Items tagged with optimization optimization Tagged Items Feed

Hi

I am trying to optimize a 39, 1 MATLAB matrix, but cannot seem to get a result beyond a 6, 1 matrix. I am getting "Warning, cannot resolve types, reassigning t##'s type" where t## varies from each time I run it, and can show multiple of these warnings. It also says "Warning, cannot translate list".

 

I found a pretty similar problem posted here earlier, where the user "Carl Love" suggested to replace a command from the original code with

Matlab(
     subsop([-1,1]= J, eval([codegen:-optimize](tmp, tryhard), pow= `^`)),
     output = string, defaulttype = numeric
);

 

I was wondering what exactly this command does? Can I apply it to my code to solve my problem? It yielded a result that looks (on the surface) as an optimized code, but I don't feel completely comfortable using it without being certain.

What I have done is simply to replace Matlab(tmp, optimize) with the suggested code above. My code is attached. Thanks in advance for any help.

OptiMatrix.mw

Hello maple users,

I have 2 functions and each functions has 8 variables. I run a matlab code and get outputs for different values of these variables. I assumed 3 of them as constant because the combinations are too many. Anyway, I plot the results and I can see that one function is much better than the other. But I need to compare these functions mathematically. I need to show some proofs. Has anyone any idea what should I do? I wrote the functions on maple and take derivative with respect to one variable and try to see the reaction of the functions to that variable. i am confused.

 

Thanks

 

I've got

f(x,y)= a.exp(1+xy) +( a^2 )*sin(x)+1

for which I've shown that there exists an implicit function x=g(y). ( df/dx <>0)

and df/dx = a*exp(1+xy) +( a^2 )*cosx now in the neighborhood of P=(0,0) for the implicit function to exist I'd need a*exp(1+xy)*y <>0 but at P, wouldn't this be 0?

Given, g(y)=x, how do I find the max,min,saddle points?

I want to know with what x,y, z,  function f is minimum, whereas function g is constant.

 

regards

Hi

I want to know with what x/y, z,  function f is minimum, whereas function g is constant.

regards

 

So i got a procedure test, she is kind of numeric, i whant to optimaze test([.5, .5, .5], 1, 3, 100, 100, true, [x, 0, 0, 0, 0])=0 by x. But optinization substitutes x like a symbol, i tryed all methods but they all do the same.

f := proc (x) options operator, arrow; abs(test([.5, .5, .5], 1, 3, 100, 100, true, [x, 0, 0, 0, 0])-.4) end proc; Minimize(f(x));
%;
Error, (in test) cannot determine if this expression is true or false: 0 < -43.0+100*x


Can i some how use optinization on such procedure?

file link  - >    primset.mw

 

 

 

 

Hello

I wqant to minimize a function that has som parameters (here number of parameters are two). how can i do that?

I have attache a picture from my target function. Could you please help me?

Tahnk you.

 

 

Hello,

      I would like to solve a system of 9 nonlinear equations, with the constraints on all 9 variables to be that they are nonnegative. How can I do this?

My code is below - I am trying NLPSolve and have tried solve, but am getting stuck.

with(Optimization);

restart; eq1 := 531062-S/(70*365)-(.187*(1/365))*(H+C+C1+C2)*S/N = 0;eq2 := (4/365*(T+C))*S/N-(.187*(1/365))*(H+C+C1+C2)*T/N-(1/(70*365)+1/(5*365))*T = 0; eq3 := (.187*(1/365))*(H+C+C1+C2)*S/N-(4/365)(T+C)*H/N-(1/(70*365)+1/(4*365))*H = 0; eq4 := (.187*(1/365))*(H+C+C1+C2)*T/N+(4/365*(T+C))*H/N-(1/(70*365)+3/(8*365)+.2*(1/365)+.1)*C = 0; eq5 := .1*C-(1/(70*365)+1/(4*365)+1/60+.5)*C1 = 0; eq6 := (1/60)*C1-(1/(70*365)+1/(4*365)+1/210+.5)*C2 = 0; eq7 := .5*C1-(1/(70*365)+1/60+0.1e-2)*CT1 = 0; eq8 := .5*C2-(1/(70*365)+1/210+(1/9)*(0.1e-2*7))*CT2+(1/60)*CT1 = 0; eq9 := N-S-T-H-C-C1-C2-CT1-CT2 = 0; soln := NLPSolve({eq1, eq2, eq3, eq4, eq5, eq6, eq7, eq8, eq9}, {C, C1, C2, CT1, CT2, H, N, S, T}, assume = nonnegative);

Hi all

The aim of following program is minimization but it is unable to produce it. where is the mistake?

