8 Barrier Methods for Constrained Optimization

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1 IOE 519: NL, Winter 2012 c Marina A. Epelman 55 8 Barrier Methods for Constrained Optimization In this subsection, we will restrict our attention to instances of constrained problem () that have inequality constraints only: S = {x 2 R n : g i (x) apple 0, i =1,...,n}. The algorithm we develop can be easily extended to problems that also have linear equality constraints, in the spirit of the previous section. Barrier methods can also be applied to problems with general equality constraints, however, an appropriate problem transformation is required. Di erent implementations of barrier methods deal with this issue di erently. Barrier methods (also known as Interior oint Methods) are designed to solve () by instead solving a sequence of specially constructed unconstrained optimization problems. In a barrier method, we presume that we are given a point x 0 that lies in the interior of the feasible region S, and we impose a very large cost on feasible points that lie ever closer to the boundary of S, thereby creating a barrier to exiting the feasible region. A barrier function for () is any continuous function b(x) defined on the the interior of the feasible set S such that b(x)! +1 as max,...,m g i (x)! 0. Two common examples of barrier functions are 1 b(x) = ln( g i (x)) and b(x) = g i (x). The idea in a barrier method is to dissuade points x from ever approaching the boundary of the feasible region. We consider solving: B( ) : minimize f(x)+ b(x) s.t. g(x) < 0 x 2 R n. for a sequence of k > 0 such that k! 0. Note that the problem B( ) is unconstrained, with inequalities g(x) < 0 defining the open domain of the objective function. The next result concerns convergence of the barrier method. Recall that N (x) denote the ball of radius centered at the point x. Theorem 8.1 (Barrier Convergence Theorem). Suppose f(x), g(x), andb(x) are continuous functions, and suppose B( ) hasasolutionforany >0. Let the sequence { k } satisfy 0 < k+1 < k and k! 0 as k!1, and let x k denote the exact solution to B( k ). Suppose there exists an optimal solution x? of () for which N (x? ) \{x : g(x) < 0} 6= ; for every >0. Then any limit point x of {x k } solves (). roof: Let x be any limit point of the sequence {x k }. Without loss of generality, assume that x is in fact the limit of the sequence. From the continuity of f(x) and g(x), lim k!1 f(x k )=f( x) and g( x) =lim k!1 g(x k ) apple 0. Thus, x is feasible for (). If g( x) < 0, then lim k!1 k b(x k ) = 0. If x lies on the boundary of S, then, by our assumption, lim k!1 b(x k )=+1. In either case, lim inf k!1 kb(x k ) 0,

2 IOE 519: NL, Winter 2012 c Marina A. Epelman 56 which implies that lim inf k!1 n o f(x k )+ k b(x k ) = f( x)+liminf kb(x k ) f( x). k!1 If x were not a global minimizer, there would exist a feasible vector x? such that f(x? ) <f( x) that can be approached arbitrarily closely through the interior of the feasible set. That is, there exists a point x in the domain of the barrier function such that f( x) <f( x). By the definition of x k, f(x k )+ k b(x k ) apple f( x)+ k b( x) for all k, which by taking the limit at k!1implies that f( x) apple f( x). This is a contradiction, therefore proving that x is a global minimizer of (). Advantages of the barrier methods: Avoid the combinatorial issues associated with finding the active constraints Converge under mild conditions rovide estimates of KKT/Lagrangian multipliers at the optimum (see below) For many barrier functions (including the two examples above) the barrier function is convex if () is a convex problem. Barrier problems are typically solved via Newton s method, which converges rapidly if initialized close to the solution (of the barrier problem). Since for su ciently small the solution of B( ) is su ciently close to the optimal solution of (), one may wonder why you should bother with solving a sequence of barrier problems instead of solving just one, with a low value of. However, the Hessian r 2 xxb(x, ) of the barrier function tends to become ill-conditioned when is very small, making the problem di cult to solve. Solving a sequence of barrier problems provides a remedy: when the solution x k 1 of the previous barrier problem is used as an initial point for solving the current barrier problem, the problem tends to be easier to solve if k is not much smaller than k 1, since then, under reasonable assumptions, x k 1 is close to x k. Suppose more generally, b(x) = (g(x)), where (y) :R m! R, and assume that (y) is continuously di erentiable for all y<0. Note that to be a valid barrier function, (y) should (y) 0 for values of y<0 such that y i is close to 0. (g(x)) rb(x) = rg i (x), and if x k solves B( k ), then rf(x k )+ k rb(x k ) = 0, that is, rf(x k )+ k m (g(x k )) rg i (x k )=0. (7) Let us define u k i = (g(x k )). (8)

