Implementation of Infinite Dimensional Interior Point Method for Solving Multi-criteria Linear Quadratic Control Problem L. Faybusovich Department of Mathematics, University of Notre Dame, Notre Dame, IN 46556 USA. T. Mouktonglang Department of Mathematics, University of Notre Dame, Notre Dame, IN 46556 USA. T. Tsuchiya The Institute of Statistical Mathematics, 4-6-7 Minami-Azabu, Minato-Ku, Tokyo, 06-8569, Japan. May 2004 Abstract We describe an implementation of an infinite-dimensional primal-dual algorithm based on the Nesterov-Todd direction. Several applications to both continuous and discrete-time multi-criteria linear-quadratic control problems and linear-quadratic control problem with quadratic constraints are described. Numerical results show a very fast convergence (typically, within 3-4 iterations) to optimal solutions. Introduction One of the current trends in the development of interior-point algorithms of optimization is the extension of the domain of their applicability. While at first the linear programming problem was the major target, subsequent developments led to creation of algorithms and software for much broader class of symmetric programming problems. This class includes semidefinite programming and second-order cone programming problems. See e.g. [2] for numerous This author was supported in part by National Science Foundation Grant No. 002628. This author was supported in part by National Science Foundation Grant No. 002628. This author was supported in part by the Grant-in-aid for scientific Research (C) 55044 of the Ministry of Education, culture, Sport, Science, and Technology of Japan.
engineering applications. Of course, these extensions are not without a price. While the general theory (very similar to the linear programming case) has been developed and cast in a very elegant form with the help of the technique of Euclidean Jordan algebras (see e.g. [, 3, 4, 5, 6, 3, 4, 6, 8] ), the computational cost of implementation grew quite significantly (mostly due to the cost of implementation of the major computational Newton-like step). In the present paper we make the next natural step along this road. We consider a class of infinite-dimensional optimization problems and a primal-dual algorithm based on the Nesterov-Todd direction. We implemented this algorithm for a class of control problems: multi-criteria linear-quadratic control problem and linearquadratic control problem with quadratic constraints (in both continuous and discrete-time settings). A computation of Nesterov-Todd direction is an infinitedimensional (linear) problem. Thus, the cost of the major computational step is even higher in comparison with previous essentially finite-dimensional schemes. What is crucial here is a careful consideration of the structure of the problem. In particular, in control applications mentioned above the infinite-dimensional part of the Newton-like step is reduced to solving the linear-quadratic control problem with extra linear term in the cost function. As we show in this paper, the presence of this extra linear term does not lead to any significant complications and (as in the standard control-theoretic setting) the solution of this problem is essentially reduced to solving one of the standard Riccati equations (differential, difference or algebraic depending on the concrete setting of the problem). The conceptual difference in implementation of the algorithm for continuous-time and discrete-time systems is quite minimal: it is essentially in the organization of data, different schemes for computation of scalar products in various functional spaces, and solving the linear-quadratic problem briefly described above. A Newton-like step for continuous-time problem is computationally expensive operation. Good news here is that it is much cheaper for discrete-time systems. Combining this with an experimental fact (which can be partially explained by t heoretical complexity estimates [0] ) that three-four iterations of the algorithm lead to very reasonable approximations of optimal solutions, one can hope t hat on-line versions of the algorithm can be implemented. This, in turn, may lead to entirely new and exciting methodologies for the multi-criteria regulation of control systems. 2 Primal-dual interior-point algorithms In this section, following [0], we describe a general set up for a class of infinitedimensional optimization problems. We then proceed with the optimality criterion and describe a primal-dual algorithm based on Nesterov-Todd (NT) direction. Let (H, <, >) be a Hilbert space, V be a vector space: V = IR H. Let, further, V = V V (m-times), X V be a closed vector subspace in V. 2
