Computational Optimization. Constrained Optimization Part 2

Transcription

1 Computational Optimization Constrained Optimization Part

2 Optimality Conditions Unconstrained Case X* is global min Conve f X* is local min SOSC f ( *) = SONC

3 Easiest Problem Linear equality constraints min f( ) f R s.. t A = b A R, b R n m n m

4 Assume world is smooth Objective and constraints are twice continuously differentiable for the purpose of this lecture.

5 Net Easiest Problem Linear equality constraints min f( ) f R s.. t A b A R, b R n m n m Constraints form a polyhedron

6 Only some constraints are active. a = b a = b At *, only the constraint a >=b is at bound/active. Polyhedron A>=b a a 3 = b f (*) a 4 = b a * a = b contour set of function constrained minimum unconstrained minimum At optimal solution, *, only one constraint is active.

7 In this eample, the problem is equivalent to minimizing over a = b Polyhedron A>=b active set (if you knew it) a = b a Equality FONC: a 3 = b3 * min f( ) st.. a= b a 4 = b4 a f (*) * a = b Equality FONC: f (*) =λ a a ' * = b Note λ i >=

8 But in this case * doesn t solve the inequality problem Polyhedron A>=b a 5 = b 5 a = b a Equality FONC: a 3 = b3 3 * min f( ) st.. a= b Equality FONC: a 4 = b 4 a * f a = b (*) f (*) =λ a a ' * = b Now λ <. SIGN OF λ matters!

9 Polyhedron A>=b a 4 = b 4 Inequality Case a 5 = b 5 a =b 5 a f (*) * a a 3 = b 3 a = b Inequality problem * min f( ) st.. a b =... d Inequality FONC: T f (*) = A λ* = λ a A* b λ ( A* b) = i i i λ Nonnegative Multipliers imply gradient points to the greater than Side of the constraint. i i i i i i

10 Complementarity Condition a 5 = b 5 a =b 5 Inequality FONC: Polyhedron A>=b a a 3 = b 3 f (*) a 4 = b 4 a * a = b λ ( a* b) = i i i λ > a* = b ie.. constraint is active i i i λ = constraint doesn't matter. It can be dropped without i changing result.

11 Lagrangian Multipliers A if i i i i λ represents sensitivity i b λ > then a = b i If λ > then increasing a ˆ > b causes the objective to increase. If b is changed to make feasible region bigger then the objective will decrease. i i

12 First Order Necessary Conditions for Linear inequality Constraints If * is a local min of f over { A b}, then for some vector λ* f(*) A' λ* = A* b λ* λ *'( A * b) =. KKT

13 Second Order Necessary Conditions for Linear Constraints If * is a local min of f over { A b}, and Z is a null-space matri for active constraints then for some vector λ* f(*) A' λ* = A* b λ* λ *'( A * b) = ' ( *)... and Z f Z is p s d KKT

14 Z for FONC Z is null space matri for matri of active constraints at * Inde set of Active Constraints I = { i ai* = bi} Matri of active constraints A I is A restricted to rows Z = null( A I ) of Ain I

15 Every d satisfying Intuition f( )' d < Ad is a feasible descent direction at. The FONC make sure that this set is empty. f ( )' d < if and only if f ( ) = A' λ Ad Due to Farkas s Lemma λ has has no solution d a solution

16 Second Order Sufficient Conditions for Linear Inequalities If (*,λ*) satisfies A* b Primal feasibility f ( *) = A' λ * λ* Dual feasibility λ *'( A * b) = Complementarity and SOSC Z+ ' f ( *) Z+ is pd.. Then * is a strict local minimizer

17 Sufficient Conditions for Linear Inequalities where Z + is a basis matri for Null(A + ) and A corresponds to + nondegenerate active constraints) i.e. A + = A J { j A * = b, λ > } j j * j

18 Sufficient Eample Find solution and verify SOSC min /( + ) + / st.. * = [,]' λ* = [,]

19 Linear Inequality Constraints - I min s.t. Put min s.t. in ( + ) standard ( + ) form + + : [ ] ( *) f (*) + f (*) = = = Active constraint is A = - + * = by inspection

20 Linear Inequality Constraints - II ( ) [ ] - A is Active constraint *) ( * *) ( = = = + = + f f

21 Linear Inequality Constraints - III KKT Conditions A* = b f(*) = A T λ λ * λ *( A* b) =? λ = since second constraint is inactive = λ = λ λ = T f(*) A K K T Point: * = λ * =

