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I'm using cplex linear programming in matlab (cplexlp) to solve the problem
min f'u s.t. Au>=b, u>=lb
using
[u,minima,flag] = cplexlp(f,-A,-b,[],[],lb);
but I need a solution tolerance of below 1e-9, which is the minimal tolerance in documentation.
I figured I can just scale the problem (e.g. by 10000) and achieve an effective tolerance of 1e-13.
scale=10000;
tolerance=1e-9;
options = cplexoptimset('cplex');
options.simplex.tolerances.feasibility = tolerance;
options.simplex.tolerances.optimality = tolerance;
[u,minima,flag] = cplexlp(scale*f,-scale*A,-scale*b,[],[],scale*lb,[], [], options);
minima = minima / scale;
It does not work, and the tolerance is improved to 1e-11 but not less. The image below show in log10 scale the 'real solution' (found using a different method) and the solution returned by the algorithm for different parameters (each color is different A,b and the x axes is some parameter of the problem which control the solution). As you can see, the real solution is achieved as long as it is above 1e-11.
Any suggestion for why is it so or how to avoid this problem?
CPLEX, like most other codes, uses double precision arithmetic. Within this environment you usually can trust your results only up to the first 9 or 10, maybe 11, digits. That's also the reason why you cannot set a smaller tolerance in CPLEX.
To get more accurate results you would have to use a solver that uses rational arithmetic. QSopt_ex and SoPlex (with iterative refinement) are two possibilities. I do not know how this is possible from within Matlab.
Additional answers offered in the cplex forum, but none solved it.
I am processing a set of data using ridge regression. I found a very interesting phenomenon when apply the learned function to data. Namely, when the ridge parameter increases from zero, the test error keeps increasing. But if we penalize small coefficients(set the parameter <0), the test error can even be smaller.
This is my matlab code:
for i = 1:100
beta = ridgePolyRegression(ty_train,tX_train,lambda(i));
sqridge_train_cost(i) = computePolyCostMSE(ty_train,tX_train,beta);
sqridge_test_cost(i) = computePolyCostMSE(ty_valid,tX_valid,beta);
end
plot(lambda,sqridge_test_cost,'color','b');
lambda is the ridge parameter. ty_train is the output of the training data, tX_train is the input of training data. Also, we use a quadratic function regression here.
function [ beta ] = ridgePolyRegression( y,tX,lambda )
X = tX(:,2:size(tX,2));
tX2 = [tX,X.^2];
beta = (tX2'*tX2 + lambda * eye(size(tX2,2))) \ (tX2'*y);
end
The plotted picture is:
Why the error is minimal when lambda is negative? Is it a sign of under-fitting?
You should not use negative lambdas.
From (probabilistic) theoretic point of view, lambda relates to the inverse of variance of parameter prior distribution, and variance can't be negative.
From computational point of view, it can (given it's less that the smallest eigenvalue of the covariance matrix) turn your positive-definite form into an indefinite form, which means you'll have not a maximum, but a saddle point. It also means there are points where your target function is as small (or as big) as you want, so you can reduce loss indefinitely and no minimum / maximum exists at all.
Your optimization algorithm gives you just a stationary point, which will be a global maximum if and only if the form is positive definite.
Short Answer: When lambda is negative, you're actually overfitting your data. Hence, it's reasonable to get much less error.
Long Answer:
The regularization term (or the penalty term as described by many statisticians) aims to penalize the weights (or the betas as written in the coming Eq.) for going too high (overfitting) and going too low (underfitting). Giving you the power to control how your model behaves, and you usually aim the "right fitting" model.
For mathematical intuition, you can check the following Eq. (P. S. Equation is screenshotted from Elements of Statistical Learning by Trevor Hastie et. al)
When you decide to make your lambda negative, the penalty term is indeed turned into a utility term that helps to increase the weights (i.e., overfitting).
Overfitting is, simply, understanding your data along with the features more than you should, because you do not have the whole population yet; therefore, what you understood so far is possibly wrong on a different dataset.
So, you should never be using negative values of lambdas.
I'm implementing AdaBoost on Matlab. This algorithm requires that in every iteration the weights of each data point in the training set sum up to one.