Consider the following standard optimization problem: < a, z > V min (2.) z (b + X) Ω (2.2) and its dual < b, w > V min (2.3) b (a + X ) Ω (2.4) where our cone Ω is a second order cone(open convex set in V ): Ω = {(s, y) IR H : s > y H } (sometimes also called the cone of squares) and Ω = Ω Ω (m-times). It is known that Suppose z, w V, Ω = {(s, y) IR H : s y H }. z = {(s, x ),, (s m, x m )} w = {(t, y ),, (t m, y m )}. Then we define an inner product on V as follows: The cone Ω is self-dual, i.e. < z, w > V = (s i t i + < x i, y i > H ). i= Ω = {z V :< w, z > 0, w Ω} = Ω. Denoted by X the orthogonal complement of X in V with respect to <, > V. Let, further, F = [(x + X) Ω] [(a + X ) Ω ]. We assume that int(f) = [(x + X) int( Ω)] [(a + X ) int( Ω )] 0. (2.5) It is easy to see that if a pair of z, w satisfies (2.2),(2.4) and < z, w >= 0, then z is an optimal solution to (2.)-(2.2) and w is an optimal solution to (2.3)-(2.4). 3
Definition 2.. Given (z, w) then the duality gap µ(z, w) is defined as follows: where r is a positive constant. µ(z, w) = < z, w > 2m A typical primal-dual algorithm generates a sequence of pairs primal and dual feasible points (z (k), w (k) ) int(f), k =, 2,... such that µ(z (k+), w (k+) ) ( δ r ω )µ(z(k), w (k) ), (2.6) for some positive δ and ω (see e.g. [9]). The following theorem (see [0] ) provides necessary and sufficient conditions for a pair (z, w ) to be an optimal solution to (2.),(2.2) and (2.3),(2.4). Theorem 2.2. Suppose the conditions (2.5) holds. Then both (2.),(2.2) and (2.3),(2.4) have optimal solutions. Moreover, z is an optimal solution to(2.),(2.2) and w is an optimal solution to (2.3),(2.4) if and only if < z, w > V = 0. One of the most important steps in the implementation of primal-dual algorithms is a computation of a descent direction which drives duality gap µ to zero. One of these directions (the so-called NT-direction) is described below. Let z i = (t i, x i ), w i = (s i, y i ) V. We define a determinant of z i as follows: det(z i ) = t 2 i x i 2. And the multiplication in V is defined as follows: (t i, x i ) (s i, y i ) = (t i s i + < x i, y i > H, s i x i + t i y i ) Observe that the above multiplication is commutative but not associative. This defines the structure of a Jordan algebra (see e.g. [0] for more details). Then the inverse z i of z i is given by the formula Next we consider a function z i = det(z i ) (t i, x i ). f(z i ) = ln det(z i ) = ln(t 2 i x i 2 ). The quadratic representation P (z i ) is defined as follows: where H(z i ) is the Hessian of f(z i ). P (z i ) = H(z i ) 4
Remark: Since V is a product of m copies of V. We define a quadratic representation P (z) for z = (z,, z m ) V as follows: P (z) = (P (z ),, P (z m )). One of the main ingredients in the construction of NT-direction is the socalled scaling point. Proposition 2.3. Given (z, z 2 ) Ω Ω. Then there exists a unique z 3 Ω such that P (z 3 )z = z 2. (2.7) The following Corollary gives an explicit formula for the scaling point. Corollary 2.4. Let z, z 2 Ω, z = (s, y), z 2 = (t, x). Then consider z 3 = (r, u) with µ µ s + µ 2 t r = µ 2 2 + µ µ 2 < (s, y), (t, x) > µ µ 2 x µ y u = µ 2 2 + µ µ 2 < (s, y), (t, x) > µ i =, i =, 2. det(zi ) Then z 3 Ω is a unique solution to (2.7) for V = IR H. Proposition 2.5. Every element z V admits the following spectral decomposition z = λ e + λ 2 e 2 where λ i IR, e i e j = δ ij e i for i, j =, 2. The following proposition gives us an explicit formula for the spectral decomposition of an element z V. Proposition 2.6. Let z = (s, y) V, y 0. Consider Then e = 2 (, and e i e j = δ ij e i for i, j =, 2. y y ), e 2 = 2 (, y y ) λ = s + y, λ 2 = s y. (s, y) = λ e + λ 2 e 2 Proposition 2.7. Let Ω be the cone of squares of V and (s, y) Ω. Consider z = ( µ 2, y µ ), µ = s + y + s y. 5
Then z Ω and z 2 = (s, y). Moreover if z 2 = λ e + λ 2 e 2 is the spectral decomposition of z 2, then Denote z by (s, y) 2. Proposition 2.8. We have Hence, z = λ e + λ 2 e 2. P (z 2 ) 2 = P (z). P (z 2 ) = P (z) 2. (2.8) For proofs of the above propositions and corollary, see [0]. We are now in a position to introduce Nesterov-Todd direction (see e.g. [5]). Given z, z 2 Ω, let z 3 Ω be the scaling point of z and z 2 defined in (2.7). Then the NTdirection (ξ, η) X X is defined to be the solution to the following system of linear equations: P (z 3 )ξ + η = γµ(z, z 2 )z z 2, (2.9) ξ X, η X. (2.0) Here 0 < γ < is a real parameter. One can show (see [0]) that (2.9),(2.0) admits a unique solution. Proposition 2.9. By following the NT-direction, the duality gap decreases. Proof Let (z (k), z(k) 2 ) Ω Ω, (ξ(k), η (k) ) be NT-direction at (z (k), z(k) 2 ), and t (k) > 0. Define z (k+) = z (k) + t (k) ξ (k), z (k+) 2 = z (k) 2 + t (k) η (k). Then, consider the following equality: < z (k+), z (k+) 2 > = < z (k) + t (k) ξ (k), z (k) 2 + t (k) η (k) > = < z (k), z(k) 2 > +t (k) (< z (k), η(k) > + < z (k) 2, ξ(k) >) So it suffices to show that < z (k), η(k) > + < z (k) 2, ξ(k) >< 0. < z (k), η(k) > + < z (k) 2, ξ(k) > = < z (k), η(k) > + < P (z 3 )z (k), ξ(k) > = < z (k), η(k) + P (z 3 )ξ (k) > = < z (k), γµ(z(k), z(k) 2 )(z(k) ) z (k) 2 > = (γ ) < z (k), z(k) 2 > < 0 ( since: 0 < γ < ). 6
