 Classic List Threaded 11 messages Open this post in threaded view
|

 Hi! I was wondering if someone could help me out. I'm minimizing a following function: \begin{equation} $$\sum_{j=1}^{J}(m_{j} -\hat{m_{j}})^2,$$ \text{subject to} $$m_{j-1}\leq m_{j}-\delta_{1}$$ $$\frac{1}{Q_{j-1}-Q_{j-2}} (m_{j-2}-m_{j-1}) \leq \frac{1}{Q_{j}-Q_{j-1}} (m_{j-1}-m_{j})-\delta_{2}$$ \end{equation} I have tried quadratic programming, but something is off. Does anyone have an idea how to approach this? Thanks in advance! 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]] ______________________________________________ [hidden email] mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-helpPLEASE do read the posting guide http://www.R-project.org/posting-guide.htmland provide commented, minimal, self-contained, reproducible code.
Open this post in threaded view
|

 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 <[hidden email]> wrote: > > Hi! > > I was wondering if someone could help me out. I'm minimizing a following > function: > > \begin{equation} > $$\sum_{j=1}^{J}(m_{j} -\hat{m_{j}})^2,$$ > \text{subject to} > $$m_{j-1}\leq m_{j}-\delta_{1}$$ > $$\frac{1}{Q_{j-1}-Q_{j-2}} (m_{j-2}-m_{j-1}) \leq \frac{1}{Q_{j}-Q_{j-1}} > (m_{j-1}-m_{j})-\delta_{2}$$ > \end{equation} > > I have tried quadratic programming, but something is off. Does anyone have > an idea how to approach this? > > Thanks in advance! > > 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]] > > ______________________________________________ > [hidden email] mailing list -- To UNSUBSCRIBE and more, see > https://stat.ethz.ch/mailman/listinfo/r-help> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html> and provide commented, minimal, self-contained, reproducible code. ______________________________________________ [hidden email] mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-helpPLEASE do read the posting guide http://www.R-project.org/posting-guide.htmland provide commented, minimal, self-contained, reproducible code.
Open this post in threaded view
|

 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 <[hidden email]> 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 > <[hidden email]> wrote: > > > > Hi! > > > > I was wondering if someone could help me out. I'm minimizing a following > > function: > > > > \begin{equation} > > $$\sum_{j=1}^{J}(m_{j} -\hat{m_{j}})^2,$$ > > \text{subject to} > > $$m_{j-1}\leq m_{j}-\delta_{1}$$ > > $$\frac{1}{Q_{j-1}-Q_{j-2}} (m_{j-2}-m_{j-1}) \leq \frac{1}{Q_{j}-Q_{j-1}} > > (m_{j-1}-m_{j})-\delta_{2}$$ > > \end{equation} > > > > I have tried quadratic programming, but something is off. Does anyone have > > an idea how to approach this? > > > > Thanks in advance! > > > > 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]] > > > > ______________________________________________ > > [hidden email] mailing list -- To UNSUBSCRIBE and more, see > > https://stat.ethz.ch/mailman/listinfo/r-help> > PLEASE do read the posting guide http://www.R-project.org/posting-guide.html> > and provide commented, minimal, self-contained, reproducible code. ______________________________________________ [hidden email] mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-helpPLEASE do read the posting guide http://www.R-project.org/posting-guide.htmland provide commented, minimal, self-contained, reproducible code.
Open this post in threaded view
|

 One more thing, is bvec supposed to be a matrix? Note you may need to provide a reproducible example, for better help... On Mon, Sep 21, 2020 at 10:09 PM Abby Spurdle <[hidden email]> wrote: > > 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 <[hidden email]> 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 > > <[hidden email]> wrote: > > > > > > Hi! > > > > > > I was wondering if someone could help me out. I'm minimizing a following > > > function: > > > > > > \begin{equation} > > > $$\sum_{j=1}^{J}(m_{j} -\hat{m_{j}})^2,$$ > > > \text{subject to} > > > $$m_{j-1}\leq m_{j}-\delta_{1}$$ > > > $$\frac{1}{Q_{j-1}-Q_{j-2}} (m_{j-2}-m_{j-1}) \leq \frac{1}{Q_{j}-Q_{j-1}} > > > (m_{j-1}-m_{j})-\delta_{2}$$ > > > \end{equation} > > > > > > I have tried quadratic programming, but something is off. Does anyone have > > > an idea how to approach this? > > > > > > Thanks in advance! > > > > > > 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]] > > > > > > ______________________________________________ > > > [hidden email] mailing list -- To UNSUBSCRIBE and more, see > > > https://stat.ethz.ch/mailman/listinfo/r-help> > > PLEASE do read the posting guide http://www.R-project.org/posting-guide.html> > > and provide commented, minimal, self-contained, reproducible code. ______________________________________________ [hidden email] mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-helpPLEASE do read the posting guide http://www.R-project.org/posting-guide.htmland provide commented, minimal, self-contained, reproducible code.
Open this post in threaded view
|

