Right upfront: this is an issue I encountered when submitting an R package to CRAN. So I
dont have control of the stack size (as the issue occured on one of CRANs platforms)
I cant provide a reproducible example (as I dont know the exact configurations on CRAN)
Problem
When trying to submit the cSEM.DGP package to CRAN the automatic pretest (for Debian x86_64-pc-linux-gnu; not for Windows!) failed with the NOTE: C stack usage 7975520 is too close to the limit.
I know this is caused by a function with three arguments whose body is about 800 rows long. The function body consists of additions and multiplications of these arguments. It is the function varzeta6() which you find here (from row 647 onwards).
How can I adress this?
Things I cant do:
provide a reproducible example (at least I would not know how)
change the stack size
Things I am thinking of:
try to break the function into smaller pieces. But I dont know how to best do that.
somehow precompile? the function (to be honest, I am just guessing) so CRAN doesnt complain?
Let me know your ideas!
Details / Background
The reason why varzeta6() (and varzeta4() / varzeta5() and even more so varzeta7()) are so long and R-inefficient is that they are essentially copy-pasted from mathematica (after simplifying the mathematica code as good as possible and adapting it to be valid R code). Hence, the code is by no means R-optimized (which #MauritsEvers righly pointed out).
Why do we need mathematica? Because what we need is the general form for the model-implied construct correlation matrix of a recursive strucutral equation model with up to 8 constructs as a function of the parameters of the model equations. In addition there are constraints.
To get a feel for the problem, lets take a system of two equations that can be solved recursivly:
Y2 = beta1*Y1 + zeta1
Y3 = beta2*Y1 + beta3*Y2 + zeta2
What we are interested in is the covariances: E(Y1*Y2), E(Y1*Y3), and E(Y2*Y3) as a function of beta1, beta2, beta3 under the constraint that
E(Y1) = E(Y2) = E(Y3) = 0,
E(Y1^2) = E(Y2^2) = E(Y3^3) = 1
E(Yi*zeta_j) = 0 (with i = 1, 2, 3 and j = 1, 2)
For such a simple model, this is rather trivial:
E(Y1*Y2) = E(Y1*(beta1*Y1 + zeta1) = beta1*E(Y1^2) + E(Y1*zeta1) = beta1
E(Y1*Y3) = E(Y1*(beta2*Y1 + beta3*(beta1*Y1 + zeta1) + zeta2) = beta2 + beta3*beta1
E(Y2*Y3) = ...
But you see how quickly this gets messy when you add Y4, Y5, until Y8.
In general the model-implied construct correlation matrix can be written as (the expression actually looks more complicated because we also allow for up to 5 exgenous constructs as well. This is why varzeta1() already looks complicated. But ignore this for now.):
V(Y) = (I - B)^-1 V(zeta)(I - B)'^-1
where I is the identity matrix and B a lower triangular matrix of model parameters (the betas). V(zeta) is a diagonal matrix. The functions varzeta1(), varzeta2(), ..., varzeta7() compute the main diagonal elements. Since we constrain Var(Yi) to always be 1, the variances of the zetas follow. Take for example the equation Var(Y2) = beta1^2*Var(Y1) + Var(zeta1) --> Var(zeta1) = 1 - beta1^2. This looks simple here, but is becomes extremly complicated when we take the variance of, say, the 6th equation in such a chain of recursive equations because Var(zeta6) depends on all previous covariances betwenn Y1, ..., Y5 which are themselves dependend on their respective previous covariances.
Ok I dont know if that makes things any clearer. Here are the main point:
The code for varzeta1(), ..., varzeta7() is copy pasted from mathematica and hence not R-optimized.
Mathematica is required because, as far as I know, R cannot handle symbolic calculations.
I could R-optimze "by hand" (which is extremly tedious)
I think the structure of the varzetaX() must be taken as given. The question therefore is: can I somehow use this function anyway?
