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This function takes number of FTE and annual $/FTE and determines labor cost
solvecost_labor(fte, cost, time = "day")
A numeric value for labor $/time.
Number of FTEs. Can be decimal.
$/year per FTE
Desired output units, one of c("day", "month", "year"). Defaults to "day".
laborcost <- solvecost_labor(1.5, 50000) library(dplyr) cost_data <- tibble( fte = seq(1, 10, 1) ) %>% mutate(costs = solvecost_labor(fte = fte, cost = .08))
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