how can we increase a value until we get answer

Question

arian hoseini 2021-12-28

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1618755-how-can-we-increase-a-value-until-we-get-answer

评论： DGM 2021-12-29

clc
clear all
% -------------------------------------data input-----------------------------------
alpha = [510; 310; 78; 120; 350; 490; 53; 150; 170; 149];    
beta = [7.20; 7.85; 7.98; 7.10; 7.30; 6.90; 8.01; 7.30; 7.42; 7.16];
gamma = [0.00142; 0.00194; 0.00482; 0.00321; 0.00210; 0.00178; 0.00212; 0.00350; 0.00267; 0.00390];
delta = [0.0000001; 0.0000002; 0.0000003; 0.0000001; 0.0000002; 0.0000003; 0.0000001; 0.0000002; 0.0000003; 0.0000001];
% Pmin =[150; 100; 50; 50; 100; 100; 100; 100; 100; 100];
% Pmax =[600; 400; 200; 200; 350; 500; 300; 300; 300; 300];
n = 10;
Pr = 2500;
L = 8.6;
%--------------------------------------Program start here--------------------------------
while eps >= 0.001
    L = 
    syms P
    % P = sym('P');
    % L = sym('L');
    F = sym('F');
    Lambda = sym('Lambda');
    % n = length(alpha);
    for i = 1:n
        F(i) = alpha(i) + beta(i).*P + gamma(i).*P.^2 + delta(i).*P.^3;
        disp(F(i))
        Lambda = diff(F(i),P) - L == 0;
        disp(Lambda);
        S = solve(Lambda, P);
        disp(S)
        S1 = max(S)
        % disp(S1)
        X(i) = S1
        % W = vpa(X(i),6)
        double(X)
    end
    Pt = sum(X)
    disp(Pt)
    double(Pt)
    eps = 2500 - double(Pt)
    double(eps)
end

i want to increase L until i get eps >=0.01

for example i wanna increase L 0.01.....l=8.01,8.02,8.03

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Image Analyst 2021-12-28

eps is a built-in function that gives the smallest number that can be represented. It's value depends on the value of the variable. Like eps around 2 is different than eps around 10^30. That is why eps() can take a input number.

You should not use eps as the name of your own custom variable because it is a built-in function.

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Answer 1

DGM 2021-12-28

1
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1618755-how-can-we-increase-a-value-until-we-get-answer#answer_863770

编辑：DGM 2021-12-28

在 MATLAB Online 中打开

This can be simplified, but let's start with the given approach.

format long
alpha = [510; 310; 78; 120; 350; 490; 53; 150; 170; 149];    
beta = [7.20; 7.85; 7.98; 7.10; 7.30; 6.90; 8.01; 7.30; 7.42; 7.16];
gamma = [0.00142; 0.00194; 0.00482; 0.00321; 0.00210; 0.00178; 0.00212; 0.00350; 0.00267; 0.00390];
delta = [0.0000001; 0.0000002; 0.0000003; 0.0000001; 0.0000002; 0.0000003; 0.0000001; 0.0000002; 0.0000003; 0.0000001];
n = 10;
Pr = 2500;
L = 8.6;
dL = 0.01;
maxep = 0.001;
syms P
F = sym('F',[1 n]);
X = sym('X',[1 n]);
ep = 1; % eps() is a built-in function; avoid overloading it
while ep >= maxep
    for k = 1:n
        F(k) = alpha(k) + beta(k).*P + gamma(k).*P.^2 + delta(k).*P.^3;
        Lambda = diff(F(k),P) - L == 0;
        S = solve(Lambda, P);
        X(k) = max(S);
    end
    Pt = sum(X);
    ep = 2500 - double(Pt)
    L = L+dL;
end
ep = 
     1.025620698183243e+02
ep = 
  83.166759617834032
ep = 
  63.781164518970400
ep = 
  44.405265975430211
ep = 
  25.039045507277024
ep = 
   5.682484700585064
ep = 
 -13.664434792915927

I doubt that negative eps is intended, so if not, you'll have to adjust the step size accordingly. Also, it should suffice to solve for P first instead of doing it every time. The following is an example of both.

% note that the solution doesn't depend on alpha
beta = [7.20; 7.85; 7.98; 7.10; 7.30; 6.90; 8.01; 7.30; 7.42; 7.16];
gamma = [0.00142; 0.00194; 0.00482; 0.00321; 0.00210; 0.00178; 0.00212; 0.00350; 0.00267; 0.00390];
delta = [0.0000001; 0.0000002; 0.0000003; 0.0000001; 0.0000002; 0.0000003; 0.0000001; 0.0000002; 0.0000003; 0.0000001];
n = 10;
Pr = 2500;
L = 8.6;
dL = 0.01;
maxep = 0.001;
ep = 1; % eps() is a built-in function; avoid overloading it
while ep >= maxep
    X1 = (sqrt(gamma.^2 + 3*L*delta - 3*beta.*delta) - gamma)./(3*delta);
    X2 = -(sqrt(gamma.^2 + 3*L*delta - 3*beta.*delta) + gamma)./(3*delta);
    ep = 2500 - double(sum(max(X1,X2)));
    
    if ep < 0
        L = L-dL;
        dL = dL/2;
        L = L+dL;
        ep = 1; % reset to dummy value
    elseif ep > maxep
        L = L+dL;
    end
    
