Static and Dynamic Optimization (42111)

Static and Dynamic Optimization (42111)Niels Kjølstad PoulsenSection for Mathematical StatisticsDept. of Informatics and Mathematical ModellingThe Technical University of Denmarkbuild. 305, room 213email: nkp@imm.dtu.dkphone: +45 4525 3356mobile: +45 2890 3797Lecture 13: Time Optimal Problems1 / 27

Outline of lectureTime optimal problemsReading: 47-54, 67-692 / 27

Motion controlThe problem is to bring the systemẋ = u t x 0 = x 0from the initial position, x 0 , to the final position, x T , in minimum time,while the control action (or the decision function) is bounded to|u t| ≤ uThe performance index is∫ T ∫ TJ = T T + 0 dt = 0+ 1 dt0 0Notice, either is (φ T = T , L = 0) or (φ = 0, L = 1)The Hamiltonian function is:H = λ tu t5 / 27

Motion controlThe Hamiltonian function is:H = λ tu tand the optimality conditions are simplyẋ = u t−˙λ = 0u t= arg min uHwith the boundary conditions:x 0 = x 0 x T = x T λ T = νThe Transversality condition is1+λ T u T = 06 / 27

Motion control (Costate)The Hamiltonian function is:H = λ tu tand the optimality conditions are simplyẋ = u tλ = νu t= arg min uHwith the boundary conditions:x 0 = x 0 x T = x T λ T = νThe Transversality condition is1+λ T u T = 07 / 27

Motion control (Control)The Hamiltonian function is:H = λ tu tand the optimality conditions are simplyẋ = u tλ = νu t = −u sign(λ t) = −u sign(ν)with the boundary conditions:x 0 = x 0 x T = x T λ T = νThe Transversality condition is1+λ T u T = 08 / 27

Motion control (Transversality)The Hamiltonian function is:H = λ tu tand the optimality conditions are simplyẋ = u tλ = νu t = −u sign(ν)with the boundary conditions:x 0 = x 0 x T = x T λ T = νThe Transversality condition is1−νu sign(ν) = 0 → ν = ± 1 u9 / 27

Motion control (State equation)The Hamiltonian function is:H = λ tu tand the optimality conditions are simplyx t = x 0 −u sign(ν) t and x T = x 0 −u sign(ν) Tλ = νu t = −u sign(ν)with the boundary conditions:x 0 = x 0 x T = x T → T = |x T −x 0 |uThe Transversality condition issign(ν) = −sign(x T −x 0 )λ T = νν = ± 1 u→ ν = − 1 u sign(x T −x 0 )10 / 27

Motion control (Conclusion)This results in the control strategyu t = u sign(x T −x 0 )andx t = x 0 +u sign(x T −x 0 ) t11 / 27

Pause12 / 27

Bang Bang Control.Consider a mass affected by a force. This is a second order system given by[ ] [ ] [ ] [ ]d z v z z0==dt v u v v0 0z position v velocity u control action, specific forceAssume the control action, i.e. the specific force is limited to|u| ≤ 1while the objective is to take the system from its original state to the origin[ ] [ ] zT 0x T = = vT 0in minimum time. The performance index is accordinglyJ = TThe Hamilton function isH = λ z v +λ v u♠13 / 27

Bang Bang Control.The Hamilton function and the terminal contribution areH = λ z v +λ v uφ = TWe can now write the conditions as the state equation,[ ] [ ]d z v=dt v uthe costate equations− d dt[ λ z λ v ] = [ 0 λ z ] ⇐the optimality condition (Pontryagins maximum principle)u t = −sign(λ v t )and the boundary conditions[ ] [ ] [ ] [ z0 z0 zT 0= =v 0 v 0 vT 0The transversality condition is:] [ λ zTλ v T] [ ] ν z=ν v1+H T = 1+λ z T v T +λ v T u T = 1+λ v T u T = 0 ⊲⊳ ♠⊲⊳14 / 27

Bang Bang Control.The Hamilton function isH = λ z v +λ v uWe can now write the conditions as the state equation,[ ] [ ]d z v=dt v uthe costate equationsλ z t = νz λ v t = νv +ν z (T −t) ⊲⊳the optimality condition (Pontryagins maximum principle)u t = −sign(λ v t)⇐and the boundary conditions[ ] [ ] [ ] [ z0 z0 zT 0= =v 0 v 0 vT 0The transversality condition is:] [ λ zTλ v T] [ ] ν z=ν v1+λ v T u T = 0 → λ v T = νv = ±1 ⊲⊳15 / 27