3 IOE 519: NL, Winter 2012 c Marina A. Epelman 57 Then (7) becomes: rf(x k )+ u k i rg i (x k )=0. (9) Therefore we can interpret the u k as a sort of vector of Karush-Kuhn-Tucker multipliers. In fact, we have: Lemma 8.2 (Karush-Kuhn-Tucker Multipliers in Barrier Methods) Let () satisfy the conditions of the Barrier Convergence Theorem. Suppose (y) is continuously di erentiable, and let u k be defined by (8). Then if x k! x, and x satisfies the linear independence condition for gradient vectors of active constraints, then u k! ū, where ū is a vector of Karush-Kuhn-Tucker multipliers for the optimal solution x of (). roof: Let x k! x and let I = {i : g i ( x) =0} and N = {i : g i ( x) < 0}. For all i 2 N, u k i = (g(x k ))! 0 as k!1, since k! 0 and g i (x k )! g i ( x) < 0, (g( x)) is finite (since ( ) is continuously di erentiable). Also u k i 0 for all i 2 I, and k su ciently large. Suppose u k! ū as k!1. Then ū functions involved, (9) implies that rf( x)+ 0, and ū i = 0 for all i 2 N. From the continuity of all ū i rg i ( x) =0, ū 0, ū T g( x) =0. Thus ū is a vector of Kuhn-Tucker multipliers. It remains to show that u k! ū for some unique ū. Suppose {u k } 1 k=1 has no accumulation point. Then kuk k!1.butthendefine v k = uk ku k k, and then kv k k = 1 for all k, and so the sequence {v k } 1 k=1 has some accumulation point v satisfying k vk = 1. Notice that for all i 2 N, wehave v i = 0. Using (9), we have u vi k rg i (x k k )= i ku k rg i (x k )= rf(xk ) k ku k k for all k. As k!1,wehavex k! x, v k! v (with v i = 0 for i 2 N and k vk = 1), and ku k k!1, and so the above becomes X v i rg i ( x) =0, i2i which violates the linear independence condition. Therefore {u k } is a bounded sequence, and so has at least one accumulation point. Now suppose that {u k } has two accumulation points, ũ and ū. Note that ū i = 0 and ũ i = 0 for i 2 N, and so X ū i rg i ( x) = rf( x) = X ũ i rg i ( x), i2i i2i

4 IOE 519: NL, Winter 2012 c Marina A. Epelman 58 so that X (ū i ũ i )rg i ( x) =0. i2i But by the linear independence condition, ū i ũ i = 0 for all i 2 I, and so ū i =ũ i. This then implies that ū =ũ.

