5.6 Penalty method and augmented Lagrangian method

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1 5.6 Penalty method and augmented Lagrangian method Consider a generic NLP problem min f (x) s.t. c i (x) 0 i I c i (x) = 0 i E (1) x R n where f and the c i s are of class C 1 or C 2, and I and E are the sets of indices of the inequality and equality constraints. Notation, examples and proofs: see Chapter 17 of J. Nocedal, S. Wright, Numerical Optimization, Springer, 1999, p Edoardo Amaldi (PoliMI) Optimization Academic Year / 16

2 5.6.1 Quadratic penalty method Idea: Delete the constraints, add terms to the objective function which penalize the constraint violation, and solve a sequence of resulting unconstrained optimization problems. We describe the method for problems with only equality constraints: min f (x) s.t. c i (x) = 0 i E = {1,..., m} (2) x R n. Definition: The quadratic penalty function problem associated to problem (2) is where µ > 0 is the penalty parameter. 1 min Q(x, µ) = f (x) + x Rn 2µ ci 2 (x) (3) We consider {µ k } k 1 with lim k µ k = 0 and, for each k, we determine an approximate solution x k of (3) by using an unconstrained optimization method. i E Edoardo Amaldi (PoliMI) Optimization Academic Year / 16

3 Example: min x 1 + x 2 s.t. x x = 0 with optimal solution ( 1, 1) t. Quadratic penalty problem: 1 min Q(x, µ) = x1 + x2 + x R2 2µ (x x2 2 2) 2. For µ = 1 the minimizer of Q(x, 1) is close to ( 1.1, 1.1) t. For µ = 0.1 the minimizer of Q(x, 0.1) is much closer to ( 1, 1) t. Figures with the contours of Q(x, µ) for differenr values of µ: From J. Nocedal, S. Wright, Numerical Optimization, Springer, 1999, p Edoardo Amaldi (PoliMI) Optimization Academic Year / 16

4 General scheme 0) Select µ 0, sequence of tolerances {τ k } k 0 with τ k > 0 and lim k τ k = 0. Choose initial x s 0 and set k = 0. 1) Determine an approximate minimizer x k of Q(x, µ k ) starting from x s k when Q(x, µ k ) τ k. and terminate 2) If termination condition is satisfied (e.g, f (x k 1 ) f (x k ) < ε) Then return solution x k Else choose µ k+1 (0, µ k ) and starting x s k+1, set k = k + 1 and Goto 1) Choices: For the convergence results, it suffices that lim k τ k = 0. Sequence {µ k } k 0 generated adaptively starting from µ 0: if minimization of Q(x, µ k ) is difficult set for instance µ k+1 = 0.7µ k, otheriwse set µ k+1 = 0.1µ k. Judicious choice of the starting solution x s k when solving the unconstrained optimization problem at each iteration: x s k+1 := x k. Edoardo Amaldi (PoliMI) Optimization Academic Year / 16

5 Convergence Theorem 1: Suppose each x k is a global minimizer of Q(x, µ k ) and lim k µ k = 0, then every limit point x of the sequence {x k } k 0 generated according to the above scheme (with τ k = 0 for every k 0) is a global minimum of problem (2). Proof: Let x be an optimal solution of (2). Since x k is a global minimizer of Q(x, µ k ) and x is feasible for (2), Q(x k, µ k ) Q(x, µ k ) namely Thus f (x k ) + 1 2µ k m c 2 i (x k ) f (x) + 1 2µ k ci 2 (x) = f (x) k. (4) c 2 i (x k ) 2µ k [f (x) f (x k )] k. (5) Consider any convergent subsequence of {x k } k 0 with k K, such that lim k K x k = x. By taking the limit as k with k K in (5), we obtain (since µ k 0) ci 2 (x ) = lim ci 2 (x k ) lim 2µ k [f (x) f (x k )] = 0 (6) k K k K which implies that c i (x ) = 0 for each i E, that is x is feasible. Edoardo Amaldi (PoliMI) Optimization Academic Year / 16