22 Linear Inequality Constraints - IV Now look at SOSC A = [ ] Z = λ * = f(*) = Z + + ' + [] f(*) Z = a p.d. matri + Therefore SOSC are satistified. X* is a strict local minima

23 Why Necessary and Sufficient? Sufficient conditions are good for? Way to confirm that a candidate point is a minimum (local) But not every min satisfies any given SC Necessary tells you: If necessary conditions don t hold then you know you don t have a minimum. Under appropriate assumptions, every point that is a min satisfies the necessary cond. Good stopping criteria Algorithms look for points that satisfy Necessary conditions

24 You Try Solve the problem using above theorems: Verify *=[8/9,4/8] is optimal by checking KKT. Also check SOSC min / + st.. +

25 General Constraints min f( ) st.. g ( ) = i E i g ( ) i I i

26 Active Set The active set at a point consists of Equality constraints Inequality constraints at bound A( ) = E { i i I, g ( ) = i }

27 Lagrangian Function Optimality conditions epressed using Lagrangian function L(, λ) = f ( ) λigi( ) = f ( ) λ ' g( ) i = and Jacobian matri m g( )' were each row is a gradient of a constraint

28 Lagrangian Function Lagrangian Function L(, λ) = f ( ) λigi( ) = f ( ) λ ' g( ) i = Lagragian Gradient Lagragian Hessian m m L(, λ) f ( ) λ g ( ) = i i i = m λ λi i i = L(, ) = f ( ) g ( )

29 Feasible descent directions A point is not optimal at if we can take a small step s that is feasible and decrease the function min f (, ) s. t.

30 Feasible descent/improvement directions Case : point in interior min f (, ) s. t. f ( ) s Can only be optimal if f( ) =

31 Feasible descent/improvement directions Case : point at bound s f () g () min s. t. f (, Any point in cone is an Improvement directions g ( ) s Can only be optimal if Has no solution f ( ) f ( )' s< g ( )' s )

32 Feasible descent/improvement directions min f (, ) This has no as solution s s. t. f (*)' s< g (*)' s Same as saying there eists λ* λ Or equivalently f (*) = * g (*) λ* L (*, λ*) = λ* s g ( ) f ( )

33 General Nonlinear Constraints Want to solve min f ( ) st.. g ( ) = g ( ) Take FOTSE about a neighborhood of to pick feasible descent directions at assuming both constraints active g ( + d) g ( ) + g ( )' d i i i f( + d) f( ) + f( )' d The set of g ( )' d =, g ( )' d, and f( )' d < is the set of feasible descent directions for the linearized problem.

34 The set Rough idea g ( )' d =, g ( )' d, f( )' d < has no solution d, if and only if f()= g ( ) λ + g ( ) λ, λ or equivalently L(, λ, λ )=, λ has a solution So same KKT as for linear case should work as long as linearization is okay. (waving hands)

35 Sufficient Conditions Nonlinear Equality If (*,λ*) satisfies g( *) = Primal feasibility f ( *) = g( *)' λ * ( equivalently L(*, λ*)=) Dual feasibility and SOSC Z ' L( *) Z is pd.. Then * is a strict local minimizer

36 Sufficient Conditions Nonlinear Equality where Z is a basis matri for Null(A ) and A is Jacobian of active constraints i.e. For the jth row of A g j j ( *) = A ctive C onstraint A = g (*)' j

37 Second Order Sufficient Conditions Nonlinear If (*,λ*) satisfies Inequality g( *) Primal feasibility f ( *) = g( *)' λ * ( equivalently L(*, λ*)=) Dual feasibility λ* (for inequalities only) λ *' g( *) = Complementarity and SOSC Z+ ' L( *) Z+ is pd.. Then * is a strict local minimizer

38 Second Order Sufficient Conditions Nonlinear Inequality where Z + is a basis matri for Null(A i.e. + ) and A corresponds to Jacobian + of nondegenerate active constraints) For the jth row of Jacobian g λ j * j ( *) = A ctive C onstraint > ( ) + Nondegenerate A = g ( *) ' j j

39 Sufficient Eample Find solution and verify SOSC min /( + ) / + st.. / / / * = [,]' λ* =

40 = = + = = + + *) ( ) ( ) ( * Guess s.t. ) ( min - f f Nonlinear Inequality Constraints - I