Here we used (2.9). Hence we have < z (k+), z (k+) 2 > < < z (k), z(k) 2 > µ(z (k+), z (k+) 2 ) < µ(z (k), z(k) 2 ). We conclude this section by a brief description of a primal-dual algorithm based on NT-direction. For a fix ɛ > 0, suppose that (z (0), w (0) ) int(ω) (b + X) int(ω) (a + X ). Let (ξ k, η k ) be the descent NT-direction obtained by solving the system of linear equations (2.9),(2.0), t > 0 be the largest value such that (z (k) + tξ k, w (k) + tη k ) int(ω) (b + X) int(ω) (a + X ). Then we set (z (k+), w (k+) ) = (z (k) + tξ k, w (k) + tη k ). We stop the iteration when µ(z (k), w (k) ) ɛ. In [0], it is shown that by choosing the right stepsize carefully, the algorithm runs in polynomial-time, i.e. it generates a sequence of primal and dual feasible iterates satisfying (2.6) with ω = 0.5 or and with δ being an independent positive constant. 3 Abstract Formulation of the Problem and Implementation Let H be a Hilbert space. Consider the following optimization problem: t min, (3.) W i (y y i ) H t, i =,.., m, (3.2) y c + Z. (3.3) Here H is the norm induced by the scalar product < > H, y i H are fixed ; Z is a closed vector subspace in H, W i : H H is a bounded linear operator for i =,, m. The problem (3.)-(3.3) is an example of an infinitedimensional second-order cone programming problem which has been analyzed in detail in [0]. We have implemented a version of a primal-dual algorithm based on Nestreov- Todd direction described in [0]. The problem (3.)-(3.3) can be easily rewritten in the conic form: Let Ω = {(s, y) R H : s y H }; V = IR H; Ω = Ω Ω... Ω (m-times) ; V = V... V (m-times). Let, further, Λ : V V be a linear operator such that: 7
Λ(s, y) = {(s, W y),..., (s, W m y)}, a = ((, 0), (0, 0),..., (0, 0)) V, b = ((0, W (c y ),..., (0, W m (c y m ))), and X = Λ(IR Z). The scalar product <, > V in V is defined as follows: < (t, x ),..., (t m, x m ), (s, y ),..., (s m, y m ) > V = {t i s i + < x i, y i > H }b. With these notations we can rewrite the problem (3.)-(3.3) in the conic form (2.),(2.2). Proposition 3.. We have X = {(r, u ),..., (r m, u m ) V ; r +... + r m = 0, i= Wi u i Z } (3.4) i= where Z is the orthogonal complement of Z and Wi each i. is the adjoint of W i for A conic dual to (2.),(2.2) will have the following form ([0]): < W i (c y i ), u i > H min, (3.5) i= Wi u i Z, u i r i, i =,..., m, (3.6) i= Here W i is the adjoint of W i. r i =. (3.7) i= 3. Calculation of NT-direction Given (z, z 2 ) a pair of feasible solutions to the problem (2.),(2.2), and (2.3),(2.4). To obtain NT-direction for each iteration, we need to solve the following equation for (ξ, η), P (z 3 )ξ X, ξ X (3.8) where z 3 is the scaling point of z and z 2,and = γµ(z, z 2 )z z 2. (Compare with (2.9), (2.0).) The equation (3.8) is equivalent to : < P (z 3 )ξ, ξ > < ξ, > min, 2 ξ X. 8
Let z 3 = ((t, x 2 ), (t 2, x 2 ),, (t m, x m )) and = ((r, u ),, (r m, u m )). For (µ, ζ) IR Z, ξ = Λ(µ, ζ), i= ρ(µ, ζ) = < P (z 3)ξ, ξ > < ξ, > 2 = (t 2 i x i 2 ) W i ζ 2 + < x i, W i ζ > 2 < u i, W i ζ > + ν µ 2 + ν 2 µ, 2 2 where i= i= i= i= ν = (t 2 i + x i 2 ), ν 2 = 2 t i < x i, W i ζ > r i. Hence, if we denote by φ(ζ) = min{ρ(µ, ζ) : µ IR} where φ(ζ) = < ζ, Mζ > 2 + 2 M = m+ i= i= i= ε i < υ i, ζ > 2 + < υ 0, ζ > ( m i= r i) 2 [ ] (t 2 i x i 2 Qi 0 ) 0 R i i= 2ν, (3.9) (3.0) υ 0 = ( r i )υ m+ W i u i. (3.) ν i= υ i = 2W i x i, i =, 2,..., m, υ m+ = 2 ν m i= t i W i x i, (3.2) ε i =, i =,..., m, ε m+ =, (3.3) As a result, to obtain NT-direction, it suffices to solve the following optimization problem: φ(ζ) min, which is equivalent to solving: < ζ, Mζ > 2 + 2 m+ i= ζ Z. ε i < υ i, ζ > 2 + < υ 0, ζ > min, (3.4) ζ Z. The following theorem further simplifies the problem (3.4). 9
Theorem 3.2. Let ζ 0 be the optimal solution to the problem: ζ, Mζ 2 + v 0, ζ min, (3.5) ζ Z, (3.6) and ζ i, i =, 2,, m +, be optimal solutions to the problems: ζ, Mζ 2 + V i, ζ min, (3.7) ζ Z, (3.8) where V i = ε i v i. Let S = (s ij ), s ij = v i, ζ j, i, j =, 2,, m +. Then the set of optimal solutions to the problems (3.4) is in one-to-one correspondence with the set of the solutions of the system of linear equations (I S) δ.. δ m+ = v, ζ 0. v m+, ζ 0 More precisely, if (δ,..., δ m+ ) is a solution to (3.9), then (3.9) m+ ζ(δ) = ζ 0 + δ i ζ i (3.20) i= is the optimal solution to the problem (3.4). For a proof see [8, 0]. Remark: The procedure of reduction of (3.4), (3.5), (3.7) is a version of Sherman-Morrison-Woodbury formula. Solving (3.4) is therefore equivalent to solve m+2 problems of the following type < Mζ, ζ > + < υ, ζ > min, ζ Z (3.2) 2 and a system of (m + ) (m + ) linear algebraic equations. Let ζ be an optimal solution to (3.4). Since dρ ν2 dµ = 0. Hence µ = ν. Therefore, we obtain ξ = Λ(µ, ζ) = ((µ, W ζ),..., (µ, W m ζ)) which is the primal descent direction. A dual descent direction is simply η = P (z)ξ. 0
