I'm working with the Simulink block MATLAB FUNCTION and I'm having problems with the bounds of the variables that I define in there.
This is the part of the code where I’m getting troubles
function P_S1_100= fcn(SOC_S1_100,S1_AGENTS_10,time_CAP_100)
assert(time_CAP_100(1)<100)
tcharging_a1_1=[0:0.05:time_CAP_100(1)]
tcharging_a1_2=[time_CAP_100(1):0.05:time_CAP_100(1)*2]
tcharging_a1=[0:0.05:time_CAP_100(1)]
(Where time_CAP_100 is a vector [1x6])
And this is the error that I'm getting:
Computed maximum size of the output of function 'colon' is not bounded.
Static memory allocation requires all sizes to be bounded.
The computed size is [1 x :?].
Function 'Subsystem1/Slow Charge/S1/MATLAB Function5' (#265.262.302), line 8, column 16:
"[time_CAP_100(1):0.05:time_CAP_100(1)*2]"
Could anyone give me an idea of how to solve this error?
Thanks in advance.
For each of your variable-size data inputs/outputs, you need to define what the upper bound is. See http://www.mathworks.co.uk/help/simulink/ug/declare-variable-size-inputs-and-outputs.html for more details.
Only work around I can think of is to manually write a loop with fixed loop bounds to expand [time_CAP_100(1):0.05:time_CAP_100(1)*2]. That expression is what is causing the problem. You need to know the bounds of this vector. Then you can write a loop something like
% max_size is the maximum length possible for tcharging_a1_2
tcharging_a1_2 = zeros(1,max_size);
tcharging_a1_2(1) = time_CAP_100(1);
for ii=2:max_size
if tcharging_a1_2(ii) < time_CAP_100(1)*2
tcharging_a1_2(ii) = tcharging_a1_2(ii) + .05;
end
end
Related
I'm trying to solve a nonlinear system of 7 equations with least_squares, as some variables are molar fractions and go between 0 and 1 (fsolve doesnt allow constraints as far as I've read). Heres the main part of the code:
def fun_t(w,x_2):
x_1=1-x_2;
z_A=w[0];
z_B=w[1];
z_AB=w[2];
gamma_A=w[3];
gamma_B=w[4];
gamma_AB=w[5];
eps=w[6]
K_gamma=gamma_AB/(gamma_A*gamma_B)
K=K_gamma*z_AB/(z_A*z_B)
phi_A=z_A*v_A/(z_A*v_A+z_B*v_B+z_AB*v_AB);
phi_B=z_B*v_B/(z_A*v_A+z_B*v_B+z_AB*v_AB);
phi_AB=z_AB*v_AB/(z_A*v_A+z_B*v_B+z_AB*v_AB);
return [z_AB-eps/(1-eps),
z_A-(x_2-eps)/(1-eps),
z_B-(x_1-eps)/(1-eps),
eps-(1-(1-4*K/(K+K_gamma)*x_1*x_2)**0.5)/2,
R*T*np.log(gamma_A)-v_A*(alphaT_AB*phi_B**2+alphaT_AAB*phi_AB**2+(alphaT_AB+alphaT_AAB-alphaT_BAB)*phi_B*phi_AB),
R*T*np.log(gamma_B)-v_B*(alphaT_AB*phi_A**2+alphaT_BAB*phi_AB**2+(alphaT_AB+alphaT_BAB+alphaT_AAB)*phi_A*phi_AB),
R*T*np.log(gamma_AB)-v_AB*(alphaT_AAB*phi_A**2+alphaT_BAB*phi_B**2+(alphaT_AAB+alphaT_BAB-alphaT_AB)*phi_A*phi_B)
]
x_2=np.linspace(0,1,11)
guess=[0.5,0.5,0.5,1,1,1,0.25];
roots_t=np.zeros(shape=(len(x_2),len(guess)))
for i in range(len(x_2)):
roots_t[i] = least_squares(fun_t, guess, bounds = ((0,0,0,-5,-5,-5,0),(1,1,1,5,5,5,0.5)), args=x_2[i])
I get the error TypeError: fun_t() argument after * must be an iterable, not numpy.float64. My goal is to give the elements of the array x_2 as the fuction's second argument in order to get the rest of the variables for a set of compositions. I dont understand why it cant iterate over x_2, as it worked when I was using fsolve.
I'm just starting using python for this kind of calculations, so any help will be much appreciated!
I am currently implementing an optimization problem with pyomo and since now some hours I get the message that my problem is unbounded. After searching for the issue, I came along one term which seems to be unbounded. I excluded this term from the objective function and it shows that it takes a very high negative value, which supports the assumption that it is unbounded to -Inf.