41 Nonlinear Inequality Constraints - II Has one active constraint: + T Jacobian: g() = [ ] = [ ] L (, λ) = f( ) + λ' g ( ) L (, λ) = f( ) λ g( ) - = λ =

42 Nonlinear Inequality Constraints - III L (, λ) = f( ) λ ' g ( ) λ λi i i L (, ) = f( ) g( ) - = = positive definite = = [ ] T Z +, since g() - T Z+ L(, λ) Z+ = [ ] positive definite = so SOSC satisifed, * is a strict local minimum

43 Sufficient Eample Find solution and verify SOSC min st.. = * = [,]' λ* =

44 Nonlinear Inequality Constraints - V min s.t. - = * = by observation L(, λ) = f( ) λ' g( ) = - λ( - ) ' L(, λ) f( ) λ g( ) = = λ = L (*, λ*) = = λ* = λ* =

45 Nonlinear Inequality Constraints - VI T g ( ) = [ ] Z+ = λ λi i i L (, ) = f( ) g( ) = ( ) = T Z+ L(, λ) Z+ = [ ] positive definite = So SOSC are satisfied, and * is a strict local minimum.

46 CAREFUL Sometimes things aren t nice and the KKT conditions don t eist even when a point is a min. We must impose additional conditions (constraint qualifications) to make sure the KKT conditions eist.

47 Eample with no KKT point min s. t. f (,) = y y g(,) = g(,) = f(,) + λ g (,) + λ g (,) for any λ

48 What s wrong Set of linearized feasible directions do not match the real set of feasible directions that can be achieved in the limit (called the tangent cone). (Take NLP in fall to learn more about tangent cones)

49 Constraint Qualification Linear Independence CQ (LICQ) gradients of active constraints are linearly independent (also called regular point) Constraints are linear Conve program with strictly interior feasible points More eists that make sure linearization works out okay.

50 First Order Necessary Conditions- Equality If * is a local min of f over { g()=}, and LICQ (or other CQ) then there eists λ * L (*, λ*) = g (*) =

51 Second Order Necessary Conditions- Equality If * is a local min of f over { g()=}, Z is a null-space matri of the Jacobian g(*), and LICQ then there eists λ * L ( *, λ*) = or equivalently Z' f( *) = g (*) = ' ( *)... and Z L Z is p s d

52 Necessary Conditions Inequality If * is a local min of f over { g()>=}, Z is a null-space matri of the Jacobian g(*) of the active constraints, and * is a regular point (LICQ) then there eists λ * L ( *, λ*) = or equivalently Z' f( *) = g (*) λ g = ' * ( *) and Z L Z is p s d ' ( *)...

53 LICQ=Regular point If * is a regular point with respect to the constraints g(*) if the gradient of the active constraints are linearly independent. For equality constraints, all constraints are active so g (*)' I should have linearly independent rows.

54 Necessary Eample Show optimal solution *=[,] is regular and find KKT point ma + 3 st.. min ( ) 3 st.. ( + ) + ( ) + KKT point * = [, ] ' λ * = [ /, ] '

55 Constraint Qualifications Regularity or LICQ is an eample of a constraint qualification CQ. The KKT conditions are based on linearizations of the constraints. CQ guarantees that this linearization is not getting us into trouble. Problem is KKT point might not eist. There are many other CQ,e.g., for conve inequalities Slater is there eists g()<. Note CQ not needed for linear constraints.

56 Necessary Conditions General If * satisfies LICQ and is a local min of f over { g()>=,h()=}, there eists λ, λ * * I E * * = L ( *, λi*, λe*) = f( *) + λi gi( *) + λi hi( *) i i g (*), h (*) = λ g ' * ( *) = and Z L Z is p s d where Z constraints) ' ( *).. is the null space matri for A( *) (active

57 General SOSC Consider problem min of f over { g()>=,h()=}, * * If there eists *, λi, λe such that * * = L ( *, λi*, λe*) = f( *) + λi gi( *) + λi hi( *) i i g (*), h (*) = ' λ * ( *) = + ' ( *) +.. g andz L Z ispd where Z is the null space matri for A ( *) (active constraints with + + positive multipliers) then * is a strict local minimizer

58 KKT Summary X* is global min X* is local min Conve f Conve constraints CQ SOSC KKT Satisfied