EE364 Review Session 4
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1 EE364 Review Session 4 EE364 Review Outline: convex optimization examples solving quasiconvex problems by bisection exercise
2 Convex optimization problems we have seen linear programming (LP) quadratic programming (QP) quadratically constrained QP (QCQP) second-order cone programming (SOCP) geometric programming (GP) semidefinite programming (SDP) EE364 Review Session 4 2
3 Force/moment generation with thrusters rigid body with center of mass at origin p = 0 R 2 n forces with magnitude u i, acting at p i = (p ix, p iy ), in direction θ i (p ix, p iy ) θ i u i PSfrag replacements EE364 Review Session 4 3
4 resulting horizontal force: F x = n i=1 u i cos θ i resulting vertical force: F y = n i=1 u i sin θ i resulting torque: T = n i=1 (p iyu i cos θ i p ix u i sin θ i ) force limits: 0 u i 1 (thrusters) fuel usage: u u n problem: find thruster forces u i that yield given desired forces and torques Fx des, Fy des, T des, and minimize fuel usage (if feasible) EE364 Review Session 4 4
5 can be expressed as LP: minimize subject to 1 T u F u = f des 0 u i 1, i = 1,..., n where F = cos θ 1 cos θ n sin θ 1 sin θ n p 1y cos θ 1 p 1x sin θ 1 p ny cos θ n p nx sin θ n, f des = (Fx des, Fy des, T des ), 1 = ( 1, 1, 1 ) EE364 Review Session 4 5
6 clear all; close all; % input data % % thrusters x-coordinates px = [ ]; % thrusters y-coordinates py = [ ]; % angles thetas = [ ]*pi/180; F = [ cos(thetas); sin(thetas); py.*cos(thetas) - px.*sin(thetas)]; % different problem specified by each column of f_des f_des = [ ; ; ]; EE364 Review Session 4 6
7 % problem solution thrus = []; for i=1:6 end cvx_begin variable u(5) minimize ( sum ( u ) ) F*u == f_des(:,i) u >= 0 u <= 1 cvx_end thrus = [thrus u]; EE364 Review Session 4 7
8 for F des x = 0, F des y = 0.5, T des = 0: EE364 Review Session 4 8
9 for F des x = 0, F des y = 1, T des = 0: EE364 Review Session 4 9
10 for F des x = 1, F des y = 0, T des = 0: EE364 Review Session 4 10
11 for F des x = 0.5, F des y = 0, T des = 0: EE364 Review Session 4 11
12 for F des x = 0, F des y = 0, T des = 2: EE364 Review Session 4 12
13 for F des x = 0, F des y = 0, T des = 2: EE364 Review Session 4 13
14 Extensions of thruster problem opposing thruster pairs: minimize subject to u 1 = n i=1 u i F u = f des u i 1, i = 1,..., n can express as LP more accurate fuel use model: minimize subject to n i=1 φ i(u i ) F u = f des 0 u i 1, i = 1,..., n φ i are piecewise linear increasing convex functions can express as LP EE364 Review Session 4 14
15 minimize maximum force/moment error: minimize subject to F u f des 0 u i 1, i = 1,..., n can express as LP minimize number of thrusters used: minimize subject to # thrusters on F u = f des 0 u i 1, i = 1,..., n can t express as LP (but we could check feasibility of each of the 2 n subsets of thrusters) EE364 Review Session 4 15
16 Optimizing structural dynamics linear elastic structure PSfrag replacements f 1 f 2 f 3 f 4 dynamics (ignoring damping): M d + Kd = 0 d(t) R k : vector of displacements M = M T 0 is mass matrix; K = K T 0 is stiffness matrix EE364 Review Session 4 16
17 Fundamental frequency solutions have form d i (t) = k α ij cos(ω j t φ j ) j=1 where 0 ω 1 ω 2 ω k are the modal frequencies, i.e., positive solutions of det(ω 2 M K) = 0 fundamental frequency: ω 1 = λ 1/2 min (K, M) = λ1/2 min (M 1/2 KM 1/2 ) structure behaves like mass at frequencies below ω 1 gives stiffness measure (the larger ω 1, the stiffer the structure) ω 1 Ω Ω 2 M K 0 so ω 1 is quasiconcave function of M, K EE364 Review Session 4 17
18 design variables: x i, cross-sectional area of structural member i (geometry of structure fixed) M(x) = M 0 + i x im i, K(x) = K 0 + i x ik i structure weight w = w 0 + i x iw i problem: minimize weight s.t. ω 1 Ω, limits on cross-sectional areas as SDP: minimize w 0 + i x iw i subject to Ω 2 M(x) K(x) 0 l i x i u i EE364 Review Session 4 18
19 Solving quasiconvex problems via bisection minimize subject to f i convex, f 0 quasiconvex f 0 (x) f i (x) 0, i = 1,..., m Ax = b idea: express sublevel set f 0 (x) t as sublevel set of convex function: f 0 (x) t φ t (x) 0 where φ t : R n R is convex in x for each t now solve quasiconvex problem by bisection on t, solving convex feasibility problem φ t (x) 0, f i (x) 0, i = 1,..., m, Ax = b (with variable x) at each iteration EE364 Review Session 4 19
20 bisection method for quasiconvex problem: given l < p ; feasible x; ɛ > 0 u := f 0 (x) repeat t := (u + l)/2 solve convex feasibility problem φ t (x) 0, f i (x) 0, Ax = b if feasible, u := t x := any solution of feas. problem else l := t until u l ɛ reduces quasiconvex problem to sequence of convex feasibility problems EE364 Review Session 4 20
21 4.47 Maximum determinant positive semidefinite matrix completion. We consider a matrix A S n, with some entries specified, and the others not specified. Say A = The positive semidefinite matrix completion problem is to determine values of the unspecified entries of the matrix so that A 0 (or to determine that such a completion does not exist). EE364 Review Session 4 21
22 (a) Why can we assume (w.l.o.g) that A ii are specified? (b) Formulate this problem as an SDP feasibility problem? EE364 Review Session 4 22
23 (c) Assume that A has at least one completion that is positive definite, and the diagonal entries of A are specified (i.e., fixed). The positive definite completion with largest determinant is called the maximum determinant completion. Show that the maximum determinant completion is unique. Show that if A is the maximum determinant completion, then (A ) 1 has zeros in all the entries of the original matrix that were not specified. (d) Suppose A is specified on its tridiagonal part, i.e., we are given A 11,..., A nn and A 12,..., A n 1,n. Show that if there exists a positive definite completion of A, then there is a positive definite completion whose inverse is tridiagonal. EE364 Review Session 4 23
24 Matlab code: n = 4; % create and solve the problem cvx_begin sdp % A is a PSD symmetric matrix (n-by-n) variable A(n,n) symmetric; maximize( det_rootn( A ) ) A >= 0; % constrained matrix entries. A(1,1) == 3; A(2,2) == 2; A(3,3) == 1; A(4,4) == 5; A(1,2) ==.5; A(1,4) ==.25; A(2,3) ==.75; cvx_end EE364 Review Session 4 24
25 Matrix A with maximum determinant (20.578) is: A = Its eigenvalues are: eigs = The inverse of matrix A is: EE364 Review Session 4 25
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