Spectroscopic Follow-up:

Average follow-up time for a volume-limited sample is about 3/7 = 0.43 of the
time to do the faintest object. If N(z) flattens, this factor might be even
better, about 0.3

Here I calculate the follow-up for Bmax = Vmax = 18.8, corresponding to z = 0.1
needed to get S/N = 10 per 3A pixel. For K-corrections this is overkill.
S/N = 10 per 50A should be sufficient, which requires 16.6x less exposure; this
would still give S/N ~ 45 per 1000A filter.

** Exposure Schedule for s=1 SNe ar z = 0.1 ***

Day    m(B)   Exptime    Exptime      Exptime
                4-m        4-m        2.5-m
               (min)      (min)       (min)
             S/N=10 3A  S/N=10 50A   S/N=10 3A
------------------------------------------------
disc    19.8    25        ----         60
-10     19.1     6        ----         15
  0     18.8     4        ----         10
 10     19.7    20        ----         50
 20     20.9   175*       ----        450*
 30     21.4   440*         25        ---
 40     21.8   900*         54        ---
 50     22.0  1325*         80        ---
 60     22.3  2300*        140*       ---

(* exposure times over 2 hr cap)

So this says we need 3/7*8.8 hrs of 4-m time per SN, which is 3/7*33 = 14 (8-hr)
nights on a 4-m. I would just set all exposures past day 15 to 2 hrs for
z = 0.1. This gives 3/7 * 11 hrs of 4-m time per SNe, or 18 nights of 4-m time
for 30 SNe. This cap means that S/N per 3A would be about 2.5 at day 60; still
good enough for K-corrections I think.

 Using 2.5-m telescopes only saves about 3/7*1 hr of 4-m time per SNe, or about
2 nights. So, we want ~16 nights of 4-m time and 4 nights of 2.5-m telescope.
We also need discovery screening time, which the 2.5-m telescopes can be used
for (S/N doesn't have to be nearly as good for screening).

Note that if the 2 hr cap is applied to all SNe, regardless of brightness, then
more time is needed and the calculation is more complicated. For example at
z = 0.05, exposures on a 4-m are 2 hrs or less out to about day 50. Once we
have a prediction of the redshift distribution, a better estimate can be made
(although there will still be a range of 1 mag in peak brightness at a given
redshift - really need the distribution of peak magnitudes, which Peter is
generating).