Taylor2.mws

thanks a lot.

Mahmood   Dadkhah

Ph.D Candidate

Applied Mathematics Department

According to this site,"It is known that every even number can be written as a sum of at most six primes". 

http://www.theage.com.au/national/education/christians-goldbachs-magic-sum-20140903-3es2t.html

i wanted to test this using maple.

restart:
> PF := proc (a::integer)

> local cst,obj,res;
> cst := add(x[i], i = 1 .. numtheory:-pi(prevprime(a))) <= 6;
> obj := add(x[i]*ithprime(i), i = 1 .. numtheory:-pi(prevprime(a)))-a;
> res := Optimization:-LPSolve(obj, {cst ,obj>=0}, assume={nonnegative,integer}); end proc:
> PF(30);
[0, [x[1] = 0, x[2] = 0, x[3] = 6, x[4] = 0, x[5] = 0, x[6] = 0,x[7] = 0, x[8] = 0, x[9] = 0, x[10] = 0]]

the third prime is 5 and 6 of them make 30. as an aside, it would be nice to know how to get maple to output "30 = 6x5".

this is obviously pretty limited, because 30 can be written as the sum of two primes (7+23 and 11+19) [GOLDBACH], but using DS's GlobalSearch for all solutions takes a long time to compute. also I have to nominate the highest prime.

any suggestions?

Dear all

is it possible to solve bilevel optimization problems in maple?

            min F(x,y)

     s.t.    min G(x,y)

        s.t.   k(x,y)<=0

As I am trying to solve this integration:

restart; with(linalg); with(stats); with(plots); with(Statistics); with(LinearAlgebra); with(Optimization);
lambda0 := proc (t) options operator, arrow; gamma0+gamma1*t+gamma2*t^2 end proc;
lambda := lambda0(t)*exp(beta*s);
t1 := 145; t3 := 250; t2 := (t1+t3)*(1/2);
s := 1/(273.16+50); s1 := 1/(273.16+t1); s3 := 1/(273.16+t3); s2 := 1/(273.16+t2); gamma0 := 0.1e-3; gamma1 := .5; gamma2 := 0; beta := -3800;
c := 300; n := 200;
Theta := solve(1-exp(-(gamma0*tau1+(1/2)*gamma1*tau1^2+(1/3)*gamma2*tau1^3)*exp(beta*s1)) = 1-exp(-(gamma0*a+(1/2)*gamma1*a^2+(1/3)*gamma2*a^3)*exp(beta*s2)), a);

a := Theta[1];

Delta := solve(1-exp(-(gamma0*(a+tau2-tau1)+(1/2)*gamma1*(a+tau2-tau1)^2+(1/3)*gamma2*(a+tau2-tau1)^3)*exp(beta*s2)) = 1-exp(-(gamma0*b+(1/2)*gamma1*b^2+(1/3)*gamma2*b^3)*exp(beta*s3)), b);

b := Delta[1];

