from:"Priyank Dwivedi"

Re: [R] Fwd: Matrix Constraints in R Optim

2016-06-21 Thread Priyank Dwivedi

   }
>
> 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
> }
>
> # RETURN CALCULATED MATRIX TO THE ERROR FUNCTION
> qjk.cal.matrix
> }
>
> Error.func <- function( my.data.var ) {
> #First convert vector(my.data.var) to MATRIX
> # and 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
>   )
>
> #FIND DIFFERENCE BETWEEN CAL. MATRIX AND ORIGINAL MATRIX
> diff.values <- my.data.matrix.prod - Calc.Qjk.Value
>
> #sum of square root of the diff
> Error <- ( ( colSums( ( diff.values^2 )
> , na.rm = FALSE
> , dims = 1
> )
>) / nrow( my.data.matrix.inj )
>  )^0.5
> #print(paste(Error))
>
> # total avg error
> Error_total <- sum( Error
>   , na.rm=FALSE
>   ) / ncol( my.data.matrix.prod )
>
> Error_total
> }
>
> n <- ncol( my.data.matrix.prod )
> m <- ncol( my.data.matrix.inj )
> k <- ( 1 + n ) * m
> ciA <- numeric( k )
> uiA <- array( 0, dim = c( m+1, n, k ) )
> ia <- 0
> #Aoff2 <- (m+1) * n
> for ( i in seq.int( m ) ) {
> ia <- ia + 1L
> # sum of columns <= 1
> uiA[ i+1, , i ] <- -1
> ciA[ ia ] <- -1
> }
> for ( i in ( 1 + seq.int( m ) ) ) {
> for ( j in seq.int( n ) ) {
> ia <- ia + 1L
> # elements > 0
> uiA[ i, j, ia ] <- 1
> ciA[ ia ] <- 0
> }
> }
> uiA <- matrix( uiA, nrow = k, byrow = TRUE )
>
> my.data.varA <- c( 10, 0.25, 0.25, 0.25, 0.25, 0.25
>  , 10, 0.25, 0.25, 0.25, 0.25, 0.25
>  , 10, 0.25, 0.25, 0.25, 0.25, 0.25
>  , 10, 0.24, 0.24, 0.24, 0.24, 0.24
>  )
> # interior?
> all( uiA %*% my.data.var1 - (ciA) > 0 )
>
> sols <- constrOptim( my.data.varA
>, Error.func
>, NULL
>, ui = uiA
>, ci = ciA
>, method = "SANN"
>)
> # meets constraint?
> all( uiA %*% sols$par - (ciA) >= 0 )
>
> sols
> ##
>
>
> On Sun, 19 Jun 2016, Priyank Dwivedi wrote:
>
>> All,
>> Here are the dput files of the input data to the code.
>>
>> Thanks for any advice.
>>
>> I am adding the entire code below too just in case.
>>
>>
>> file <- file.path("Learning R","CRM_R_Ver4.xlsx")
>> file
>> my.data <- readWorksheetFromFile(file,sheet=1,startRow=1)
>> str(my.data)  # DATA FRAME
>> my.data.matrix.inj <- as.matrix(my.data)  #convert DATA FRAME to MATRIX
>> my.data.matrix.inj
>>
>> dput(my.data.matrix.inj,"my.data.matrix.inj.txt")
>>
>>
>> my.data.2 <- readWorksheetFromFile(file,sheet=2,startRow=1)
>> str(my.data.2)  # DATA FRAME
>> my.data.matrix.time <- as.matrix(my.data.2)  #convert DATA FRAME to MATRIX
>> my.data.matrix.time
>>
>> dput(my.data.matrix.time,"my.data.matrix.time.txt")
>>
>> my.data <- readWorksheetFromFile(

Re: [R] Fwd: Matrix Constraints in R Optim

2016-06-20 Thread Priyank Dwivedi

All,
Here are the dput files of the input data to the code.

Thanks for any advice.

I am adding the entire code below too just in case.


file <- file.path("Learning R","CRM_R_Ver4.xlsx")
file
my.data <- readWorksheetFromFile(file,sheet=1,startRow=1)
str(my.data)  # DATA FRAME
my.data.matrix.inj <- as.matrix(my.data)  #convert DATA FRAME to MATRIX
my.data.matrix.inj

dput(my.data.matrix.inj,"my.data.matrix.inj.txt")


my.data.2 <- readWorksheetFromFile(file,sheet=2,startRow=1)
str(my.data.2)  # DATA FRAME
my.data.matrix.time <- as.matrix(my.data.2)  #convert DATA FRAME to MATRIX
my.data.matrix.time

dput(my.data.matrix.time,"my.data.matrix.time.txt")

my.data <- readWorksheetFromFile(file,sheet=3,startRow=1)
str(my.data)  # DATA FRAME
my.data.matrix.prod <- as.matrix(my.data)  #convert DATA FRAME to MATRIX
my.data.matrix.prod

dput(my.data.matrix.prod,"my.data.matrix.prod.txt")

 # my.data.var <- vector("numeric",length = 24)
 # my.data.var

my.data.var <- c(10,0.25,0.25,0.25,0.25,0.25,
 10,0.25,0.25,0.25,0.25,0.25,
 10,0.25,0.25,0.25,0.25,0.25,
 10,0.25,0.25,0.25,0.25,0.25)
my.data.var

dput(my.data.var,"my.data.var.txt")


my.data.qo <- c(5990,150,199,996)   #Pre-Waterflood Production
my.data.timet0 <- 0 # starting condition for time

#FUNCTION
Qjk.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

sols<-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),
  lower=c(0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0))

sols

On 17 June 2016 at 16:55, Jeff Newmiller <jdnew...@dcn.davis.ca.us> wrote:
> Your code is corrupt because you failed to send your email in plain text
> format.
>
> You also don't appear to have all data needed to reproduce the problem. Use
> the dput function to generate R code form of a sample of your data.
> --
> Sent from my phone. Please excuse my brevity.
>
> On June 17, 2016 1:07:21 PM PDT, Priyank Dwivedi <dpriyan...@gmail.com>
> wrote:
>>
>> By mistake, I sent it earlier to the wrong address.
>>
>> -- Forwarded message --
>> From: Priyank Dwivedi <dpriyan...@gmail.com>
>> Date: 17 June 2016 at 14:50
>> Subject: Matrix Constraints in R Optim
>> To: r-help-ow...@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 matri

[R] Fwd: Matrix Constraints in R Optim

2016-06-17 Thread Priyank Dwivedi

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

-- Forwarded message --
From: Priyank Dwivedi <dpriyan...@gmail.com>
Date: 17 June 2016 at 14:50
Subject: Matrix Constraints in R Optim
To: r-help-ow...@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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Re: [R] Fwd: Matrix Constraints in R Optim

Re: [R] Fwd: Matrix Constraints in R Optim

[R] Fwd: Matrix Constraints in R Optim

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