how to optimize a sweep for solving inequalities?

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Alex Muniz on 16 Jan 2023
Commented: Alex Muniz on 17 Jan 2023
Solution of inequalities. For a project I need to solve systems of inequalities. The script below works, however it is very slow, it takes several minutes to resolve. How can I make it more efficient?
% ki < 0 && -kd + ki + 1 > 0 && -15*kd + ki -55 < 0
syms ki kd
i0 = 1
i1 = -1
i2 = -1
eqn1 = ki*i0
eqn2 = (-kd + ki + 1)*i1
eqn3 = (-15*kd + ki -55)*i2
x_min = -100;
x_max = 100;
y_min = -100;
y_max = 100;
% Define o passo de busca para x e y
step = 1;
for ki_ = x_min:step:x_max
for kd_ = y_min:step:y_max
eqn1_subs_ki_kd = subs(eqn1, {ki}, {ki_});
eqn2_subs_ki_kd = subs(eqn2, {ki kd}, {ki_ kd_});
eqn3_subs_ki_kd = subs(eqn3, {ki kd}, {ki_ kd_});
if eqn1_subs_ki_kd > 0 && eqn2_subs_ki_kd > 0 && eqn3_subs_ki_kd > 0
disp("solution found!")
ki_
kd_
end
end
end

KSSV on 16 Jan 2023
You need to use syms. You can straight away substitute the values in the equation and get the lligcal indexing.
% ki < 0 && -kd + ki + 1 > 0 && -15*kd + ki -55 < 0
i0 = 1 ;
i1 = -1 ;
i2 = -1 ;
x_min = -100;
x_max = 100;
y_min = -100;
y_max = 100;
% Define o passo de busca para x e y
thestep = 1 ;
[ki_,kd_] = meshgrid(x_min:thestep:x_max,y_min:thestep:y_max) ;
eqn1 = ki_*i0 ;
eqn2 = (-kd_ + ki_ + 1)*i1 ;
eqn3 = (-15*kd_ + ki_ -55)*i2 ;
idx = eqn1 > 0 & eqn2 > 0 & eqn3 > 0 ;
sol = [ki_(idx) kd_(idx)]
sol = 4851×2
1 3 1 4 1 5 1 6 1 7 1 8 1 9 1 10 1 11 1 12

Sulaymon Eshkabilov on 16 Jan 2023
One of the possible ways to speed up your code is to get rid of display of resuts. Instead storing them, e.g.:
syms ki kd
i0 = 1;
i1 = -1;
i2 = -1;
eqn1 = ki*i0;
eqn2 = (-kd + ki + 1)*i1 ;
eqn3 = (-15*kd + ki -55)*i2;
x_min = -100;
x_max = 100;
y_min = -100;
y_max = 100;
% Define o passo de busca para x e y
step = 1;
ii=1;
for ki_ = x_min:step:x_max
for kd_ = y_min:step:y_max
eqn1_subs_ki_kd = subs(eqn1, {ki}, {ki_});
eqn2_subs_ki_kd = subs(eqn2, {ki kd}, {ki_ kd_});
eqn3_subs_ki_kd = subs(eqn3, {ki kd}, {ki_ kd_});
if eqn1_subs_ki_kd > 0 && eqn2_subs_ki_kd > 0 && eqn3_subs_ki_kd > 0
%disp("solution found!")
KI(ii) = ki_;
KD(ii)=kd_;
ii=ii+1;
end
end
end
Alex Muniz on 17 Jan 2023
In this way the solution is found very quickly.
The problem is that I am using a symbolic equation and when I use the symbolic equation it gives an error.
V_jw_even = (- kd - 7)*w^4 + (ki - 9*kd + 17)*w^2 + 9*ki
This is part of the code. V_jw_even is calculated elsewhere in the code and the variables w, ki and kd are symbolic. Could you use your solution with this symbolic equation?
i0 = -1
i1 = 1
i2 = -1
% Define o passo de busca para x e y
thestep =0.2;
[ki,kd] = meshgrid(x_min:thestep:x_max,y_min:thestep:y_max);
%
% eqn1 = ki_;
% eqn2 = (-kd_ + ki_ + 1) ;
% eqn3 = (-15*kd_ + ki_ -55);
eqn1 = subs(V_jw_even,w,0)*i0
eqn2 = subs(V_jw_even,w,w_roots)*i1
eqn3 = subs(V_jw_even,w,inf)*i2
idx = eqn1 > 0 & eqn2 > 0 & eqn3 > 0;
sol = [ki(idx) kd(idx)]
Error displayed:
Error using symengine
Unable to prove '0 < -9*ki & 0 < 10*ki - 10*kd + 10 & 0 < Inf*sign(kd + 7) - Inf*(ki - 9*kd + 17)' literally. Use
'isAlways' to test the statement mathematically.
Error in sym/subsindex (line 849)
Error in Calculo_Ki_Kd_dotM (line 218)
sol = [ki(idx) kd(idx)]

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