Once conceivable approach is to try to convince the CRAN maintainers that there's no easy way for you to fix the problem. This is a NOTE, not a WARNING; The CRAN repository policy says
In principle, packages must pass R CMD check without warnings or significant notes to be admitted to the main CRAN package area. If there are warnings or notes you cannot eliminate (for example because you believe them to be spurious) send an explanatory note as part of your covering email, or as a comment on the submission form
So, you could take a chance that your well-reasoned explanation (in the comments field on the submission form) will convince the CRAN maintainers. In the long run it would be best to find a way to simplify the computations, but it might not be necessary to do it before submission to CRAN.
This is a bit too long as a comment, but hopefully this will give you some ideas for optimising the code for the varzeta* functions; or at the very least, it might give you some food for thought.
There are a few things that confuse me:
All varzeta* functions have arguments beta, gamma and phi, which seem to be matrices. However, in varzeta1 you don't use beta, yet beta is the first function argument.
I struggle to link the details you give at the bottom of your post with the code for the varzeta* functions. You don't explain where the gamma and phi matrices come from, nor what they denote. Furthermore, seeing that beta are the model's parameter etimates, I don't understand why beta should be a matrix.
As I mentioned in my earlier comment, I would be very surprised if these expressions cannot be simplified. R can do a lot of matrix operations quite comfortably, there shouldn't really be a need to pre-calculate individual terms.
For example, you can use crossprod and tcrossprod to calculate cross products, and %*% implements matrix multiplication.
Secondly, a lot of mathematical operations in R are vectorised. I already mentioned that you can simplify
1 - gamma[1,1]^2 - gamma[1,2]^2 - gamma[1,3]^2 - gamma[1,4]^2 - gamma[1,5]^2
as
1 - sum(gamma[1, ]^2)
since the ^ operator is vectorised.
Perhaps more fundamentally, this seems somewhat of an XY problem to me where it might help to take a step back. Not knowing the full details of what you're trying to model (as I said, I can't link the details you give to the cSEM.DGP code), I would start by exploring how to solve the recursive SEM in R. I don't really see the need for Mathematica here. As I said earlier, matrix operations are very standard in R; analytically solving a set of recursive equations is also possible in R. Since you seem to come from the Mathematica realm, it might be good to discuss this with a local R coding expert.
If you must use those scary varzeta* functions (and I really doubt that), an option may be to rewrite them in C++ and then compile them with Rcpp to turn them into R functions. Perhaps that will avoid the C stack usage limit?
Related
I have been working for a couple of months with OpenMDAO and I find myself struggling with my code when I want to impose conditions for trying to replicate a physical/engineering behaviour.
I have tried using sigmoid functions, but I am still not convinced with that, due to the difficulty about trading off sensibility and numerical stabilization. Most of times I found overflows in exp so I end up including other conditionals (like np.where) so loosing linearity.
outputs['sigmoid'] = 1 / (1 + np.exp(-x))
I was looking for another kind of step function or something like that, able to keep linearity and derivability to the ease of the optimization. I don't know if something like that exists or if there is any strategy that can help me. If it helps, I am working with an OpenConcept benchmark, which uses vectorized computations ans Simpson's rule numerical integration.
Thank you very much.
PD: This is my first ever question in stackoverflow, so I would like to apologyze in advance for any error or bad practice commited. Hope to eventually collaborate and become active in the community.
Update after Justin answer:
I will take the opportunity to define a little bit more my problem and the strategy I tried. I am trying to monitorize and control thermodynamics conditions inside a tank. One of the things is to take actions when pressure P1 reaches certein threshold P2, for defining this:
eval= (inputs['P1'] - inputs['P2']) / (inputs['P1'] + inputs['P2'])
# P2 = threshold [Pa]
# P1 = calculated pressure [Pa]
k=100 #steepness control
outputs['sigmoid'] = (1 / (1 + np.exp(-eval * k)))
eval was defined in order avoid overflows normalizing the values, so when the threshold is recahed, corrections are taken. In a very similar way, I defined a function to check if there is still mass (so flowing can continue between systems):
eval= inputs['mass']/inputs['max']
k=50
outputs['sigmoid'] = (1 / (1 + np.exp(-eval*k)))**3
maxis also used for normalizing the value and the exponent is added for reaching zero before entering in the negative domain.
