Constrained optimization: indirect methods

Transcription

1 Constrained optimization: indirect methods Jussi Hakanen Post-doctoral researcher

2 On constrained optimization We have seen how to characterize optimal solutions in constrained optimization KKT optimality conditions include the balance of forces ( f x, g i x, i I and h j (x )) and complementarity conditions (μ i g i x = 0 i) Regularity of x need to be assumed Now, we are interested in how to find such solutions

3 Methods for constrained optimization Many methods utilize knowledge about the constraints Linear inequalities or linear equalities Nonlinear inequalities or equalities For example, if a linear constraint is active at some point, you know that by taking steps along the direction of the constraint, it remains active For nonlinear constraints, you don t have such a direction Methods for constrained optimization can be characterized based on how they treat constraints

4 Classification of the methods Indirect methods: the constrained problem is converted into a sequence of unconstrained problems whose solutions will approach to the solution of the constrained problem, the intermediate solutions need not to be feasible Direct methods: the constraints are taking into account explicitly, intermediate solutions are feasible

5 Transforming the optimization problem Constraints of the problem can be transformed if needed g i x 0 g i x + y i 2 = 0, where y i is a slack variable; constraint is active if y i = 0 By adding y i 2 no need to add y i 0 If g i (x) is linear, then linearity is preserved by g i x + y i = 0, y i 0 g i x 0 g i x 0 h i x = 0 h i x 0 & h i x 0

6 Examples of indirect methods Penalty function methods Lagrangian methods

7 Penalty function methods Include constraints into the objective function with the help of penalty functions that penalize constraint violations or even approaching the boundary of S Different types Penalty function: penalize for constraint violations Barrier function: prevents leaving the feasible region Exact penalty function Resulting unconstrained problems can be solved by using the methods presented earlier in the course

8 Penalty function methods Generate a sequence of points that approach the feasible region from outside Constrained problem is converted into min f x + r α(x), x Rn where α(x) is a penalty function and r is a penalty parameter Requirements: α x 0 x R n and α x = 0 if and only if x S

9 On convergence When r, the solutions x r of penalty function problems converge to a constrained minimizer (x r x and rα x r 0) All the functions should be continuous For each r, there should exist a solution for penalty functions problem and {x r } belongs to a compact subset of R n

10 Examples of penalty functions Can you give an example of a penalty function α(x)? For equality constraints l i=1 or h i x = 0 α x = h i x 2 α x = l i=1 h i (x) p, p 2 For inequality constraints g i x 0 α x = i=1 max 0, g i (x) or m α x = max 0, g i x p, p 2 m i=1

11 How to choose r? Should be large enough in order for the solutions be close enough to the feasible region If r is too large, there could be numerical problems in solving the penalty problems For large values of r, the emphasis is on finding feasible solutions and, thus, the solution can be feasible but far from optimum Typically r is updated iteratively Different parameters can be used for different constraints (e.g. g i r i, g j r j ) For the sake of simplicity, same parameter is used here for all the constraints

12 Algorithm 1) Choose the final tolerance ε > 0 and a starting point x 1. Choose r 1 > 0 (not too large) and set h = 1. 2) Solve min f x + x Rn rh α(x) with some method for unconstrained problems (x h as a starting point). Let the solution be x h+1 = x(r h ). 3) Test optimality: If r h α x h+1 < ε, stop. Solution x h+1 is close enough to optimum. Otherwise, set r h+1 > r h (e.g. r h+1 = κr h, where κ can be initialized to be e.g. 10). Set h = h + 1 and go to 2).

13 From Miettinen: Nonlinear optimization, 2007 (in Finnish) Example min x s. t. x Let α x = max [0, ( x + 2)] 2 Then α x = 0, if x 2 x + 2 2, if x < 2 Minimum of f + rα is at 2 1 2r

14 Barrier function method Prevents leaving the feasible region Suitable only for problems with equality constraints Set x g i x < 0 i} should not be empty Problem to be solved is min Θ r s. t. r 0, r where Θ r = inf f x + rβ x g i x < 0 i] x β is a barrier function: β x 0 when g i x < 0 i and β x when x approaches boundary of S Constraints g i x < 0 can be omitted since β in the boundary of S

15 On convergence Denote Θ r = f x r + rβ(x r ) Under some assumptions, the solutions x r of barrier problems converge to a constrained minimizer (x r x and rβ x r 0) when r 0 + All functions should be continuous x g i x < 0 i}

16 Properties of barrier functions Nonnegative and continuous in x g i x < 0 i} Approaches when the boundary of the feasible region is approached from inside Ideally: β = 0 in x g i x < 0 i} and β = in the boundary Guarantees staying in the feasible region This kind of discontinuity causes problems for any numerical method Examples of barrier functions m β x = 1 i=1 g i x m i=1 β x = ln (min[1, g i (x)])

17 Algorithm 1) Choose the final tolerance ε > 0 and a starting point x 1 s.t. g i x < 0 i. Choose r 1 > 0, not too small (and a parameter 0 < τ < 1 for reducing r). Set h = 1. 2) Solve min f x + r h β(x) s. t. g i x < 0 i x by using the starting point x h. Let the solution be x h+1. 3) Test optimality: If r h β x h+1 < ε, stop. Solution x h+1 is close enough to optimum. Otherwise, set r h+1 < r h (e.g. r h+1 = τr h ). Set h = h + 1 and go to 2).

