how to optimize a sweep for solving inequalities?

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

Answer 1

[KSSV]

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

Answer 2

[Sulaymon Eshkabilov]

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]

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)

X = find(mupadmex('symobj::logical',A.s,9)) - 1;

Error in Calculo_Ki_Kd_dotM (line 218)

sol = [ki(idx) kd(idx)]