3.2 Computation of the step size The Jordan algebraic techniques significantly simplifies the computation of the step-size. Observe that z + tξ = P (z 2 )(e + tp (z 2 )ξ). Hence, z + tξ Ω, is equivalent to e + tp (z 2 )ξ Ω. If tp (z 2 )ξ = µ f + µ 2 f 2 is the spectral decomposition, then z + tξ Ω is equivalent to + tµ 0, + tµ 2 0. (3.22) Since the primal-dual cone is the product of several copies of Ω, we obtain step size t as the maximal value of t for which all inequalities (3.22) are satisfied. We use 0.9t for the actual step size in out algorithm. The stopping rule we used is quite standard: the duality gap µ < ɛ for a given ɛ > 0. Recall (see [0]) that Nesterov-Todd direction (ξ, η) is determined by the following conditions ξ + P (z) η = γµ(z, z 2 )z 2 z, ξ X, η X, where (z, z 2 ) Ω Ω and z 3 Ω is the scaling point uniquely determined by the equation: P (z 3 )z = z 2. As a result of numical experiments,we arrived at γ = 0.5 as a value which minimizes the number of iterations. 4 Examples of concrete problems We now consider several concrete examples of the problem (3.)-(3.3). 4. Multi-criteria linear quadratic control problem Denote by L n 2 [0, T ] the Hilbert space of square integrable functions f : [0, T ] IR n, T > 0. Let (x, u) H = L n 2 [0, T ] L l 2[0, T ]. Consider the following optimization problem: J(x, u) = max i [,m] T 0 (x x i ) T Q i (x x i ) + (u u i ) T R i (u u i )dt min (4.) ẋ(t) = Ax(t) + Bu(t), x(0) = x 0. (4.2)
Here (x i, u i ) H, i =, 2,..., m; A, B, Q i, R i, i =, 2,..., m are given matrices of appropriate dimensions. We will assume that Q i, R i are symmetric positive definite matrices. The problem (4.),(4.2) can be easily rewritten in the second order cone programming problem of the form (3.)- (3.3) Here H is the norm induced by the scalar product < > H, y i = (x i, u i ), i =, 2,..., m; Z is a closed vector subspace in H described as follows: Z = {(x, u); ẋ(t) = Ax(t) + Bu(t), x(0) = 0}; (4.3) [ ] L T W i = Qi 0 0 L T, i =,..., m, R i where L Qi, L Ri are lower-triangular matrices obtained by Cholesky factorization of Q i and R i respectively. Then by following the scheme described in the previous section, we have to show that the condition (2.5) is satisfied, and then compute NT-direction. Constructing feasible solutions Primal problem: After some numerical experiments, we end up with the following construction. Let K 0 be a unique stabilizing solution to the algebraic Riccati equation: KBB T K A T K KA I = 0. Here stabilizing means that the close-loop matrix A BB T K 0 is stable. Let x be the solution to the following system of linear differential equations: ẋ = (A BB T K 0 )x; x(0) = x 0, u = B T K 0 x. It is quite obvious that y = (x, u) c + Z. and hence if we choose s = max i [,m] W i(y y i ) + δ for some δ > 0, the initial point for primal problem is Λ(s, y). Dual problem: We simply take u i = 0, r i = m, i =, 2,..., m. Computation of NT-direction For T <, the problem (3.2) is simply classical LQ problem with a linear term on a finite interval (see e.g. [8, ]). Hence on each iteration, we need to solve a differential Riccati equation, system of linear differential equations and a system of (m + ) (m + ) linear algebraic equations (see [0]). For the case when T =, the problem (3.2) is LQ problem with a linear term on a semi-infinite interval. The complete solution to such a problem is described in the following section. 2
4.. Solution to LQ problem with a linear term on a semi-infinite interval Let (x, u) H = L n 2 [0, ) L m 2 [0, ), Σ = Σ T. Consider the following LQcontrol problem with a linear term on a semi-infinite J(x, u) = 2 0 [ x < u ] [ x, Σ u ] [ x > dt + < 0 u ] [ y, v ] > dt min (4.4) By using block partition, denote [ Σ Σ Σ = 2 Σ T 2 Σ 22 For proofs of the theorems in this section, see [9]. ẋ = Ax + Bu, x(0) = x 0 (4.5) Theorem 4.. Let A be an antistable n by n matrix ( i.e. real parts of all eigenvalues of A are positive). Consider the following system of linear differential equations: ẋ = Ax + f, (4.6) where f L n 2 [0, ). There exists a unique solution L(f) of (4.6) such that L(f) L n 2 [0, ). Moreover the map f L(f) is linear and bounded. Explicitly: L(f)(t) = + 0 ]. e Aτ f(τ + t)dτ (4.7) Theorem 4.2. The following conditions are equivalent: i) Σ 22 is a positive definite (symmetric) matrix and the following Riccati equation has a stabilizing solution. where KLK + KÃ + ÃT K Q = 0 (4.8) Ã = A BΣ 22 Σ 2, Q = Σ Σ 2 Σ 22 Σ 2, L = BΣ 22 BT. ii) The pair (A, B) is stabilizable and there exists ɛ > 0 such that Γ(x, u) = for all (x, u) Z. + 0 < [ x u ] [ x, Σ u Let further Z H be as follows: ] > dt ɛ + 0 ( x 2 + u 2 )dt Z = {(x, u) H; ẋ = Ax + Bu, x is absolutely continuous, x(0) = 0}. 3
Theorem 4.3. Suppose that the pair (A, B) is stabilizable. Then Z is a closed vector subspace in H. Then [ ] Z ṗ + A = { T p B T : p L n p 2 [0, + ), p is absolutely continuous, ṗ L n 2 [0, + )} We are now in the position to describe the optimal solution to (4.4). Theorem 4.4. Suppose that the conditions of theorem 4.2 are satisfied. Then the problem (4.4)-(4.5) has a unique solution which can be described as follows. There exists a stabilizing solution K 0 to the Riccati equation (4.8). Then the matrix C = (Ã + LK 0) is antistable, (K 0 B Σ 2 )Σ 22 v + y Ln 2 [0, + ). Let ρ be a unique solution from L n 2 [0, + ) of the system of differential equations ρ = C T ρ + (K 0 B Σ 2 )Σ 22 v + y (4.9) (which exists according to Theorem 4.), x is the solution to the system of differential equations ẋ = (Ã + LK 0)x + Lρ BΣ 22 v, x(0) = x 0, (4.0) p = K 0 x + ρ, u = Σ 22 (BT p v Σ 2 x). Now back to the problem of finding NT-direction for T =. Solving (3.2) is reduced to solving an algebraic Riccati equation (4.8), systems of differential equations (4.9),(4.0) and a system of (m+) (m+) linear algebraic equations on each iteration. 