similar to: Quadratic Constraints

I was playing around with a simple example using solve.qp ( function is in the quadprog package ) and the code is below. ( I'm not even sure there if there is a reasonable solution because I made the problem up ). But, when I try to use solve.QP to solve it, I get the error that D in the quadratic function is not positive definite. This is because Dmat is zero because I don't have a

Hi R, A quick question.... How can I optimize the objective function constrained to quadratic constraints? Which function of R is useful for quadratic constraints? Many Thanks, Shubha This e-mail may contain confidential and/or privileged i...{{dropped:13}}

Hi Bert, I won't post any more messages on this thread as problem has shifted from Optimization in R to Graph Algorithms. Rest fine Khris. On Aug 2, 2012, at 9:13 PM, Bert Gunter [via R] wrote: > This discussion needs to be taken off (this) list, as it appears to > have nothing to do with R. > > -- Bert > > On Thu, Aug 2, 2012 at 2:27 AM, khris <[hidden email]>

Hi, I need to maximize a quadratic function under constraints in R. For minimization I used solve.QP but for maximization it is not useful since the matrix D of the quadratic function should be positive definite hence I cannot simply change the sign. any suggestion ? thanks -- View this message in context:

Hi R experts, Could you please help me to fit a linear, cubic and quadratic curve in a figure? I was trying to show all these three fitting curves with different colour in one figure. I spent substantial time to figure it out, but I could not. I have given here a example and what I did for linear, but no idea for cubic and quadratic fitting curve > dput(test) structure(list(sp = c(4L, 5L,

-----Original Message----- Message: 6 Date: Mon, 20 Jan 2003 01:11:24 +0100 From: Martin Michlmayr <tbm at cyrius.com> To: r-help at stat.math.ethz.ch Subject: [R] quadratic trends and changes in slopes I'd like to use linear and quadratic trend analysis in order to find out a change in slope. Basically, I need to solve a similar problem as discussed in

Dear list users How is it possible to visualise both a linear trend line and a quadratic trend line on a plot of two variables? Here my almost working exsample. data(Duncan) attach(Duncan) plot(prestige ~ income) abline(lm(prestige ~ income), col=2, lwd=2) Now I would like to add yet another trend line, but this time a quadratic one. So I have two trend lines. One linear trend line

Hello, I have a system of quadratic equations (results of a Hamiltonian optimization) which I need to find the roots for. Is there a package and/or function which will find the roots for a quadratic system? Note that I am not opimizing, but rather solving the first order conditions which come from a Hamiltonian. I am basically looking for something in R that will do the same thing as fsolve in

Good day for all, I'm a beginner aRgonaut, thus I'm having a problem to plot a quadratic model of regression in a plot. First I wrote: >plot(Y~X) and then I tried: >abline(lm(Y~X+I(X^2)) but "abline" only uses the first two of three regression coefficients, thus I tried: >line(lm(Y~X+I(X^2)) but a message error is showed ("insufficient observations").

Hello. I'm trying to solve a quadratic programming problem of the form min ||Hx - y||^2 s.t. x >= 0 and x <= t using solve.QP in the quadprog package but I'm having problems with Dmat not being positive definite, which is kinda okay since I expect it to be numerically semi-definite in most cases. As far as I'm aware the problem arises because the Goldfarb and Idnani method first

How to define bounds for a semi continous variable in lp_solve. Min 5x1 +9x2 +7.15x3 +0.1x4 subject to x1+x2+x3+x4=6.7 x1+x4 <= 6.5 And x3 can be 0 or greater than 3.6 hence x3 is a semi continous variable how to define bounds as well as semicontinous function because using set.semicont and set. bound simantaneously doesn't seem to work.Thanks in advance for the help -- View this

Below. -- Bert Bert Gunter On Thu, Jul 13, 2017 at 3:07 AM, Luigi Biagini <luigi.biagini at gmail.com> wrote: > I have two ideas about it. > > 1- > i) Entering variables in quadratic form is done with the command I > (variable ^ 2) - > plsr (octane ~ NIR + I (nir ^ 2), ncomp = 10, data = gasTrain, validation = > "LOO" > You could also use a new variable

Hello, I want to know if two quadratic regressions are significantly different. I was advised to make the test using step 1 bootstrapping both quadratic regressions and get their slope coefficients. (Let's call the slope coefficient *â*^1 and *â*^2) step 2 use the slope difference *â*^1-*â*^2 and bootstrap the slope coefficent step 3 find out the sampling distribution above and

Hello, I have a linear problem with quadratic constraints but i have no idea about what package to use... any help? Thanks in advance

Hi, Can I use "constrOptim" for quadratic constraints? If not, which optimizer can I go for? BR, Shubha This e-mail may contain confidential and/or privileged i...{{dropped:13}}

I have an equation describing the best-fit model for a set of points (just 2 axes) that is in the form: y=b+mx+px^2 Where b is the intercept, m is the slope describing a linear term, and p is a slope of the quadratic term. I would like to plot this equation on a curve (I know the equation is y=(.1766x^2)+(.171x)+.101) on the original scatterplot. Any easy way to plot this equation and

Hello , I want to compare two quadratic regression models with non-parametric bootstrap. However, I do not know which R package can serve the purpose, such as boot, rms, or bootstrap, DeltaR. Please kindly advise and thank you. Elaine The two quadratic regression models are y1=a1x^2+b1x+c1 y1= observed migration distance of butterflies() y2=a2x^2+b2x+c2 y2= predicted migration distance of

Hi, I need to solve an optimization problem in R having linear objective function and quadratic constraints(number of variables is around 80). What are the possible choices to do this in R. optim() function only allows box constrained problems. Is it possible in nlm()? Or please tell me if there is any other routine. Thanks Amit

I have two ideas about it. 1- i) Entering variables in quadratic form is done with the command I (variable ^ 2) - plsr (octane ~ NIR + I (nir ^ 2), ncomp = 10, data = gasTrain, validation = "LOO" You could also use a new variable NIR_sq <- (NIR) ^ 2 ii) To insert a square variable, use syntax I (x ^ 2) - it is very important to insert I before the parentheses. iii) If you want to

?? If I haven't misunderstood, they are completely different! 1) NIR must be a matrix, or poly(NIR,...) will fail. 2) Due to the previously identified bug in poly, degree must be explicitly given as poly(NIR, degree =2,raw = TRUE). Now consider the following example: > df <-matrix(runif(60),ncol=3) > y <- runif(20) > mdl1 <-lm(y~df*I(df^2)) > mdl2