A1 := `assuming`([unapply(int(exp(beta*s1)*exp(-(gamma0*t+(1/2)*gamma1*t^2+(1/3)*gamma2*t^3)*exp(beta*s1))/(gamma0+gamma1*t+gamma2*t^`2`), t = N .. M), N, M)], [N > 0, M > 0]);
A2 := unapply(int(exp(beta*s2)*exp(-(gamma0*(a+t-tau1)+(1/2)*gamma1*(a+t-tau1)^2+(1/3)*gamma2*(a+t-tau1)^3)*exp(beta*s2))/(gamma0+gamma1*(a+t-tau1)+gamma2*(a+t-tau1)^2), t = N .. M), N, M);
A3 := unapply(int(exp(beta*s3)*exp(-(gamma0*(b+t-tau2)+(1/2)*gamma1*(b+t-tau2)^2+(1/3)*gamma2*(b+t-tau2)^3)*exp(beta*s3))/(gamma0+gamma1*(b+t-tau2)+gamma2*(b+t-tau2)^2), t = N .. M), N, M);
B1 := `assuming`([unapply(int(t^2*exp(beta*s1)*exp(-(gamma0*t+(1/2)*gamma1*t^2+(1/3)*gamma2*t^3)*exp(beta*s1))/(gamma2*t^2+gamma1*t+gamma0), t = N .. M), N, M)], [N > 0, M > 0]);
B2 := unapply(int((a+t-tau1)^2*exp(beta*s2)*exp(-(gamma0*(a+t-tau1)+(1/2)*gamma1*(a+t-tau1)^2+(1/3)*gamma2*(a+t-tau1)^3)*exp(beta*s2))/(gamma0+gamma1*(a+t-tau1)+gamma2*(a+t-tau1)^2), t = N .. M), N, M);
B3 := unapply(int((b+t-tau2)^2*exp(beta*s3)*exp(-(gamma0*(b+t-tau2)+(1/2)*gamma1*(b+t-tau2)^2+(1/3)*gamma2*(b+t-tau2)^3)*exp(beta*s3))/(gamma0+gamma1*(b+t-tau2)+gamma2*(b+t-tau2)^2), t = N .. M), N, M);

F0 := A1(0, tau1)+A2(tau1, tau2)+A3(tau2, c);
F1 := B1(0, tau1)+B2(tau1, tau2)+B3(tau2, c);

NLPSolve(1/(n^3*(F0*F1-F1)), tau1 = 115 .. 201, tau2 = 237 .. 273);

I need to have tau1 tau2 as varibles to get there optimal values ..

But this error keeps coming :


Error, (in Optimization:-NLPSolve) integration range or variable must be specified in the second argument, got HFloat(1.0) = HFloat(158.0) .. HFloat(255.0)

Please Help ..

As am trying to solve this integration:


A(c,n,m):=evalf(int(1/(y)*exp(-c*y^(2)),y=n..m))

B(c,n,m):=evalf(int(exp(-c*y^(2)),y=n..m))

C(c,n,m):=evalf(int(y*exp(-c*y^(2)),y=n..m))

d(c,n,m):=evalf(int(y^(2)*exp(-c*y^(2)),y=n..m))

E(c,n,m):=evalf(int(y^(3)*exp(-c*y^(2)),y=n..m))

F0 := exp(beta*s1)*exp(gamma0^2*exp(beta*s1)/(2*gamma1))*A((1/2)*gamma1*exp(beta*s1), gamma0/gamma1,gamma0/gamma1+tau1)/gamma1+exp(beta*s2)*exp(gamma0^2*exp(beta*s2)/(2*gamma1))*A((1/2)*gamma1*exp(beta*s2), gamma0/gamma1+a, gamma0/gamma1+tau2-tau1+a)/gamma1+exp(beta*s3)*exp(gamma0^2*exp(beta*s3)/(2*gamma1))*A((1/2)*gamma1*exp(beta*s3), gamma0/gamma1+b, gamma0/gamma1+c-tau2+b)/gamma1

F1 := exp(beta*s1)*exp(gamma0^2*exp(beta*s1)/(2*gamma1))*(gamma0^2*A((1/2)*gamma1*exp(beta*s1), gamma0/gamma1, gamma0/gamma1+tau1)/gamma1^2-2*gamma0*B((1/2)*gamma1*exp(beta*s1), gamma0/gamma1, gamma0/gamma1+tau1)/gamma1+C((1/2)*gamma1*exp(beta*s1), gamma0/gamma1, gamma0/gamma1+tau1))/gamma1+exp(beta*s2)*exp(gamma0^2*exp(beta*s2)/(2*gamma1))*(gamma0^2*A((1/2)*gamma1*exp(beta*s2), gamma0/gamma1+a, gamma0/gamma1+tau2-tau1+a)/gamma1^2-2*gamma0*B((1/2)*gamma1*exp(beta*s2), gamma0/gamma1+a, gamma0/gamma1+tau2-tau1+a)/gamma1+C((1/2)*gamma1*exp(beta*s2), gamma0/gamma1+a, gamma0/gamma1+tau2-tau1+a))/gamma1+exp(beta*s3)*exp(gamma0^2*exp(beta*s3)/(2*gamma1))*(gamma0^2*A((1/2)*gamma1*exp(beta*s3), gamma0/gamma1+b, gamma0/gamma1+c-tau2+b)/gamma1^2-2*gamma0*B((1/2)*gamma1*exp(beta*s3), gamma0/gamma1+b, gamma0/gamma1+c-tau2+b)/gamma1+C((1/2)*gamma1*exp(beta*s3), gamma0/gamma1+b, gamma0/gamma1+c-tau2+b))/gamma1

