Optimization problem with non-decision variable boundary limits

1 Ansicht (letzte 30 Tage)
sastra university
sastra university am 26 Okt. 2020
Hello everyone,
Anyone please clarify my doubt.
I have 2 decsion variables (p1, p2) with boundary limits. I can optimize the function with these values.
But i have one more variable which depends the one of the decision variable said above.This 3rd varaible also having boundary limits.
Based on the optimized result of p1 and p2, i need to ensure this 3rd variable also within the boundary limits.
What can i do for this?
Please help me out from this.
Thanks in advance.
  7 Kommentare
Walter Roberson
Walter Roberson am 26 Okt. 2020
p3_lower = b - a * p1 implies a*p1 = b + p3_lower implies p1 = (b+p3_lower)/a . This is either an upper bound or a lower bound depending on sign(a)
Likewise p1 = (b+p3_upper)/a is either an upper bound or lower bound depending on sign(a)
If your existing lower bound for p1 is less than the lower bound implied by the above, raise it to the value calculated. Likewise if your existing upper bound for p1 is greater than the upper bound implied by above, lower it to the value calculated.
Now you have restricted p1 to values that cannot possibly cause p3 to exceed the established bounds.
This general technique depends upon the relationship between p1 and p3 being monotonic within the segment of interest, but does not depend upon the relationship being linear. As long as you can solve() the p3 upper and lower bounds for p1 and get unique solutions within the range of interest, this is all you need.
You would have additional challenges, though, if there are multiple solutions within the range, especially discontinuous. For example if the restriction was that p3 = sin(a*p1)+b had to be within a particular range, and p1 could span more than pi/a then you could end up in situations where internal sub-ranges of p1 were prohibitted by the p3 restriction.
sastra university
sastra university am 26 Okt. 2020
Thank you so much for the response.

Melden Sie sich an, um zu kommentieren.

Antworten (0)

Produkte

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by