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how can i fix subscript indices must be either real positive integers or logicals
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I am new to matlab and i am tring to do multi objective optimization using gamultiobj , i have multi functions but when i try to run each function it says "Not enough input argument " plus when i try to run the main fun it says subscript indices must be either real positive integer or logicals" i have attached all of my functions!
댓글 수: 5
KSSV
2021년 6월 23일
You cannot staright away run the function, you need to input the varibles it wants and then call it. So the error "Not enough input argument " is legit.
KSSV
2021년 6월 23일
Specify the line number where you are getting the error: subscript indices must be either real positive integer or logicals
Walter Roberson
2021년 6월 23일
disp('Total Function Count:'+gaoutput1.funccount);
When you use that code, it takes the number in funccount and it adds it (numerically) to the character codes for every element of the vector 'Total Function Count:' --- which is, recall, the vector ['T' 'o' 't' 'a' 'l' and so on]
If you instead were to use
disp("Total Function Count:"+gaoutput1.funccount);
then the outputcome would be completely different. The + operator for string() objects (such as those created with double-quotes) is the "append" operator, and a number as the other operand will be automatically converted to character representation before the appending.
metasebia dabi
2021년 6월 25일
When I try to use
disp("Total Function Count:"+gaoutput1.funccount); It will make " red and tell me it is invalid characteristics
Walter Roberson
2021년 6월 25일
편집: Walter Roberson
2021년 6월 25일
Then you are using R2015b or earlier, and should have mentioned that. You should switch to using fprintf()
fprintf('Total Function Count: %d\n', gaoutput1.funccount);
When someone says that they are new to MATLAB, we do not expect them to be using a version that is 5 years old.
채택된 답변
Walter Roberson
2021년 6월 23일
At a location approximately half way through the third line, you have
(x(60^3*(x(1) - x(2)))/6
You might possibly have intended something like (x(6)^3*(x(1)-x(2)))/6 . But watch out for just replacing the 0 with a ) -- you would have had to add an extra ) to get the ) to balance and there is no telling where you added it.
You appear to have used some kind of code generation to generate your code from a symbolic expression. I can tell that you did not use matlabFunction() just by itself -- but it is possible that you used matlabFunction() and then hand-editted pieces together.
I recommend that you go back to the code generation and tell it to optimize. Not so much that you want the optimization (though that would not hurt), but rather that you want the expressions to be split into multiple lines for readability.
And when you create the symbolic expressions that you are going to do code generation for, you should wrap every numeric constant with a sym() call, except that you can leave powers alone. Seeing numbers such as
(19515681011313314790985107421875*x(5))/73786976294838206464 in your code makes it clear that you just let it convert floating point constants to whatever was most convenient... that divisor is 2^66 exactly.
So for example,
syms t
c = 2.99792458e8
arfreq = 0.000000488;
obj = arfreq/c*t
you would instead do something like
syms t
Q = @(v) sym(v);
c = Q(299792458);
arfreq = Q(488)/sym(10)^9; %avoid floating point
obj = arfreq/c*t
In this case the pure numeric constants would lead to generating 2304514843640387*t/4722366482869645213696 but the more careful constants would lead to 61*t/37474057250000000 . Which is still a bit "funny money", magic constants with no obvious meaning to look at them, but they would be a lot more comprehendable, without the problems like 0.000000488 being converted to 2304514843640387/4722366482869645213696
댓글 수: 45
metasebia dabi
2021년 6월 23일
THE other thing is can u help me with how can i simiplify the numeric form ? IT is a long equation!
Walter Roberson
2021년 6월 23일
Go back to the expressions and wrap all of the numeric constants with sym()
Better yet, when you have floating point numbers such as 0.5831 express them as sym() of an integer divided by 10 to a power, such as sym(5831)/10000 .
And when you use matlabFunction() to generate the code, tell it to write to a file, and tell it to turn on optimization.
Unless, that is, you are using R2019b or R2020a... matlabFunction optimize option is broken for those releases.
metasebia dabi
2021년 6월 25일
Again after fixing (x(60^3*(x(1) - x(2)))/6 it again says the subscript indices must be either real positive integer or logicals!
Walter Roberson
2021년 6월 25일
S = fileread('objval.m');
unique(regexp(S, 'x\(.*?\)', 'match'))
Now look for an output that is not x(NUMBER)
metasebia dabi
2021년 7월 4일
thank you and sir i have another issue no i have result in the pareto front but the result is a single point where could be my mistek?
Walter Roberson
2021년 7월 4일
Sometimes there is only a single global minima. Sometimes the local minima are just hard to find.
metasebia dabi
2021년 7월 4일
so should i diduct the variable ..... what is the solution to get real pareto front solution
rakeb yonas
2021년 7월 4일
greetings Walter
i am Metasebia
this is my current code
pls help me with this!
metasebia dabi
2021년 7월 4일
Greetings Sir i have attached the code by another account!
please help me with this problem!