F01 := exp(beta*s1)*exp(gamma0^2*exp(beta*s1)/(2*gamma1))*(B((1/2)*gamma1*exp(beta*s1), gamma0/gamma1, gamma0/gamma1+tau1)-gamma0*A((1/2)*gamma1*exp(beta*s1), gamma0/gamma1, gamma0/gamma1+tau1)/gamma1)/gamma1+exp(beta*s2)*exp(gamma0^2*exp(beta*s2)/(2*gamma1))*(B((1/2)*gamma1*exp(beta*s2), gamma0/gamma1+a, gamma0/gamma1+tau2-tau1+a)-gamma0*A((1/2)*gamma1*exp(beta*s2), gamma0/gamma1+a, gamma0/gamma1+tau2-tau1+a)/gamma1)/gamma1+exp(beta*s3)*exp(gamma0^2*exp(beta*s3)/(2*gamma1))*(B((1/2)*gamma1*exp(beta*s3), gamma0/gamma1+b, gamma0/gamma1+c-tau2+b)-gamma0*A((1/2)*gamma1*exp(beta*s3), gamma0/gamma1+b, gamma0/gamma1+c-tau2+b)/gamma1)/gamma1

`F&beta;` := int((s1^2*(gamma0*t+(1/2)*gamma1*t^2)*exp(beta*s1)*(gamma1*t+gamma0)*exp(beta*s1))*exp(-(gamma0*t+(1/2)*gamma1*t^2)*exp(beta*s1)), t = 0 .. tau1)+int((s2^2*(gamma0*(a+t-tau1)+(1/2)*gamma1*(a+t-tau1)^2)*exp(beta*s2)*(gamma0+gamma1*(a+t-tau1))*exp(beta*s2))*exp(-(gamma0*(a+t-tau1)+(1/2)*gamma1*(a+t-tau1)^2)*exp(beta*s2)), t = tau1 .. tau2)+int((s3^2*(gamma0*(b+t-tau2)+(1/2)*gamma1*(b+t-tau2)^2)*exp(beta*s3)*(gamma0+gamma1*(b+t-tau2))*exp(beta*s3))*exp(-(gamma0*(b+t-tau2)+(1/2)*gamma1*(b+t-tau2)^2)*exp(beta*s3)), t = tau2 .. c)+int((s3^2*(gamma0*(b+t-tau2)+(1/2)*gamma1*(b+t-tau2)^2)*exp(beta*s3)*(gamma0+gamma1*(b+t-tau2))*exp(beta*s3))*exp(-(gamma0*(b+t-tau2)+(1/2)*gamma1*(b+t-tau2)^2)*exp(beta*s3)), t = c .. infinity)

I need to have tau2 as varibles to get there optimal values ..

Minimize(1/((F0*F1-F01^2)*n^3*`F&beta;`), tau2 = 237..273})

But this error keeps coming :

Error, (in Optimization:-NLPSolve) integration range or variable must be specified in the second argument, got HFloat(1.0) = 121.0828419 .. HFloat(193.0828419)

Please Help ..

Hi all

 

I am trying to maximize a function f(x,y,z,w) in terms of x. (Only x is treated as a variable, and the others are treated as parameters).

However, all I know is that y,z,w they are parameters and they are non-negative. I have already tried with the "optmization help page" from maplesoft's website, and it looks like it will search the range of x,y,z,w, and it will return numerical values at which this function is maximized. 

 

 

However, what I want is instead a close-form solution of x=g(y,z,w) that will maximize the function.   In other words, I would like to keep the parameter in symbolic forms. 

 

Can Maple do that?

1 2 3 4 5 6 7 Last Page 1 of 12