MATH Dr. Pedro V squez UPRM. P. V squez (UPRM) Conferencia 1/ 17

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1 Dr. Pedro V squez UPRM P. V squez (UPRM) Conferencia 1/ 17

2 Quadratic programming MATH 6026 Equality constraints A general formulation of these problems is: min x 2R nq (x) = 1 2 x T Qx + x T c (1) subjec to Ax = b where A is the m " n Jacobian of constraints (with m # n) whose rows ai T, i 2 # and b is the vector in R m. We assume that A has full row rank (rank m).so that the constraints are consistent. P. V squez (UPRM) Conferencia 2/ 17

3 First order necessary conditions The FONC for x $ to be a solution of (1) state that there is a vector l $ such that the following system of equations is satisöed:! "! " Q %A T x $ A 0 l $ =! " %c b The system can be rewritten in a form that is useful for computation by expressing x $ as x $ = x + p, where x is some estimate of the solution and p is the desired step, so the above system can be written:! Q A T A 0 "! " %p l $ =! " g h where h = Ax % b, g = c + Qx, p = x $ % x (2) (3) P. V squez (UPRM) Conferencia 3/ 17

4 The matrix in (2) is called the Karush-Kuhn-Tucker (KKT) matrix, and the following result gives conditions under which is non singular. We use Z as the n " (n % m) matrix whose columns are the basic null space of A, that is Z has full rank and satisöes AZ = 0. Lemma Let A have full row rank, and assume that the reduced-hessian matrix Z T QZ! is postive " deönite. Then the KKT matrix Q A T K = (4) A 0 is nonsingular and hence there is a unique vector pair (x $, l $ ) satisfying (2). P. V squez (UPRM) Conferencia 4/ 17

5 Example P. V squez (UPRM) Conferencia 5/ 17

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7 Theorem Let A have full row rank and assume that the reduced-hessian matrix Z T QZ is positive deönite. Then the vector x $ satisfying (2) is the unique global solution of (1). P. V squez (UPRM) Conferencia 7/ 17

8 Direct solution of the KKT system One important observation is that if m & 1, the KKT matrix is always indeönite. We deöne the inertia of a symmetric matrix K to be the triple that indicates the numbers n +, n %, and n 0 of positive, negative and zero eigenvalues, respectively, that is, inertia(k )=(n +, n %, n 0 ) Theorem Let K be deöned by (4), and suppose that A has rank m. inertia(k )=inertia(z T QZ )+(m, m, 0). Then Therefore, if Z T QZ is positive deönite, inertia(k )=(m, m, 0) P. V squez (UPRM) Conferencia 8/ 17

9 Factoring the full KKT system One option to solve (3) is to perform a triangular factorization on the full KKT matrix and then perform backward and forward substitution with the triangular factors. Because of indeöniteness, we cannot use the Cholesky factorization. We would use Gaussian elimination with partial pivoting to obtain the L and U factors, but this approach has the disadvantage that ignores symmetry.. The most e ective strategy in this case is to use a symmetric indeönite factorization. For a general symmetric matrix K, the factorization has the form: P T KP = LBL T where P is a permutation matrix, L is unit lower triangular, and B is blocked-diagonal with either 1 " 1 or 2" 2 blocks. P. V squez (UPRM) Conferencia 9/ 17

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11 Null- space method The null space method does not require nonsingularity of Q and therefore has wider applicability. It assumes only that the conditions of lemma, that A has full row rank and that Z T QZ is positive deönite. However requires knowledge of the null-space basis matrix Z Suppose that the vector p has a partition into two components: p = Yp Y + Zp Z where Z is the n " (n % m) null-space matrix, Y is any n " n matrix such that [Y jz ] is nonsingular p Y is an n-vector p Z is an (n % m)-vector Yx Y is a particular solution of Ax = b Zx Z is a displacement along the constraints P. V squez (UPRM) Conferencia 11 / 17

12 Then we obtain: (AY ) p Y = %h Since A has rank m and [Y jz ] is n " n nonsingular, the product A [Y jz ] = [AY j0] has rank m. Therefore, AY is a nonsingular m " m matrix, and p Y is well determined, substituting : %QYp Y % QZp Z + A T l $ = g and multiply by Z T : # Z T QZ $ p Z = %Z T QYp Y % Z T g The system can be solved by performing a Cholesky factorization of the reduced Hessian matrix Z T QZ to determine p Z. We therefore compute the total step p = Yp Y + Zp Z. To obtain the lagrange multiplier, we use: which can be solved for l $. (AY ) T l $ = Y T (g + Qp) P. V squez (UPRM) Conferencia 12 / 17

13 Example P. V squez (UPRM) Conferencia 13 / 17

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15 Iterative solution for the KKT system CG applied to the reduced system Assuminf that the solution of the QP is: x $ = Yx Y + Zx Z for some vectors x Z 2 R n%m, x Y 2 R m, the constraints Ax = b yield: AYx Y = b whixh determines the vector x. To obtain x Z solves the unsconstrained reduced problem: where 1 min x 2 x Z T Z T QZx Z + xz T c Z, Z c Z = Z T QYx Y + Z T c The solution x Z satisöes the linear system Z T QZx Z = %c Z Since Z T QZ is positive deönite, we can apply the CG method: P. V squez (UPRM) Conferencia 15 / 17

16 Algorithm 16.1 (preconditioned CG for reduced systems Choose an initial point x Z ; Compute r Z = Z T QZx Z + c Z, g Z = W ZZ %1 r Z, and d Z = %g Z repeat a r Z g Z /dz T Z T QZd Z ; x Z x Z + ad Z ; r + Z r Z + az T QZd Z ; g + Z W ZZ %1 r + # $ Z ; b r + T Z g + Z /rz T g Z ; d Z %g + Z + bd Z ; g Z g + Z ; r Z r + Z until a termination test is satisöed. P. V squez (UPRM) Conferencia 16 / 17

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