ORIGINAL: kmitahj
This careful estimate matches well with the naive estimate:
days to full repair = days to availability * (1 - 1/e)
Interesting result! Did you reach that DOA*(1-exp(-1)) in analytical way? If so you seem to be one of that few human beeings left which are adept in an ancient art of mathematics [&o]. When I was working on it I limited myself to running series of simulations on large number of input data (DOA & fab sizes) and was happy to see that no matter what was the starting point all results seem to converge to single factor which I estimated at about 0.63 of DOA. (or in other words 37% of DOA time left for r/d pts generation when starting from fully damaged factory). Your results seem to confirm my rough estimations which made me happy [:)]
Let me see if I can reconstruct the logic...
The formula for R&D factory repair is that the probability of repair is
such that you get a fraction 1/(days_to_nominal_arrival) of the
factory's total fully-repaired size repaired each day.
100 days to arrival, factory size 1 = 1% chance of 1 repair
100 days to arrival, factory size 30 = 30% chance of 1 repair
1000 days to arrival, factory size 30 = 3% chance of 1 repair
In general, if you go 10% of the way to the scheduled arrival
date, you get 10% of the capacity repaired. Or if you go
1% of the way to the schedule arrival date, you get 1% of
the capacity repaired.
So the first 10% of the repairs will take the time remaining
down to 90% of the original time remaining. [As a crude
estimate]
Or the first 1% of the repairs will take the time remaining
down to 99% of the original time remaining. [As a better
estimate]
And the next 1% of the repairs will take the time remaining
down to 99% of _that_.
If you count it out in 10% increments then the formula is
(1 - 1/10) ** 10
If you count it out in 1% increments then the formula is
(1 - 1/100) ** 100
As the number of days to nominal availability increases
without bound, you get
lim x->oo (1 - 1/x) ** x
Now for x very large there is an approximation that applies
here:
1 - 1/x ~= 1 / (1 + 1/x)
[e.g. .99 ~= 1/1.01]
so the above limit is equal to:
lim x->oo 1/(1 + 1/x) ** x
Which is equal to
1 / lim x->oo (1 + 1/x) ** x
Which is clearly equal to 1/e
If that's the fraction of time remaining left at the moment of
full repairs then the time to full repair is 1 - 1/e of the way
through the interval.
Q.E.D.