Provides the approximate cumulative time or cost required for units m through n (inclusive) using the Crawford unit model. Provides nearly the exact output as unit_cum_exact(), usually only off by 1-2 units but reduces computational time drastically if trying to calculate cumulative hours (costs) for over a million units.
unit_cum_appx(t, n, r, m = 1, na.rm = FALSE)
NA
values be removed?library(learningCurve) # An estimator believes that the first unit of a product will # require 100 labor hours. How many total hours will be required # for 125 units given the organization has historically experienced # an 85% learning curve? unit_cum_appx(t = 100, n = 125, r = .85)#> [1] 5202.988## [1] 5202.998 # Computational difference between unit_cum_exact() and unit_cum_appx() # for 1 million units system.time(unit_cum_exact(t = 100, n = 1000000, r = .85))#> user system elapsed #> 0.25 0.00 0.25## user system elapsed ## 0.105 0.004 0.109 system.time(unit_cum_appx(t = 100, n = 1000000, r = .85))#> user system elapsed #> 0 0 0## user system elapsed ## 0 0 0