5 IOE 519: NL, Winter 2012 c Marina A. Epelman 59 9 enalty Methods for Constrained Optimization Consider the constrained optimization problem (): : min x f(x) s.t. g i (x) apple 0 i =1,...,m x 2 R n. We will write g(x) :=(g 1 (x),...,g m (x)) T for convenience. Definition: A function p(x) :R n! R is called a penalty function for () if p(x) satisfies: p(x) =0ifg(x) apple 0 and p(x) > 0ifg(x) 6apple 0. Example: p(x) = (max{0,g i (x)}) 2. We then consider solving the following penalty program: (c) : min x f(x)+cp(x) s.t. x 2 R n for an increasing sequence of constants c as c! +1. Note that in the program (c) we are assigning a penalty to the violated constraints. The scalar quantity c is called the penalty parameter. Let c k 0,k =1,...,1, be a sequence of penalty parameters that satisfies c k+1 >c k for all k and lim k!1 c k =+1. Let q(c, x) =f(x)+cp(x), and let x k be the exact solution to the program (c k ), and let x? denote any optimal solution of (). The following Lemma presents some basic properties of penalty methods: Lemma 9.1 (enalty Lemma) (i) q(c k,x k ) apple q(c k+1,x k+1 ) (ii) p(x k ) p(x k+1 ) (iii) f(x k ) apple f(x k+1 ) (iv) f(x? ) q(c k,x k ) f(x k ) roof: (i) We have q(c k+1,x k+1 )=f(x k+1 )+c k+1 p(x k+1 ) f(x k+1 )+c k p(x k+1 ) f(x k )+c k p(x k )=q(c k,x k ) (ii) and f(x k )+c k p(x k ) apple f(x k+1 )+c k p(x k+1 ) f(x k+1 )+c k+1 p(x k+1 ) apple f(x k )+c k+1 p(x k )

6 IOE 519: NL, Winter 2012 c Marina A. Epelman 60 Thus (c k+1 c k )p(x k ) (c k+1 c k )p(x k+1 ), whereby p(x k ) p(x k+1 ). (iii) We have f(x k+1 )+c k p(x k+1 ) f(x k )+c k p(x k ). But p(x k ) p(x k+1 ) from (ii), which implies that f(x k+1 ) f(x k ). (iv) f(x k ) apple f(x k )+c k p(x k ) apple f(x? )+c k p(x? )=f(x? ). The next result concerns convergence of the penalty method. Theorem 9.2 (enalty Convergence Theorem). Suppose that f(x), g(x), andp(x) are continuous functions. Let {x k },k =1,...,1, be a sequence of solutions to (c k ). Then any limit point x of {x k } solves (). roof: Let x be a limit point of {x k }. From the continuity of the functions involved, lim k!1 f(x k )= f( x). Also, from the enalty Lemma, Thus q? := lim k!1 q(c k,x k ) apple f(x? ). h i lim c kp(x k )= lim q(c k,x k ) f(x k ) = q? f( x). k!1 k!1 But c k!1, which implies from the above that lim k!1 p(xk )=0. Therefore, from the continuity of p(x) and g(x), p( x) = 0, and so g( x) apple 0, that is, x is a feasible solution of (). Finally, from the enalty Lemma, f(x k ) apple f(x? ) for all k, and so f( x) apple f(x? ), which implies that f( x) =f(x? ) and therefore x is an optimal solution of (). An often-used class of penalty functions is: We note the following: p(x) = [max{0,g i (x)}] q, where q 1. (10) If q = 1, p(x) in (10) is called the linear penalty function. This function may not be di erentiable at points where g i (x) = 0 for some i. Setting q = 2 is the most common form of (10) that is used in practice, and is called the quadratic penalty function. Notice that the penalty function p(x) is often a function only of g + (x), where g + (x) = max{0,g i (x)} (the nonnegative part of g i (x)), i =1,...,m. Then we can write p(x) = (g + (x)) where (y) is a function of y 2 R m +. Two examples of this type of penalty function are (y) = y i,