Bang Bang Control.To summarize,we have 3 unknown quantities, ν z , ν v and T and 3 conditions to meet,z T = 0 v T = 0 and λ v T = ±1.If the control has a constant values u = ±1, then the solution is simplyv t= v 0 +utz t = z 0 +v 0 t+ 1 2 ut2 16 / 27

Bang Bang Control.u = 1 (Accelration)v t = v 0 +tz t= z 0 +v 0 t+ 1 2 t22Trajectories for u=11.510.5v0−0.5−1−1.5−2−4 −3 −2 −1 0 1 2 3 4 5z17 / 27

Bang Bang Control.u = −1 (Deacceleration)v tz t= v 0 −t= z 0 +v 0 t− 1 2 t22Trajectories for u=−11.510.5v0−0.5−1−1.5−2−5 −4 −3 −2 −1 0 1 2 3 4z18 / 27

Bang Bang Control - On the switching curve.No switching if the terminal point (origin) and the starting point is on the same parabola (and inthe directionis of arrows), i.e. satisfy the equations0 = v 0 +uT f u = ±1 (1)0 = z 0 +v 0 T f + 1 2 uT2 fwhere T f = T . Solution:T f = − v 0u ≥ 0 z 0 = 1 v022 ufor either u = 1 or u = −1. Since T f ≥ 0:(must be satisfied) (2)u = −sign(v 0 ) T f = |v 0 |z0 = − 1 2 sign(v 0)v 2 0(The Switching curve)If v 0 > 0 then u = −1 and z 0 = − 1 2 v2 0 .If v 0 < 0 then u = 1 and z 0 = 1 2 v2 0 . 19 / 27

Bang Bang Control.The switching (or the acceleration/deaccelleration) curve:{ −1z 0 = 2 v2 0 deacc. u = −1 for v 0 > 012 v2 0 acc. u = 1 for v 0 < 0= − 1 2 v2 0 sign(v 0)2Trajectories for u=± 11.510.5v0−0.5−1−1.5−2−5 −4 −3 −2 −1 0 1 2 3 4 5z20 / 27

Bang Bang Control.Let us return to the general case.From Pontryagin and the solution to the costate equation:( )u t = −sign ν v +ν z (T −t)There will be a switching unless the initial point lies on switching curve. We will have 2 possibilities.u t ={ 1 for 0 ≤ t ≤ Ts−1 for T s < t ≤ Tz 0 < − 1 2 v2 0sign(v 0 )Below the switching curve.{ −1 for 0 ≤ t ≤ Tsu t =1 for T s < t ≤ Tz 0 > − 1 2 v2 0 sign(v 0)Above the switching curve.21 / 27

Bang Bang Control.v2Trajectories for u=± 11.510.50−0.5−1−1.5−2−5 −4 −3 −2 −1 0 1 2 3 4 5zAs a state feedback law⎧1 for z < − 1 2 v2 sign(v) (Below switching curve)⎪⎨ −1 for z = − 1 2 v2 and v > 0 (Down along switching curve)u t =−1 for z > − 1 2⎪⎩v2 sign(v) (Above switching curve)1 for z = 1 2 v2 and v < 0 (Up along switching curve)22 / 27

Bang Bang Control.Searching for T and T s. Assume z 0 > − 1 2 v2 0 sign(v 0) (above switching curve)2Trajectories for u=± 11.510.5v0−0.5−1−1.5−2−5 −4 −3 −2 −1 0 1 2 3 4 5zFor t < T s u t = −1 and T s < t ≤ T u t = 123 / 27

Bang Bang Control.For t < T s u t = −1 andv t = v 0 −t z t = z 0 +v 0 t− 1 2 t2 (1)which is valid until we reach the switching curve given byz = − 1 2 v2 sign(v) = 1 2 v2 (because v < 0) (2)This happens at T s wherez 0 +v 0 T s − 1 2 T2 s = 1 2 (v 0 −T s) 2orT s = v 0 +√z 0 + 1 2 v2 024 / 27

Bang Bang Control.The velocity at the switching point isv s = v 0 −T s (≤ 0)The time left for the origin is˜T = |v s| = T s −v 0For t > T s we accelerate (u t = 1) and in total√T = ˜T +T s = T s −v 0 +T s = v 0 +2 z 0 + 1 2 v2 025 / 27

6Contour areas for T=2, 4 and 642v0−2−4−6−20 −15 −10 −5 0 5 10 15 20z26 / 27

Project 2Name and ID on frontName(s) and ID(s) of group mates on front (separated from ovn id)Only electronic upload to CampusNet.Deadline 21.12.2012helpdesk: nkp (305/213)27 / 27