6 By taking the limit as k with k K in (4), we obtain (since µ k 0 and c i (x k ) 2 0) f (x ) f (x 1 ) + lim ci 2 (x k K 2µ k ) f (x), k namely x is an optimal solution of (2). Since in general the unconstrained penalty problems are solved approximately, the following result is more relevant in practice. Theorem 2: If in the above scheme the tolerances τ k satisfy lim k τ k = 0 the penalty prameters satisfy lim k µ k = 0, then every limit point x of the sequence {x k } k 0 at which all c i (x ), with i E, are linearly independent is a KKT point of problem (2). For such points, the subsequence defined by K with lim k K x k = x satisfies lim c i(x k ) = ui i E, (7) k K µ k where u is the multiplier vector satisfying with x the KKT conditions for problem (2). Edoardo Amaldi (PoliMI) Optimization Academic Year / 16

7 Observation: (7) implies that i) The minimizer x k of Q(x, µ k ) does not satisfy c i (x) = 0 exactly, for all i E, but approximately, namely c i (x k ) = µ k ui for all i E. Thus µ k must be driven to 0 in order to obtain a feasible solution. ii) In some circumstances c i (x k ) µ k may be used as estimates of the Lagrange multipliers u i. Recall that the Lagrange function associated to the problem (2) is L(x, u) = f (x) u i c i (x) (8) and that the KKT conditions require that, apart from c i (x) = 0 for each i E, By comparing xl(x, u) = f (x) xq(x, µ) = f (x) + 1 µ and (9), it appears that c i (x) µ has been substituted with u i. u i c i (x) = 0. (9) c i (x) c i (x) = 0 (10) Edoardo Amaldi (PoliMI) Optimization Academic Year / 16

8 It can be proved that if τ k 0 then x k x and c i (x k ) µ k u i i = 1, 2,..., m. Simple approach but unfortunately Observation: When µ k 0 the quadratic penalty problem (3) becomes ill conditioned. 2 xxq(x, µ k ) = 2 f (x) + 1 µ k A t (x) A(x) + 1 µ k m c i (x) 2 c i (x) (11) where A t (x) = [ c 1(x),..., c m(x)] and A is m n and of full rank m n, with usually m < n. When x is close to the minimizer of Q(x, µ k ) and the assumptions of Theorem 2 are satisfied, (7) implies that 2 xxq(x, µ k ) 2 xxl(x, u ) + 1 µ k A t (x) A(x). (12) Since 2 xxl(x, u ) does not depend on µ k and 1 µ k A t (x) A(x) has n m eigenvalues of value 0 and m eigenvalues of value O(1/µ k ), numerical issues arise when µ k 0. Edoardo Amaldi (PoliMI) Optimization Academic Year / 16

9 Problems with both equality and inequality constraints: Consider the quadratic penalty problem 1 min Q(x, µ) = f (x) + x Rn 2µ where [y] denotes max( y, 0). i E c 2 i (x) + 1 2µ ([c i (x)] ) 2 (13) i I Other penalty functions are also available. In the case with only equality constraints c i (x) = 0, i E, the exact penalty problem is 1 min Q(x, µ) = f (x) + x Rn 2µ N.B.: Q is not everywhere differentiable. c i (x). (14) i E Edoardo Amaldi (PoliMI) Optimization Academic Year / 16

10 5.6.2 Augmented Lagrangian method Idea: Reduce ill-conditioning issues of the unconstrained subproblems (in the quadratic penalty method) by introducing explicit estimates of the Lagrange multipliers. We describe the method for problems with only equality constraints: min f (x) s.t. c i (x) = 0 i E = {1,..., m} (15) x R n. Definition: The augmented Lagrange function associated to problem (15) is L A (x, u, µ) = f (x) u i c i (x) + 1 ci 2 (x) = L(x, u) + 1 ci 2 (x), (16) 2µ 2µ where u is the multiplier vector and µ is the penalty parameter. N.B.: L A (x, u, µ) is a combination of L(x, u) and Q(x, µ). Since the KKT conditions of (15) require that xl(x, u ) = 0 and c i (x ) = 0 for every i E, at optimality L A coincides with the Lagrange function L, and µ does not need to tend to 0. Edoardo Amaldi (PoliMI) Optimization Academic Year / 16

