[R] Fwd: Matrix Constraints in R Optim

[Priyank Dwivedi]

By mistake, I sent it earlier to the wrong address.

---------- Forwarded message ----------
From: Priyank Dwivedi <dpriyank23 at gmail.com>
Date: 17 June 2016 at 14:50
Subject: Matrix Constraints in R Optim
To: r-help-owner at r-project.org


Hi,

Below is the code snippet I wrote in R:

The basic idea is to minimize error by optimizing set of values (in this
scenario 12) in the form of a matrix. I defined the matrix elements as
vector "*my.data.var" * and then stacked it into a matrix called
"*my.data.var.mat"
in the error function. *

The only part that I can't figure out is "what if the column sum in
the *my.data.var.mat
needs to be <=1"; that's the constraint/s.. Where do I introduce it in the
OPTIM solver or elsewhere?*






*my.data.matrix.inj* <- as.matrix(my.data)  #convert DATA FRAME to MATRIX
my.data.matrix.inj


*my.data.matrix.time* <- as.matrix(my.data.2)  #convert DATA FRAME to MATRIX
my.data.matrix.time


*my.data.matrix.prod* <- as.matrix(my.data)  #convert DATA FRAME to MATRIX
my.data.matrix.prod


*my.data.var* <-
c(2,0.8,0.5,0.2,0.2,0.1,10,0.01,0.02,0.2,0.1,0.01,2,0.8,0.5,0.2,0.2,0.1,10,0.01,0.02,0.2,0.1,0.01,2,0.8,0.5,0.2,0.2,0.1,10,0.01,0.02,0.2,0.1,0.01)
my.data.var

*my.data.qo* <- c(5990,150,199,996)   #Pre-Waterflood Production

*my.data.timet0* <- 0 # starting condition for time


*#FUNCTIONQjk.Cal.func* <-
function(my.data.timet0,my.data.qo,my.data.matrix.time,
                         my.data.matrix.inj,
my.data.matrix.prod,my.data.var,my.data.var.mat)
{

  qjk.cal.matrix <- matrix(,nrow = nrow(my.data.matrix.prod),
ncol=ncol(my.data.matrix.prod))

  count <- 1
  number <- 1
  for(colnum in 1:ncol(my.data.matrix.prod))   # loop through all PROD
wells columns
  {
    sum <-0
    for(row in 1:nrow(my.data.matrix.prod)) #loop through all the rows
    {
      sum <-0
      deltaT <-0
      expo <-0


        for(column in 1:ncol(my.data.matrix.inj)) #loop through all the
injector columns to get the PRODUCT SUM
         {
            sum = sum +
my.data.matrix.inj[row,column]*my.data.var.mat[colnum,number+column]
         }

      if(count<2)
      {
        deltaT<- my.data.matrix.time[row]
      }
      else
      {deltaT <- my.data.matrix.time[row]-my.data.matrix.time[row-1]}


      expo <- exp(-deltaT/my.data.var.mat[colnum,1])                  #
change here too

      if(count<2)
      {
        qjk.cal.matrix[row,colnum] = my.data.qo[colnum]*expo + (1-expo)*sum
      }
      else
      {
        qjk.cal.matrix[row,colnum]=qjk.cal.matrix[row-1,colnum]*expo +
(1-expo)*sum
      }
      count <- count+1
    }

    count <-1
  }

  qjk.cal.matrix      # RETURN CALCULATED MATRIX TO THE ERROR FUNCTION

}


*# ERROR FUNCTION* - FINDS DIFFERENCE BETWEEN CAL. MATRIX AND ORIGINAL
MATRIX. Miminize the Error by changing my.data.var

*Error.func* <- function(my.data.var)
{
  #First convert vector(my.data.var) to MATRIX aand send it to calculate
new MATRIX
  *my.data.var.mat* <- matrix(my.data.var,nrow =
ncol(my.data.matrix.prod),ncol = ncol(my.data.matrix.inj)+1,byrow = TRUE)

*  Calc.Qjk.Value* <-
Qjk.Cal.func(my.data.timet0,my.data.qo,my.data.matrix.time,
                                 my.data.matrix.inj,
my.data.matrix.prod,my.data.var,my.data.var.mat)


  diff.values <- my.data.matrix.prod-Calc.Qjk.Value    #FIND DIFFERENCE
BETWEEN CAL. MATRIX AND ORIGINAL MATRIX


  Error <- ((colSums ((diff.values^2), na.rm = FALSE, dims =
1))/nrow(my.data.matrix.inj))^0.5    #sum of square root of the diff
  print(paste(Error))

  Error_total <- sum(Error,na.rm=FALSE)/ncol(my.data.matrix.prod)   # total
avg error


 * Error_total*
}

# OPTIMIZE

*optim*(*my.data.var*
,Error.func,method="L-BFGS-B",upper=c(Inf,1,1,1,1,1,Inf,1,1,1,1,1,Inf,1,1,1,1,1,Inf,1,1,1,1,1))



-- 
Best Regards,
PD



-- 
Best Regards,
Priyank Dwivedi

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