Effacer les filtres
Effacer les filtres

Loop "for" optimization

2 vues (au cours des 30 derniers jours)
Aleksei
Aleksei le 10 Déc 2023
Commenté : Dyuman Joshi le 16 Déc 2023
Hello.
I have a code and need to optimize loop "for". I need make one "for" instead of two. How to make less then 99*99 itaration?
f =[6; 6; 30; 2; 13; 8; 21;
14; 1; 24; 28; 25; 24; 7;
18; 9; 26; 30; 25; 8; 13;
24; 25; 11; 18; 23; 12; 29;
26; 19; 29; 16; 4; 27; 24;
19; 14; 7; 15; 24; 6; 21 ];
g =[43; 28; 21; 15; 20; 47; 48;
47; 32; 15; 39; 31; 20; 19;
42; 23; 33; 14; 35; 26; 32;
10; 43; 24; 37; 39; 24; 31;
17; 31; 9; 28; 46; 25; 12;
17; 43; 19; 20; 49; 16; 22];
AA = zeros(1, 99);
a_k=1;
BB = zeros(1, 99);
b_k=1;
for v = 0.01:0.01:0.99
for u = 0.01:0.01:0.99
if (v+u == 1)
D=(v*f)+(u*g);
B = round(D);
E = reshape(B,7,6);
[X,K] = linprog(E,[],[],Aeq,beq,lb);
A = round(X);
H = reshape(A,7,6);
Optim_plan_A = H';
N =A'*f;
L =A'*g;
AA(a_k)=N;
BB(b_k)=L;
a_k=a_k+1;
b_k=b_k+1;
end
end
end
figure
plot (AA,BB, 'g')

Réponses (1)

Atsushi Ueno
Atsushi Ueno le 10 Déc 2023
After 99*99 iterations, v+u==1 is satisfied only 99 times.
The same thing can be done without repeating the double nested loop by calclating u as u=1-v.
f = [6; 6; 30; 2; 13; 8; 21; 14; 1; 24; 28; 25; 24; 7; 18; 9; 26; 30; 25; 8; 13; 24; 25; 11; 18; 23; 12; 29; 26; 19; 29; 16; 4; 27; 24; 19; 14; 7; 15; 24; 6; 21 ];
g = [43; 28; 21; 15; 20; 47; 48; 47; 32; 15; 39; 31; 20; 19; 42; 23; 33; 14; 35; 26; 32; 10; 43; 24; 37; 39; 24; 31; 17; 31; 9; 28; 46; 25; 12; 17; 43; 19; 20; 49; 16; 22];
AA = zeros(1,99); a_k = 1;
BB = zeros(1,99); b_k = 1;
for v = 0.01:0.01:0.99
u = 1 - v;
D = (v*f)+(u*g);
B = round(D);
E = reshape(B,7,6);
[X,K] = linprog(E,[],[],Aeq,beq,lb);
A = round(X);
H = reshape(A,7,6);
Optim_plan_A = H';
N = A'*f;
L = A'*g;
AA(a_k) = N;
BB(b_k) = L;
a_k = a_k + 1;
b_k = b_k + 1;
end
figure
plot (AA,BB,'g')
  2 commentaires
Aleksei
Aleksei le 10 Déc 2023
Thank you very much!
Dyuman Joshi
Dyuman Joshi le 16 Déc 2023
Hello @Aleksei, if this answer solved your problem, please consider accepting the answer.

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