I'm trying to minimize a function which takes a vector as input and is subjected to some non linear constraints. I'm very new to Julia. I’m trying to implement pseudospectral methods using Ipopt.My isssue is Optimizer which i'm using takes gradient of cost function and constraints. Functions like "ForwardDiff , ReverseDiff" are not helping in finding the gradient of my vector function.
I found that similar issue has been face by #acauligi. So far I haven't found any solution.
using LinearAlgebra, DiffEqOperators, ForwardDiff, ApproxFun, FFTW, ToeplitzMatrices
using ModelingToolkit,DifferentialEquations,NLPModels,ADNLPModels,NLPModelsIpopt
using DCISolver,JSOSolvers
# Number of collocation points
N=31 # This number can go up to 200
function Dmatrix(N::Integer)
h=2*pi/N;
ns=range(1,N-1,step=1);
col1(nps)=0.5*((-1)^nps)/sin(nps*h/2);
col=[0,col1.(ns)...];
row=[0,col[end:-1:2]...];
D=Toeplitz(col,row)
end
Dmat=Dmatrix(N);
function dzdt(x,y,t,a)
u=(1-(x^2)/4)-y^2;
dx=-4*y+x*u+a*x;
dy=x+y*u+a*y;
[dx,dy]
end
# initial guess
tfinal=1.1*pi;
tpoints=collect(range(1,N,step=1))*tfinal/N;
xguess=sin.((2*pi/tfinal)*tpoints)*2.0
yguess=-sin.((2*pi/tfinal)*tpoints)*0.5
function dxlist(xs,ys,tf,a)
nstates=2
ts=collect(range(1,N,step=1))*tf/N;
xytsZip=zip(xs,ys,ts);
dxD0=[dzdt(x,y,t,a) for (x,y,t) in xytsZip];
dxD=reduce(hcat, dxD0)';
xlyl=reshape([xs;ys],N,nstates);
dxF=(Dmat*xlyl)*(2.0*pi/tf);
err=dxD-dxF;
[vcat(err'...).-10^(-10);-vcat(err'...).+10^(-10)]
end
function cons(x)
tf=x[end-1];
a=x[end];
xs1=x[1:N];
ys1=x[N+1:2*N];
dxlist(xs1,ys1,tf,a)
end
a0=10^-3;
x0=vcat([xguess;yguess;[tfinal,a0]]);
obj(x)=0.0;
xlower1=push!(-3*ones(2*N),pi);
xlower=push!(xlower1,-10^-3)
xupper1=push!(3*ones(2*N),1.5*pi);
xupper=push!(xupper,10^-3)
consLower=-ones(4*N)*Inf;
consUpper=zeros(4*N)
# println("constraints vector = ",cons(x0))
model=ADNLPModel(obj,x0,xlower,xupper,cons,consLower,consUpper; backend =
ADNLPModels.ReverseDiffAD)
output=ipopt(model)
xstar=output.solution
fstar=output.objective
I got the solution for this same problem in 3 minutes in MatLab.(solution to this problem is . Time period of system is "pi" when a=0.).
I was hoping I could get the same result much faster in Julia. I have asked in Julia discourse so far I have got any suggestion. Any suggestion on how fix this issue highly appreciated. Thank you all.
I think there was two issues with your code. First,
xupper1=push!(3*ones(2*N),1.5*pi);
xupper=push!(xupper1,10^-3)
and then for some reason the product of the Toeplitz matrix by another matrix gives an error with the automatic differentiation. However, the following works:
function dxlist(xs,ys,tf,a)
nstates=2
ts=collect(range(1,N,step=1))*tf/N;
xytsZip=zip(xs,ys,ts);
dxD0=[dzdt(x,y,t,a) for (x,y,t) in xytsZip];
dxD=reduce(hcat, dxD0)';
xlyl=reshape([xs;ys],N,nstates);
dxF=vcat((Dmat*xlyl[:,1])*(2.0*pi/tf), (Dmat*xlyl[:,2])*(2.0*pi/tf));
err=vcat(dxD...) - dxF;
[err.-10^(-10);-err.+10^(-10)]
end
At the end, Ipopt returns the right results
model=ADNLPModel(obj,x0,xlower,xupper,cons,consLower,consUpper)
output=ipopt(model)
xstar=output.solution
fstar=output.objective
I also noticed that using Percival.jl is faster
using Percival
output=percival(model, max_time = 300.0)
xstar=output.solution
fstar=output.objective
Note that ADNLPModels.jl is receiving some attention and will improve significantly.
