Worthwhile task shaving

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.

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.

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

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.

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:

               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