 Thank you for your response! Bvec is supposed to be a matxit. I'm following the solve.QP ( https://www.rdocumentation.org/packages/quadprog/versions/1.5-8/topics/solve.QP). I'm not sure what would be the best way to solve a quadratic programming problem in R. ma 21. syysk. 2020 klo 13.20 Abby Spurdle ([hidden email]) kirjoitti: > One more thing, is bvec supposed to be a matrix? > > Note you may need to provide a reproducible example, for better help... > > On Mon, Sep 21, 2020 at 10:09 PM Abby Spurdle <[hidden email]> wrote: > > > > 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 <[hidden email]> > 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 > > > <[hidden email]> wrote: > > > > > > > > Hi! > > > > > > > > I was wondering if someone could help me out. I'm minimizing a > following > > > > function: > > > > > > > > \begin{equation} > > > > $$\sum_{j=1}^{J}(m_{j} -\hat{m_{j}})^2,$$ > > > > \text{subject to} > > > > $$m_{j-1}\leq m_{j}-\delta_{1}$$ > > > > $$\frac{1}{Q_{j-1}-Q_{j-2}} (m_{j-2}-m_{j-1}) \leq > \frac{1}{Q_{j}-Q_{j-1}} > > > > (m_{j-1}-m_{j})-\delta_{2}$$ > > > > \end{equation} > > > > > > > > I have tried quadratic programming, but something is off. Does > anyone have > > > > an idea how to approach this? > > > > > > > > Thanks in advance! > > > > > > > > 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]] > > > > > > > > ______________________________________________ > > > > [hidden email] mailing list -- To UNSUBSCRIBE and more, see > > > > https://stat.ethz.ch/mailman/listinfo/r-help> > > > PLEASE do read the posting guide > http://www.R-project.org/posting-guide.html> > > > and provide commented, minimal, self-contained, reproducible code. >         [[alternative HTML version deleted]] ______________________________________________ [hidden email] mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-helpPLEASE do read the posting guide http://www.R-project.org/posting-guide.htmland provide commented, minimal, self-contained, reproducible code.
Open this post in threaded view
|

 In reply to this post by Maija Sirkjärvi Hi, Sorry, for my rushed responses, last night. (Shouldn't post when I'm about to log out). I haven't used the quadprog package for nearly a decade. And I was hoping that an expert using optimization in finance in economics would reply. Some comments: (1) I don't know why you think bvec should be a matrix. The documentation clearly says it should be a vector (implying not a matrix). The only arguments that should be matrices are Dmat and Amat. (2) I'm having some difficulty following your quadratic program, even after rendering it. Perhaps you could rewrite your expressions, in a form that is consistent with the input to solve.QP. That's a math problem, not an R programming problem, as such. (3) If that fails, then you'll need to produce a minimal reproducible example. I strongly recommend that the R code matches the quadratic program, as closely as possible. On Mon, Sep 21, 2020 at 9:28 PM Maija Sirkjärvi <[hidden email]> wrote: > > Hi! > > I was wondering if someone could help me out. I'm minimizing a following > function: > > \begin{equation} > $$\sum_{j=1}^{J}(m_{j} -\hat{m_{j}})^2,$$ > \text{subject to} > $$m_{j-1}\leq m_{j}-\delta_{1}$$ > $$\frac{1}{Q_{j-1}-Q_{j-2}} (m_{j-2}-m_{j-1}) \leq \frac{1}{Q_{j}-Q_{j-1}} > (m_{j-1}-m_{j})-\delta_{2}$$ > \end{equation} > > I have tried quadratic programming, but something is off. Does anyone have > an idea how to approach this? > > Thanks in advance! > > 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]] > > ______________________________________________ > [hidden email] mailing list -- To UNSUBSCRIBE and more, see > https://stat.ethz.ch/mailman/listinfo/r-help> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html> and provide commented, minimal, self-contained, reproducible code. ______________________________________________ [hidden email] mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-helpPLEASE do read the posting guide http://www.R-project.org/posting-guide.htmland provide commented, minimal, self-contained, reproducible code.
Open this post in threaded view
|

