On Sat, Jun 5, 2021, 3:42 PM Karl <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.