Lower and upper bounds in optimization technique
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Dear all
I have 28 variables, I want to put 14 variables with (lower bound (lb) and upper bound(ub)) and the 14 other without (lower bound (lb) and upper bound(ub))... How can I do that?
thank you in advance.
8 个评论
tahseen alshmary
2021-10-7
Almost. It is necessary to supply the function names (and spell the function names correctly), and concatenate them using square brackets [] —
lb = [0.05*ones(1,14) zeros(1,14)]
I assume that the first term is supposed to call the ones function, and I completed this by supplying the right square bracket.
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tahseen alshmary
2021-10-7
I am not certain that I understand what you want, since ‘not zeroes’ can be anything.
Perhaps substituting Inf in the upper bound or -Inf in the lower bound could do what you want. So for example if you wanted the last three elements of ‘lb’ (in this example) to be -Inf do this —
lb = [0.05*ones(1,14) zeros(1,11) -Inf(1,3)]
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tahseen alshmary
2021-10-7
I was a bit mystified.
The same general idea —
lb = [0.05*ones(1,14) 3*ones(1,14)]
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tahseen alshmary
2021-10-7
Star Strider
2021-10-7
As always, my pleasure!
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采纳的回答
Star Strider
2021-10-6
0 个投票
Create appropriate vectors for each bound. Specify the bounds for the parameters that have bounds, and +Inf and -Inf for those allowed to vary without bound.
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4 个评论
tahseen alshmary
2021-10-6
Star Strider
2021-10-6
A representative example (not for all 14 parameters) would be something like this —
lb = [2 3 2 1 -Inf -Inf]
ub = [10 4 5 7 +Inf +Inf]
So long as the upper bounds are greater than the lower bounds, this should work. The elements of each vector must correspond to the order of the parameters in the model, so here ‘p(1)’ would be allowed to vary between 2 and 10, while ‘p(6)’ would be essentially unbounded.
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tahseen alshmary
2021-10-6
First, it is not possible to use parentheses to create the vector. It is necessary to use square brackets to concatenate the elements of the vector into a single expression.
The correct way to write those would be something like this —
lb = 0.05 * ones(1,14)
ub = ones(1,14)
If the other 14 are not bounded:
lb = [0.05 * ones(1,14) -Inf(1,14)]
ub = [ones(1,14) Inf(1,14)]
However if they have specific bounds that could not be defined by multiples of ones, additions to zeros, or something else, they would have to be entered individually, as I demonstrated earlier.
Also, optimising 28 parameters could be difficult.
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