4.2 Discrete-time Multi-criteria linear quadratic control problem In this section, we consider a discrete-time formulation for the problem (4.)- (4.2). For simplicity, let us introduce some useful notations. Let x denote a sequence {x k } IR n for k = 0,,. We say that x l2 n (IN) if i= x i 2 < where is a norm induced by an inner product <, > in IR n. Let (x, u) l2 n (IN) l2 m (IN). Then the Discrete-time Multi-criteria Linear Quadratic Control Problem takes the form: max i=,...,m k=0 {< (x k ϕ (i) k ), Q i(x k ϕ (i) k ) > + < (u k φ (i) k x k+ = Ax k + Bu k, k = 0,,, x 0 = γ 0, ), R i(u k φ (i) k ) >} min (4.) where (ϕ (i), φ (i) ) H = l n 2 (IN) l m 2 (IN) for i =,, m, A, B, Q i, R i for i =,, m are matrices of appropriate sizes. We assume that Q i s and R i s are positive definite. It is easy to see that the problem (4.) can be rewritten into the secondorder cone programming problem (3.)- (3.3). 4
Observe now the inner product in H has the following form: < (x, y), (u, v) > H = {< x k, u k > + < y k, v k >}. (4.2) k=0 The vector subspace Z now takes the form: Z = {(x, u) H : x k+ = Ax k + Bu k, k = 0,,, x 0 = 0} The constructions of a initial feasible point for primal and dual problems are similar to the constructions of a feasible points in the continuous case. In the primal, we solve a discrete LQ problem via discrete Algebraic Riccati equation. In the dual, we take the same choice as in the continuous case. Now the problem (3.2) is discrete-time linear-quadratic control problem with linear term. 4.2. Discrete-time linear-quadratic control problem with linear term In this section, we describe a complete solution to discrete linear-quadratic control problem with a linear term. In order to derive the solution, we need the following auxiliary results. Theorem 4.5. Let A be a d-stable matrix (ie. all eigenvalues of A lie inside a unit circle on a on a complex plane. ) Consider the following system of linear difference equation. x k = Ax k+ + f k, k = 0,, (4.3) where f = {f k } l2 n (IN). Then there exists a unique solution L(f) of (4.3) such that L(f) l2 n (IN). Moreover the map f L (f) is linear and bounded. Explicitly, x k = L(f) k = A r f r+k, k = 0,,. (4.4) Proof Since A be a stable matrix, there exists S = S T 0, such that A T SA S = I (4.5) (see e.g. [2]). Let w k =< x k, Sx k >. Obviously, w k 0, k. Then using (4.3) and (4.5), we obtain the following: w k+ = w k+ w k = < x k+, Sx k+ > < x k, Sx k > = < x k+, Sx k+ > < (Ax k+ + f k ), S(Ax k+ + f k ) > = < x k+, Sx k+ > < x k+, A T SAx k+ > 2 < Ax k+, Sf k > < f k, Sf k > = < x k+, x k+ > 2 < Ax k+, Sf k > < f k, Sf k > x k+ 2 2 x k+ A T S f k S f k 2. 5
But then it is easy to see that the following inequality is true. 2 x k+ A T S f k 2 x k+ 2 + 2 A T S 2 f k 2 Then w k+ 2 x k+ 2 (2 A T S 2 + S ) f k 2 Then it follows that r= k+ w k+ w 0 = w r r= k+ ( 2 x r 2 (2 A T S 2 + S ) f r 2 ). r= Hence, k+ k+ x r 2 2 (2 A T S 2 + S ) f r 2 + 2(w k+ w 0 ). (4.6) Let now then we have that r= x k = A r f r+k, k = 0,,, (4.7) Ax k+ + f k = A = = = A r f r+k+ + f k A r+ f r+k+ + f k A r f r+k + f k r= A r f r+k = x k. Hence, (4.7) satisfies (4.3). Let us now consider the following inequilities: x k = A r f r+k A r f r+k A r f r+k. 6
Then by Cauch-Schawrz inequality we have x k 2 A r 2 A 2r s=0 f r+k 2 f s 2 for all k. But since A is stable and all norms are equivalent, by Householder s Theorem [7] we can assume without lost of generality that Therefore, A <. x k 2 < for all k. Hence {x k } is bounded and as a result {w k } is also bounded. But since x 0 = Ar f r, x 0 2 Then from (4.6) we have, A 2r A 2 f r 2 f r 2. k+ k+ x r 2 2(2 A T S 2 + S + A 2 ) f r 2 + 2(w k+ w 0 ) for all k. Since {w k } is bounded, we conclude from (4.8): {x k } l2 n (IN). Hence As a result, From (4.8), we now have lim x k = 0. k lim w k = 0. k x r 2 2(2 A T S 2 + S + A 2 ) f r 2 2w 0. (4.8) But since w 0 =< x 0, Sx 0 > 0, 7
we have x r 2 2(2 A T S 2 + S + A 2 ) f r 2 C f r 2 (4.9) for some constant C. But now suppose that { x k } l n 2 (IN) be another solution to (4.3). Then y k = x k x k = A(x k+ x k+ ) = Ay k+ for k = 0,. Since A is d-stable and {y k } l n 2 (IN), one can easily see that y k = 0 for all k = 0,. Therefore the system (4.3) may have only one solution {x k } l n 2 (IN). Hence we can conclude that the linear map f L (f) is correctly defined and bounded. Let now consider the following classical discrete-time LQ problem 2 < k=0 [ xk u k ] [ xk, Σ u k ] > min, (4.20) where x k+ = Ax k + Bu k, x 0 = x 0 IR n and [ Σ Σ Σ = 2 Σ T 2 Σ 22 ]. It is well known the the following discrete algebraic Riccati equation plays a very important role in finding the optimal solution to the problem (4.20) [2]: K = A T KA (A T KB + Σ 2 )(Σ 22 + B T KB) (A T KB + Σ 2 ) + Σ. (4.2) Observe that (4.2) is defined under assumption that the matrix Σ 22 + B T KB is invertible. Let F = (A T KB + Σ 2 )(Σ 22 + B T KB). Then K 0 a symmetric solution to (4.2) is called a stabilizing if Σ 22 +B T KB > 0 and A T F B T is d-stable. Theorem 4.6. Suppose Σ 22 is a positive definite matrix, (4.2) has a stabilizing solution. Then [ ] [ ] xk xk <, Σ > > 0. (4.22) 2 u k u k k=0 8