7 IOE 519: NL, Winter 2012 c Marina A. Epelman 61 which corresponds to the linear penalty function, and (y) = yi 2, which corresponds to the quadratic penalty function. Note that even if (y) is continuously di erentiable, p(x) might not be continuously di erentiable, since g + (x) is not di erentiable at points x where g i + (x) = 0 for some i. However, if we assume the (y) = 0 at y i =0, i =1,...,m, (11) then p(x) is di erentiable whenever the functions g i (x) are di erentiable, i =1,...,m, and we can (g + (x)) rp(x) = rg i (x). (12) Now let x k solve (c k ). Then x k will satisfy that is, Let us define Then rf(x k )+c k rf(x k )+c k rp(x k )=0, m (g + (x k )) rg i (x k )=0. u k i = c (g + (x k )). (13) rf(x k )+ u k i rg i (x k )=0, and so we can interpret the u k as a sort of vector of Karush-Kuhn-Tucker multipliers. In fact, we have: Lemma 9.3 Suppose (y) is monotone in y, is continuously di erentiable, and satisfies (11), and that f(x) and g(x) are di erentiable. Let u k be definied by (13). Then if x k! x, and x satisfies the linear independence condition for gradient vectors of active constraints, then u k! ū, where ū is a vector of Karush-Kuhn-Tucker multipliers for the optimal solution x of (). roof: From the enalty Convergence Theorem, x is an optimal solution of (). Let I = {i : g i ( x) =0} and N = {i : g i ( x) < 0}. For i 2 N, g i (x k ) < 0 for all k su ciently large, so u k i = 0 for all k su ciently large, whereby ū i = 0 for i 2 N. From (13) and the definition of a penalty function, it follows that u k i su ciently large. 0 for i 2 I, for all k Suppose u k! ū as k! 1. Then ū i = 0 for i 2 N. From the continuity of all functions involved, rf(x k )+ u k i rg(x k )=0

8 IOE 519: NL, Winter 2012 c Marina A. Epelman 62 implies rf( x)+ ū i rg( x) =0. From the above remarks, we also have ū 0 and ū i = 0 for all i 2 N. Thus ū is a vector of Karush-Kuhn-Tucker multipliers. It therefore remains to show that u k! ū for some unique ū, which is done along the same lines as in Lemma 8.2. Remark. The quadratic penalty function satisfies the condition (11), but the linear penalty function does not satisfy (11). 9.1 Exact enalty Methods The idea in an exact penalty method is to choose a penalty function p(x) and a constant c so that the optimal solution x of (c) is also an optimal solution of the original problem (). Theorem 9.4 Suppose () is a convex program for which the Karush-Kuhn-Tucker conditions are necessary. Suppose that p(x) := (g i (x)) +. Then as long as c is chosen su ciently large, the sets of optimal solutions of (c) and () coincide. In fact, it su cies to choose c>max i {u? i }, where u? is a vector of Karush-Kuhn-Tucker multipliers. roof: Suppose ˆx solves (). For any x 2 R n we have: q(c, x) =f(x)+c m (g i (x)) + f(x)+ m (u? i g i(x)) + f(x)+ m u? i g i(x) f(x)+ m u? i (g i(ˆx)+rg i (ˆx) T (x = f(x)+ m u? i rg i(ˆx) T (x Thus q(c, ˆx) apple q(c, x) for all x, and therefore ˆx solves (c). Next suppose that x solves (c). Then if ˆx solves (), we have: ˆx) ˆx)) = f(x) rf(ˆx) T (x ˆx) f(ˆx) =f(ˆx)+c m (g i (ˆx)) + = q(c, ˆx). and so f( x)+c (g i ( x)) + apple f(ˆx)+c (g i (ˆx)) + = f(ˆx), f( x) apple f(ˆx) c (g i ( x)) +. (14)