PLot (sorry it seems I cannot post images yet for my reputation)
It may be important to highlight that both mass and pressure are calculated from coupled ODE integration, in which this activation functions take part. I guess OpenConcept nature 'explore' a lot of possible values before arriving the solution, so most of the times giving negative infeasible values for massand pressure and creating overflows. For that sometimes I try to include:
eval[np.where(eval > 1.5)] = 1.5
eval[np.where(eval < -1.5)] = -1.5
That is not a beautiful but sometimes effective solution. I try to avoid using it since I taste that this bounds difficult solver and optimizer work.
I could give you a more complete answer if you distilled your question down to a specific code example of the function you're wrestling with and its expected input range. If you provide that code-sample, I'll update my answer.
Broadly, this is a common challenge when using gradient based optimization. You want some kind of behavior like an if-condition to turn something on/off and in many cases thats a fundamentally discontinuous function.
To work around that we often use sigmoid functions, but these do have some of the numerical challenges you pointed out. You could try a hyberbolic tangent as an alternative, though it may suffer the same kinds of problems.
I will give you two broad options:
Option 1
sometimes its ok (even if not ideal) to leave the purely discrete conditional in the code. Lets say you wanted to represent a kind of simple piecewise function:
y = 2x; x>=0
y = 0; x < 0
There is a sharp corner in that function right at 0. That corner is not differentiable, but the function is fine everywhere else. This is very much like the absolute value function in practice, though you might not draw the analogy looking at the piecewise definition of the function because the piecewise nature of abs is often hidden from you.
If you know (or at least can check after the fact) that your final answer will no lie right on or very near to that C1 discontinuity, then its probably fine to leave the code the way is is. Your derivatives will be well defined everywhere but right at 0 and you can simply pick the left or the right answer for 0.
Its not strictly mathematically correct, but it works fine as long as you're not ending up stuck right there.
Option 2
Apply a smoothing function. This can be a sigmoid, or a simple polynomial. The exact nature of the smoothing function is highly specific to the kind of discontinuity you are trying to approximate.
In the case of the piecewise function above, you might be tempted to define that function as:
2x*sig(x)
That would give you roughly the correct behavior, and would be differentiable everywhere. But wolfram alpha shows that it actually undershoots a little. Thats probably undesirable, so you can increase the exponent to mitigate that. This however, is where you start to get underflow and overflow problems.
So to work around that, and make a better behaved function all around, you could instead defined a three part piecewise polynomial:
y = 2x; x>=a
y = c0 + c1*x + c2*x**2; -a <= x < a
y = 0 x < -a
you can solve for the coefficients as a function of a (please double check my algebra before using this!):
c0 = 1.5a
c1 = 2
c2 = 1/(2a)
The nice thing about this approach is that it will never overshoot and go negative. You can also make a reasonably small and still get decent numerics. But if you try to make it too small, c2 will obviously blow up.
In general, I consider the sigmoid function to be a bit of a blunt instrument. It works fine in many cases, but if you try to make it approximate a step function too closely, its a nightmare. If you want to represent physical processes, I find polynomial fillet functions work more nicely.
It takes a little effort to derive that polynomial, because you want it to be c1 continuous on both sides of the curve. So you have to construct the system of equations to solve for it as a function of the polynomial order and the specific relaxation you want (0.1 here).
My goto has generally been to consult the table of activation functions on wikipedia: https://en.wikipedia.org/wiki/Activation_function
I've had good luck with sigmoid and the hyperbolic tangent, scaling them such that we can choose the lower and upper values as well as choosing the location of the activation on the x-axis and the steepness.
Dymos uses a vectorization that I think is similar to OpenConcept and I've had success with numpy.where there as well, providing derivatives for each possible "branch" taken. It is true that you may have issues with derivative mismatches if you have an analysis point right on the transition, but often I've had success despite that. If the derivative at the transition becomes a hinderance then implementing a sigmoid or relu are more appropriate.
