On Sat, Jun 5, 2021, 3:42 PM Karl <[1]gmkarl@gmail.com> wrote:

   S_n and a 1 minute delay.
   St(n): time person n arrives.
   Sts(n): time person n begins being served
   Stf(n): time person n is done being served and leaves
   Stf(n) = Sts(n) + 1
   Sts(n) = max(St(n), Stf(n -1))

   Sts(n) = max(St(n), Sts(n - 1) + 1)
   Each iteration, the final value is either one more than the previous,
   or a newly defined value.

References

   1. mailto:gmkarl@gmail.com