9 IOE 519: NL, Winter 2012 c Marina A. Epelman 63 However, if x is not feasible for (), then f( x) f(ˆx)+rf(ˆx) T ( x ˆx) m = f(ˆx) u? i rg i(ˆx) T ( x f(ˆx)+ m u? i (g i(ˆx) g i ( x)) m = f(ˆx) u? i g i( x) >f(ˆx) ˆx) c m (g i ( x)) +, which contradicts (14). Thus x is feasible for (). That being the case, f( x) apple f(ˆx) f(ˆx) from (14), and so x solves (). c m (g i ( x)) + = 9.2 enalty Methods for Inequality and Equality Constraints The presentation of penalty methods has assumed that the problem () has no equality constraints. If the problem has equality constraints h(x) = 0, we could convert them into inequality constraints by writing h(x) apple 0 and h i (x) apple 0, but such conversion usually violates good judgement in that it unnecessarily complicates the problem. Furthermore, it can cause the linear independence condition to be automatically violated for every feasible solution. Therefore, instead let us consider the constrained optimization problem () with both inequality and equality constraints: : min x f(x) s.t. g(x) apple 0 h(x) =0 x 2 R n, where g(x) and h(x) are vector-valued functions, that is, g(x) :=(g 1 (x),...,g m (x)) T and h(x) := (h 1 (x),...,h l (x)) T for notational convenience. Definition: A function p(x) :R n! R is called a penalty function for () if p(x) satisfies: p(x) =0ifg(x) apple 0 and h(x) =0 p(x) > 0ifg(x) 6apple 0 or h(x) 6= 0. The main class of penalty functions for this general problem are of the form: p(x) = [max{0,g i (x)}] q + lx h i (x) q, where q 1. All of the results of this section extend naturally to problems with equality constraints and for penalty functions of the above form. For example, in the analogous result of Theorem 9.4, it su ces to choose c>max{u? 1,...,u? m, v 1?,..., v? l }. 9.3 Augmented Lagrangian methods A modification of the classical penalty methods involves explicitly incorporating the KKT multipliers into the method. This complicates the algorithm, but has the potential of ameliorating the ill-conditioning encountered in the penalty method, and improving rates of convergence.

10 IOE 519: NL, Winter 2012 c Marina A. Epelman 64 Consider the problem of the form The Lagrangian of this problem is defined to be min f(x) s.t. h(x) =0. L(x, v) =f(x)+v T h(x). If KKT conditions are necessary at a local minimizer x?, then for v = v? (the set of KKT multipliers) x? is a stationary point of the Lagrangian with v = v?, but usually not a minimizer. It is a minimizer of the problem min f(x)+v T h(x) s.t. h(x) =0. Applying the penalty approach to the latter problem, consider min A(x, v, ) =f(x)+v T h(x)+ 1 x 2 h(x)t h(x) to obtain x?. The above function is referred to as the augmented Lagrangian. Intuitively, a minimizer of A(x, v, ) is close to the solution x? if v is close to v? and >0 is su is very large. Example Consider ciently large. min f(x) = 1 2 (x2 1 + x 2 2)subjecttox 1 =1. The optimal solution is x? =(1, 0) T and v? = 1. The augmented Lagrangian is min x A(x, v, ) = 1 2 (x2 1 + x 2 2)+v(x 1 1) (x 1 1) 2. By setting its gradient to zero, the unique unconstrained minimizer is Thus, for all >0, On the other hand, for all v, x 1 (v, ) = c v c +1, x 2(v, ) =0. lim x 1(v, ) =1, v!v? lim x 1(v, ) =0. v!v? lim x 1(v, ) =1, lim x 1(v, ) =0.!1!1 An augmented Lagrangian algorithm is initialized with some x 0,v 0, and 0 > 0, and proceeds as follows: 1. If rl(x k,v k ) = 0, then stop 2. Compute x k+1 by solving min x A(x, v k, k ) 3. Determine v k+1 and k+1. For example, take v k+1 = v k + k h(x k+1 ), and k+1 k.

11 IOE 519: NL, Winter 2012 c Marina A. Epelman 65 It can be shown that if v k = v? and is su ciently large, so that the Hessian of A(x?,v?, )is positive definite, then the unconstrained minimizer of the augmented Lagrangian is a minimizer of the original problem. Thus, if we knew the Lagrangian multipliers at the solution, the problem can be solved in one iteration. Moreover, the closer v k estimates v?, the closer the next iterate x k+1 will be to x? (assuming k is su ciently large).

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