Related
I am trying to set up a model for a dynamic optimisation problem with pyomo.DAE.
I defined my state variable as well as it's corresponding Derivative (both indexed by m.Time). I then set up a simple constraint that expresses the relationship between state and derivative variable in the most simple terms. Solving the problem with a dummy objective (so just testing the constraint), I get the following error:
ERROR: Model contains an expression (calc_my_state[0])
that contains a variable (derivative_var[0]) that is not
attached to an active block on the submodel being written
Here's an excerpt of what I wrote:
(.....)
m.state_var = Var(m.Time, initialize=0)
m.derivative_var = DerivativeVar(m.state_var, wrt=m.Time)
def calc_my_state(m,i):
return m.derivative_var[i] == m.state_var[i]*2
m.calc_my_state = Constraint(m.Time, rule=calc_my_state)
m.obj = Objective(expr=1)
opt = SolverFactory("glpk")
results = opt.solve(m)
I tried to reproduce the simple setup of an DAE in pyomo, more or less copied and pasted lines from the pyomoDAE docu.
I printed derivative_var.get_state_var() and it gives me the right state variable without error.
I also tried solving simple DAE examples that I found on the internet and solving them with my solver settings worked fine as well.
What am I missing? I am grateful for any input!!! Thanks!
I found the missing link: I did not specify the "Discretization Transformation". Once something like the following was added, the script ran without error!
discretizer = TransformationFactory('dae.finite_difference')
discretizer.apply_to(m, wrt=m.Time)
I am still a beginnner in R. For a project I am trying to fit a gam model on a simple dataset with a timeset and year. I am doing it in R and I keep getting an error message that claims an argument is unused, even though I specify it in the code.
It concerns a dataset which includes a categorical variable of "Year", with only two levels. 2020 and 2022. I want to investigate if there is a peak in the hourly rate of visitors ("H1") in a nature reserve. For each observation period the average time was taken, which is the predictor variable used here ("T"). I want to use a Gam model for this, and have the smoothing applied differently for the two years.
The following is the line of code that I tried to use
`gam1 <- gam(H1~Year+s(T,by=Year),data = d)`
When I try to run this code, I get the following error message
`Error in s(T, by = Year) : unused argument (by = Year)`
I also tried simply getting rid of the "by" argument
`gam1 <- gam(H1~Year+s(T,Year),data = d)`
This allows me to run the code, but when trying to summon the output using summary(gam1), I get
Error in [<-(tmp, snames, 2, value = round(nldf, 1)) : subscript out of bounds
Since I feel like both errors are probably related to the same thing that I'm doing wrong, I decided to combine the question.
Did you load the {mgcv} package or the {gam} package? The latter doesn't have factor by smooths and as such the first error message is what I would expect if you did library("gam") and then tried to fit the model you showed.
To fit the model you showed, you should restart R and try in a clean session:
library("mgcv")
# load you data
# fit model
gam1 <- gam(H1 ~ Year + s(T, by = Year), data = d)
It could well be that you have both {gam} and {mgcv} loaded, in which case whichever you loaded last will be earlier on the function search path. As both packages have functions gam() and s(), R might just be finding the wrong versions (masking), so you might also try
gam1 <- mgcv::gam(H1 ~ Year + mgcv::s(T, by = Year), data = d)
But you would be better off only loading {mgcv} if you wan factor by smooths.
#Gavin Simpson
I did have both loaded, and I tried just using mgcv as you suggested. However, then I get the following error.
Error in names(dat) <- object$term :
'names' attribute [1] must be the same length as the vector [0]
I am assuming this is simply because it's not actually trying to use the "gam" function, but rather it attempts to name something gam1. So I would assume I actually need the package of 'gam' before I could do this.
The second line of code also doesn't work. I get the following error
Error in model.frame.default(formula = H1 ~ Year + mgcv::s(T, by = Year), :
invalid type (list) for variable 'mgcv::s(T, by = Year)'
This happens no matter the order I download the two packages in. And if I don't download 'gam', I get the error as described above.
