search for: inequalities

Dear R-users, in the last days I have been trying to estimate a normal linear model with equality and inequality constraints. Please find below a simple example of my problem. Of course, one could easily see that, though the constraints are consistent, there is some redundancy in the specific constraints. Nevertheless my actual applications can get much larger and I would not like to manually

Dear All, I would like to minimise a nonlinear function subject to linear inequality constraints as part of an R program. I have been using the constrOptim function. I have tried all of the methods that come with Optim, but nothing finds the correct solution. If I use the correct solution as the vector of starting values, though, my program does output the correct solution and optimum - the

Dear All, Which package/function could i use to solve following linear least square problem? A over determined system of linear equations is given. The nnls-function may would be a possibility BUT: The solving is constrained with a inequality that all unknowns are >= 0 and a equality that the sum of all unknowns is 1 The influence of the equations according to the solving process is

Dear R Users, I would like to make a lesast squares regression similar to that what is done by the command "lm". But additionally, I would like to impose some restrictions: 1) The sum of all regression coefficients should be equal to 1. 2) Each coefficient should assume a value between 0 and 1. (inequality restrictions) Which command is the best to use in order to solve this problem

Hello, I was interested in trying to write an R script to calculate a UCL for a lognormal distribution using the Chebyshev Inequality — MVUE Approach (based on EPA’s guidance found in http://www.epa.gov/oswer/riskassessment/pdf/ucl.pdf). This looks like it should be straight forward, but I am need to calculate an MVUE for the population mean and an MVUE for the population variance, which requires

Hello, I'm fairly new to R and I can't work out how to produce a double inequality like (LaTeX) $0 \leq x \leq 1$ in the legend of a graph. If I try > legend(50, 0.1, legend = c(expression(0 <= x <= 1), c(2 <= x <= 3)), pch = c(1,1), col = c(2, 3)) then I get an error message "unexpected '<=' in ...". I've checked the help files for plotmath and

...") print(1/(1+exp(-predict(model)))) The last expression should be a nondecreasing sequence, as fitted probabilities make no sense to decrease in this case. Is there any elegant way to constrain the model fit? Right now, I am trying to do this by defining a "large" set of linear inequalities and using constrOptim() on the log-likelihood. However, this seems to be very sensitive to the starting values, and I am not really happy with the results. BTW: yes, working with lrm() would probably be more natural, but lrm() breaks down on my data because of a rank-deficient information matri...

Dear R users, I?m trying to optimize simultaneously two binomials inequalities used to acceptance sampling, which are nonlinear solution, so there is no simple direct solution. Please, let me explain shortly the the problem and the question as following. The objective is to obtain the smallest value of 'n' (sample size) satisfying both inequalities: (1-alpha) <=...

Hi,, In src/lib/mountpoint.c:53 the following had to be commented out: // point_r->type = p_strdup(pool, buf.f_fstypename); because f_fstypename is undeclared on this OS. Also the following errors occur with the DEC C compiler (which compiles the other files just fine): source='file-cache.c' object='file-cache.o' libtool=no DEPDIR=.deps depmode=none

Dear R users, I?m trying to optimize simultaneously two binomials inequalities (used in acceptance sampling) which are nonlinear solution, so there is no simple direct solution. Please, let me explain shortly the the problem and the question as following. The objective is to obtain the smallest value of 'n' (sample size) satisfying both inequalities: (1-alpha) <...

Dear everybody! take a<-c(5,3,NA,6). if(a[1]!=NA){b<-7} if(a[3]!=5){b<-7} if(a[3]!=NA){b<-7} if(a[3]==NA){b<-7} will alltogeather return Fehler in if (a[1] != NA) { : Fehlender Wert, wo TRUE/FALSE n?tig ist (or simularly). Somehow this is logical. But how else should I get out, whether a certain vector-component has an existing value? Thank you in advance! Yours, Mag. Ferri

Hello, I watched your YouTube videos for explanation for R package. They are really helpful for new beginner for using R. Please I need your help for solving inequalities that contain beta functions using R studio. I attached the file that contains this inequalities. I appreciate your help and looking forward to hear from you. Akram -------------- next part -------------- A non-text attachment was scrubbed... Name: Inequalities.pdf Type: application/pdf Size: 2892...

