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Covering the ceiling with foil would give it about US R10, ie 1.76 mK/W,
with E = 0.03 and a Tc = 20 C ceiling temp and a Tf = 14 C floor temp
and a large air gap. This would reduce the radiation from the ceiling
to the floor, es((Tc+273)^4-Tf+273)^4) W/m^2, with s = 5.6697x10^-8
W/m^2-K^4. If the upper foil surface is perfectly clean, with no dust
(you are German, right? :-), this may work even better. The linearized
radiation conductance is 4esTm^3 W/m^2-K, where Tm is the approximate
mean absolute temp.

Su, what the **** does all that gibberish mean? Oh, I know, it means Nick
can show off he knows a couple of equations and has a calculator. So,
rather than impress us with all of your learning, why not take the time to
explain if this is good or bad and translates to dollars (or Euros) saved.

Yes, that closed cell EPS will be a good insulator.