Timeline for Minimizing $\mathrm{trace}(S)+\mathrm{trace}(S^{-2})$ using CVX
Current License: CC BY-SA 4.0
20 events
| when toggle format | what | by | license | comment | |
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| May 17, 2018 at 6:15 | history | edited | Johan Löfberg | CC BY-SA 4.0 |
edited body
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| May 6, 2013 at 8:20 | vote | accept | user2987 | ||
| Mar 13, 2013 at 12:10 | comment | added | Johan Löfberg | Well, then your problem formulation does not capture what you really want to compute, i.e., it is really not an issue about the implementation of your problem, but the relevance of the model to begin with. | |
| Mar 13, 2013 at 7:06 | comment | added | user2987 | it works and it does satisfy the constraint but when I compare $S^{Estimated}$ to $S^{true}$ they are totally different.. | |
| Mar 13, 2013 at 6:58 | comment | added | user2987 | here is my whole problem math.stackexchange.com/questions/329040/… | |
| Mar 13, 2013 at 6:49 | comment | added | Johan Löfberg | You mean the particular problem above? I don't see that. Running 100 examples shows that it is active in roughly 50% of those cases. | |
| Mar 13, 2013 at 1:39 | comment | added | user2987 | Do you have any idea why this problem leads always to the unconstrained solution. Actually the unconstrained solution for $f(S)$ is $\propto I$ but given my constraint $\|S-C\|_1\leq \alpha$, where $\alpha$ depends on $C$ and $\| \dot \|_1$, the estimated $S_*$ should not be any more diagonal. | |
| Mar 12, 2013 at 10:06 | comment | added | Johan Löfberg | Based on email correspondence, the zero solution was due to other mistakes and not related to the question here. The zero solution is not possible (S zero would lead to X infinite which would lead to Z infinite) | |
| Mar 12, 2013 at 8:27 | comment | added | user2987 | Is that normal to find always $Z=0_{m,m}$ and $S\sim 0_{m,m}$?!! | |
| Mar 12, 2013 at 7:43 | history | edited | Johan Löfberg | CC BY-SA 3.0 |
clarifying role of X and Z
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| Mar 12, 2013 at 7:41 | comment | added | user2987 | Yes you're right! | |
| Mar 12, 2013 at 7:38 | comment | added | Johan Löfberg | $Z$ should be equal to $S^{-2}$, i.e., $S = Z^{-1/2}$ | |
| Mar 12, 2013 at 7:35 | comment | added | Johan Löfberg | The BoundSquare is not equivalent to $X^TZ^{-1}X\geq I$, you have reversed the inequality. | |
| Mar 12, 2013 at 7:32 | history | edited | Johan Löfberg | CC BY-SA 3.0 |
Adding comments in code
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| Mar 12, 2013 at 7:25 | comment | added | user2987 | Actually I tried your code using cvx for not fixed S and I found that $S\ne Z^2$. | |
| Mar 12, 2013 at 7:25 | history | edited | Johan Löfberg | CC BY-SA 3.0 |
Adding comments in code
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| Mar 12, 2013 at 7:14 | comment | added | Johan Löfberg | The case with a fixed S was just to create a simple test to check correctness. If it wouldn't work for a fixed S, it would obviously not work when you optimize over S (i.e., when you have S linearly parameterized in some decision variabables). You never add an explicit constraint $Z=S^2$ or something like that. It would kill convexity. The epigraph-based code above is complete, except that you have some more code to define S, what ever it might be. Note that it only works if you minimize the $trace(S^{-2})$ term, as it is based on an upper, in optimality tight, bound. | |
| Mar 12, 2013 at 2:22 | comment | added | user2987 | what do you mean by for a given S? My objective function to minimize is $f(S)=\mathrm{trace}(S)+\mathrm{trace}(S^{-2})$ where S is the unknown that we are looking to find. What you have suggested is equivalent to have $X\geq S^{-1}$ and $X^TZ^{-1}X\geq I$. So this constraints would impose an upper bound for $S^{-1}$ and an another lower bound for the quadratic form of $Z^{-1}$. Do you mean that by adding these two constraints I can avoid writing $Z=S^2$ which cannot be accepted in a convex problem because it's not an affine equality? | |
| Mar 11, 2013 at 20:36 | vote | accept | user2987 | ||
| Mar 17, 2013 at 21:45 | |||||
| Mar 11, 2013 at 12:58 | history | answered | Johan Löfberg | CC BY-SA 3.0 |