Nested for loops to optimize price

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Yousif Alhammadi
Yousif Alhammadi 2019-12-7
评论： Walter Roberson 2019-12-7
Testing various parameters to find the one that gives the most profit, through price equation:
Price = (0.12*P)-(Price of Reactor +0.04*M+1400000*Q). How to test all possible combinations of P, price of reactor, M, and Q to find the largest price/i.e. most profitable.
function Project = optimize(T)
T=[298.15:873.15];
length(T);
K=zeros(1,length(T));
myanswer=nan(2,length(T));
M=linspace(1,100,length(T));
Q=zeros(1,length(T));
Price=zeros(1,length(T));
for i=1:length(T)
K(i)=exp(-20000/(8.314*T(i)));
end
for j=1:length(T);
syms P
eqn = K(j)*M(j)^2-(2*M(j)*P*K(j))+(K(j)*P^2)-P;
myanswer(:,j)=solve(eqn);
myanswer(:,j)=double(myanswer(:,j));
end
P1=myanswer(1,:);
P2=myanswer(2,:);
r=linspace(2,100,length(T));
A=zeros(1,length(r));
for k=1:length(r)
A(k)=10*3.14*(r(k))^2;
Q(k)=25*3.14*r(k)^2*(T(k)-298.15);
end
Priceofreactor=A*150000;
Q=Q/0.12; %% accounts for 12% efficiency
P1=double(P1);
for d=1:length(T)
Price(d) = 0.12*P1(d)-(Priceofreactor(d)+0.04*M(d)+140*Q(d));
end
max(Price)
end
  1 个评论
Walter Roberson
Walter Roberson 2019-12-7
for j=1:length(T);
syms P
eqn = K(j)*M(j)^2-(2*M(j)*P*K(j))+(K(j)*P^2)-P;
myanswer(:,j)=solve(eqn);
myanswer(:,j)=double(myanswer(:,j));
end
That can be done much more efficiently.
syms k m P
eqn = k*m^2-(2*m*P*k)+(k*P^2)-P;
sol = solve(eqn,P);
myanswer = double( subs(sol, {k,m}, {K,M}) );

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