k 0.5 t i 0 else. q(t) 1 p(t) t t Sebastian Sager Combinatorial Integral Approximation 25

49 Sum Up Rounding [Sager, 2005] Assume continuous solution q j,i [0, 1]. Goal: p j,i {0, 1}. For t i = t i+1 t i and all j = 1... n ω, i = 1... n int set p j,i = { 1 if i k=0 q j,k t k i 1 k=0 p j,k t k 0.5 t i 0 else. q(t) 1 p(t) t t Sebastian Sager Combinatorial Integral Approximation 25

50 Sum Up Rounding [Sager, 2005] Assume continuous solution q j,i [0, 1]. Goal: p j,i {0, 1}. For t i = t i+1 t i and all j = 1... n ω, i = 1... n int set p j,i = { 1 if i k=0 q j,k t k i 1 k=0 p j,k t k 0.5 t i 0 else. Sebastian Sager Combinatorial Integral Approximation 25

51 Approximating the control [S., Bock, Diehl, Mathematical Programming, to appear] LEMMA. Let functions α : [0, t f ] [0, 1] nω, and ω : [0, t f ] {0, 1} nω, α j (t) = q j,i, t [t i, t i+1 ] ω j (t) = p j,i, t [t i, t i+1 ] { 1 if i p j,i = k=0 q j,k t k i 1 k=0 p j,k t k 0.5 t i 0 else. be given. Then it holds t α(τ) ω(τ) dτ max t i i (Similar strategy and lemma allow to include SOS1 constraints for nonlinear case) Sebastian Sager Combinatorial Integral Approximation 26

52 Implications 2 lemmata together: if SUR is used then y(t) x(t) C t for any (relaxed) solution x( ), control grid size t Objective function is a continuous function, thus Φ(x(t f )) Φ(y(t f )) γ t Sebastian Sager Combinatorial Integral Approximation 27

53 Implications 2 lemmata together: if SUR is used then y(t) x(t) C t for any (relaxed) solution x( ), control grid size t Objective function is a continuous function, thus Φ(x(t f )) Φ(y(t f )) γ t Extension to continuous constraints straightforward Sebastian Sager Combinatorial Integral Approximation 27

54 Implications 2 lemmata together: if SUR is used then y(t) x(t) C t for any (relaxed) solution x( ), control grid size t Objective function is a continuous function, thus Φ(x(t f )) Φ(y(t f )) γ t Extension to continuous constraints straightforward We know the exact lower bound from α( ) solution Sum Up Rounding constructive way to get integer solution Sebastian Sager Combinatorial Integral Approximation 27

55 MS MINTOC algorithm [Sager 2005; Sager, J Proc Control, 2009] 1. k = 0. Input: control discretization grid G 0, TOL R If necessary, reformulate and convexify problem. Relax this problem to α( ) [0, 1] nω. 3. REPEAT 3.1 Solve relaxed problem on G k. Obtain T k = (x k ( ), u k ( ), α k ( )) and the grid dependent optimal value φ RC. 3.2 If T k on G k fulfills ω k ( ) := α k ( ) {0, 1} nω then STOP. 3.3 Apply Sum Up Rounding to α k ( ). Fix u k ( ). Obtain y k ( ) and upper bound φ STO by simulation. 3.4 If φ STO < φ RC + TOL then STOP. 3.5 Refine the control grid G k. 3.6 k = k Bijection to obtain solution Φ = φ STO for original problem. Algorithm terminates in finite number of iterations [S., 2005; S. 2009] Sebastian Sager Combinatorial Integral Approximation 28

56 Intermediate summary Main trick: binary controls propagated to continuous states Most applications: no explicit constraints on binary controls Sebastian Sager Combinatorial Integral Approximation 29

