x i + y i cos ⁡ ( p i k − a i )...

pouzdrotf

pouzdrotf

Answered

2022-06-30

x i + y i cos ( p i k a i ) z i q i k > x j + y j cos ( p j k a j ) z j q j k
x i + y i cos ( p i n a i ) z i q i n < x m + y m cos ( p m n a m ) z m q m n
All p and q are parameters, I need to solve for x i , y i , z i , a i for all i.

Answer & Explanation

Ronald Hickman

Ronald Hickman

Expert

2022-07-01Added 18 answers

For purposes of this exposition, let's presume you have a system of k inequalities in n variables, which has been put into the form (trivial to do)
f i ( x ) 0 , i = 1.. k
where x is the n by 1 vector of variables.
Solve (using an optimizer which can handle the nonlinear inequalities) the following nonlinearly-constrained optimization problem: (but see the generalization below).
minimize Σ i = 1 k M i with respect to x,M subject to
f i ( x ) + M i 0 , i = 1.. k
M i 0 , i = 1.. kIf the optimal objective of this problem is zero, then you have found a feasible solution to the original system of nonlinear inequalities. If it is positive, then the original system of nonlinear inequalities is infeasible (does not have an exact solution), but you will have found a solution which is close as possible to feasible, as measured per the objective function of the optimization problem.
You can use a different objective function in order to prioritize different levels of errors and relative importance of compliance among the various constraints. For instance, more generally consider an objective function such as
Σ i = 1 k w i M i q
where w i 0, which allows a nonlinear "non-compliance" penalty and allows different weights for the different inequalities. The simple version presented above corresponds to all w i = 1 and q = 1. If you want to get really fancy, q need not be the same for all k inequalities.
bandikizaui

bandikizaui

Expert

2022-07-02Added 7 answers

It is also possible to linearize the inequalities using substitutions b i = y i cos a i :
x i x j + b i cos p i k b j cos p j k + c i sin p i k c j sin p j k z i q i k + z j q j k 0

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