quadprog constraints with minimum value

I would like to set up the quadprog constraint for minimum position in each asset as 0 or 0.5%.
Because the solver returns a whole bunch of tiny positions between 0-0.5% which are unrealistic to implement.
Thanks.

2 个评论

If "position" are solution variables, set their lower bounds (lb in the call to quadprog) to 0.005, e.g.
Thanks Torsten,
If the lower bounds of asset are set to 0.5%, then all assets in the optimization universe would need to have at least 0.5% weights.
Here's an example to illustrate the desired setting:
I have 1000 assets in the optimization universe, a sub-group of which (determined by optimizer) would have a minimum weights of 0.5%, where the rest would stay at 0% as the final solution.
Basically, I would like to have a discontinuous constraint.

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 采纳的回答

Matt J
Matt J 2022-5-10
编辑:Matt J 2022-5-10
One possible alternative would be to use minL1lin,
instead of quadprog. An L1-norm cost function will tend to have sparser solutions than a quadratic cost. So, that may be a way of reducing small residual non-zeros in the solution.

1 个评论

Thanks Matt,
This is probably the only way we can set up such a discontinuous constraint.

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更多回答(2 个)

Matt J
Matt J 2022-5-9
编辑:Matt J 2022-5-9
If the lower bounds of asset are set to 0.5%, then all assets in the optimization universe would need to have at least 0.5% weights.
lb is a vector-valued argument. Set each lb(i) to the appropriate lower bound.

2 个评论

Thanks Matt,
In this case, I don't need to differentiate constraints for each asset, but rather a one-size-fits-all is applied.
The tricky thing is we don't know ex-ante which asset is sitting at 0% and which asset has a non-zero weight and need to be greater than 0.5%, due to the discontinuous nature of constaint.
If it's a discontinuous constraint, it's not a quadratic program anymore.

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Matt J
Matt J 2022-5-10
If you have the Global Optimization Toolbox, you could also try using ga() with additional binary variables b(i). Denoting your originally set of continuous variables as x(i), you could impose linear inequality constraints x(i)-0.5*b(i)>=0 to enforce your bounds.

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