フィルターのクリア

Optimization to minimize output

1 回表示 (過去 30 日間)
Anand Ra
Anand Ra 2021 年 10 月 6 日
コメント済み: Anand Ra 2021 年 10 月 6 日
Hello,
I am trying to optimize the lengths to minimize the power ( details in the code)
I am unable to proceed with the optimization part. ( Iam not getting the expected result of optimized power)
Any help would be greatly appreciated.
% Objective:Optimize the lengths to minimze the power using a new variable.
%Constants
k= 1;
w=1;
%Variable lengths
l1 = 3*(5)^0.5;
l2 = 5*(2)^0.5
l2 = 7.0711
l3 = sqrt(5);
v1 = sqrt(5);
v2 = 2*sqrt(2);
v3 = 2*sqrt(5);
% Expressing the l and v in terms of a b and c. Open to expressing differently
% if it helps the objective below.
a = l1*v1;
b = l2*v2;
c = l3*v3;
%Actual Power
P = (l1*v1 + l2*v2 + l3*v3)*k*w; % Power
% Objective
%To search and find the values for a, b and c to minimize power
%Considering x to represent a,b and c.
fun = @(x)(x(:,1) + x(:,2) + x(:,3)).*k.*w;
[X1, X2, X3] = ndgrid(0:.1:2); % Should I give this condition in a nested loop?
X = [X1(:), X2(:), X3(:)];
P = fun(X);
[bestP, idx] = min(P(:))
bestP = 0
idx = 1
best_X = X(idx,:)
best_X = 1×3
0 0 0
  1 件のコメント
Matt J
Matt J 2021 年 10 月 6 日
Your post now contains the results of running the code (I took the liberty...). The answer looks correct to me.

サインインしてコメントする。

回答 (1 件)

Walter Roberson
Walter Roberson 2021 年 10 月 6 日
% Objective:Optimize the lengths to minimze the power using a new variable.
%Constants
k= 1;
w=1;
%Variable lengths
l1 = 3*(5)^0.5;
l2 = 5*(2)^0.5
l2 = 7.0711
l3 = sqrt(5);
v1 = sqrt(5);
v2 = 2*sqrt(2);
v3 = 2*sqrt(5);
% Expressing the l and v in terms of a b and c. Open to expressing differently
% if it helps the objective below.
a = l1*v1;
b = l2*v2;
c = l3*v3;
%Actual Power
P = (l1*v1 + l2*v2 + l3*v3)*k*w; % Power
% Objective
%To search and find the values for a, b and c to minimize power
%Considering x to represent a,b and c.
fun = @(x)(x(:,1) + x(:,2) + x(:,3)).*k.*w;
[X1, X2, X3] = ndgrid(0:.1:2); % Should I give this condition in a nested loop?
X = [X1(:), X2(:), X3(:)];
P = fun(X);
[bestP, idx] = min(P(:))
bestP = 0
idx = 1
best_X = X(idx,:)
best_X = 1×3
0 0 0
This is obviously the correct output for the formula.
k is positive. w is positive. k*w is positive. k*w is being multiplied by (x1 + x2 + x3)
To get the smallest result of a multiplication of a positive number and another number, you want the second number to be as small as possible.
Your x1, x2, x3 are all bounded by 0 below. Increasing x1 or x2 or x3 would increase the sum x1+x2+x3 . So the smallest x1+x2+x3 is at the lower bound, where x1 = x2 = x3 = 0.
So your minima is clearly at x1 = x2 = x3 = 0 with power 0 .
  3 件のコメント
Walter Roberson
Walter Roberson 2021 年 10 月 6 日
fun = @(x)(x(:,1) + x(:,2) + x(:,3) + x(:4) + x(5).*k.*w;
missing close bracket
has same issue as before with minima at 0
Anand Ra
Anand Ra 2021 年 10 月 6 日
I think I missed the relationship constaints. I have included that now. I am not sure how to construct the minimization part, given my relationshipo constrains
% Objective:Optimize the lengths to minimze the power using a new variable.
a = 0.1:20;
b = 0.1:20;
d = a + 0.1:20;
c = b + 0.1:20;
e = d + 0.1:20
%Constants
k= 2;
w=1;
v=1.5
%Variables
AB = sqrt(a.^2 + b.^2);
BC = sqrt( c.^2 + ((e-d)/2).^2 );
CS = sqrt( c.^2 + ((e-d)/2).^2 );
VAB = sqrt(((((a.*v).^2/(((b.^2).*4))) + (v^2)/2 )));
% VBS = sqrt(((a*v)^2/((4*b*b)) + (v^2)/2 ));
VCS = ((2*c)./(e-d)).*sqrt(AB.^2);
VBC= CS.^2 + BC.^2;
%Actual Power
%P = (AB*VAB + BC*VBC + CS*VCS)*k*w; % Power
%
% Objective
%To search and find the values for a, b, c d and eto minimize power
% Not sure how to write the function for above
% fun = @(x)(x(:,1) + x(:,2) + x(:,3) + x(:,4) + x(:,5)).*k.*w;
%
% [X1, X2, X3, X4, X5] = ndgrid(0:.1:2); % Should I give this condition in a nested loop?
%
% X = [X1(:), X2(:), X3(:), X4(:), X5(:)];
%
% P = fun(X);
% [bestP, idx] = min(P(:))
% best_X = X(idx,:)

サインインしてコメントする。

カテゴリ

Help Center および File ExchangeSolver Outputs and Iterative Display についてさらに検索

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by