Walter Roberson
2021년 7월 4일
You have 250 lines of nonlinear constraints, and you have a linear equality constraint (and a linear inequality constraint too.)
It looks like it is struggling a lot just to find any points that satisfy the constraints.
Walter Roberson
2021년 7월 4일
In over 60000 function calls, it is able to get objective #2 to 0, but the lowest it can get for objective #1 is about 2.84E+14 . It is not able to satisfy the constraints even once.
Checking in more detail, it is not possible to satisfy the nonlinear constraints. You define
fc=50000000;
fc is a large positive constant
B1=0.85-((0.05).*(fc-28)./7);
small positive constant multiplied by roughly 1% of a large positive constant, gives a large negative value
and you have the constraint
0.65-B1
small positive constant minus large negative constant gives large positive constant. But inequality constraints must be negative to be satisfied.
This constraint is independent of the inputs, and therefore should not be used.
I am also having trouble with the constraint
1-Mn./Mr;
but that is going to take more investigation to determine whether there is any chance of satisfying the constraint. So far it has never been satisfied over a million random locations.
Metasebia Gelan
2021년 7월 4일
i only perform the optimization for some constrantes and fix the things that you said but still there is an error saying
Error using matlab.graphics.chart.primitive.Line/set
Complex values are not supported.
Error in gaplotpareto>plotFirstFront (line 84)
set(plotHandle,'Xdata',xy(:,1), 'Ydata',xy(:,2));
Error in gaplotpareto (line 28)
plotFirstFront(state.Score(range,:),state.Rank(range),markers{1 +
mod(i,5)},objectivesToPlot(:)',flag,tag);
Error in gadsplot>callOnePlotFcn (line 222)
optimvalues = plotfcn(varargin{1:end});
Error in gadsplot (line 169)
[state,optimvalues] =
callOnePlotFcn(fname,plotNames{i},state,options.OutputPlotFcnOptions,optimvalues,flag,args{i}{:});
Error in gamultiobjsolve (line 115)
gadsplot(options,state,currentState,'Genetic Algorithm');
Error in gamultiobj (line 303)
[x,fval,exitFlag,output,population,scores] = gamultiobjsolve(FitnessFcn,nvars, ...
Metasebia Gelan
2021년 7월 4일
편집: Metasebia Gelan
2021년 7월 4일
greetings walter
i have attached my current code to!!
best regards
metasebia
Walter Roberson
2021년 7월 5일
You did not fix the B1 problem. Your constraints cannot be met.
I have attached a cleaned up version of nonlcon.m . You should save your current version and rename this one to nonlcon.m . It would be easier on me if you could develop using this modified version instead of what you have been using.
This version is vectorized -- it can take an N x 8 array of sample points and calculate all the nonlinear constraints for them simultaneously. This makes it much easier to explore to see if there are any points that can satisfy the constraints. With the previous code, I sampled over 1 million points in the bounds.
This version also builds up the final constraint array in pieces, such as
cin1 = abs(sqrt(M./(0.27.*W.*fc)))-x(:,1);
cin2 = ftop-fa;
cin3 = C-x(:,2);
[...]
Cineq=[cin1, cin2, cin3, cin4, cin5, cin6, cin7, cin8, cin9, cin10, cin11, cin12, cin13, cin14, cin15, cin16, cin17, cin18, cin19];
This makes it easier to debug problems with mismatched variable sizes, and make it easier to explore why particular constraints are not satisfied.
You will notice this version has
%{
cin8 = zeros(size(M)); cin8(1) = 0.65-B1;
%}
cin8 = zeros(size(M,1),0); %should be present, but always positive
This leaves logical space for the 0.65-B1 constraint but leaves it empty, because as I demonstrated earlier A) the constraint is independent of inputs and so should not be present at this level; and B) as I showed, B1 is a quite negative number so 0.65-B1 is always positive so the constraint cannot ever be satisfied.
I also put in
%{
cin14 = abs((Vu-Vp).*dv)-Mu;
cin15 = Vs-VS;
cin16 = Vu./q-Vn;
cin17 = 0.5.*q.*(Vc+Vp)-Vu;
cin18 = Vu./q-Vn;
cin19 = fpe-0.8.*fpy;
%}
cin14 = zeros(size(M,1),0); %removed
cin15 = zeros(size(M,1),0); %removed
cin16 = zeros(size(M,1),0); %removed
cin17 = zeros(size(M,1),0); %removed
cin18 = zeros(size(M,1),0); %removed
cin19 = zeros(size(M,1),0); %removed
to make it easier if you want to restore those constraints (that you used to have) later.
You absolutely cannot get anywhere until you either fix your B1 constraint or you decide to leave it disabled (like I do here.)
If you put in new code, please be sure to use .* and ./ and .^ to vectorize the constraints.