 I was wondering if you're trying to fit a curve, subject to monotonicity/convexity constraints... If you are, this is a challenging topic, best of luck... On Tue, Sep 22, 2020 at 8:12 AM Abby Spurdle <[hidden email]> wrote: > > Hi, > > Sorry, for my rushed responses, last night. > (Shouldn't post when I'm about to log out). > > I haven't used the quadprog package for nearly a decade. > And I was hoping that an expert using optimization in finance in > economics would reply. > > Some comments: > (1) I don't know why you think bvec should be a matrix. The > documentation clearly says it should be a vector (implying not a > matrix). > The only arguments that should be matrices are Dmat and Amat. > (2) I'm having some difficulty following your quadratic program, even > after rendering it. > Perhaps you could rewrite your expressions, in a form that is > consistent with the input to solve.QP. That's a math problem, not an R > programming problem, as such. > (3) If that fails, then you'll need to produce a minimal reproducible example. > I strongly recommend that the R code matches the quadratic program, as > closely as possible. > > > On Mon, Sep 21, 2020 at 9:28 PM Maija Sirkjärvi > <[hidden email]> wrote: > > > > Hi! > > > > I was wondering if someone could help me out. I'm minimizing a following > > function: > > > > \begin{equation} > > $$\sum_{j=1}^{J}(m_{j} -\hat{m_{j}})^2,$$ > > \text{subject to} > > $$m_{j-1}\leq m_{j}-\delta_{1}$$ > > $$\frac{1}{Q_{j-1}-Q_{j-2}} (m_{j-2}-m_{j-1}) \leq \frac{1}{Q_{j}-Q_{j-1}} > > (m_{j-1}-m_{j})-\delta_{2}$$ > > \end{equation} > > > > I have tried quadratic programming, but something is off. Does anyone have > > an idea how to approach this? > > > > Thanks in advance! > > > > 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]] > > > > ______________________________________________ > > [hidden email] mailing list -- To UNSUBSCRIBE and more, see > > https://stat.ethz.ch/mailman/listinfo/r-help> > PLEASE do read the posting guide http://www.R-project.org/posting-guide.html> > and provide commented, minimal, self-contained, reproducible code. ______________________________________________ [hidden email] mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-helpPLEASE do read the posting guide http://www.R-project.org/posting-guide.htmland provide commented, minimal, self-contained, reproducible code.
Open this post in threaded view
|