If I simply use the following normalization v = v / sum(v) I get a vector whose 1-norm is 1 except some numerical error which later leads to the failure of the algorithm.
Is there a matlab function for normalizing a vector so that it's 1-norm is EXACTLY 1?
Assuming you want identical values to be normalised with the same factor, this is not possible. Simple counter example:
v=ones(21,1);
v = v / sum(v);
sum(v)-1
One common way to deal with it, is enforce values sum(v)>=1 or sum(v)<=1 if your algorithm can deal with a derivation to one side:
if sum(v)>1
v=v-eps(v)
end
Alternatively you can try using vpa, but this will drastically increase your computation time.
I am performing logistic regression in MATLAB with L2 regularization on text data. My program works well for small datasets. For larger sets, it keeps running infinitely.
I have seen the potentially duplicate question (matlab fminunc not quitting (running indefinitely)). In that question, the cost for initial theta was NaN and there was an error printed in the console. For my implementation, I am getting a real valued cost and there is no error even with verbose parameters being passed to fminunc(). Hence I believe this question might not be a duplicate.
I need help in scaling it to larger sets. The size of the training data I am currently working on is roughly 10k*12k (10k text files cumulatively containing 12k words). Thus, I have m=10k training examples and n=12k features.
My cost function is defined as follows:
function [J gradient] = costFunction(X, y, lambda, theta)
[m n] = size(X);
g = inline('1.0 ./ (1.0 + exp(-z))');
h = g(X*theta);
J =(1/m)*sum(-y.*log(h) - (1-y).*log(1-h))+ (lambda/(2*m))*norm(theta(2:end))^2;
gradient(1) = (1/m)*sum((h-y) .* X(:,1));
for i = 2:n
gradient(i) = (1/m)*sum((h-y) .* X(:,i)) - (lambda/m)*theta(i);
end
end
I am performing optimization using MATLAB's fminunc() function. The parameters I pass to fminunc() are:
options = optimset('LargeScale', 'on', 'GradObj', 'on', 'MaxIter', MAX_ITR);
theta0 = zeros(n, 1);
[optTheta, functionVal, exitFlag] = fminunc(#(t) costFunction(X, y, lambda, t), theta0, options);
I am running this code on a machine with these specifications:
Macbook Pro i7 2.8GHz / 8GB RAM / MATLAB R2011b
The cost function seems to behave correctly. For initial theta, I get acceptable values of J and gradient.
K>> theta0 = zeros(n, 1);
K>> [j g] = costFunction(X, y, lambda, theta0);
K>> j
j =
0.6931
K>> max(g)
ans =
0.4082
K>> min(g)
ans =
-2.7021e-05
The program takes incredibly long to run. I started profiling keeping MAX_ITR = 1 for fminunc(). With a single iteration, the program did not complete execution even after a couple of hours had elapsed. My questions are:
Am I doing something wrong mathematically?
Should I use any other optimizer instead of fminunc()? With LargeScale=on, fminunc() uses trust-region algorithms.
Is this problem cluster-scale and should not be run on a single machine?
Any other general tips will be appreciated. Thanks!
This helped solve the problem: I was able to get this working by setting the LargeScale flag to 'off' in fminunc(). From what I gather, LargeScale = 'on' uses trust region algorithms, while keeping it 'off' uses quasi-newton methods. Using quasi-newton methods and passing the gradient worked a lot faster for this particular problem and gave very nice results.
I was able to get this working by setting the LargeScale flag to 'off' in fminunc(). From what I gather, LargeScale = 'on' uses trust region algorithms, while keeping it 'off' uses quasi-newton methods. Using quasi-newton methods and passing the gradient worked a lot faster for this particular problem and gave very nice results.
Here is my advise:
-Set the Matlab flag to show debug output during run. If not just print out in your cost function the cost, which will allow you to monitor iteration count and error.
And second, which is very important:
Your problem is illposed, or so to say underdetermined. You have a 12k feature space and provide only 10k examples, which means for an unconstrained optimization the answer is -Inf. To make a quick example why this is, your problem is like:
Minimize x+y+z given that x+y-z = 2. Feature space dim 3, spanned vector space - 1d. I suggest use PCA or CCA to reduce the dimensionality of the the text files by retaining their variation up to 99%. This will probably give you a feature space ~100-200dim.
