```
##' Balancing algorithm via Maximum Likelihood with all Normal Distributions
##'
##' This function has more or less the same inputs as balancing(), with the
##' exception of a handful of the tuning options that don't apply. If all
##' distributions are normal and unbounded, there is a simple analytical
##' solution to the balancing equation. Even if we have bounds, we can
##' iteratively find the optimal solution by 1. optimizing analytically and
##' ignoring bounds, 2. adjust elements that fall outside of their bounds, 3.
##' repeat. We will have at most as many iterations as elements and thus it
##' will be very fast, and we are also guaranteed to find the true optimum.
##'
##' Minimizing the negative log-likelihood of the sum of the normal densities is
##' equivalent to minimizing the weighted sums of squares (where the weights are
##' 1/sigma_i^2). Thus, the analytical solution is to just proportion the
##' difference according to those weights (sigma_i^2).
##'
##' @param param1 A vector of the first parameter for each of the elements. For
##' a normal distribution, this is the mean.
##' @param param2 A vector of the second parameter for each of the elements. For
##' a normal distribution, this is the standard deviation.
##' @param sign A vector of the sign of each element. These values should all
##' be +1 or -1, and they indicate if Delta_1, Delta_2, ... should be
##' pre-multiplied by a negative or not. Usually, these will all be +1.
##' @param lbounds A vector of the lower bounds for each element. These values
##' should all be numeric
##' @param ubounds A vector of the upper bounds for each element. These values
##' should all be numeric
##' @param failed Logic value indicating if the balancing occurs after the failure of some cases
##' when imbalance>0
##' @return A vector of the final balanced values
##'
##' @examples
##'
##' # Production, Imports, Exports, Stock, Food, Processing, Feed, Waste,
##' # Seed, Industrial, Tourist
##' param1 = c(54418808, 1999076, 32789894, -230630, 0, 26331060, 4898000,
##' 560306, 1904246, 0, 67)
##' param2 = c(544188, 0, 0, 89854, 0, 1749, 244900, 56031, 1129, 0, 7)
##' sign = c(1, 1, -1, -1, -1, -1, -1, -1, -1, -1, -1)
##' lbounds = c(0, 0, 0, -Inf, 0, 0, 0, 0, 0, 0, 0)
##' ubounds = rep(Inf, 11)
##'
balancingProportional = function(param1, param2, sign,
lbounds = rep(0, length(param1)),
ubounds = rep(Inf, length(param1)),
failed = FALSE){
## Data quality checks
## (None implemented as this function is a helper function and should not be
## called directly (but rather called from balancing()).
sign2 =sign
sign2[sign2==0]=-1
imbalance = param1 %*% sign2
signSupply= sign
signSupply[signSupply<0]=0
supply= param1 %*% signSupply
maxResidual=supply*0.05
solution = param1
if(imbalance>0&!failed)
{
signResidual=sign
signResidual[signResidual==0]=100
signResidual[signResidual<100]=0
signResidual[signResidual==100]=1
if(imbalance<maxResidual){
solution= solution + imbalance *(signResidual)
imbalance=0
}else{
solution= solution+ maxResidual *(signResidual)
imbalance= imbalance-maxResidual
}
}
if(all(param2 == 0) & imbalance != 0){
stop("Balance was not possible because of constraints.")
}
if(any(is.infinite(param1)) | any(is.infinite(param2))){
stop("Balance was not possible because presence of infinite values in means or param1 or param2")
}
## Initialize solution
weight = abs(param2)
if(all(weight==0)){
weight=weight
} else {weight = weight / sum(weight)}
solution = solution + imbalance * weight * (-sign2)
failedCases = solution < lbounds | solution > ubounds
if(any(failedCases)){
## Assign to boundary if it crossed boundary
solution[solution < lbounds] = lbounds[solution < lbounds]
solution[solution > ubounds] = ubounds[solution > ubounds]
## Resolve (i.e. reallocate new residual to elements that did not
## violate bounds).
solution = balancingProportional(param1 = solution,
param2 = ifelse(failedCases, 0, param2),
sign = sign, lbounds = lbounds,
ubounds = ubounds,failed = TRUE)
}
return(solution)
}
```

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