Walter Roberson
2021년 7월 5일
Example of reason to vectorize constraints:
%sample many points across the entire defined range
N = rand(1e6, 8);
rval = lb + (ub-lb).*N;
%calculate constraints at each point, all at once
val = nonlcon(rval);
%see if it is even hypothetically possible to satisfy the constraints
min(val)
if the min() is positive in any entry, then that is a constraint that was not satisfied on any of the 1 million points, and you need to chase through the code to figure out why that constraint was never negative (never satisfied). You might, for example, discover that to get negative for that constraint would require that one of the inputs is outside the range you set in lb and ub. Or you might find a logic error in calculating the constraint.
Once you get to a state where min(val) is 0 or positive for all entries, then that does not mean that there is definitely a point that can satisfied all constraints. There is the possibility that some constraints are mutually exclusive, that the range of values that can satisfy one of them can never satisfy another, and vice versa. But at least once min(val) is <= 0 then you have a fighting chance; if any min(val) > 0 then you have no chance.
metasebia dabi
2021년 7월 5일
Thank you for u support Walter Theother question that I have is how can I vectorize the objective and constraint function without manual metoude b/c the function is to long , like Mn in my function Best regards
Walter Roberson
2021년 7월 5일
I already posted the vectorized nonlinear constraint, above.
Because you already are using x(:,INDEX), vectorizing is mostly a matter of doing some search and replace
* -> .*
/ -> ./
^ -> .^
then search for
..
to find any cases where you might have ended up changing an operationg that was already .* into ..* .
Mr=min(Mcr,1.33.*Mu);
That is already vectorized.
F=[f1,f2];
and that is already fine for the case where your rows correspond to different input samples and your columns correspond to the input variables.
Your previous nonlcon function required additional changes to handle S and SMax in vectorized form. It also needed a small adjustment to the way that min() is calculated for dva .
Vectorized objective attached.
metasebia dabi
2021년 7월 6일
편집: metasebia dabi
2021년 7월 6일
gteatings walter
thank you in advance for ur support .
i have some changes in my objective fuction and it became longer , when i try to vectors it manualy it is hard and out of the wendow , i coun't be able ro simplify as well as fectorize it , i have already attached it in hear can you pls help me with it.
best regards
metasebia dabi
2021년 7월 6일
편집: metasebia dabi
2021년 7월 6일
and i fix the B1 case it was rhe errror in the inpute thst is why i want to change the objective function fc in line 8 of the nonlinear.m is 50 not 50000000. so should i write the constraint of B1 like others?
Walter Roberson
2021년 7월 6일
Your x(3) is not used in either your objective function or your constraints function. It is, however, used in your Aeq, B, and ub / lb, and your Aineq, bineq.
Effectively, it will be adjusted to whatever value is needed to work out the constraints, and is useless on its own. Ideally it should be replaced algebraically to reduce the number of variables being searched over, as the system will continually be wasting time just adjusting the useless variable.
Your x(7) is not used in either your objective function or your constraints function or your inequalities. Your code makes it look like it is used, but instead its input value is ignored and a calculated value is written into it. It should be removed as an input, and the variables renamed in the code to a meaningful name.
Your x(8) is used in your objective function, but is not used in your constraints functions or your inequalities. Your constraints function makes it look like it is used, but instead its input value is ignored and a calculated value is written into it. This is confusing, and the variable should be renamed in the constraints function to a more meaningful name.
I have attached current working version of the code.
If you execute it you will notice that it does not run an optimization, and instead complains about unsolved constraints.
The unsolved constraint it is refering to is the 1-Mn/Mr constraint -- cin13 in the code. Your Mn is always negative and your Mr is always positive, so the ratio of the two is always negative, and subtracting a negative from 1 gives a value that is always positive; therefore the constraint cannot be met.
Your change for fc to fc=50 allows the B1 constraint to pass. However, it is still a constraint that is independent of the inputs, so you should not have it in the code: either check the ratio before you start any optimization, or else put it in as an assert() statement to abort processing if it cannot be met. A constraint that is either always true or always false regardless of inputs is a poor constraint.
metasebia dabi
2021년 7월 10일
편집: metasebia dabi
2021년 7월 10일
greeting walter u have helped me so much in this but i came up with additional chalege can eyou please see it and help me with it
i have solved the issue with the unsolved constraint which is 1-Mn/Mr but mister wolter the resultes in the pareto chart are not good i want the result of safty or my seconde function f2 to be positive no. and must be greater than one but see the figer , the parito front what should i do??
i atached the new code and the results from pareto front please see it and help me with this?
best regards
Metasebia Dabi
metasebia dabi
2021년 7월 10일
편집: metasebia dabi
2021년 7월 10일
the results in the left side of the chart are negative as well as are almost zero why did this happen.
