Worthwhile task shaving

Worthwhile task shaving is a draft programming task. It is not yet considered ready to be promoted as a complete task, for reasons that should be found in its talk page.

Recreate https://xkcd.com/1205/ which shows a (humorous) table of how long you can work on making a routine task more efficient before spending more time than saved, for various s(h)avings against how often the task is run.

There are of course several possible interpretations of "day" and "week" in this context. The Phix implementation assumes 8 hour days and 5 day weeks might be more realistic, whereas it seems the original author worked with 24 hour days and 7 day weeks, and, tbh, my interest is piqued to see what built-in facilities other languages might have for handling such non-standard terms, if any. Extra kudos awarded for getting into the mind of the original author and reproducing their results exactly, or drumming up non-trivial (but still elegant) and potentially actually useful routines. This task can be made as trivial or as convoluted as you please, and should aim more for a little playfulness than rigid scientific accuracy.

Phix

with javascript_semantics
constant SEC = 1,
         MIN = 60,
         HOUR = 60*MIN,
         DAY = 8*HOUR,      -- (allow some sleepage)
         WEEK = 5*DAY,      -- (omit weekends)
         MONTH = 4*WEEK,
         YEAR = 12*MONTH,   -- (as 48 weeks/omit holidays)
         shavings = {1,5,30,MIN,5*MIN,30*MIN,HOUR,6*HOUR,DAY},
         frequencies = {{50,DAY},{5,DAY},{1,DAY},{1,WEEK},{1,MONTH},{1,YEAR}},
         roundto = {SEC, MIN, HOUR, DAY, WEEK, MONTH, YEAR},
         ts = {"sec", "min", "hour", "day", "week", "month", "year"}

function duration(atom a)
    string es
    for rdx=1 to length(roundto) do
        atom t = trunc(a/roundto[rdx])
        if rdx>1 and t<1 then exit end if
        es = sprintf("%d %s%s",{t,ts[rdx],iff(t=1?"":"s")})
    end for
    return es
end function

printf(1,"               50/day       5/day       daily      weekly     monthly      yearly\n")
for s=1 to length(shavings) do
    integer si = shavings[s]
    string line = sprintf("%10s ",duration(si))
    for f=1 to length(frequencies) do
        integer {per,slot} = frequencies[f]
        if si*per > slot then
            line &= sprintf("%10s  ","n/a")
        else
            atom shaving = (5*YEAR/slot * per) * si
            line &= sprintf("%10s  ",duration(shaving))
        end if
    end for
    printf(1,"%s\n",line)
end for

Output:

One outlier here is 1hr 5/day ==> 3 years vs original 10 months: as per notes above for 5/8ths the cutoff is indeed 3 years.
Note that the standard builtins such as elapsed() have no facilities for non-standard terms such as 8 hour working days.

               50/day       5/day       daily      weekly     monthly      yearly
     1 sec     2 days      1 hour     20 mins      4 mins       1 min      5 secs
    5 secs    2 weeks       1 day      1 hour     20 mins      5 mins     25 secs
   30 secs   3 months      1 week       1 day     2 hours     30 mins      2 mins
     1 min   6 months     2 weeks      2 days     4 hours      1 hour      5 mins
    5 mins    2 years    3 months     2 weeks      2 days     5 hours     25 mins
   30 mins        n/a      1 year    3 months     3 weeks      3 days     2 hours
    1 hour        n/a     3 years    7 months     1 month      1 week     5 hours
   6 hours        n/a         n/a     3 years    9 months    2 months      3 days
     1 day        n/a         n/a     5 years      1 year    3 months      1 week