But I have checked the problem further and it is impossible that the term is unbounded, as following code and results show:
model.nominal_cap_storage = Var(model.STORAGE, bounds=(0,None)) #lower bound is 0
#I assumed very high CAPEX for each storage (see print)
dict_capex_storage = {'battery': capex_battery_storage,
'co2': capex_co2_storage,
'hydrogen': capex_hydrogen_storage,
'heat': capex_heat_storage,
'syncrude': capex_syncrude_storage}
print(dict_capex_storage)
>>> {'battery': 100000000000000000, 'co2': 100000000000000000,
'hydrogen': 1000000000000000000, 'heat': 1000000000000000, 'syncrude': 10000000000000000000}
From these assumptions I already assume that it is impossible that the one term can be unbounded towards -Inf as the capacity has the lower bound of 0 and the CAPEX is a positive fixed value. But now it gets crazy. The following term is has the issue of being unbounded:
model.total_investment_storage = Var()
def total_investment_storage_rule(model):
return model.total_investment_storage == sum(model.nominal_cap_storage[storage] * dict_capex_storage[storage] \
for storage in model.STORAGE)
model.total_investment_storage_con = Constraint(rule=total_investment_storage_rule)
If I exclude the term from the objective function, I get following value after the optimization. It seems, that it can take high negative values.
>>>>
Variable total_investment_storage
-1004724108.3426505
So I checked the term regarding the component model.nominal_cap_storage to see the value of the capacity:
model.total_cap_storage = Var()
def total_cap_storage_rule(model):
return model.total_cap_storage == sum(model.nominal_cap_storage[storage] for storage in model.STORAGE)
model.total_cap_storage_con = Constraint(rule=total_cap_storage_rule)
>>>>
Variable total_cap_storage
0.0
I did the same for the dictionary, but made a mistake: I forgot to delete the model.nominal_cap_storage. But the result is confusing:
model.total_capex_storage = Var()
def total_capex_storage_rule(model):
return model.total_capex_storage == sum(model.nominal_cap_storage[storage] * dict_capex_storage[storage] \
for storage in model.STORAGE)
model.total_capex_storage_con = Constraint(rule=total_capex_storage_rule)
>>>>
Variable total_capex_storage
0.0
So my question is why is the term unbounded and how is it possible that model.total_investment_storage and model.total_capex_storage have different solutions though both are calculated equally? Any help is highly appreciated.
I think you are misinterpreting "unbounded." When the solver says the problem is unbounded, that means the objective function value is unbounded based on the variables and constraints in the problem. It has nothing to do with bounds on variables, unless one of those variable bounds prevents the objective from being unbound.
If you want help on above problem, you need to edit and post the full problem, with the objective function, and (if possible) the error. What you have now is a collection of different snippets of different variations of a problem, which isn't really informative on the overall issue.
I solved the problem by setting a lower bound to the term, which takes a negative value:
model.total_investment_storage = Var(bounds=(0, None)
I am still not sure why this term can take negative values but this solved at least my problem
I write a vector in Dymola mos script in a simple manner like this:
x_axis = cell.spatialSummary.x_cell;
output: x_axis={1,2,3,4,5} // row vector
I want to do the same thing in a function.'x_cell' has 5 values which I want to store in a row vector. I use DymolaCommands.Trajectories.readTrajectory function to read x_cell values one by one in for loop (I use for loop because, readTrajectory throws an error when I try to read entire x_cell)
Real x_axis[:],axis_value[:,:];
Integer len=5;
for i in 1:len loop
axis_value:=readTrajectory(result,{"cell.spatialSummary.x_cell["+String(i)+"]"},1); //This intermediate variable returns [1,1] matrix
x_axis[i]:=scalar(axis_value);
end for;
I get an error:
Assignment failed x_axis[i] = scalar(axis_value);
what's wrong here? All I want to do is read all values of x_cell and write it into a vector. How can I do this in dymola function?
Thank you!
Solution: Initialize the vector with a certain value. In this case,
x_axis :=fill(0, len);
This solved the above problem for me.
Pre filling as in the other solution works, and is generally the best solution. However, in some cases you might have to append to the vector as follows:
x_axis=fill(0.0, 0);
for i in 1:len loop
axis_value:=readTrajectory(result,{"cell.spatialSummary.x_cell["+String(i)+"]"},1); //This intermediate variable returns [1,1] matrix
x_axis:=cat(1, x_axis, {scalar(axis_value)});
end for;
(This takes x_axis and concatenates a new element at the end. It is generally slower.)