i have attached my design code in here i used every thing in the nonlcon from my design code
Walter Roberson
2021년 7월 10일
편집: Walter Roberson
2021년 7월 10일
Mcr=(((31.*2.^(1./2))./10 - ((x1 - (10000.*x2 + (100.*x5 - 100.*x6).*(x2 + 100./3) + 1750.*x2.^2 - x4.*(x1 - x4./2).*(2.*x5 + x6 - 3500) + 2.*x6.*(x1./2 + x2./2).*(x1 - x2) + 1000000./3)./(3500.*x2 + 100.*x5 - 100.*x6 + 2.*x6.*(x1 - x2) - x4.*(2.*x5 + x6 - 3500) + 10000)).*((12354898479057801913548821935370803857421875.*x2)./309485009821345068724781056 + (352997099401651483244252055296308681640625.*x5)./309485009821345068724781056 - (352997099401651483244252055296308681640625.*x6)./309485009821345068724781056 + (14119883976066059329770082211852347265625.*x6.*(x1 - x2))./618970019642690137449562112 - (14119883976066059329770082211852347265625.*x4.*(2.*x5 + x6 - 3500))./1237940039285380274899124224 + 454739836963295317117559435156108837041015625./77371252455336267181195264))./((500000.*x5)./3 - (500000.*x6)./3 + 10000.*(x2 - (10000.*x2 + (100.*x5 - 100.*x6).*(x2 + 100./3) + 1750.*x2.^2 - x4.*(x1 - x4./2).*(2.*x5 + x6 - 3500) + 2.*x6.*(x1./2 + x2./2).*(x1 - x2) + 1000000./3)./(3500.*x2 + 100.*x5 - 100.*x6 + 2.*x6.*(x1 - x2) - x4.*(2.*x5 + x6 - 3500) + 10000) + 100./3).^2 + 3500.*x2.*(x2./2 - (10000.*x2 + (100.*x5 - 100.*x6).*(x2 + 100./3) + 1750.*x2.^2 - x4.*(x1 - x4./2).*(2.*x5 + x6 - 3500) + 2.*x6.*(x1./2 + x2./2).*(x1 - x2) + 1000000./3)./(3500.*x2 + 100.*x5 - 100.*x6 + 2.*x6.*(x1 - x2) - x4.*(2.*x5 + x6 - 3500) + 10000)).^2 + 2.*(50.*x5 - 50.*x6).*(x2 - (10000.*x2 + (100.*x5 - 100.*x6).*(x2 + 100./3) + 1750.*x2.^2 - x4.*(x1 - x4./2).*(2.*x5 + x6 - 3500) + 2.*x6.*(x1./2 + x2./2).*(x1 - x2) + 1000000./3)./(3500.*x2 + 100.*x5 - 100.*x6 + 2.*x6.*(x1 - x2) - x4.*(2.*x5 + x6 - 3500) + 10000) + 100./3).^2 + (875.*x2.^3)./3 + (x6.^3.*(x1 - x2))./6 - (x4.^3.*(2.*x5 + x6 - 3500))./12 + 2.*x6.*(x1 - x2).*(x1./2 + x2./2 - (10000.*x2 + (100.*x5 - 100.*x6).*(x2 + 100./3) + 1750.*x2.^2 - x4.*(x1 - x4./2).*(2.*x5 + x6 - 3500) + 2.*x6.*(x1./2 + x2./2).*(x1 - x2) + 1000000./3)./(3500.*x2 + 100.*x5 - 100.*x6 + 2.*x6.*(x1 - x2) - x4.*(2.*x5 + x6 - 3500) + 10000)).^2 - x4.*(x4./2 - x1 + (10000.*x2 + (100.*x5 - 100.*x6).*(x2 + 100./3) + 1750.*x2.^2 - x4.*(x1 - x4./2).*(2.*x5 + x6 - 3500) + 2.*x6.*(x1./2 + x2./2).*(x1 - x2) + 1000000./3)./(3500.*x2 + 100.*x5 - 100.*x6 + 2.*x6.*(x1 - x2) - x4.*(2.*x5 + x6 - 3500) + 10000)).^2.*(2.*x5 + x6 - 3500) + 8947848533333333./536870912) + ((1836769977535370803857421875.*x2)./73786976294838206464 + (52479142215296308681640625.*x5)./73786976294838206464 - (52479142215296308681640625.*x6)./73786976294838206464 + (2099165688611852347265625.*x6.*(x1 - x2))./147573952589676412928 - (2099165688611852347265625.*x4.*(2.*x5 + x6 - 3500))./295147905179352825856 - (5.