 I really appreciate you helping me with this! I just don't seem to figure it out. (1) I don't know why you think bvec should be a matrix. The documentation clearly says it should be a vector (implying not a matrix). - I've written it in a form of a matrix with one row and 2*J-3 columns. (0,2*J-3,1). I thought that it would pass as a vector. I made it a vector now. The thing is that I'm trying to replicate a C++ code with R. The C++ code imposes shape restrictions on the function and works perfectly. The C++ code is attached and after that the same for R. You said you haven't used the QP package for a decade. Is there a better/another package for these types of problems? Thanks again! C++ code: int main() { Print("Begin"); /* Bootstrap Parameters */ long RandomSeed = -98345; // Set random seed const int S = 100; // Number of bootstrap replications /* Housing Demand Parameters */ const double Beta =  1.1613; const double v    =  0.7837; const double Eta  = -0.5140; /* Quadratic Programming Tolerance Parameters */ const double Delta1 =  0.000001; const double Delta2 =  0.0001; /* Read in Dataset */ DataFileDelim d("K:\\Data\\Local Jurisdictions\\AEJ Data.dat",'\t'); d.SortBy("AfterTaxPrice"); int J = d.NumObs(); Vector Price = d.Get("AfterTaxPrice"); Vector Educ  = d.Get("EducAll"); Vector Crime = d.Get("CrimeIndex"); Vector Dist  = d.Get("RushTrav"); /* Adjust Crime Data */ for(int j=0; j Rank1(J); for(int j=0; j X(J); Matrix Y(J,2); Vector Z(J); for(int j=0; j RhoGrid1 = Grid(-1.2,-0.1,RhoK1); Matrix gTrans(J,RhoK1); for(int rk=0; rk hSmooth(J); for(int j=0; j Q(J); for(int j=0; j H(J,Zero); Vector c(J,Zero); Matrix Aeq(0,J); Vector beq(0); Matrix Aneq(2*J-3,J,Zero); Vector bneq(2*J-3); Vector lb(J,-Inf); Vector ub(J,Inf); for(int j=0; j I was wondering if you're trying to fit a curve, subject to > monotonicity/convexity constraints... > If you are, this is a challenging topic, best of luck... > > > On Tue, Sep 22, 2020 at 8:12 AM Abby Spurdle <[hidden email]> wrote: > > > > Hi, > > > > Sorry, for my rushed responses, last night. > > (Shouldn't post when I'm about to log out). > > > > I haven't used the quadprog package for nearly a decade. > > And I was hoping that an expert using optimization in finance in > > economics would reply. > > > > Some comments: > > (1) I don't know why you think bvec should be a matrix. The > > documentation clearly says it should be a vector (implying not a > > matrix). > > The only arguments that should be matrices are Dmat and Amat. > > (2) I'm having some difficulty following your quadratic program, even > > after rendering it. > > Perhaps you could rewrite your expressions, in a form that is > > consistent with the input to solve.QP. That's a math problem, not an R > > programming problem, as such. > > (3) If that fails, then you'll need to produce a minimal reproducible > example. > > I strongly recommend that the R code matches the quadratic program, as > > closely as possible. > > > > > > On Mon, Sep 21, 2020 at 9:28 PM Maija Sirkjärvi > > <[hidden email]> wrote: > > > > > > Hi! > > > > > > I was wondering if someone could help me out. I'm minimizing a > following > > > function: > > > > > > \begin{equation} > > > $$\sum_{j=1}^{J}(m_{j} -\hat{m_{j}})^2,$$ > > > \text{subject to} > > > $$m_{j-1}\leq m_{j}-\delta_{1}$$ > > > $$\frac{1}{Q_{j-1}-Q_{j-2}} (m_{j-2}-m_{j-1}) \leq > \frac{1}{Q_{j}-Q_{j-1}} > > > (m_{j-1}-m_{j})-\delta_{2}$$ > > > \end{equation} > > > > > > I have tried quadratic programming, but something is off. Does anyone > have > > > an idea how to approach this? > > > > > > Thanks in advance! > > > > > > 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]] > > > > > > ______________________________________________ > > > [hidden email] mailing list -- To UNSUBSCRIBE and more, see > > > https://stat.ethz.ch/mailman/listinfo/r-help> > > PLEASE do read the posting guide > http://www.R-project.org/posting-guide.html> > > and provide commented, minimal, self-contained, reproducible code. >         [[alternative HTML version deleted]] ______________________________________________ [hidden email] mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-helpPLEASE do read the posting guide http://www.R-project.org/posting-guide.htmland provide commented, minimal, self-contained, reproducible code.
Open this post in threaded view
|

 > I'm trying to replicate a C++ code with R. Notes: (1) I'd recommend you make the code more modular. i.e. One function for initial data prep/modelling, one function for setting up and solving the QP, etc. This should be easier to debug. (However, you would probably have to do it to the C++ code first). (2) Your R code is not completely reproducible. i.e. AEJData (3) For the purposes of a reproducible example, your code can be simplified. i.e. Only one contributed R package should be attached. Regardless of (1) above, you should be able to identify at what point the C++ and R code becomes inconsistent. The simplest approach is to add print-based functions into both the C++ and R code, and print out state data, at each major step. Then all you need to do is compare the output for both. > Is there a better/another package for these types of problems? I'm not sure. After a quick search, this is the best I found: scam::scam scam::shape.constrained.smooth.terms ______________________________________________ [hidden email] mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-helpPLEASE do read the posting guide http://www.R-project.org/posting-guide.htmland provide commented, minimal, self-contained, reproducible code.