Lower and upper bounds in optimization technique

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.

[tahseen alshmary]

How can I write two different number like

...

[0.05* (1,14) zero (1,14)

Is that coorect

Star Strider

[Star Strider]

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)]

lb = 1×28

0.0500 0.0500 0.0500 0.0500 0.0500 0.0500 0.0500 0.0500 0.0500 0.0500 0.0500 0.0500 0.0500 0.0500 0 0 0 0 0 0 0 0 0 0 0 0 0 0

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]

Thanks

If I want to use three not zeroes in right square bracket ? How?

[Star Strider]

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)]

lb = 1×28

0.0500 0.0500 0.0500 0.0500 0.0500 0.0500 0.0500 0.0500 0.0500 0.0500 0.0500 0.0500 0.0500 0.0500 0 0 0 0 0 0 0 0 0 0 0 -Inf -Inf -Inf

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[tahseen alshmary]

I mean

lb = [0.05* ones (1,14) three (1,14)]

how can I write the number of three

[Star Strider]

I was a bit mystified.

The same general idea —

lb = [0.05*ones(1,14)  3*ones(1,14)]

lb = 1×28

0.0500 0.0500 0.0500 0.0500 0.0500 0.0500 0.0500 0.0500 0.0500 0.0500 0.0500 0.0500 0.0500 0.0500 3.0000 3.0000 3.0000 3.0000 3.0000 3.0000 3.0000 3.0000 3.0000 3.0000 3.0000 3.0000 3.0000 3.0000

.

[tahseen alshmary]

Thank you so much

[Star Strider]

As always, my pleasure!

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Answer 1

[Star Strider]

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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[tahseen alshmary]

Can you write example ? please

[Star Strider]

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]

If I want to write the bounded variable as below

lb = 0.05 * (1,14)

ub = 1 * (1,14)

so how can I write the 14 other bound

[Star Strider]

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)

lb = 1×14

0.0500 0.0500 0.0500 0.0500 0.0500 0.0500 0.0500 0.0500 0.0500 0.0500 0.0500 0.0500 0.0500 0.0500

ub = ones(1,14) 

ub = 1×14

1 1 1 1 1 1 1 1 1 1 1 1 1 1

If the other 14 are not bounded:

lb = [0.05 * ones(1,14) -Inf(1,14)]

lb = 1×28

0.0500 0.0500 0.0500 0.0500 0.0500 0.0500 0.0500 0.0500 0.0500 0.0500 0.0500 0.0500 0.0500 0.0500 -Inf -Inf -Inf -Inf -Inf -Inf -Inf -Inf -Inf -Inf -Inf -Inf -Inf -Inf

ub = [ones(1,14) Inf(1,14)]

ub = 1×28

1 1 1 1 1 1 1 1 1 1 1 1 1 1 Inf Inf Inf Inf Inf Inf Inf Inf Inf Inf Inf Inf Inf Inf

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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