*2.^(1./2).*((500000.*x5)./3 - (500000.*x6)./3 + 10000.*(x2 - (10000.*x2 + (100.*x5 - 100.*x6).*(x2 + 100./3) + 1750.*x2.^2 - x4.*(x1 - x4./2).*(2.*x5 + x6 - 3500) + 2.*x6.*(x1./2 + x2./2).*(x1 - x2) + 1000000./3)./(3500.*x2 + 100.*x5 - 100.*x6 + 2.*x6.*(x1 - x2) - x4.*(2.*x5 + x6 - 3500) + 10000) + 100./3).^2 + 3500.*x2.*(x2./2 - (10000.*x2 + (100.*x5 - 100.*x6).*(x2 + 100./3) + 1750.*x2.^2 - x4.*(x1 - x4./2).*(2.*x5 + x6 - 3500) + 2.*x6.*(x1./2 + x2./2).*(x1 - x2) + 1000000./3)./(3500.*x2 + 100.*x5 - 100.*x6 + 2.*x6.*(x1 - x2) - x4.*(2.*x5 + x6 - 3500) + 10000)).^2 + 2.*(50.*x5 - 50.*x6).*(x2 - (10000.*x2 + (100.*x5 - 100.*x6).*(x2 + 100./3) + 1750.*x2.^2 - x4.*(x1 - x4./2).*(2.*x5 + x6 - 3500) + 2.*x6.*(x1./2 + x2./2).*(x1 - x2) + 1000000./3)./(3500.*x2 + 100.*x5 - 100.*x6 + 2.*x6.*(x1 - x2) - x4.*(2.*x5 + x6 - 3500) + 10000) + 100./3).^2 + (875.*x2.^3)./3 + (x6.^3.*(x1 - x2))./6 - (x4.^3.*(2.*x5 + x6 - 3500))./12 + 2.*x6.*(x1 - x2).*(x1./2 + x2./2 - (10000.*x2 + (100.*x5 - 100.*x6).*(x2 + 100./3) + 1750.*x2.^2 - x4.*(x1 - x4./2).*(2.*x5 + x6 - 3500) + 2.*x6.*(x1./2 + x2./2).*(x1 - x2) + 1000000./3)./(3500.*x2 + 100.*x5 - 100.*x6 + 2.*x6.*(x1 - x2) - x4.*(2.*x5 + x6 - 3500) + 10000)).^2 - x4.*(x4./2 - x1 + (10000.*x2 + (100.*x5 - 100.*x6).*(x2 + 100./3) + 1750.*x2.^2 - x4.*(x1 - x4./2).*(2.*x5 + x6 - 3500) + 2.*x6.*(x1./2 + x2./2).*(x1 - x2) + 1000000./3)./(3500.*x2 + 100.*x5 - 100.*x6 + 2.*x6.*(x1 - x2) - x4.*(2.*x5 + x6 - 3500) + 10000)).^2.*(2.*x5 + x6 - 3500) + 8947848533333333./536870912))./(2.*(x1 - (10000.*x2 + (100.*x5 - 100.*x6).*(x2 + 100./3) + 1750.*x2.^2 - x4.*(x1 - x4./2).*(2.*x5 + x6 - 3500) + 2.*x6.*(x1./2 + x2./2).*(x1 - x2) + 1000000./3)./(3500.*x2 + 100.*x5 - 100.*x6 + 2.*x6.*(x1 - x2) - x4.*(2.*x5 + x6 - 3500) + 10000))) + 182273709256141679345841015625./18446744073709551616)./(((19.*x1)./20 + ((500000.*x5)./3 - (500000.*x6)./3 + 10000.*(x2 - (10000.*x2 + (100.*x5 - 100.*x6).*(x2 + 100./3) + 1750.*x2.^2 - x4.*(x1 - x4./2).*(2.*x5 + x6 - 3500) + 2.*x6.*(x1./2 + x2./2).*(x1 - x2) + 1000000./3)./(3500.*x2 + 100.*x5 - 100.*x6 + 2.*x6.*(x1 - x2) - x4.*(2.*x5 + x6 - 3500) + 10000) + 100./3).^2 + 3500.*x2.*(x2./2 - (10000.*x2 + (100.*x5 - 100.*x6).*(x2 + 100./3) + 1750.*x2.^2 - x4.*(x1 - x4./2).*(2.*x5 + x6 - 3500) + 2.*x6.*(x1./2 + x2./2).*(x1 - x2) + 1000000./3)./(3500.*x2 + 100.*x5 - 100.*x6 + 2.*x6.*(x1 - x2) - x4.*(2.*x5 + x6 - 3500) + 10000)).^2 + 2.*(50.*x5 - 50.*x6).*(x2 - 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3500) + 8947848533333333./536870912))./(2.*(x1 - (10000.*x2 + (100.*x5 - 100.*x6).*(x2 + 100./3) + 1750.*x2.^2 - x4.*(x1 - x4./2).*(2.*x5 + x6 - 3500) + 2.*x6.*(x1./2 + x2./2).*(x1 - x2) + 1000000./3)./(3500.*x2 + 100.*x5 - 100.*x6 + 2.*x6.