for all (x k, u k ) satisfying x k+ = Ax k + Bu k, (4.23) x 0 = 0, (x, u) l n 2 (IN) l m 2 (IN). Proof Let K be a stabilizing solution to (4.2), using (4.22) we obtain the following equalities: < x k+, Kx k+ > < x k, Kx k > = < (Ax k + Bu k ), K(Ax k + Bu k ) > < x k, Kx k > = < x k, (A T KA K)x k > + < x k, A T KBu k > + < u k, B T KAx k > + < u k, (B T KB + Σ 22 )u k > < u k, Σ 22 u k >. Let now R K = Σ 22 + B T KB, and S K = A T KB + Σ 2 then using (4.2), we obtain: < x k+, Kx k+ > < x k, Kx k > = < x k, (S K R K S K Σ )x k > + < x k, A T KBu k > + < u k, B T KAx k > + < u k, R K u k > < u k, Σ 22 u k > = < x k, (S K R K S K)x k > + < x k, S K u k > + < u k, SKx T k > + < u k, R K u k > < x k, Σ 2 u k > < u k, Σ T 2x k > < u k, Σ 22 u k > < x k, Σ x k > = < (R K S Kx k + u k ), R K (R K S Kx k + u k ) > < [ xk u k ], Σ [ xk ] >. u k Then take summation with respect to k from k = 0 to k = N to obtain: N < k=0 [ xk u k ] [ xk, Σ u k ] > = N k=0 < (R K S Kx k + u k ), R K (R K S Kx k + u k ) > + < x 0, Kx 0 > < x N+, Kx K+ >. Since we have x 0 = 0 and lim x N = 0, N < k=0 [ xk u k ], Σ [ xk ] > = u k k=0 < (R K S Kx k + u k ), R K (R K S Kx k + u k ) >. Hence, since R K > 0, (4.22) is satisfied. Theorem is proved. Remark Suppose that (A, B) is d-stabilizable and Σ = I (identity matrix). Then (4.2) has a stabilizing solution (see e.g. [2]). Theorem 4.7. Let H = l n 2 (IN) l m 2 (IN). Suppose the pair (A, B) is d-stabilizable. Let X be defined as follows: 9
X = {(x, u) = ({x k }, {u k }) H; x k+ = Ax k + Bu k, x 0 = 0 IR n k = 0,, }. Let also Z = {({ξ k }, {η k }); ξ 0 = A T p 0, ξ k = A T p k p k for k =, 2,, η k = B T p k, for k = 0,,, {p k } l n 2 (IN)}. Then X is a closed vector subspace and Z is an orthogonal complement of X in H, i.e. Z = X. Proof Let (x, u) X and ({ξ k }, {η k }) Z. Then < (x, u), ({ξ k }, {η k }) > H = < x 0, A T p 0 > + < u 0, B T p 0 > + lim = < x, p 0 > + lim = < x, p 0 > + lim n k= n k= = lim n < x n+, p n > = 0. n k= n (< x k, A T p k p k > + < u k, B T p k >) n (< Ax k + Bu k, p k > < x k, p k >) n (< x k+, p k > < x k, p k >) Next we show that given ({ψ k }, {φ k }) H, it admits the following representation: ψ 0 = x 0 [A T p 0 ] (4.24) and for k =, 2, ψ k = x k [A T p k p k ] (4.25) φ k = u k B T p k (4.26) for k = 0,,, where ({x k }, {u k }) X and {p k } l n 2 (IN). This will imply easily that Z = X and both X and Z are closed. We look for p k in the form Then by (4.25), we have p k = Kx k + ρ k (4.27) p k = ψ k x k + A T p k where K = K T is a symmetric matrix. Substituting this into (4.27), we get ρ k = A T p k + ψ k x k + Kx k. (4.28) 20
But then (4.26) implies: u k = B T p k + φ k. (4.29) Our goal now is to eliminate p k. By using (4.27) p k = Kx k+ + ρ k+ K(Ax k + Bu k ) + ρ k+. By substituting this into (4.29), we have Hence, Hence, we have Using u k = B T ( K(Ax k + Bu k ) + ρ k+ ) + φ k = B T KAx k B T KBu k + B T ρ k+ + φ k. (I + B T KB)u k = B T KAx k + B T ρ k+ + φ k. u k = (I + B T KB) B T KAx k + (I + B T KB) (B T ρ k+ + φ k ). (4.30) we obtain x k+ = Ax k + Bu k x k+ = Ax k B(I + B T KB) B T KAx k +B(I + B T KB) (B T ρ k+ + φ k ). (4.3) But then since p k = Kx k+ + ρ k+, we have p k = KAx k + KB(I + B T KB) B T KAx k KB(I + B T KB) (B T ρ k+ + φ k ) + ρ k+. Then from (4.28) we have the following ρ k = A T p k + ψ k x k + Kx k = ψ k x k + Kx k A T KAx k + A T KB(I + B T KB) B T KAx k A T KB(I + B T KB) (B T ρ k+ + φ k ) + A T ρ k+. Recall, now, if Σ is the identity matrix then according to Remark which follows Theorem 4.6 the corresponding discrete algebraic Riccati (4.2) equation has a stabilizing solution. In this case, Σ = I n, Σ 22 = I m, Σ 2 = 0 i.e. K = A T KA A T KB(I + B T KB) A T KB + I (4.32) 2
If we choose K to be stabilizing solution, we obtain ρ k = A T (I KB(I+B T KB) B T )ρ k+ A T KB(I+B T KB) φ k +ψ k (4.33) It is known that if K 0 is a stabilizing solution to DARE (4.32), then the matrix A T (I K 0 B(I + B T K 0 B) B T ) is d-stable. Then by the Theorem 4.5, there exists a unique solution {ρ k } l2 n (IN) to (4.33). Hence by using (4.3) where x 0 = 0, (4.30) and then (4.27) we have shown that any ({ψ k }, {φ k }) H admits a unique decomposition as a sum of elements in X and X. Hence the Theorem is proved. Now we are in the position to describe a solution of the LQ-problem on a semi-infinite interval. 2 where < k=0 [ xk u k ], Σ [ xk ] > + < u k k=0 [ xk u k ], [ ψk ] > min, (4.34) φ k x k+ = Ax k + Bu k + g k, x 0 = x 0 R n, {g k } l n 2 (IN), (x, u) H. (4.35) Theorem 4.8. Suppose there exists a stabilizing solution K to DARE (4.2): Then {f k } l n 2 (IN) where f k = (A T KB + Σ 2 )(Σ 22 + B T KB) φ k + Σ 2 (Σ 22 + B T KB) B T Kg k ψ k. Furthermore, let {ρ k } be a unique solution from l n 2 (IN) of the system where ρ k = Cρ k+ + f k C = A T (A T KB + Σ 2 )(Σ 22 + B T KB) B T. Then the solution ({x k }, {u k }) to LQ-problem on a semi-infinite interval (4.34) can be described by the following recurrent relations: x k+ = C T x k + B(Σ 22 + B T KB) (B T ρ k+ B T Kg k φ k ) (4.36) u k = (Σ 22 + B T KB) (B T KA + Σ 2 )x k +(Σ 22 + B T KB) (B T ρ k+ B T Kg k φ k ). (4.37) Proof By Theorem 4.5, the functional (4.34) restricted to subspace X is convex. Hence necessary and sufficient optimality condition for (4.34),(4.35)takes the form [ ] [ ] [ ] xk ψk A Σ + = T p k p k u k φ k B T p k 22