If x is of a magnitude such that it can cause overflows, consider applying units or using scaling to put it within reasonable limits if you cannot bound it directly.
Introduction
I'm doing research in computational contact mechanics, in which I try to solve a PDE using a finite difference method. Long story short, I need to solve a linear system like Ax = b.
The suspects
In the problem, the matrix A is sparse, and so I defined it accordingly. On the other hand, both x and b are dense arrays.
In fact, x is defined as x = A\b, the potential solution of the problem.
So, the least one might expect from this solution is to satisfy that Ax is close to b in some sense. Great is my surprise when I find that
julia> norm(A*x-b) # Frobenius or 2-norm
5018.901093242197
The vector x does not solve the system! I've tried a lot of tricks discover what is going on, but no clues as of now. My first candidate is that I've found a bug, however I need more evidence to make this assertion.
The hints
Here are some tests that I've done to try to pinpoint the error
If you convert A to dense, the solution changes completely, and in fact it returns the correct solution.
I have repeated the proccess above in matlab, and it seems to work well with both sparse and dense matrices (that is, the sparse version does not agree with that of Julia's)
Not all sparse matrices cause a problem. I have tried other initial conditions and the solver seems to work quite well. I am not able to predict what property of the matrix can be causing this discrepancy. However;
A has a condition number of 120848.06, which is quite high, although matlab doesn't seem to complain. Also, the absolute error of the solution to the real solution is huge.
How to reproduce this "bug"
Download the .csv files in the following link
Run the following code in the folder of the files (install the packages if necessary
using DelimitedFiles, LinearAlgebra, SparseArrays;
A = readdlm("A.csv", ',');
b = readdlm("b.csv", ',');
x = readdlm("x.csv", ',');
A_sparse = sparse(A);
println(norm(A_sparse\b - x)); # You should get something close to zero, x is the solution of the sparse implementation
println(norm(A_sparse*x - b)); # You should get something not close to zero, something is not working!
Final words
It might easily be the case that I'm missing something. Are there any other implementations apart from the usual A\b to test against?
To solve a sparse square system Julia chooses to do a sparse LU decomposition. For the specific matrix A in the question, this decomposition is numerically ill-conditioned. This is evidenced by the cond(lu(A_sparse).U) == 2.879548971708896e64. This causes the solve routine to make numerical errors in turn.
A quick solution is to use a QR decomposition instead, by running x = qr(A_sparse)\b.
The solve or LU routines might need to be fixed to handle this case, or at least maintainers need to know of this issue, so opening an issue on the github repo might be good.
(this is a rewrite of my comment on question)
This is a continuation of my questions:
Declaring a functional recursive sequence in Matlab
Is there a more efficient way of nesting logarithms?
Nesting a specific recursion in Pari-GP
But I'll keep this question self contained. I have made a coding project for myself; which is to program a working simple calculator for a tetration function I've constructed. This tetration function is holomorphic, and stated not to be Kneser's solution (as to all the jargon, ignore); long story short, I need to run the numbers; to win over the nay-sayers.
As to this, I have to use Pari-GP; as this is a fantastic language for handling large numbers and algebraic expressions. As we are dealing with tetration (think numbers of the order e^e^e^e^e^e); this language is, of the few that exist, the best for such affairs. It is the favourite when doing iterated exponential computations.
Now, the trouble I am facing is odd. It is not so much that my code doesn't work; it's that it's overflowing because it should over flow (think, we're getting inputs like e^e^e^e^e^e; and no computer can handle it properly). I'll post the first batch of code, before I dive deeper.
The following code works perfectly; and does everything I want. The trouble is with the next batch of code. This produces all the numbers I want.
\\This is the asymptotic solution to tetration. z is the variable, l is the multiplier, and n is the depth of recursion
\\Warning: z with large real part looks like tetration; and therefore overflows very fast. Additionally there are singularities which occur where l*(z-j) = (2k+1)*Pi*I.