57 Intermediate summary Main trick: binary controls propagated to continuous states Most applications: no explicit constraints on binary controls At the price of chattering, relaxed solution can always be reached no integer gap!!! However: for any fixed control grid there may be a gap black box MINLP performs badly! Sum Up Rounding constructive way to get integer solution Needs to be combined with iterative refinement of grid Is Outer Convexification crucial? Sebastian Sager Combinatorial Integral Approximation 29

58 Example: MIOCP nonlinear in v min x,v x 2 (t f ) s.t. ẋ 0 = x 0 sin(v 1 ) sin(1) + (x 0 + x 1 ) v (x 0 x 1 ) v 3 3 ẋ 1 = (x 0 + 2x 1 ) v 1 + (x 0 2x 1 ) v 2 + (x 0 + x 1 ) v 3 + (x 0 x 1 x 2 ) v 2 2 (x 0 x 1 x 2 ) v 2 3, (1) ẋ 2 = x0 2 + x 1 2, x(0) = (0.5, 0.5, 0) T, x 1 0.4, v {(1, 0, 0), (0, 1, 0), (0, 0, 1)} with t [t 0, t f ] = [0, 1]. This problem can be relaxed by requiring v 1, v 2, v 3 [0, 1], 3 v i = 1 i=1 Sebastian Sager Combinatorial Integral Approximation 30

59 Example: partially outer convexified with ω min x,ω x 2 (t f ) s.t. ẋ 0 = x 0 ω 1 + (x 0 + x 1 ) ω 2 + (x 0 x 1 ) ω 3, ẋ 1 = (x 0 + 2x 1 ) ω 1 + (x 0 2x 1 ) ω 2 + (x 0 + x 1 ) ω 3, ẋ 2 = x x 2 1, x(0) = (0.5, 0.5, 0) T, x 1 0.4, ω i {0, 1}, 3 ω i = 1 i=1 (2) with t [t 0, t f ] = [0, 1]. Relaxation: α i [0, 1] Is (by construction) identical to the one in (?) and (?). Sebastian Sager Combinatorial Integral Approximation 31

60 OCP (1) - nonlinear relaxed: Φ = Sebastian Sager Combinatorial Integral Approximation 32

61 OCP (2) - outer convexification relaxed: Φ = Sebastian Sager Combinatorial Integral Approximation 33

62 MIOCPs (1) and (2) - Sum Up Rounding: Φ = Sebastian Sager Combinatorial Integral Approximation 34

63 Elchtest benchmark problem [Gerdts, OCAM, 2005] Elchtest in min. time with gear shifts, N discretization points. N t f CPU Time t f CPU Time :23: :00: :25: :00: :04:19 Left: Branch and Bound, Inner Convexification [Gerdts, OCAM, 2005] Pentium III with 750 MHz Right: MS MINTOC, Outer Convexific. [Kirches, Bock, Schlöder, S., OCAM, 2009] AMD Athlon XP with 2166 MHz Similar speedup for free switching times [Gerdts OCAM 2006, Kirches et al. 2009] Sebastian Sager Combinatorial Integral Approximation 35

64 Why is this formulation so beneficial?!? v(t) = 1 m ( (F µ lr F ) Ax cos β(t) + Flf cos ( δ(t) + β(t) ) ( F sr F Ay ) sin β(t) Fsf sin ( δ(t) + β(t) )). F µ lr is a function of φ, F B, v, and i µ g. Maximum acceleration φ := 1, F B := 0. Sebastian Sager Combinatorial Integral Approximation 36

65 Why is this formulation so beneficial?!? v(t) = 1 m ( (F µ lr F ) Ax cos β(t) + Flf cos ( δ(t) + β(t) ) ( F sr F Ay ) sin β(t) Fsf sin ( δ(t) + β(t) )). F µ lr is a function of φ, F B, v, and i µ g. Maximum acceleration φ := 1, F B := 0. Sebastian Sager Combinatorial Integral Approximation 36