Anyone know if NLopt works with univariate optimization. Tried to run following code:
using NLopt
function myfunc(x, grad)
x.^2
end
opt = Opt(:LD_MMA, 1)
min_objective!(opt, myfunc)
(minf,minx,ret) = optimize(opt, [1.234])
println("got $minf at $minx (returned $ret)")
But get following error message:
> Error evaluating untitled
LoadError: BoundsError: attempt to access 1-element Array{Float64,1}:
1.234
at index [2]
in myfunc at untitled:8
in nlopt_callback_wrapper at /Users/davidzentlermunro/.julia/v0.4/NLopt/src/NLopt.jl:415
in optimize! at /Users/davidzentlermunro/.julia/v0.4/NLopt/src/NLopt.jl:514
in optimize at /Users/davidzentlermunro/.julia/v0.4/NLopt/src/NLopt.jl:520
in include_string at loading.jl:282
in include_string at /Users/davidzentlermunro/.julia/v0.4/CodeTools/src/eval.jl:32
in anonymous at /Users/davidzentlermunro/.julia/v0.4/Atom/src/eval.jl:84
in withpath at /Users/davidzentlermunro/.julia/v0.4/Requires/src/require.jl:37
in withpath at /Users/davidzentlermunro/.julia/v0.4/Atom/src/eval.jl:53
[inlined code] from /Users/davidzentlermunro/.julia/v0.4/Atom/src/eval.jl:83
in anonymous at task.jl:58
while loading untitled, in expression starting on line 13
If this isn't possible, does anyone know if a univariate optimizer where I can specify bounds and an initial condition?
There are a couple of things that you're missing here.
You need to specify the gradient (i.e. first derivative) of your function within the function. See the tutorial and examples on the github page for NLopt. Not all optimization algorithms require this, but the one that you are using LD_MMA looks like it does. See here for a listing of the various algorithms and which require a gradient.
You should specify the tolerance for conditions you need before you "declare victory" ¹ (i.e. decide that the function is sufficiently optimized). This is the xtol_rel!(opt,1e-4) in the example below. See also the ftol_rel! for another way to specify a different tolerance condition. According to the documentation, for example, xtol_rel will "stop when an optimization step (or an estimate of the optimum) changes every parameter by less than tol multiplied by the absolute value of the parameter." and ftol_rel will "stop when an optimization step (or an estimate of the optimum) changes the objective function value by less than tol multiplied by the absolute value of the function value. " See here under the "Stopping Criteria" section for more information on various options here.
The function that you are optimizing should have a unidimensional output. In your example, your output is a vector (albeit of length 1). (x.^2 in your output denotes a vector operation and a vector output). If you "objective function" doesn't ultimately output a unidimensional number, then it won't be clear what your optimization objective is (e.g. what does it mean to minimize a vector? It's not clear, you could minimize the norm of a vector, for instance, but a whole vector - it isn't clear).
Below is a working example, based on your code. Note that I included the printing output from the example on the github page, which can be helpful for you in diagnosing problems.
using NLopt
count = 0 # keep track of # function evaluations
function myfunc(x::Vector, grad::Vector)
if length(grad) > 0
grad[1] = 2*x[1]
end
global count
count::Int += 1
println("f_$count($x)")
x[1]^2
end
opt = Opt(:LD_MMA, 1)
xtol_rel!(opt,1e-4)
min_objective!(opt, myfunc)
(minf,minx,ret) = optimize(opt, [1.234])
println("got $minf at $minx (returned $ret)")
¹ (In the words of optimization great Yinyu Ye.)
So, I have this project for school in which I have to write code in SciLab to solve a puzzle (Tents). The code is getting longer and longer as it gets better and better, but I suddenly got an error stating "stack size exceeded".
Error log:
!--error 17
stack size exceeded!
Use stacksize function to increase it.
Memory used for variables: 28875
Intermediate memory needed: 59987764
Total memory available: 10000000
I tried using this line
stacksize('max')
And this one
stacksize(1e8)
Neither of which works, all that happens is SciLab shutting itself down without any warning at all.
How did I exceed my stacksize? Is there a way to prevent this? How can I continue further?
I figured out myself how to solve this problem. Here's what I did wrong for people with the same problem:
Within a function I used the line
[m,n] = [x,y]
to save the coordinates of an object from a matrix. This was called within a loop using x and y to browse through the matrix.
Apparently this caused the stacksize exceeded error and here's how I wrote it afterwards:
m = x
n = y
I have no idea why this line caused this error, but this is how I've solved it.