\\j,k are integers
beta_function(z,l,n) =
{
my(out = 0);
for(i=0,n-1,
out = exp(out)/(exp(l*(n-i-z)) +1));
out;
}
\\This is the error between the asymptotic tetration and the tetration. This is pretty much good for 200 digit accuracy if you need.
\\modify the 0.000000001 to a bigger number to make this go faster and receive less precision. When graphing 0.0001 is enough
\\Warning: This will blow up at some points. This is part of the math; these functions have singularities/branch cuts.
tau(z,l,n)={
if(1/real(beta_function(z,l,n)) <= 0.000000001, //this is where we'll have problems; if I try to grab a taylor series with this condition we error out
-log(1+exp(-l*z)),
log(1 + tau(z+1,l,n)/beta_function(z+1,l,n)) - log(1+exp(-l*z))
)
}
\\This is the sum function. I occasionally modify it; to make better graphs, but the basis is this.
Abl(z,l,n) = {
beta_function(z,l,n) + tau(z,l,n)
}
Plugging this in, you get the following expressions:
Abl(1,log(2),100)
realprecision = 28 significant digits (20 digits displayed)
%109 = 0.15201551563214167060
exp(Abl(0,log(2),100))
%110 = 0.15201551563214167060
Abl(1+I,2+0.5*I,100)
%111 = 0.28416643148885326261 + 0.80115283113944703984*I
exp(Abl(0+I,2+0.5*I,100))
%112 = 0.28416643148885326261 + 0.80115283113944703984*I
And so on and so forth; where Abl(z,l,n) = exp(Abl(z-1,l,n)). There's no problem with this code. Absolutely none at all; we can set this to 200 precision and it'll still produce correct results. The graphs behave exactly as the math says they should behave. The problem is, in my construction of tetration (the one we actually want); we have to sort of paste together the solutions of Abl(z,l,n) across the value l. Now, you don't have to worry about any of that at all; but, mathematically, this is what we're doing.
This is the second batch of code; which is designed to "paste together" all these Abl(z,l,n) into one function.
//This is the modified asymptotic solution to the Tetration equation.
beta(z,n) = {
beta_function(z,1/sqrt(1+z),n);
}
//This is the Tetration function.
Tet(z,n) ={
if(1/abs(beta_function(z,1/sqrt(1+z),n)) <= 0.00000001,//Again, we see here this if statement; and we can't have this.
beta_function(z,1/sqrt(1+z),n),
log(Tet(z+1,n))
)
}
This code works perfectly for real-values; and for complex values. Some sample values,
Tet(1+I,100)
%113 = 0.12572857262453957030 - 0.96147559586703141524*I
exp(Tet(0+I,100))
%114 = 0.12572857262453957030 - 0.96147559586703141524*I
Tet(0.5,100)
%115 = -0.64593666417664607364
exp(Tet(0.5,100))
%116 = 0.52417133958039107545
Tet(1.5,100)
%117 = 0.52417133958039107545
We can also effectively graph this object on the real-line. Which just looks like the following,
ploth(X=0,4,Tet(X,100))
Now, you may be asking; What's the problem then?
If you try and plot this function in the complex plane, it's doomed to fail. The nested logarithms produce too many singularities near the real line. For imaginary arguments away from the real-line, there's no problem. And I've produced some nice graphs; but the closer you get to the real line; the more it misbehaves and just short circuits. You may be thinking; well then, the math is wrong! But, no, the reason this is happening is because Kneser's tetration is the only tetration that is stable about the principal branch of the logarithm. Since this tetration IS NOT Kneser's tetration, it's inherently unstable about the principal branch of the logarithm. Of course, Pari just chooses the principal branch. So when I do log(log(log(log(log(beta(z+5,100)))))); the math already says this will diverge. But on the real line; it's perfectly adequate. And for values of z with an imaginary argument away from zero, we're fine too.
So, how I want to solve this, is to grab the Taylor series at Tet(1+z,100); which Pari-GP is perfect for. The trouble?