when (x, u) satisfy (4.35) for some {p k } l n 2 (IN). Moreover, we are looking for {p k } in the form p k = Kx k+ + ρ k+ where K is a stabilizing solution to DARE (4.2). We finish the proof exactly as in Theorem 4.7. 5 Some extensions 5. Linear-quadratic control problem with quadratic constraints Consider the following optimization problem: W 0 (y y 0 ) H min, (5.) W i (y y i ) H t i, i =,..., m (5.2) y c + Z. (5.3) With our choice of H, Z and other parameters, we arrive at the linearquadratic control problem with quadratic constraints ( see e.g [7, 20]). We first consider the problem of checking the feasibility of (5.) - (5.3): t min, (5.4) W i (y y i ) H t + t i, i =,..., m (5.5) To cast (5.4) - (5.6) in the conic form, consider y c + Z. (5.6) Λ : IR Z (IR H) (IR H)...(IR H), m-times such that Λ(t, y) = (t, W y), (t, W 2 y),..., (t, W m y), a = ((, 0), (0, 0),..., (0, 0)), b = ((t, W (c y )),..., (t m, W m (c y m ))) and X = Λ(IR Z). Then we can rewrite (5.4)-(5.6) in the form of (2.),(2.2). Observe that the only difference between (2.),(2.2) and our conic formulation is the choice of b. It is quite obvious that the dual will be of the form t i r i + < W i (c y i ), u i > H min, (5.7) i= i= Wi u i Z, u i r i, i =,..., m, (5.8) i= r i =. (5.9) i= 23
Thus, the whole numerical scheme remain the same for (5.4)-(5.6)( the only difference is in the choice of initial feasible solution). Returning to (5.)-(5.3) consider the following; Λ : IR Z (IR H) (IR H)...(R H), (m+)-times such that Λ(t, y) = (t, W 0 y), (0, W y),..., (0, W m y), X = Λ(IR Z), a = ((, 0), (0, 0),..., (0, 0)), b = ((0, W 0 (c y 0 )), (t, W (c y )),..., (t m, W m (c y m ))). We can rewrite (5.)-(5.3) in the conic form < a, z > min, z (b + X) Ω. One can easily see that the dual will have the following form t i r i + < W i (c y i ), u i > H min, (5.0) i= i=0 Wi u i Z, u i r i, i =,..., m, (5.) i=0 u 0. (5.2) It is quite easy to work out a scheme for the computation of the Nesterov- Todd direction following the ideas developed in [0]. 6 Numerical results In this section we consider several examples for both continuous and discretetime formulations. The major difference between the two methods is in solving linear-quadratic control problem with linear term. For continuous-time case for finite interval (respectively, infinite interval), solving linear-quadratic control problem with linear term requires solving differential Riccati equation (continuous algebraic Riccati equation) and systems of linear differential equations. For discrete-time case, we have to solve discrete Riccati equation and systems of difference equations. The numerical experiments confirm fast convergence to optimal solution and good approximation of the continuous formulations by discrete formulations. The algorithms are implemented in MATLAB and integrations of the ordinary differential equations are carried out with ODE45, a MATLAB ODE solver function for nonstiff problems. 24
6. Example : Multi-criteria Linear Quadratic Control Problem We consider Multi-criteria Linear Quadratic Control Problem in both continuous and discrete cases. For continuous formulations, we first consider the problem (4.),(4.2) (i.e. problem on the finite interval [0, T ]) with m = n = 3 and the following data: 2 3 4 5 6 x i e 3t e 3t sin(t)e 4t e 2t cos(t)e 4t cos(t)e 3t x 2i cos(t)e 4t sin(t)e 4t e 3t sin(t)e 5t e 3t e 4t x 3i cos(t)e 4t cos(t)e 4t cos(t)e 4t sin(t)e 5t e 2t e 3t u i t 2 +5 u 2i t 2 +8 u 3i t 2 +6 A = t 2 +3 t 2 +8 t 2 +6 2 5 2 4 2 3 2 3 2 t 2 +3 t 2 +8 t 2 +6 t 2 +5 t 2 +8 t 2 +6 7 8 x i cos(t)e 4t sin(t)e 3t x 2i e 4t e 4t x 3i e 3t e 3t u i t 2 +7 u 2i t 2 +8 u 3i t 2 +6, B = t 2 +8 t 2 +6 t 2 +6 3 2 4 5 2 2 3 4 2, α 0 = t 2 +8 t 2 +6 t 2 +6 2 5 t 2 +8 t 2 +8 t 2 +6 Q = diag(69, 96, 225), Q 2 = diag(69, 69, 9), Q 3 = diag(96, 2, 25), Q 4 = diag(225, 8, 49), Q 5 = diag(96, 49, 8), Q 6 = diag(289, 25, 2), Q 7 = diag(324, 9, 69), Q 8 = diag(36, 2, 225), R = diag(96, 44, 96), R 2 = diag(44, 9, 6), R 3 = diag(69, 225, 36), R 4 = diag(96, 49, 64), R 5 = diag(225, 8, 00), R 6 = diag(256, 2, 44), R 7 = diag(289, 69, 96), R 8 = diag(324, 225, 36), Then by applying a primal-dual algorithm based on NT-direction, we obtain the following table of solutions for different T for ɛ = 0.0 T 2 3 4 5 Opt V alue 2035 207 2075 2076 2077 N umber of Iterations 6 6 6 6 6 T 6 7 8 9 0 Opt V alue 2077 2077 2077 2077 2077 N umber of Iterations 6 6 6 6 6 25
