Rquefts (version 1.2-3)

fertApp: Optimal fertilizer application

Description

Compute the optimal fertilizer application rates given a target nutrient application and the available products (fertilizer blends) and their prices.

Usage

fertApp(nutrients, fertilizers, price, exact=TRUE, retCost=FALSE)

Arguments

nutrients

data.frame with columns "N", "P", "K" in kg (per unit area)

fertilizers

data.frame with fertilizer products (see examples)

price

numeric. Vector with fertilizer product prices. Should have length of nrow(fertilizers)

exact

logical. If FALSE the cheapest solution is returned that includes at least as much of each nutrient as desired, but possibly more, if that is cheaper than the exact solution; or when there is no exact solution

retCost

logical. If FALSE the optimal solution is returned (the amounts of fertilizers). If TRUE, the price of the optimal solution is returned

Examples

# fertilizer product list
fert <- fertilizers()
# shortening some of the names for display
fert[,2] = substr(fert[,2], 1, 20)
# contents are expressed as a percentage.
ferts <- fert[c(8,15:17), 2:5] 
ferts

x <- fertApp(data.frame(N=100, P=50, K=50), ferts, c(1, 1.5, 1.25, 1))
# show that it is correct
nutrientRates(ferts, x[,2])

fertApp(data.frame(N=seq(0,200,50), P=50, K=50), ferts, c(1, 1.5, 1.25, 0.75))
fertApp(data.frame(N=seq(0,200,50), P=50), ferts[,-3], c(1, 1.5, 1.25, 0.75))
fertApp(data.frame(N=seq(0,200,50), P=50), ferts[,-3], c(1, 1.5, 1.25, 5.75))