Tet(1+z,100)
*** at top-level: Tet(1+z,100)
*** ^------------
*** in function Tet: ...unction(z,1/sqrt(1+z),n))<=0.00000001,beta_fun
*** ^---------------------
*** _<=_: forbidden comparison t_SER , t_REAL.
The numerical comparison I've done doesn't translate to a comparison between t_SER and t_REAL.
So, my question, at long last: what is an effective strategy at getting the Taylor series of Tet(1+z,100) using only real inputs. The complex inputs near z=0 are erroneous; the real values are not. And if my math is right; we can take the derivatives along the real-line and get the right result. Then, we can construct a Tet_taylor(z,n) which is just the Taylor Series expansion. Which; will most definitely have no errors when trying to graph.
Any help, questions, comments, suggestions--anything, is greatly appreciated! I really need some outside eyes on this.
Thanks so much if you got to the bottom of this post. This one is bugging me.
Regards, James
EDIT:
I should add that a Tet(z+c,100) for some number c is the actual tetration function we want. There is a shifting constant I haven't talked about yet. Nonetheless; this is spurious to the question, and is more a mathematical point.
This is definitely not an answer - I have absolutely no clue what you are trying to do. However, I see no harm in offering suggestions. PARI has a built in type for power series (essentially Taylor series) - and is very good at working with them (many operations are supported). I was originally going to offer some suggestions on how to get a Taylor series out of a recursive definition using your functions as an example - but in this case, I'm thinking that you are trying to expand around a singularity which might be doomed to failure. (On your plot it seems as x->0, the result goes to -infinity???)
In particular if I compute:
log(beta(z+1, 100))
log(log(beta(z+2, 100)))
log(log(log(beta(z+3, 100))))
log(log(log(log(beta(z+4, 100)))))
...
The different series are not converging to anything. Even the constant term of the series is getting smaller with each iteration, so I am not entirely sure there is even a Taylor series expansion about x = 0.
Questions/suggestions:
Should you be expanding about a different point? (say where the curve
crosses the x-axis).
Does the Taylor series satisfy some recursive relation? For example: A(z) = log(A(z+1)). [This doesn't work, but perhaps there is another way to write it].
I suspect my answer is unlikely to be satisfactory - but then again your question is more mathematical than a practical programming problem.
So I've successfully answered my question. I haven't programmed in so long; I'm kind of shoddy. But I figured it out after enough coffee. I created 3 new functions, which allow me to grab the Taylor series.
\\This function attempts to find the number of iterations we need.
Tet_GRAB_k(A,n) ={
my(k=0);
while( 1/real(beta(A+k,n)) >= 0.0001, k++);
return(k);
}
\\This function will run and produce the same results as Tet; but it's slower; but it let's us estimate Taylor coefficients.
\\You have to guess which k to use for whatever accuracy before overflowing; which is what the last function is good for.
Tet_taylor(z,n,k) = {
my(val = beta(z+k,n));
for(i=1,k,val = log(val));
return(val);
}
\\This function produces an array of all the coefficients about a value A.
TAYLOR_SERIES(A,n) = {
my(ser = vector(40,i,0));
for(i=1,40, ser[i] = polcoeff(Tet_taylor(A+z,n,Tet_GRAB_k(A,n)),i-1,z));
return(ser);
}
After running the numbers, I'm confident this works. The Taylor series is converging; albeit rather slowly and slightly less accurately than desired; but this will have to do.
Thanks to anyone who read this. I'm just answering this question for completeness.
As I was working out how Epi generates the basis for its spline functions (via the function Ns), I was a little confused by how it handles the detrend argument.
When detrend=T I would have expected that Epi::Ns(...) would more or less project the basis given by splines::ns(...) onto the orthogonal complement of the column space of [1 t] and finally extract the set of linearly independent columns (so that we have a basis).
However, this doesn’t appear to be the exactly the case; I tried
library(Epi)
x=seq(-0.75, 0.75, length.out=5)
Ns(x, knots=c(-0.5,0,0.5), Boundary.knots=c(-1,1), detrend=T)
and
library(splines)
detrend(ns(x, knots=c(-0.5,0,0.5), Boundary.knots=c(-1,1)), x)
The matrices produced by the above code are not the same, however, they do have the same column space (in this example) suggesting that if plugged in to a linear model, the fitted coefficients will be different but the fit (itself) will be the same.
