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Minimizing infinity norm squared |
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| Jun6-11, 03:43 AM | #1 |
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Minimizing infinity norm squared
I have to minimize an expression of the following type:
min <a,x>-L||x-u||_inf^2 s.t.: ||x||_inf <= R, where a is a vector of coefficients, x is the vector of decision variables, <.,.> denotes the scalar product, R and L are scalars, u is some constant (known) vector, and 'inf' denotes the infinity norm. I know how to deal with the situation where the expression is of the type ||x-u||_inf. But, I do not know how to approach the expression when the infinity norm is squared. I hope someone out there can suggest some ideas, or a book that I can consult, etc. |
| Jun6-11, 12:34 PM | #2 |
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A hint that might be useful: if ||x||_inf <= R, then ||x||_inf^2 <= R^2. You should then be able to proceed just as in your other case. If not, you need to show us more ...
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| Jun7-11, 05:10 AM | #3 |
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well.. the actual problem is to find the argument which maximizes the following expression:
<a,x> - (L/2)||x-u||_inf^2, s.t.: x >= 0, ||x||_inf <= R, where a and u are known vectors, x is the vector of decision variables, and R and L are known scalars. Furthermore, the vector u satisfies the following: u >= 0, ||u||_inf <= R. I know of a simple trick for dealing with the minimization of an expression of the form ||Ax-b||_inf. But I cannot figure out how to approach the situation where the infinity norm expression is squared and where the scalar product adds a linear term as well. |
| Jun7-11, 05:54 AM | #4 |
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Minimizing infinity norm squared
i have an approach in mind.. i thought i'd get your opinion on it..
the problem is to find the maximum of the following: <a,x> - (L/2)||x-u||_inf^2 s.t.: x >= 0, ||x||_inf <= R. The is the same as finding the minimum of (L/2)||x-u||_inf^2 - <a,x> s.t.: x >= 0, ||x||_inf <= R. I use the following trick: min t s.t.: (L/2)||x-u||_inf^2 - <a,x> <= t, x >= 0, ||x_inf|| <= R. And write this as follows: min t s.t. (L/2)(xi-ui)^2 - <a,x> <= t^2, for all i = 1,...,m x >= 0, ||x||_inf <= R. This gives me a quadratic cost with quadratic and linear constraints. And the constraints are coupled. Do you think that a) the above is correct; and b) if it's possible to have a formulation where the constraints are linear or uncoupled? |
| Jun7-11, 06:47 AM | #5 |
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there is another approach.. this gives a quadratic cost with coupled linear constraints.. i'd be very grateful if you could check the working..
the problem is to find the maximum of <a,x> - (L/2)||x-u||_inf^2 s.t.: x >= 0, ||x||_inf <= R. this is the same as finding the minimum of (L/2)||x-u||_inf^2 - <a,x> s.t.: x >= 0, ||x||_inf <= R. I express this as follows: min t0^2 + t1 + t2 + ... + tm s.t.: (L/2)||x-u||_inf^2 <= t0^2, -ai*xi <= ti, i = 1,...,m, x >= 0, ||x||_inf <= R. I express the first constraint as follows: (L/2)(xi - ui)^2 <= t0^2, i = 1,...,m. I express these constraints as follows: ui - sqrt(2/L)*t0 <= xi <= ui + sqrt(2/L)*t0, i = 1,...,m. The optimization problem is given by: min t0^2 + t1 + ... + tm s.t.: -xi - sqrt(2/L)*t0 <= - ui, i = 1,...,m xi - sqrt(2/L)*t0 <= ui, i = 1,...,m -ai*xi -ti <= 0, i = 1,...,m 0 <= xi <= R, i = 1,...,m i don't think i can uncouple the constraints somehow.. |
| Jun7-11, 09:05 AM | #6 |
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Hey, sorry for not answering earlier ... I live in a different timezone :)
s.t. 0 <= x_i <= R for all i as you did in your other example. The calculation you did seems fine to me, but I don't know either how to uncouple your constraints ... |
| Jun7-11, 10:03 AM | #7 |
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thanks..
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| Jun7-11, 11:39 AM | #8 |
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Recognitions:
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min M*z^2 - <a,x> s.t. z >= x(i)-u(i), z >= u(i)-x(i) for all i -R <= x(i) <= R for all i. This has just one single quadratic term in the objective, but is otherwise linear. Somehow, one feels there must be a better way to solve it than using a standard quadratic programming package, although that would certainly work in this case, because we have a convex quadratic programming problem. This only works because we are *minimizing* ||x-u||^2; if, instead, you wanted to maximize the objective M*||x-u||^2 - <a,x> you would have a problem with a mixed continuous-combinatorial aspect to it, and formulation/solution would be difficult. RGV |
| Jun7-11, 05:12 PM | #9 |
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perhaps, relaxing the coupling constraints using Lagrange multipliers could be an alternative to a quadratic programming package.. but i doubt the efficacy of such an approach.. it may still require global coordination.. moreover, it may be too slow for what is essentially one sub-problem in a larger algorithm requiring many iterations.. |
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