Description Usage Arguments Details Value Examples

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. adjusting elements that fall outside of their bounds

3. repeating the above steps

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.

1 2 |

`param1` |
A vector of the first parameter for each of the elements. For a normal distribution, this is the mean. |

`param2` |
A vector of the second parameter for each of the elements. For a normal distribution, this is the standard deviation. |

`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. |

`lbounds` |
A Vector of the lower bounds for each element. These values should all be numeric |

`ubounds` |
A Vector of the upper bounds for each element. These values should all be numeric |

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).

A vector of the final balanced values

1 2 3 4 5 6 7 8 | ```
# 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)
``` |

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