The first question I had was; is this true in general?
The second question is why are the two different?
Regarding the second question - when detrend is specified, Epi::Ns gives a warning that fixsl is ignored.
Diving into Epi github NS.r ... in the construction of the basis, in the call to Epi::Ns above with detrend=T, the worker ns.ld() is called (a function almost identical to the guts of splines::ns()), which passes c(NA,NA) along to splines::spline.des as the derivs argument in determining a matrix const;
const <- splines::spline.des( Aknots, Boundary.knots, 4, c(2-fixsl[1],2-fixsl[2]))$design
This is the difference between what happens in Ns(detrend=T) and the call to ns() above which passes c(2,2) to splineDesign as the derivs argument.
So that explains how they are different, but not why? Does anyone have an explanation for why fixsl=c(NA,NA) is used instead of fixsl=c(F,F) in Epi::Ns()?
And does anyone have a proof/or an answer to the first question?
I think the orthogonal complement of const's column space is used so that second (or desired) derivatives are zero at the boundary (via projection of the general spline basis) - but I'm not sure about this step as I haven't dug into the mathematics, I'm just going by my 'feel' for it. Perhaps if I understood this better, the reason that the differences in the result for const from the call to splineDesign/spline.des (in ns() and Ns() respectively) would explain why the two matrices from the start are not the same, yet yield the same fit.
The fixsl=c(NA,NA) was a bug that has been fixed since a while. See the commits on the CRAN Github mirror.
I have still sent an email to the maintainer to ask if the fix could be made a little bit more consistent with the condition, but in principle this could be closed.
I asked this question a while ago. I am not sure whether I should post this as an answer or a new question. I do not have an answer but I "solved" the problem by applying the Levenberg-Marquardt algorithm using nls.lm in R and when the solution is at the boundary, I run the trust-region-reflective algorithm (TRR, implemented in R) to step away from it. Now I have new questions.
From my experience, doing this way the program reaches the optimal and is not so sensitive to the starting values. But this is only a practical method to step aside from the issues I encounterd using nls.lm and also other optimization functions in R. I would like to know why nls.lm behaves this way for optimization problems with boundary constraints and how to handle the boundary constraints when using nls.lm in practice.
Following I gave an example illustrating the two issues using nls.lm.
It is sensitive to starting values.
It stops when some parameter reaches the boundary.
A Reproducible Example: Focus Dataset D
library(devtools)
install_github("KineticEval","zhenglei-gao")
library(KineticEval)
data(FOCUS2006D)
km <- mkinmod.full(parent=list(type="SFO",M0 = list(ini = 0.1,fixed = 0,lower = 0.0,upper =Inf),to="m1"),m1=list(type="SFO"),data=FOCUS2006D)
system.time(Fit.TRR <- KinEval(km,evalMethod = 'NLLS',optimMethod = 'TRR'))
system.time(Fit.LM <- KinEval(km,evalMethod = 'NLLS',optimMethod = 'LM',ctr=kingui.control(runTRR=FALSE)))
compare_multi_kinmod(km,rbind(Fit.TRR$par,Fit.LM$par))
dev.print(jpeg,"LMvsTRR.jpeg",width=480)
The differential equations that describes the model/system is:
"d_parent = - k_parent * parent"
"d_m1 = - k_m1 * m1 + k_parent * f_parent_to_m1 * parent"
In the graph on the left is the model with initial values, and in the middle is the fitted model using "TRR"(similar to the algorithm in Matlab lsqnonlin function ), on the right is the fitted model using "LM" with nls.lm. Looking at the fitted parameters(Fit.LM$par) you will find that one fitted parameter(f_parent_to_m1) is at the boundary 1. If I change the starting value for one parameter M0_parent from 0.1 to 100, then I got the same results using nls.lm and lsqnonlin.I have many cases like this one.
