[R] Fwd: Matrix Constraints in R Optim
Priyank Dwivedi
dpriyank23 at gmail.com
Mon Jun 20 01:17:27 CEST 2016
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 <jdnewmil at 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 <dpriyank23 at gmail.com>
> wrote:
>>
>> 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,
Priyank Dwivedi
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-------------- next part --------------
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-------------- next part --------------
c(10, 0.25, 0.25, 0.25, 0.25, 0.25, 10, 0.25, 0.25, 0.25, 0.25,
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