PS: Just to point out that the problem is very fram from cluster size requirement, which usually is 1kk+ data points and that fminunc is not at all an overkill, and LIBSVM has nothing to do with it because fminunc is just a quadratic optimizer, while LIBSVM is a classifier. To clear out LIBSVM uses something similar to fminunc just with different objective function.
Here's what I suspect to be the issue, based on my experience with this type of problem. You're using a dense representation for X instead of a sparse one. You're also seeing the typical effect in text classification that the number of terms increasing roughly linearly with the number of samples. Effectively, the cost of the matrix multiplication X*theta goes up quadratically with the number of samples.
By contrast, a good sparse matrix representation only iterates over the non-zero elements to do a matrix multiplication, which tends to be roughly constant per document if they're of appropriately constant length, causing linear instead of quadratic slowdown in the number of samples.
I'm not a Matlab guru, but I know it has a sparse matrix package, so try to use that.
This is my first post to stackoverflow, so if this isn't the correct area I apologize. I am working on minimizing a L1-Regularized System.
This weekend is my first dive into optimization, I have a basic linear system Y = X*B, X is an n-by-p matrix, B is a p-by-1 vector of model coefficients and Y is a n-by-1 output vector.
I am trying to find the model coefficients, I have implemented both gradient descent and coordinate descent algorithms to minimize the L1 Regularized system. To find my step size I am using the backtracking algorithm, I terminate the algorithm by looking at the norm-2 of the gradient and terminating if it is 'close enough' to zero(for now I'm using 0.001).
The function I am trying to minimize is the following (0.5)*(norm((Y - X*B),2)^2) + lambda*norm(B,1). (Note: By norm(Y,2) I mean the norm-2 value of the vector Y) My X matrix is 150-by-5 and is not sparse.
If I set the regularization parameter lambda to zero I should converge on the least squares solution, I can verify that both my algorithms do this pretty well and fairly quickly.
If I start to increase lambda my model coefficients all tend towards zero, this is what I expect, my algorithms never terminate though because the norm-2 of the gradient is always positive number. For example, a lambda of 1000 will give me coefficients in the 10^(-19) range but the norm2 of my gradient is ~1.5, this is after several thousand iterations, While my gradient values all converge to something in the 0 to 1 range, my step size becomes extremely small (10^(-37) range). If I let the algorithm run for longer the situation does not improve, it appears to have gotten stuck somehow.
Both my gradient and coordinate descent algorithms converge on the same point and give the same norm2(gradient) number for the termination condition. They also work quite well with lambda of 0. If I use a very small lambda(say 0.001) I get convergence, a lambda of 0.1 looks like it would converge if I ran it for an hour or two, a lambda any greater and the convergence rate is so small it's useless.
I had a few questions that I think might relate to the problem?
In calculating the gradient I am using a finite difference method (f(x+h) - f(x-h))/(2h)) with an h of 10^(-5). Any thoughts on this value of h?
Another thought was that at these very tiny steps it is traveling back and forth in a direction nearly orthogonal to the minimum, making the convergence rate so slow it is useless.
My last thought was that perhaps I should be using a different termination method, perhaps looking at the rate of convergence, if the convergence rate is extremely slow then terminate. Is this a common termination method?
The 1-norm isn't differentiable. This will cause fundamental problems with a lot of things, notably the termination test you chose; the gradient will change drastically around your minimum and fail to exist on a set of measure zero.
The termination test you really want will be along the lines of "there is a very short vector in the subgradient."
It is fairly easy to find the shortest vector in the subgradient of ||Ax-b||_2^2 + lambda ||x||_1. Choose, wisely, a tolerance eps and do the following steps:
Compute v = grad(||Ax-b||_2^2).
If x[i] < -eps, then subtract lambda from v[i]. If x[i] > eps, then add lambda to v[i]. If -eps <= x[i] <= eps, then add the number in [-lambda, lambda] to v[i] that minimises v[i].
You can do your termination test here, treating v as the gradient. I'd also recommend using v for the gradient when choosing where your next iterate should be.