Abby Spurdle @purd|e@@ @end|ng |rom gm@||@com
Mon Sep 21 12:09:16 CEST 2020

Sorry, ignore the last part.
What I should have said, is the inequality has the opposite sign.
>= bvec (not <= bvec)

On Mon, Sep 21, 2020 at 10:05 PM Abby Spurdle <spurdle.a using gmail.com> wrote:
>
> Are you using the quadprog package?
> If I can take a random shot in the dark, should bvec be -bvec?
>
>
> On Mon, Sep 21, 2020 at 9:28 PM Maija Sirkjärvi
> <maija.sirkjarvi using gmail.com> wrote:
> >
> > Hi!
> >
> > I was wondering if someone could help me out. I'm minimizing a following
> > function:
> >
> >
> > $$\sum_{j=1}^{J}(m_{j} -\hat{m_{j}})^2,$$
> > \text{subject to}
> > $$m_{j-1}\leq m_{j}-\delta_{1}$$
> > $$\frac{1}{Q_{j-1}-Q_{j-2}} (m_{j-2}-m_{j-1}) \leq \frac{1}{Q_{j}-Q_{j-1}} > > (m_{j-1}-m_{j})-\delta_{2}$$
> >
> >
> > I have tried quadratic programming, but something is off. Does anyone have
> > an idea how to approach this?
> >
> >
> > Q <- rep(0,J)
> > for(j in 1:(length(Price))){
> >   Q[j] <- exp((-0.1) * (Beta *Price[j]^(Eta + 1) - 1) / (1 + Eta))
> > }
> >
> > Dmat <- matrix(0,nrow= J, ncol=J)
> > diag(Dmat) <- 1
> > dvec <- -hs
> > Aeq <- 0
> > beq <- 0
> > Amat <- matrix(0,J,2*J-3)
> > bvec <- matrix(0,2*J-3,1)
> >
> > for(j in 2:nrow(Amat)){
> >   Amat[j-1,j-1] = -1
> >   Amat[j,j-1] = 1
> > }
> > for(j in 3:nrow(Amat)){
> >   Amat[j,J+j-3] = -1/(Q[j]-Q[j-1])
> >   Amat[j-1,J+j-3] = 1/(Q[j]-Q[j-1])
> >   Amat[j-2,J+j-3] = -1/(Q[j-1]-Q[j-2])
> > }
> > for(j in 2:ncol(bvec)) {
> >   bvec[j-1] = Delta1
> > }
> > for(j in 3:ncol(bvec)) {
> >   bvec[J-1+j-2] = Delta2
> > }
> > solution <- solve.QP(Dmat,dvec,Amat,bvec=bvec)
> >
> >         [[alternative HTML version deleted]]
> >
> > ______________________________________________
> > R-help using r-project.org mailing list -- To UNSUBSCRIBE and more, see
> > https://stat.ethz.ch/mailman/listinfo/r-help