*(x1 - x2) - x4.*(2.*x5 + x6 - 3500) + 10000))) + 182273709256141679345841015625./18446744073709551616))./(20.*((19.*x1)./20 + ((500000.*x5)./3 - (500000.*x6)./3 + 10000.*(x2 - (10000.*x2 + (100.*x5 - 100.*x6).*(x2 + 100./3) + 1750.*x2.^2 - x4.*(x1 - x4./2).*(2.*x5 + x6 - 3500) + 2.*x6.*(x1./2 + x2./2).*(x1 - x2) + 1000000./3)./(3500.*x2 + 100.*x5 - 100.*x6 + 2.*x6.*(x1 - x2) - x4.*(2.*x5 + x6 - 3500) + 10000) + 100./3).^2 + 3500.*x2.*(x2./2 - (10000.*x2 + (100.*x5 - 100.*x6).*(x2 + 100./3) + 1750.*x2.^2 - x4.*(x1 - x4./2).*(2.*x5 + x6 - 3500) + 2.*x6.*(x1./2 + x2./2).*(x1 - x2) + 1000000./3)./(3500.*x2 + 100.*x5 - 100.*x6 + 2.*x6.*(x1 - x2) - x4.*(2.*x5 + x6 - 3500) + 10000)).^2 + 2.*(50.*x5 - 50.*x6).*(x2 - (10000.*x2 + (100.*x5 - 100.*x6).*(x2 + 100./3) + 1750.*x2.^2 - x4.*(x1 - x4./2).*(2.*x5 + x6 - 3500) + 2.*x6.*(x1./2 + x2./2).*(x1 - x2) + 1000000./3)./(3500.*x2 + 100.*x5 - 100.*x6 + 2.*x6.*(x1 - x2) - x4.*(2.*x5 + x6 - 3500) + 10000) + 100./3).^2 + (875.*x2.^3)./3 + (x6.^3.*(x1 - x2))./6 - (x4.^3.*(2.*x5 + x6 - 3500))./12 + 2.*x6.*(x1 - x2).*(x1./2 + x2./2 - (10000.*x2 + (100.*x5 - 100.*x6).*(x2 + 100./3) + 1750.*x2.^2 - x4.*(x1 - x4./2).*(2.*x5 + x6 - 3500) + 2.*x6.*(x1./2 + x2./2).*(x1 - x2) + 1000000./3)./(3500.*x2 + 100.*x5 - 100.*x6 + 2.*x6.*(x1 - x2) - x4.*(2.*x5 + x6 - 3500) + 10000)).^2 - x4.*(x4./2 - x1 + (10000.*x2 + (100.*x5 - 100.*x6).*(x2 + 100./3) + 1750.*x2.^2 - x4.*(x1 - x4./2).*(2.*x5 + x6 - 3500) + 2.*x6.*(x1./2 + x2./2).*(x1 - x2) + 1000000./3)./(3500.*x2 + 100.*x5 - 100.*x6 + 2.*x6.*(x1 - x2) - x4.*(2.*x5 + x6 - 3500) + 10000)).^2.*(2.*x5 + x6 - 3500) + 8947848533333333./536870912)./((x1 - (10000.*x2 + (100.*x5 - 100.*x6).*(x2 + 100./3) + 1750.*x2.^2 - x4.*(x1 - x4./2).*(2.*x5 + x6 - 3500) + 2.*x6.*(x1./2 + x2./2).*(x1 - x2) + 1000000./3)./(3500.*x2 + 100.*x5 - 100.*x6 + 2.*x6.*(x1 - x2) - x4.*(2.*x5 + x6 - 3500) + 10000)).*(3500.*x2 + 100.*x5 - 100.*x6 + 2.*x6.*(x1 - x2) - x4.*(2.*x5 + x6 - 3500) + 10000))).*((500000.*x5)./3 - (500000.*x6)./3 + 10000.*(x2 - (10000.*x2 + (100.*x5 - 100.*x6).*(x2 + 100./3) + 1750.*x2.^2 - x4.*(x1 - x4./2).*(2.*x5 + x6 - 3500) + 2.*x6.*(x1./2 + x2./2).*(x1 - x2) + 1000000./3)./(3500.*x2 + 100.*x5 - 100.*x6 + 2.*x6.*(x1 - x2) - x4.*(2.*x5 + x6 - 3500) + 10000) + 100./3).^2 + 3500.*x2.*(x2./2 - (10000.*x2 + (100.*x5 - 100.*x6).*(x2 + 100./3) + 1750.*x2.^2 - x4.*(x1 - x4./2).*(2.*x5 + x6 - 3500) + 2.*x6.*(x1./2 + x2./2).*(x1 - x2) + 1000000./3)./(3500.*x2 + 100.*x5 - 100.*x6 + 2.*x6.*(x1 - x2) - x4.*(2.*x5 + x6 - 3500) + 10000)).^2 + 2.*(50.*x5 - 50.*x6).*(x2 - (10000.*x2 + (100.*x5 - 100.*x6).*(x2 + 100./3) + 1750.*x2.^2 - x4.*(x1 - x4./2).*(2.*x5 + x6 - 3500) + 2.*x6.*(x1./2 + x2./2).*(x1 - x2) + 1000000./3)./(3500.*x2 + 100.*x5 - 100.*x6 + 2.*x6.