I am relatively new to using Python and Pandas, and was looking for help with this line of code:
`Football.ydstogo[Football.ydstogo>='11']&[Football.ydstogo<='19']= '10-plus`'
I am working with data from the NFL, and trying to build a model to predict when a team will pass, or when a team will run the ball. One of my variables (ydstogo) measures the distance for the team, with the ball, to get a first down. I am trying to group together the observations after 10 yards for ease of visualization.
When I tried running the code above, the error in my output is "can't assign to operator". I've used this code before to change gender observations to dummy variables, so I'm confused why it is not working here.
As I understand, you want to find elements with (string)
value between '11' and '19' and set a new string there.
So probably your should change your code to:
Football.ydstogo[(Football.ydstogo >= '11') & (Football.ydstogo <= '19')] = '10-plus'
Alternative:
Football.ydstogo[Football.ydstogo.between('11', '19')] = '10-plus'
I am new to Flink and have gone through site(s)/examples/blogs to get started. I am struggling with the correct use of operators. Basically I have 2 questions
Question 1: Does Flink support declarative exception handling, I need to handle parse/validate/... errors?
Can I use org.apache.flink.runtime.operators.sort.ExceptionHandler or similar
to handle errors?
or Rich/FlatMap function my best option?
If Rich/FlatMap the only option then is there a way to get handle to Stream inside Rich/FlatMap function so Sink(s) could be attached for error processing?
Question 2: Can I conditionally attach different Sink(s)?
Based on certain field(s) in keyed split streams I need to select different sink(s), do I split the stream again or use a Rich/FlatMap to handle that?
I am using Flink 1.3.2. Here is the relevant portion of my job
.....
.....
DataStream<String> eventTextStream = env.addSource(messageSource)
KeyedStream<EventPojo, Tuple> eventPojoStream = eventTextStream
// parse, transform or enrich
.flatMap(new MyParseTransformEnrichFunction())
.assignTimestampsAndWatermarks(new EventAscendingTimestampExtractor())
.keyBy("eventId");
// split stream based on eventType as different reduce and windowing functions need to be applied
SplitStream<EventPojo> splitStream = eventPojoStream
.split(new EventStreamSplitFunction());
// need to apply reduce function
DataStream<EventPojo> event1TypeStream = splitStream.select("event1Type");
// need to apply reduce function
DataStream<EventPojo> event2TypeStream = splitStream.select("event2Type");
// need to apply time based windowing function
DataStream<EventPojo> event3TypeStream = splitStream.select("event3Type");
....
....
env.execute("Event Processing");
Am I using the correct operators here?
Update 1:
Tried using the ProcessFunction as suggested by #alpinegizmo but that didn't work as it depends upon a keyed stream which I don't have until I parse/validate input. I get "InvalidProgramException: Field expression must be equal to '*' or '_' for non-composite types. ".
It's such a common use case where your first parse/validate input and won't have keyed stream yet, so how do you solve it?
Thanks for your patience and help.
There's one key building block that you've overlooked. Take a look at side outputs.
This mechanism provides a typesafe way to produce any number of additional output streams. This can be a clean way to report errors, among other uses. In Flink 1.3 side outputs can only be used with ProcessFunction, but 1.4 will add side outputs to ProcessWindowFunction.
I want to compute a list of the terms of the sequence
(14747-40*n)/(2621440*(41-n)), n from 1 to 40
I tried entering the above in Wolfram Alpha, and it plots me a graph of the function. This isn't what I want. I've tried variations on this command, as well as guessing at various keywords to stick before it, and I either get back the same thing, or something unhelpful.
The help page on sequences suggests various things you might do with sequences, but doesn't say how to do something simple like this???
The following works:
Table[(14747-40*n)/(2621440*(41-n)) n, {n, 1, 40}]
Clicking on "approximate form" then on "copy plaintext" gives the following:
{0.000140257, 0.000286924, 0.000440507, 0.000601567, 0.000770728,
0.000948683, 0.00113621, 0.00133417, 0.00154356, 0.00176547,
0.00200115, 0.00225204, 0.00251976, 0.00280618, 0.00311345,
0.00344409, 0.00380101, 0.00418764, 0.00460803, 0.00506701,
0.00557035, 0.00612508, 0.00673974, 0.00742493, 0.00819385,
0.00906326, 0.0100547, 0.0111963, 0.0125257, 0.0140939, 0.0159728,
0.0182658, 0.0211282, 0.0248041, 0.0297003, 0.0365488, 0.0468139,
0.0639122, 0.0980936, 0.200607}