ORIGINAL: InfiniteMonkey
Your assumptions are wrong. If you look at the links I provided in my first post, they clearly state teh probability of repair is (the R&D factory total size) in (days to arrival). I have verfied this in testing prior to this thread and did another test a few minutes ago. It isn't an opinion question and it is easily verfiable:
I certainly won't test it (I don't benchmark the game, it looks too much like work), but I think I get it (and I understand where the 63% come from). Thank you.
So, bottomline would be
- the probability of repair is total size/nr of days to availability,
- so, for a 0(N) factory, D days away from availability, the expected nr of repairs after d days is
N ( 1/D + 1/(D-1) + ... + 1/(D-d) )
- the expected time of full repair is the time when this sum reaches N, or, factoring N out, the value of d for which
( 1/D + 1/(D-1) + ... + 1/(D-d) ) = 1
- this quantity is independent of N (the size of the factory), it can be tabulated in excel or solved using Euler formula for harmonic series
- more precisely : the sum 1 + 1/2 + ... + 1/D is very close to ln D + euler constant,
- so the left part of our formula is close to: ln D - ln (D-d), ie ln (D/(D-d))
- replacing and solving the above, we get D/(D-d)=exp(1), or d=(e-1)/e D, e=2.7183
- hence, on average, d = 0,63 D for a 0 (N) factory D days away from availability
- if the factory is partly repaired, the same formula applies, replacing the 1 on the right by the fraction of damaged size/total size, and e by the exponential of this fraction
- for instance for a 25(5) factory, replace 1 by 1/6, or e=exp(1/6), and d=0,15D...
The general formula would then be, for a A(B) factory, compute e = exp( B/(A+B)), on average, if you a D days from availability, you will have full repair in (e-1)/e D days.
Calculations yield:
- 63% of availability for fully damaged factories
- 41% for 75% damage
- 24% for 50% damage
- 10% for 25% damage
Francois