similar to: How to plot a Quadratic Model?

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,

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

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

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 All, I am unable to solve a optimization Problem Please Help Me out of this to solve. The Optimization problem is as follows :- My objective function is linear and one of the constraint is quadratic. Min z = 5 * X1 + 9* X2 + 7.15 *X3 + 2 * X4 subject to X1 + X2 + X3 +X4 = 9 X1 + X4 < = 6.55 X3(X3 - 3.5) >=0 X1,X2,X3,X4 >=0 Now the problem is how to solve this kind of

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

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

-----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

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

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, 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

Hello, I have been trying to take the derivative of a quadratic B-spline obtained by using the COBS library. What I would like to do is similar to what one can do by using fit<-smooth.spline(cdf) xx<-seq(-10,10,.1) predict(fit, xx, deriv = 1) The goal is to fit the spline to data that is approximating a cumulative distribution function (e.g. in my example, cdf is a 2-column matrix with x

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

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

Hi, I am using solve.QP (from quadprog) to solve a standard quadratic programming problem: min_w -0.5*w'Qw st ... I would like solve.QP to do two things: 1) to start the optimization from a user-supplied initial condition; i.e., from a vector w_0 that satisfies the constraints, and 2) to return the values of the lagrange multiplieres associated with the constraints. I did not find an obvious

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

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

Dear Listserv, Here is my latest in a series of simple-seeming questions that dog me. Consider the following data: x <- read.table(textConnection("temperature probability 0.11 9.4 0 2.3 0.38 8.7 0.43 9.2 0.6 15.6 0.47 8.7 0.09 12.8 0.11 9.4 0.01 7.7 0.83 8 0.65 9.3 0.05 7.4 0.34 10.1 0.02 4.8 0.07 9.1 0.6 15.6 0.01 8.4 0.9 9.6 0.83 8 0.12 8.4 0.01 8 0 5 0.11 9.7 0.41 7.4 0.05 9.4 0.09

I hope that someone can help me with the following question: I would like to solve the Markowitz optimization problem WITH short-sale constraints. Maybe a procedure to solve a quadratic optimization problem with convex constraints and positive variables is already implemented in R? Thank you very much, edg -.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.- r-help

On page 22 of the R-introduction guide it's written: the quadratic form x^{'} A^{-1} x which is used in multivariate computations, should be computed by something like x%*%solve(A,x), rather than computing the inverse of A. Why isn't it good to compute t(x) %*% solve(A) %*% x? Thanks a lot for help!