*(x1 - x2) - x4.*(2.*x5 + x6 - 3500) + 10000) + 100./3).^2 + (875.*x2.^3)./3 + (x6.^3.*(x1 - x2))./6 - (x4.^3.*(2.*x5 + x6 - 3500))./12 + 2.*x6.*(x1 - x2).*(x1./2 + x2./2 - (10000.*x2 + (100.*x5 - 100.*x6).*(x2 + 100./3) + 1750.*x2.^2 - x4.*(x1 - x4./2).*(2.*x5 + x6 - 3500) + 2.*x6.*(x1./2 + x2./2).*(x1 - x2) + 1000000./3)./(3500.*x2 + 100.*x5 - 100.*x6 + 2.*x6.*(x1 - x2) - x4.*(2.*x5 + x6 - 3500) + 10000)).^2 - x4.*(x4./2 - x1 + (10000.*x2 + (100.*x5 - 100.*x6).*(x2 + 100./3) + 1750.*x2.^2 - x4.*(x1 - x4./2).*(2.*x5 + x6 - 3500) + 2.*x6.*(x1./2 + x2./2).*(x1 - x2) + 1000000./3)./(3500.*x2 + 100.*x5 - 100.*x6 + 2.*x6.*(x1 - x2) - x4.*(2.*x5 + x6 - 3500) + 10000)).^2.*(2.*x5 + x6 - 3500) + 8947848533333333./536870912))).*(3500.*x2 + 100.*x5 - 100.*x6 + 2.*x6.*(x1 - x2) - x4.*(2.*x5 + x6 - 3500) + 10000).*((500000.*x5)./3 - (500000.*x6)./3 + 10000.*(x2 - (10000.*x2 + (100.*x5 - 100.*x6).*(x2 + 100./3) + 1750.*x2.^2 - x4.*(x1 - x4./2).*(2.*x5 + x6 - 3500) + 2.*x6.*(x1./2 + x2./2).*(x1 - x2) + 1000000./3)./(3500.*x2 + 100.*x5 - 100.*x6 + 2.*x6.*(x1 - x2) - x4.*(2.*x5 + x6 - 3500) + 10000) + 100./3).^2 + 3500.*x2.*(x2./2 - (10000.*x2 + (100.*x5 - 100.*x6).*(x2 + 100./3) + 1750.*x2.^2 - x4.*(x1 - x4./2).*(2.*x5 + x6 - 3500) + 2.*x6.*(x1./2 + x2./2).*(x1 - x2) + 1000000./3)./(3500.*x2 + 100.*x5 - 100.*x6 + 2.*x6.*(x1 - x2) - x4.*(2.*x5 + x6 - 3500) + 10000)).^2 + 2.*(50.*x5 - 50.*x6).*(x2 - (10000.*x2 + (100.*x5 - 100.*x6).*(x2 + 100./3) + 1750.*x2.^2 - x4.*(x1 - x4./2).*(2.*x5 + x6 - 3500) + 2.*x6.*(x1./2 + x2./2).*(x1 - x2) + 1000000./3)./(3500.*x2 + 100.*x5 - 100.*x6 + 2.*x6.*(x1 - x2) - x4.*(2.*x5 + x6 - 3500) + 10000) + 100./3).^2 + (875.*x2.^3)./3 + (x6.^3.*(x1 - x2))./6 - (x4.^3.*(2.*x5 + x6 - 3500))./12 + 2.*x6.*(x1 - x2).*(x1./2 + x2./2 - (10000.*x2 + (100.*x5 - 100.*x6).*(x2 + 100./3) + 1750.*x2.^2 - x4.*(x1 - x4./2).*(2.*x5 + x6 - 3500) + 2.*x6.*(x1./2 + x2./2).*(x1 - x2) + 1000000./3)./(3500.*x2 + 100.*x5 - 100.*x6 + 2.*x6.*(x1 - x2) - x4.*(2.*x5 + x6 - 3500) + 10000)).^2 - x4.*(x4./2 - x1 + (10000.*x2 + (100.*x5 - 100.*x6).*(x2 + 100./3) + 1750.*x2.^2 - x4.*(x1 - x4./2).*(2.*x5 + x6 - 3500) + 2.*x6.*(x1./2 + x2./2).*(x1 - x2) + 1000000./3)./(3500.*x2 + 100.*x5 - 100.*x6 + 2.*x6.*(x1 - x2) - x4.*(2.*x5 + x6 - 3500) + 10000)).^2.*(2.*x5 + x6 - 3500) + 8947848533333333./536870912))./(10000.*x2 + (100.*x5 - 100.*x6).*(x2 + 100./3) + 1750.*x2.^2 - x4.*(x1 - x4./2).*(2.*x5 + x6 - 3500) + 2.*x6.*(x1./2 + x2./2).*(x1 - x2) + 1000000./3);
Your Mcr come out as large negative numbers, around -1e19
Mr=min(Mcr,1.33.*Mu);
Your mu is a moderately large positive number, about 1E10. The min() of a negative and a positive is going to be the negative, so the min() is effectively just passing through the Mcr values.