Hi everyone, i?m a new user of R and i?m trying to translate an linear optimization problem from Matlab into r. The matlab code is as follow: options = optimset('Diagnostics','on'); [x fval exitflag] = linprog(f,A,b,Aeq,beq,lb,ub,[],options); exitflag fval x=round(x); Where: f = Linear objective function vector (vector of 45,rows) A = Matrix for linear inequality

Hi folks, Pls help me to understand follow; An Introduction to R 2.4 Logical vectors http://cran.r-project.org/doc/manuals/R-intro.html#R-and-statistics 1) > x [1] 1 2 3 4 5 > temp <- x != 1 > temp [1] FALSE TRUE TRUE TRUE TRUE > 2) > x [1] 1 2 3 4 5 > temp <- x > 1 > temp [1] FALSE TRUE TRUE TRUE TRUE Why NOT > temp [1] TRUE FALSE FALSE FALSE

Hi folks, Pls help me to understand follow; An Introduction to R 2.4 Logical vectors http://cran.r-project.org/doc/manuals/R-intro.html#R-and-statistics 1) > x [1] 1 2 3 4 5 > temp <- x != 1 > temp [1] FALSE TRUE TRUE TRUE TRUE > 2) > x [1] 1 2 3 4 5 > temp <- x > 1 > temp [1] FALSE TRUE TRUE TRUE TRUE Why NOT > temp [1] TRUE FALSE FALSE FALSE

I have to estimate the following model for several group of observations : y(1-y) = p[1]*(x^2-y) + p[2]*y*(x-1) + p[3]*(x-y) with constraints : p[1]+p[3] >= 1 p[1]+p[2]+p[3]+1 >= 0 p[3] >= 0 I use the following code : func <- sum((y(1-y) - p[1]*(x^2-y) + p[2]*y*(x-1) + p[3]*(x-y))^2) estim <- optim( c(1,0,0),func, method="L-BFGS-B" , lower=c(1-p[3], -p[1]-p[3]-1,

Hello, we want to identify breakpoints (different phases) in environmental data, algae cell counts of three years with intervals between 7 and 30 days (N=40). We found that breakpoints(cells ~1) works great and identifies 5 very good breaks, however we are uncertain about these, because the data are unequally spaced. Is there a way to include the information about the measurement intervals,

I have a relatively simple finance application of GMM. Given the moment condition: E[m*R]=0 where m=m[theta] I would like to constrain m>0. Any ideas? [[alternative HTML version deleted]]

...on(x){ all2=sum(mat%*%x) sum_pterm=sum(mat[1:N,]%*%x/all2) sum_x=sum(x) return(c(sum_pterm,sum_x)) } # the right hand side for the constraint eqB=c(1,budget) # the inequality function inequal1=function(x){ all3=sum(mat%*%x) p_b=mat[1:BR,]%*%x/all3 return(as.vector(p_b)) } #the lower limit for inequalities ineqLB=rep(0.00001,BR) #the upper limit for the inequalities ineqUB=rep(1.00001,BR) # the lower and upper bounds for the variables LB=rep(0,CELL) UB=rep(1,CELL) #the seed for variables cells0=rep(.3,9) #solving the problem sol=solnp(cells0,fun=obj,eqfun=equal1,eqB=eqB,ineqfun =inequal1,ineqL...

I have a spatial salinity field s and a model g(s) ~ Xb where the X comes from slightly modified GAM basis functions. I am trying to deal with the following set of requirements: 1. The underlying physics are linear, and plain salinity (the identity link) is the correct response to my covariates. 2. Dispersion (variance or sd) is almost certainly proportional to the mean. 3. The data s(x,y)