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Oct 10, 2023 at 22:55 comment added Josh Smith This worked perfectly, I really appreciate the help
Oct 10, 2023 at 22:32 history edited Erwin Kalvelagen CC BY-SA 4.0
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Oct 10, 2023 at 22:23 history edited Erwin Kalvelagen CC BY-SA 4.0
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Oct 10, 2023 at 22:15 comment added Josh Smith Would you say there's a good way to ensure turnover is <=0.25? If we let x be the weights from the previous day, then turnover = sum(abs(w_i,long - w_i,short) - (x_i,long - x_i,short))) The only idea I have for expressing that as linear constraints is to split the differences ((w_i,long - w_i,short) - (x_i,long - x_i,short)) into positive and negative differences and then add a constraint to check if sum(d_i,p + (-1)*d_i,n) <= 0.25
Oct 10, 2023 at 21:08 vote accept Josh Smith
Oct 10, 2023 at 21:02 history edited Erwin Kalvelagen CC BY-SA 4.0
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Oct 10, 2023 at 20:58 comment added Josh Smith Thank you for the help this is an elegant solution
Oct 10, 2023 at 20:54 history edited Erwin Kalvelagen CC BY-SA 4.0
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Oct 10, 2023 at 20:47 history edited Erwin Kalvelagen CC BY-SA 4.0
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Oct 10, 2023 at 20:37 history edited Erwin Kalvelagen CC BY-SA 4.0
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Oct 10, 2023 at 20:31 history answered Erwin Kalvelagen CC BY-SA 4.0