Plot of Approximation solution after 3 iterations ".."and optimal control " " 0 2 3 Value 4 5 6 7 8 0 0.5.5 2 2.5 3 Time Figure : Plot of approximation of optimal control and Optimal control: T = 3 Note also that the approximation of the optimal control approaches to the optimal control only after a few iterations. In fact, only after 3-4 iterations, we obtain a very good approximation to the optimal solution. Figure and 2 are graphs of the approximate optimal solution after three iterations and the optimal solution. The dotted line is the plot of the approximation and the solid line is the actual optimal solution (the actual optimal solution means the solution obtained in the last iteration). Our next goal is to use the discrete formulations to estimate the optimal solution and optimal value for the above problem. The construction is quite straightforward. First,we choose a sufficiently large K > 0, then we discretize the partition [0, T ] in to K sub-intervals uniformly. Next we let h = T K. Hence, our targets in the discrete formulation can be easily computed by the following simple procedure: for state components x (i) k x (2i) k x (3i) k = x i(kh) x 2i (kh) x 2i (kh) for i =,, m, and k = 0,,, K. Similarly, for control components u (i) k u (2i) k = u i(kh) u 2i (kh), u (3i) u 2i (kh) k for i =,, m, and i = 0,,, K., 26
5 Plot of approximation of state and the actual state optimal solution 4 3 value 2 0 0 0.5.5 2 2.5 3 Time Figure 2: Plot of approximation of optimal state and Optimal state: T = 3 The corresponding discrete-time problem will have the form max i=,...,m k= {< (x k ϕ (i) k ), Q i (x k ϕ (i) k ) > + < (u k φ (i) k x k+ = Ãx k + Bu k, x 0 = γ 0 ), R i (u k φ (i) k ) >} min (6.) where Q i = hq i, R i = hr i, Ã = (I + ha) and B = hb. We choose K = 300. Then using above formulation, we obtain a optimal solution which is very close to the optimal solution to the continuous case. As it is shown in the Figure 3-4, we take the difference pointwise of the optimal solutions obtained from the continuous and discrete formulation. The result is quite good. The two optimal solutions are almost identical. 6.2 Example 3: Discrete-time linear-quadratic control problem with quadratic constraints In this section we solve linear-quadratic control problem with quadratic constraints using discrete formulation. Let us first consider the checking feasibility problem (5.4)-(5.6) in the discrete formulation of the linear-quadratic control problem with quadratic constraints where Z, c, W i, y i for i =,..., m will be the same as in the Example, and t i be given by the following table. i 2 3 4 5 6 7 8 t i 4 20 39 30 32 38 43 45 It is easy to see that there exists a feasible solution to the problem (5.)-(5.3) if and only if the optimal value of the problem (5.4)-(5.6) is negative. 27
3.) Discrete-time optimal state 5 Optimal state obtained by using discrete formulation 3.2) Difference between discrete-time and continuous-time Optimal state 0.0 Pointwise difference of state spaces between discrete and continuous formulations 4 0.005 3 2 0 0.005 0 0.0 0 500 000 500 2000 2500 3000 0.05 0 500 000 500 2000 2500 3000 Figure 3: Plot of discrete-time optimal state for Multi-criteria Linear Quadratic Control Problem. 4.) Discrete-time optimal control Optimal control obtained by using discrete formulation 0 4.2) Difference between discrete-time and continuous-time Optimal control 0. Pointwise difference in control spaces between discrete and continuous formulations 0.05 2 3 0 4 0.05 5 0. 6 7 0.5 8 0.2 9 0 500 000 500 2000 2500 3000 0.25 0 500 000 500 2000 2500 3000 Figure 4: Plot of discrete-time optimal control for Multi-criteria Linear Quadratic Control Problem. 28
5.) Plot of optimal state for T = 20. 5 Plot of optimal state 5.2) Plot of optimal control for T = 20. Plot of optimal control 4 0 3 2 2 3 4 5 0 6 0 00 200 300 400 500 600 700 800 900 000 7 0 00 200 300 400 500 600 700 800 900 000 Figure 5: Plot of Discrete-time linear-quadratic control problem with quadratic constraints. Then by applying a primal-dual algorithm described in [0] based on NTdirection for T = 3, we obtain the following data of optimal value t after some number of iterations. #iterations 2 3 4 5 6 7 8 Opt Value.87 5.9.86 0.9 0.6 0.42 0.49 0.50 As it is shown in the above table, after the forth iteration we obtain a feasible solution to the linear-quadratic control problem with quadratic constraints(5.)- (5.3). Once we obtain a feasible solution for the optimization problem (5.)- (5.3), we then can apply primal-dual algorithms to solve the problem. The calculations for the NT-direction for this problem are quite similar to the calculation for (6.). The only differences are in the definition of a closed vector subspace X and its complement X. The optimal solution to the problem (5.)-(5.3) for T = 20 based on NT-direction is shown in the Figure 5. 29
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