66 Compare maximum acceleration Compare maximum acceleration for convex combinations, a = Inner convexification, use F lr = F lr (a i 2 g + (1 a) i 3 g ) where i 2 g and i 3 g are the transmission ratios of second resp. third gear Sebastian Sager Combinatorial Integral Approximation 37

67 Compare maximum acceleration Compare maximum acceleration for convex combinations, a = Inner convexification, use Outer convexification, use F lr = F lr (a ig 2 + (1 a) ig 3 ) F lr = a F lr (ig 2 ) + (1 a) F lr (ig 3 ) where i 2 g and i 3 g are the transmission ratios of second resp. third gear Sebastian Sager Combinatorial Integral Approximation 37

68 Diploma thesis F. Kehrle Sebastian Sager Combinatorial Integral Approximation 38

69 Problem characteristics Long optimization horizon, unstable = use Bock s direct multiple shooting method, 350 intervals = 1750 binary control variables = 1050 continuous control variables of discretized state variables Nonlinear dynamics, nasty engine speed + path constraints NMPC Sebastian Sager Combinatorial Integral Approximation 39

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71 Outline MIOC introduction Methods for Mixed-Integer Optimal Control... based on HJB / Dynamic Programming... based on Direct Methods MINLP MS MINTOC Switching Costs and Combinatorial Constraints Sebastian Sager Combinatorial Integral Approximation 41

72 Switching Costs and Combinatorial Constraints What if constraints depend explicitly on integer variables? 0 c(x(t), v(t), u(t)) Example: switching restrictions / costs on fixed control grid Problem: Sum Up Rounding solution infeasible! Sebastian Sager Combinatorial Integral Approximation 42

73 Switching Costs and Combinatorial Constraints What if constraints depend explicitly on integer variables? 0 c(x(t), v(t), u(t)) Example: switching restrictions / costs on fixed control grid Problem: Sum Up Rounding solution infeasible! One option: use MINLP solver Sebastian Sager Combinatorial Integral Approximation 42

74 Switching Costs and Combinatorial Constraints What if constraints depend explicitly on integer variables? 0 c(x(t), v(t), u(t)) Example: switching restrictions / costs on fixed control grid Problem: Sum Up Rounding solution infeasible! One option: use MINLP solver Alternative: decompose problem into NLP and MILP, making use of approximation theorem! Sebastian Sager Combinatorial Integral Approximation 42

75 Combinatorial Integral Approximation [S., Jung, Kirches, MMOR, to appear] Idea: minimize max t t 0 α(τ) ω(τ) dτ with binary controls ω( ) p that fulfill combinatorial constraints. Sebastian Sager Combinatorial Integral Approximation 43

76 Combinatorial Integral Approximation [S., Jung, Kirches, MMOR, to appear] Idea: minimize max t t 0 α(τ) ω(τ) dτ with binary controls ω( ) p that fulfill combinatorial constraints. min p max subject to k=1..n ω i=1..n t i max (q k,j p k,j ) t j j=1 σ k,max n t 1 i=1 p k,i p k,i+1, k = 1..n ω, p k,i {0, 1}, k = 1..n ω, i = 1..n t. (3) Given are: control discretization grid t j, maximum numbers of switches σ k,max, and relaxed solution α( ) q Sebastian Sager Combinatorial Integral Approximation 43

77 Combinatorial Integral Approximation [S., Jung, Kirches, MMOR, to appear] MINLP solution (Bonmin) >> 1800 sec NLP + MILP solution (CPlex) 360 sec Sebastian Sager Combinatorial Integral Approximation 44

78 Combinatorial Integral Approximation [S., Jung, Kirches, MMOR, to appear] MINLP solution (Bonmin) >> 1800 sec NLP + MILP solution (CPlex) 360 sec NLP + MILP solution (own code) 1.5 sec How do we get this MILP solution so fast? Sebastian Sager Combinatorial Integral Approximation 44