11 Approach similar to the penalty method one: At each iteration µ k > 0 and we determine an approximate minimizer x k of L A (x, u k, µ k ) by using an unconstrained optimization method, where u k is an updated estimate. Differentiating with respect to x, we obtain xl A (x, u, µ) = f (x) (u i c i(x) µ ) c i(x). Considerations similar to those used to prove Theorem 2 allow to establish that which is equivalnt to u i u k i c i(x k ) µ k i E, (17) c i (x k ) µ k (u k i u i ) i E. (18) Edoardo Amaldi (PoliMI) Optimization Academic Year / 16

12 General scheme 0) Choose µ 0 > 0, sequence of tolerances {τ k } k 0 with τ k > 0 and lim k τ k = 0, x s 0 and initial u 0, set k := 0. 1) Determine an approximate minimizer x k of L A (x, u k, µ k ) starting from x s k and terminate when xl A (x, u k, µ k ) τ k. 2) If overall termination condition is satisfied (e.g, f (x k 1 ) f (x k ) < ε) Then Stop Else set u k+1 i = u k i c i(x k ) µ k for i E (19) choose µ k+1 (0, µ k ) and next starting solution x s k+1 set k := k + 1 and Goto 1) Edoardo Amaldi (PoliMI) Optimization Academic Year / 16

13 Including in L A an additional term related to the Lagrange multipliers leads to substantial improvements with respect to the quadratic penalty method. Example: min x 1 + x 2 s.t. x x = 0 with optimal solution ( 1, 1) t, optimal Lagrange multiplier u = 0.5 and unconstrained optimization subproblem min L A(x, u, µ) = x 1 + x 2 u(x 2 x R x2 2 2) + 1 2µ (x x2 2 2) 2. Suppose that µ k = 1 and the current estimate of the multiplier u k = 0.4. The contours of L A (x, 0.4, 1) are similar to those of Q(x, 1) but the minimizer x k ( 1.02, 1.02) t of L A (x, 0.4, 1) is much closer to ( 1, 1) t than the minimizer of Q(x, 1), which is ( 1.1, 1.1) t. From J. Nocedal, S. Wright, Numerical Optimization, Springer, 1999, p Edoardo Amaldi (PoliMI) Optimization Academic Year / 16

14 Theorem 3: Let x be a local minimum of (15) at which the c i (x ), i E, are linearly independent and the second order sufficient optimality conditions are satisfied for u = u. Then there exists a threshold value µ > 0 such that for all µ (0, µ], x is a strict local minimum of L A (x, u, µ). N.B.: In general u is not known. The next result - is concerned with the more realistic case in which u u, - provides conditions under which there exist a minimizer of L A close to x and error bounds on x k and on the multiplier estimate u k+1. Edoardo Amaldi (PoliMI) Optimization Academic Year / 16

15 Theorem 4: Suppose that the assumptions of Theorem 3 are satisfied at x and u, and let µ > 0 be the corresponding threshold. Then there exist scalars δ > 0, ε > 0, and M such that i) For all u k and µ k satisfying the problem u k u δ/µ k, µ k µ, (20) min L A(x, u k, µ k ) x R n : x x ε has a unique solution x k. Moreover, we have x k x Mµ k u k u. ii) For all u k and µ k satisfying (20), we have where u k+1 is given by the formula (19). u k+1 u Mµ k u k u, iii) For all u k and µ k satisfying (20), the matrix 2 xxl A (x k, u k, µ k ) is positive definite and the c i (x k ), with i E, are linearly independent. Edoardo Amaldi (PoliMI) Optimization Academic Year / 16

16 Problems with also inequality constraints: We can introduce slack variables and substitute c i (x) 0, i I, with c i (x) s i = 0, s i 0, i I. In the LANCELOT solver, the bounds on the variables are explicitly taken into account in the subproblem min l inf x l sup L A (x, u k, µ k ). Edoardo Amaldi (PoliMI) Optimization Academic Year / 16

5.5 Quadratic programming

5.5 Quadratic programming Minimize a quadratic function subject to linear constraints: 1 min x t Qx + c t x 2 s.t. a t i x b i i I (P a t i x = b i i E x R n, where Q is an n n matrix, I and E are the