Quadratic Cost Function x^T Q x
in [Matlab]
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From: Matt J on 21 May 2010 08:03 "Bruno Luong" <b.luong(a)fogale.findmycountry> wrote in message <ht5qeo$26u$1(a)fred.mathworks.com>... > So what exactly he wants to minimize? It seems Jason confused many people, and even Roger has left the thread. ========== This is the objective function f(x) = ( x'*Q_1*x )^2 + ( x'*Q_2*x )^2 + ( x'*Q_3*x )^2 with some constraint to normalize the magnitude of x, e.g. we have looked at norm(x)=1 norm(x(1:2))=1 x(3)=1 etc...
From: Bruno Luong on 21 May 2010 08:24 "Jason" <jf203(a)ic.ac.uk> wrote in message <ht5si9$ipb$1(a)fred.mathworks.com>... > "Bruno Luong" <b.luong(a)fogale.findmycountry> wrote in message <ht5qeo$26u$1(a)fred.mathworks.com>... > > Well, actually I'm confused. The reason I think it's quadratic because I quote (from OP): > > > > 'As I said J = min x ( x' Q x).' > > > > OR in another thread > > > > [ If I have a quadratic optimization problem of the type > > > > min(x) x' Q x > > > > Where Q is symmetric but indefinite (3x3) matrix, and x is column vector of length 3. > > > > Fmincon and quadprog could solve this. ] > > > > So what exactly he wants to minimize? It seems Jason confused many people, and even Roger has left the thread. > > > > Bruno > > Bruno, it is quite clearly quadratic. > And it's quite clear that you minimize x, i.e. find the values of x that minimize x' Q x. > > If something is not clear in your mind then don't blame me. Well, now Jason, can you tell me the relation between: 1) minimizing x'*Q*x, and 2) x'*Q*x = 0, I quite [ ... common tangent to a set of ellipses which is solved if x'Qx = 0 ] And earlier on you wrote: [ ... J = SUM ||x^T Qi x||^2 ... ] This is clearly quadric functional, not quadratic. And one more quote, you gave a function handle passing to LSQNONLIN defined by: [ function F = opt_routine(x,sns,Q); for i = 1:sns F(i) = (x.')*Q{i}*x; end end ] Which LSQNONLIN, will intepret as the cost function as (see http://www.mathworks.com/access/helpdesk/help/toolbox/optim/ug/lsqnonlin.html) J(x) = sum f(i)^2 = sum ((x.')*Q{i}*x)^2 Which is QUADRIC and not QUADRATIC. But you state in the last post [ ... it is quite clearly quadratic... ] There is clearly a confusion in various posts between QUADRIC and QUADRATIC. Who should I blame Jason? > > Thanks for your help, but I don't think it's in my scope to try it out, I want something from scratch and not hacking a ready code, which I am not even sure is applicable for my purposes. If it's quadratic, it's applicable, and FYI it can find ALL the solutions in one shot. Bruno
From: Bruno Luong on 21 May 2010 08:28 "Matt J " <mattjacREMOVE(a)THISieee.spam> wrote in message <ht5spo$4oc$1(a)fred.mathworks.com>... > "Bruno Luong" <b.luong(a)fogale.findmycountry> wrote in message <ht5qeo$26u$1(a)fred.mathworks.com>... > > > So what exactly he wants to minimize? It seems Jason confused many people, and even Roger has left the thread. > ========== > > This is the objective function > > f(x) = ( x'*Q_1*x )^2 + ( x'*Q_2*x )^2 + ( x'*Q_3*x )^2 > But Jason just tells us [... it is quite clearly quadratic.]!!! And how on earth he can use QUADPROG for this function? I can't still figure out what Jason really wanted. Bruno
From: Bruno Luong on 21 May 2010 08:40 Sorry for the typo, *QUARTIC* and not QUADRIC Bruno
From: Matt J on 21 May 2010 09:34
"Jason" <jf203(a)ic.ac.uk> wrote in message <ht5si9$ipb$1(a)fred.mathworks.com>... > > Bruno, it is quite clearly quadratic. > And it's quite clear that you minimize x, i.e. find the values of x that minimize x' Q x. > > If something is not clear in your mind then don't blame me. ================= No it is not quadratic. You are trying to find the simultaneous *root* of 3 quadratic functions (one for each ellipse), not their minimum. To do this, you wrote down a cost function J in which you square all the quadratics for each ellipse and add them up. This leads to a 4th order function polynomial in x, not a 2nd order one |