f2=Mn./Mr;
Your Mn are moderately large positive numbers, about 1e11. As discussed above, your Mr are about -1e19. 1e11/-1e19 is on the order of -1e-9 -- a number that cannot be greater than 1 -- it cannot even be greater than 0.
metasebia dabi
2021년 7월 14일
Greetings Walter how can I handle maximizing one objective and minimizing the other, where do have to make change in my objective finction. One of my objective is to maximize the safety(f2) and the other is to minimize the cost(f1) , what kind of change is expected from me to change in the objective function
Walter Roberson
2021년 7월 14일
Have your objective function return the negative of the value to be maximized. Minimizing -1*x is the same as maximimzing x.
metasebia dabi
2021년 7월 15일
So should I make every value of x negative of should I multiplay the whole function by negative
metasebia dabi
2021년 7월 15일
What kind of chartshould I expect is it like one of the row start from large to small??
Walter Roberson
2021년 7월 15일
The plot will probably look like a gentle parabolic descending from maximum safety and minimum cost to the left, to less safety and higher cost towards the right.
The safety will still be about 6*e-10, no-where near the 1 you were hoping for. This is because the algorithm for computing f2 has not changed, and I showed above that f2 is only about 1e-10
metasebia dabi
2021년 7월 16일
편집: metasebia dabi
2021년 7월 16일
But I went a safety greater than 1 ! What is the problem and am still getting a chart which is concave up not descending , is the problem with my code?
metasebia dabi
2021년 7월 16일
How do I change the algorithm for computing f2 should I multiplay all x by -ve
Walter Roberson
2021년 7월 16일
I would not know, as I have not seen the equations.
Remember, though, the previous discussion:
Your Mn are moderately large positive numbers, about 1e11. As discussed above, your Mr are about -1e19. 1e11/-1e19 is on the order of -1e-9 -- a number that cannot be greater than 1 -- it cannot even be greater than 0.
We (already) take the negative of it in order to maximize it, but in order to be able to get the value greater than 1, then either Mn has to increase by 9 orders of magnitude, or else Mr has to decrease by 9 orders of magnitude (or some combination of the two)
metasebia dabi
2021년 7월 16일
편집: metasebia dabi
2021년 7월 17일
i fix those problems with Mn and Mr and in my current code Mn/Mr is positive grater than 1 , now i change my mind and i wont to maximize my second function f2 but rather that minimizing -f2 to compute the maximization it makes all results negative why ?
i have attached the code in here!
Walter Roberson
2021년 7월 17일
metasebia dabi
2021년 8월 5일
편집: metasebia dabi
2021년 8월 5일
i have another essue mr. walter
to check my results i work out the one of the solution in pareto solution, in my design code but there is defrence in results of the objective function why is this happening?? i have attached the design cod
Walter Roberson
2021년 8월 6일
elseif S/6>tf> S/18,tf = tf;
MATLAB interprets that as meaning
elseif ((S/6>tf)> S/18)
tf = tf;
S/6>tf is calculated, returning 0 (false) or 1 (true). That 0 or 1 is then compared to S/18.
In this particular case it does not matter, since the branch effectively does nothing even if it works, but the code is misleading.
And you really should deal with the question of what to do if S/6 == tf exactly, or if tf < S/18 . If those cases are meaningful, then deal with them; if they are not meaningful then just have the if without any else.
You have a number of similar cases.
I suggest that you learn how to use min() and max(). Instead of writing something like
if C > 0.65
C = 0.65;
elseif C < 0.65
C = C;
end
just write
C = min(C, 0.65);
It is a bit counter-intuitive, but use min() when you want to set a value to be the maximum level something should have, and use max() when you want to set a value to be the minimum level something should have.
But other than those... unfortunately I do not know what I am looking at. Your design code calculates some things, but I have no idea how those things link to your previous code.
metasebia dabi
2021년 8월 6일
I insert one of the final result of optimized solutions from my previous code , which are x1,x2,x3,x4,x5,x6,x7 and insert this results in the design code , the previous code is formulated using this design code so after inserting this results I run the code and calculate the result of safety factor Mn/Mr but in the optimization this result is fgal2 and the result is 11 but the out put of the original code using the optimized cross sections is 3,my question is why does it have this difference in results , is it b/c I use -f2 to maximize?
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