18 From Miettinen: Nonlinear optimization, 2007 (in Finnish) min x s. t. x Let β x = 1 when x 1 x+1 Minimum of f x + rβ x = x + r x 1 1 is at 1 + r Example

19 Summary: penalty and barrier function methods Penalty and barrier functions usually differentiable Minimum is obtained in a limit Penalty function: r h Barrier function: r h 0 Choosing the sequence r h essential for convergence If r h or r h 0 too slowly, a large number of unconstrained problems need to be solved If r h or r h 0 too fast, solutions of successive unconstrained problems are far from each other and solution time increases

20 Exact penalty function Idea is to have a method where the solution could be found with a small amount of iterations Suitable for both equality and inequality constraints An exact penalty function problem is e.g. of the form m l i=1 ) min f x + r( max[0, g x R n i x ] + i=1 h i (x)

21 Exact penalty function method Theorem: Consider a point x where the necessary KKT conditions hold. Let the corresponding Lagrange multipliers be μ and ν. Assume that objective and inequality constraint functions are convex and equality constraint functions are affine. Then x is a solution of the exact penalty function problem with r max[μ i, i = 1,, m, ν i, i = 1,, l] Solution can be obtained with a finite value for the penalty parameter r Algorithm is similar to penalty function method except for that r h is increased only if necessary E.g. when the feasible region is not approached fast enough

22 Properties of exact penalty function Not differentiable in points x where g i x = 0 or h i x = 0 Gradient based methods are not suitable If r and the starting point could be chosen appropriately, only one minimization would be required in principle If r is too large and the starting point is not close enough to the optimum, minimizing the exact penalty function problem could become difficult

23 Example min f x = x x 2 s. t. x 1 + x 2 1 = 0 Optimal solution is x = 1, 1 T, ν = 2x = 2x 2 = 1 Exact penalty function problem: min x x R n x r x 1 + x 2 1 Solution: x = r, r T when 0 r < 1 and x = 1, 1 T when r 1 (obtained by using KKT conditions of an equivalent differentiable problem where the absolute value term is replaced with a new variable and two inequality constraints) Thus, the solution can be found with r 1 (= ν )

24 From Miettinen: Nonlinear optimization, 2007 (in Finnish) Example: barrier function 2 min f x = x 1 x 2 s. t. x 2 1 +x x = , T, the constraint is active in x (a) level curves of f(x) and the boundary of S Logarithmic barrier function: (b) r = 0.2, (c) r = 0.001

25 From Miettinen: Nonlinear optimization, 2007 (in Finnish) Example: penalty function 2 min f x = x 1 x 2 s. t. x 2 1 +x x = , T, the constraint is active in x Quadratic penalty function: (a) r = 1, (b) r = 100

26 From Miettinen: Nonlinear optimization, 2007 (in Finnish) Example: exact penalty function min f x = x 1 x 2 2 s. t. x 1 2 +x x = , T, the constraint is active in x, μ = Exact penalty function: (a) r = 1.2, (b) r = 5, (c) r = 100

27 Lagrangian function Consider problem min f(x) s. t. h i x = 0, i = 1,, l Lagrangian function l L x, ν = f x + ν i h i (x) i=1 KKT conditions l f x + i=1 ν i h i (x) = 0 h i x = 0, i = 1,, l Let x be a minimizer and ν corresponding Lagrange multiplier

28 Properties of Lagrangian KKT conditions: x is a critical point of the Lagrangian function x is not necessarily a minimizer for L(x, ν ) Thus, minimizing the Lagrangian function doesn t necessarily give a minimum for f(x) Hessian xx 2 L(x, ν ) may be indefinite a saddle point Improve Lagrangian function!

29 Augmented Lagrangian function Augmented Lagrangian function: l L A x, ν, ρ = f x + i=1 ν i h i (x) + 1 ρ l h 2 i=1 i x 2, ρ > 0 Lagrangian function + quadratic penalty function A point (x, ν ) is a critical point of the augmented Lagrangian x L A x, ν, ρ = 0 and 1 ρ 2 l i=1 h i x 2 = 0 Hessian: 2 xx L A x, ν, ρ = 2 xx L x, ν + ρ h x T h(x ) It can be shown that for ρ > ρ, 2 xx L A x, ν, ρ is positive definite x is a local minimizer of L A x, ν, ρ Need to know ν

30 Properties of L A (x, ν, ρ) Differentiable if the original functions are x is a minimizer of L A x, ν, ρ for finite ρ Lagrangian function + quadratic penalty function

31 Algorithm 1) Choose the final tolerance ε > 0. Choose x 1, ν 1 i (i = 1,, l) and ρ. Set h = 1. 2) Test optimality: if optimality conditions are satisfied, stop. The solution is x h. 3) Solve (with a suitable method) min x R n A x, ν h, ρ by using x h as a starting point. Let the solution be x h+1. 4) Update Lagrange multipliers: e.g. ν h+1 = ν h + ρh x h+1. 5) Increase ρ if necessary: e.g. if h(x h ) h x h+1 < ε. 6) Set h = h + 1 and go to 2). Note: x h x only if ν h ν

32 From Miettinen: Nonlinear optimization, 2007 (in Finnish) Example 2 min f x = x 1 x 2 s. t. x 2 1 +x x = , T, the constraint is active in x Lagrangian function: saddle point in x

33 From Miettinen: Nonlinear optimization, 2007 (in Finnish) Example (cont.) Augmented Lagrangian function (a) ρ = 0.075, (b) ρ = 0.2, (c) ρ = 100

34 From Miettinen: Nonlinear optimization, 2007 (in Finnish) Example (cont.) Augmented Lagrangian function ν = , ρ = 0.2 (a) ν = 0.5, (b) ν = 0.9, (c) ν = 1.0

35 Topic of the lectures next week Mon, Feb 10 th : Constrained optimization: gradient projection, active set method Wed, Feb 12 th : Constrained optimization, SQP method & Matlab Study this before the lecture! Questions to be considered What is the basic idea of gradient projection? What is the basic idea of active set methods? What is the basic idea of Sequential Quadratic Programming (SQP)?