how can we increase a value until we get answer

2021 12 28
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更新时间：2021 12 29
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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
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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 采纳的回答

DGM
DGM 2021-12-28
编辑：DGM 2021-12-28
Ran in:

1 个投票

Ran in:
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
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
end
print last L that is used
DGM
DGM 2021-12-28
Ran in:
Ran in:
Working off of the final example above:
% 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];
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;
eptarget = 0.0005;
Ltarget = fzero(@(L) findep(beta,gamma,delta,L,Pmin,Pmax,Pr)-eptarget,L) % find L
Ltarget = 8.6749
findep(beta,gamma,delta,Ltarget,Pmin,Pmax,Pr) % test L
ans = 5.0000e-04
function ep = findep(beta,gamma,delta,L,Pmin,Pmax,Pr)
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);
P = min(max(max(X1,X2),Pmin),Pmax);
ep = Pr - double(sum(P));
end
DGM
DGM 2021-12-29
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