    disp([dL L ep])
end
   1.0e+02 *

   0.000100000000000   0.086100000000000   1.025620698183266

   0.010000000000000   8.619999999999999  83.166759617837670

   0.010000000000000   8.629999999999999  63.781164518971764

   0.010000000000000   8.639999999999999  44.405265975432485

   0.010000000000000   8.649999999999999  25.039045507278843

   0.010000000000000   8.659999999999998   5.682484700587793

   0.005000000000000   8.654999999999999   1.000000000000000

   0.002500000000000   8.652499999999998   1.000000000000000

   0.002500000000000   8.654999999999998   0.844851669389300

   0.001250000000000   8.653749999999999   1.000000000000000

   0.000625000000000   8.653124999999998   1.000000000000000

   0.000312500000000   8.652812499999998   1.000000000000000

   0.000312500000000   8.653124999999998   0.240189895270305

   0.000156250000000   8.652968749999998   1.000000000000000

   0.000078125000000   8.652890624999998   1.000000000000000

   0.000078125000000   8.652968749999998   0.089025922124165

   0.000039062500000   8.652929687499999   1.000000000000000

   0.000039062500000   8.652968749999999   0.013444156101286

   0.000019531250000   8.652949218749999   1.000000000000000

   0.000009765625000   8.652939453124999   1.000000000000000

   0.000004882812500   8.652934570312498   1.000000000000000

   0.000004882812500   8.652939453124997   0.003996445692792

   0.000002441406250   8.652937011718748   1.000000000000000

   0.000001220703125   8.652935791015622   1.000000000000000

   0.000001220703125   8.652937011718747   0.001634518447190

   0.000000610351563   8.652936401367185   1.000000000000000

   0.000000610351563   8.652936401367185   0.000453554875094

This can be simplified further using the available numerical solvers.

% note that the solution doesn't depend on alpha
beta = [7.20; 7.85; 7.98; 7.10; 7.30; 6.90; 8.01; 7.30; 7.42; 7.16];
gamma = [0.00142; 0.00194; 0.00482; 0.00321; 0.00210; 0.00178; 0.00212; 0.00350; 0.00267; 0.00390];
delta = [0.0000001; 0.0000002; 0.0000003; 0.0000001; 0.0000002; 0.0000003; 0.0000001; 0.0000002; 0.0000003; 0.0000001];
n = 10;
Pr = 2500;
L = 8.6;
eptarget = 0.0005; % some specific value instead of a maximum
Ltarget = fzero(@(L) findep(beta,gamma,delta,L)-eptarget,L) % find L
Ltarget = 
   8.652936377363181
findep(beta,gamma,delta,Ltarget) % test L
ans = 
     5.000000014661055e-04
function ep = findep(beta,gamma,delta,L)
    X1 = (sqrt(gamma.^2 + 3*L*delta - 3*beta.*delta) - gamma)./(3*delta);
    X2 = -(sqrt(gamma.^2 + 3*L*delta - 3*beta.*delta) + gamma)./(3*delta);
    ep = 2500 - double(sum(max(X1,X2)));
end

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arian hoseini 2021-12-28

do u know how much you helped me.....thank you very much.....and no negative eps is not intended.so can you help me on this one too i added Pmin and Pmax but its not working.

clc

clear all

% -------------------------------------data input-----------------------------------

alpha = [510; 310; 78; 120; 350; 490; 53; 150; 170; 149];

beta = [7.20; 7.85; 7.98; 7.10; 7.30; 6.90; 8.01; 7.30; 7.42; 7.16];

gamma = [0.00142; 0.00194; 0.00482; 0.00321; 0.00210; 0.00178; 0.00212; 0.00350; 0.00267; 0.00390];

delta = [0.0000001; 0.0000002; 0.0000003; 0.0000001; 0.0000002; 0.0000003; 0.0000001; 0.0000002; 0.0000003; 0.0000001];

Pmin =[150; 100; 50; 50; 100; 100; 100; 100; 100; 100];

Pmax =[600; 400; 200; 200; 350; 500; 300; 300; 300; 300];

n = 10;

Pr = 2500;

L = 8.6;

%--------------------------------------Program start here--------------------------------

syms P

% P = sym('P');

% L = sym('L');

F = sym('F');

Lambda = sym('Lambda');

% n = length(alpha);

% while abs(eps)>=0.01

% L=L+0.001

while abs(eps) >=0.001

for i = 1:n

F(i) = alpha(i) + beta(i).*P + gamma(i).*P.^2 + delta(i).*P.^3;

% disp(F(i))

Lambda = diff(F(i),P) - L == 0;

% disp(Lambda);

S = solve(Lambda, P);

% disp(S)

S1 = max(S)

% disp(S1)

X(i) = S1

if P(i)< Pmin(i)

P(i)=Pmin(i);

elseif P(i)> Pmax(i)

P(i)=Pmax(i);

end

% double(X)

end

Pt = sum(X)

% disp(Pt)

% double(Pt)

eps = Pr - double(Pt)

% double(eps)

% end

if eps>0

L=L+0.01;

elseif eps<0

L=L-0.01;

end

print last L that is used

arian hoseini 2021-12-29

thank u everyone

DGM 2021-12-29

If you feel that the question has been satisfactorily answered, you can click Accept. That way the question is moved to the "answered" queue where it may be more attractive to other people searching for similar information in the future.

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how can we increase a value until we get answer

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how can we increase a value until we get answer

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