newpars <- rbind(Fit.TRR$par,Fit.LM$par)
rownames(newpars)<- c("TRR(lsqnonlin)","LM(nls.lm)")
newpars
M0_parent k_parent k_m1 f_parent_to_m1
TRR(lsqnonlin) 99.59848 0.09869773 0.005260654 0.514476
LM(nls.lm) 84.79150 0.06352110 0.014783294 1.000000
Except for the above problems, it often happens that the Hessian returned by nls.lm is not invertable(especially when some parameters are on the boundary) so I cannot get an estimation of the covariance matrix. On the other hand, the "TRR" algorithm(in Matlab) almost always give an estimation by calculating the Jacobian at the solution point. I think this is useful but I am also sure that R optimization algorithms(the ones I have tried) did not do this for a reason. I would like to know whether I am wrong by using the Matlab way of calculating the covariance matrix to get standard error for the parameter estimates.
One last note, I claimed in my previous post that the Matlab lsqnonlin outperforms R's optimization functions in almost all cases. I was wrong. The "Trust-Region-Reflective" algorithm used in Matlab is in fact slower(sometimes much slower) if also implemented in R as you can see from the above example. However, it is still more stable and reaches a better solution than the R's basic optimization algorithms.
First off, I am not an expert on Matlab and Optimisation and have never used R.
I am not sure I see what your actual question is, but maybe I can shed some light into your puzzlement:
LM is slightly enhanced Gauß-Newton approach - for problems with several local minima it is very sensitive to initial states. Including boundaries typically generates more of those minima.
TRR is akin to LM, but more robust. It has better capabilities for "jumping out of" bad local minima. It is quite feasible that it will behave better, but perform worse, than an LM. Actually explaining why is very hard. You would need to study the algorithms in detail and look at how they behave in this situation.
I cannot explain the difference between Matlab's and R's implementation, but there are several extensions to TRR that maybe Matlab uses and R does not.
Does your approach of using LM and TRR alternatingly converge better than TRR alone?
Using the mkin package, you can find the parameters using the "Port" algorithm (which is also a kind of a TRR algorithm as far as I could tell from its documentation), or the "Marq" algorithm, which uses nls.lm in the background. Then you can use "normal" starting values or "bad" starting values.
library(mkin)
packageVersion("mkin")
Recent mkin version can speed up the process considerably as they compile the models from automatically generated C code if a compiler is available on your system (e.g. you have r-base-dev installed on Debian/Ubuntu, or Rtools on Windows).
This defines the model:
m <- mkinmod(parent = mkinsub("SFO", "m1"),
m1 = mkinsub("SFO"),
use_of_ff = "max")
You can check that the differential equations are correct:
cat(m$diffs, sep = "\n")
Then we fit in four variants, Port and LM, with or without M0 fixed to 0.1:
f.Port = mkinfit(m, FOCUS_2006_D)
f.Port.M0 = mkinfit(m, FOCUS_2006_D, state.ini = c(parent = 0.1, m1 = 0))
f.LM = mkinfit(m, FOCUS_2006_D, method.modFit = "Marq")
f.LM.M0 = mkinfit(m, FOCUS_2006_D, state.ini = c(parent = 0.1, m1 = 0),
method.modFit = "Marq")
Then we look at the results:
results <- sapply(list(Port = f.Port, Port.M0 = f.Port.M0, LM = f.LM, LM.M0 = f.LM.M0),
function(x) round(summary(x)$bpar[, "Estimate"], 5))
which are
Port Port.M0 LM LM.M0
parent_0 99.59848 99.59848 99.59848 39.52278
k_parent 0.09870 0.09870 0.09870 0.00000
k_m1 0.00526 0.00526 0.00526 0.00000
f_parent_to_m1 0.51448 0.51448 0.51448 1.00000
So we can see that the Port algorithm finds the best solution (to the best of my knowledge) even with bad starting values. The speed issue that one may have with more complicated models is alleviated using the automatic generation of C code.