79 Cutting Planes? Number of facets of feasible polytope [PORTA Christof, Löbel, Reinelt] Control values q 1,j fixed to: n t depends heavily on relaxed solution q Hence cutting planes of limited use Sebastian Sager Combinatorial Integral Approximation 45

80 Cutting Planes? Number of facets of feasible polytope [PORTA Christof, Löbel, Reinelt] Control values q 1,j fixed to: n t depends heavily on relaxed solution q Hence cutting planes of limited use Instead use Branch&Bound without LP relaxations, making use of specific structure of inequalities Sebastian Sager Combinatorial Integral Approximation 45

81 Algorithm 1: Combinatorial Branch and Bound Input : Relaxed controls q, time grid {t i }, i = 1..n t, max. numbers of switches σ k,max, k = 1..n ω. Result: Optimal solution (η, s, p ) of (3). begin Create empty priority queue Q ordered by a.η (non-decreasing), if equal by a.d (non-increasing). Push an empty node (0, {}, {}, 0.0) into the queue. while Q is not empty do a = top node of Q and remove the node from Q. /* 1st solution found is optimal since best-first search is used */ if a.d = n t then Return optimal solution (a.η, a.s, a.p). /* Create child nodes, use strong branching. */ else forall possible permutations φ of {0, 1} nω do Create new node n with n.d = d + 1, n.p = a.p, n.s = a.s. Set n.p k,d+1 = φ k, calculate n.s k,d+1. if n.s fulfills switching { constraint until time d + 1 then n.η = max a.η, max nω k=1 {± } d+1 j=1 (q k,j p k,j ) t j } Push n into Q. end

82 Combinatorial Integral Approximation [S., Jung, Kirches, MMOR, to appear] THEOREM. Assume p SUR to be the solution obtained by Sum Up Rounding. The following claims hold for the optimal solution (η, s, p ) of the MILP (3): (a) η < 0.5 δt = 0.5 min i=1..n t t i (b) p = p SUR (c) σ k,max σk SUR k = 1..n ω (d) η 0.5 t = 0.5 max t i i=1..n t where the solution p = p SUR in (b) is unique. Sebastian Sager Combinatorial Integral Approximation 47

83 Combinatorial Integral Approximation [S., Jung, Kirches, MMOR, to appear] THEOREM. Assume p SUR to be the solution obtained by Sum Up Rounding. The following claims hold for the optimal solution (η, s, p ) of the MILP (3): (a) η < 0.5 δt = 0.5 min i=1..n t t i (b) p = p SUR (c) σ k,max σk SUR k = 1..n ω (d) η 0.5 t = 0.5 max t i i=1..n t where the solution p = p SUR in (b) is unique. Hence: MILP solution approximates MINLP solution when n t and if σ k,max are large enough Sebastian Sager Combinatorial Integral Approximation 47

84 CIA summary replace 1 MINLP by 1 NLP + 1 MILP generic approach (not only for switching constraint) for given grid t the solutions may be different, but asymptotic convergence as t 0 MILP solution can be used as fast UB in MINLP MILP solution can be used to initialize Switching Time Optimization MILP structure (e.g., max switching) may allow for ultrafast strategies Sebastian Sager Combinatorial Integral Approximation 48

85 Conclusions Many applications (with optimization potential) are MIOCPs Open benchmark library: Many different approaches compete Speaker s personal favorite: MS MINTOC exact bounds and error estimates can be combined with state-of-the-art numerics extensions towards NMPC, global, robust, multi-objective, DAE, PDE, combinatorial constraints,... are possible Survey article [S. Sager. Reformulations and algorithms for the optimization of switching decisions in nonlinear optimal control. Journal of Process Control, 2009.] Literature: Sebastian Sager Combinatorial Integral Approximation 49

86 Acknowledgements Hans Georg Bock Moritz Diehl Matthias Gerdts Johannes P. Schlöder Graduate School MathComp Heidelberg MathOpt group Thank you for your attention! Questions? Sebastian Sager Combinatorial Integral Approximation 50