Slitherine

quote:

ORIGINAL: awaw

quote:

I hope fow is not an issue here, but a 2nd point I will like to clarify is, with the above research plan, what was the date when the first factory is FULLY REPAIRED? By that same date, how many "advances" of Shinden have you collected?

I have a program to estimate this.

From 12/41 to 12/45 (nominal arrival date of the Shinden) is roughly 1455 days.

By day 787 you should have a 10% chance of full repairs on any given factory
By day 914 you should have a 50% chance of full repairs on any given factory
By day 1032 you should have a 90% chance of full repairs on any given factory

That's February of 1943 for the 10% level, June of 1943 for the 50% level and
October of 1943 for the 90% level.

This careful estimate matches well with the naive estimate:

days to full repair = days to availability * (1 - 1/e)

Where e is Euler's number, 2.71828 = lim x->oo ( 1 + 1/x ) ** x

The program explicitly computes the probability distribution of possible
repaired factory sizes on every day from the starting day to the
arrival day and reports the probability of full repairs on each day.


< Message edited by alimentary -- 5/4/2012 10:41:32 PM >

(in reply to awaw)

quote:

This careful estimate matches well with the naive estimate:

days to full repair = days to availability * (1 - 1/e)

quote:

ORIGINAL: kmitahj

quote:

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 . 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

quote:

ORIGINAL: alimentary

quote:

ORIGINAL: kmitahj

quote:

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 . 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.

Created by the amazing Dixie

quote:

ORIGINAL: Mike Solli

quote:

ORIGINAL: alimentary

quote:

ORIGINAL: kmitahj

quote:

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 . 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.

quote:

ORIGINAL: alimentary
Let me see if I can reconstruct the logic...

[...]

Q.E.D.

quote:

# A factory generates research if it was fully repaired at the start of the day.
$p = ($maxsize+1)/30;

quote:

ORIGINAL: BigBadWolf
And than you hear AFBs complain about Japanese R&D. You need PhD in math just to comprehend the damn thing, let alone make it work

quote:

ORIGINAL: kmitahj
<snip>
Hey, it is not so bad as it looks. First practical result from alimentary work is solid estimation of time needed to repair fully damaged r/d factory: 63% of initial time till availability will be on average needed for repair with only slightly more then 1/3 of that initial time left for research points generation. It is just first approximation and you may expect serious dispersion of repair time for single factories around this value but using probability distribution so nicely generated by script provided by alimentary you can - when you really care - find precise answer on more complicated questions (like: what is an exact chance that my factory will repair after given fixed number of days). And the script is also helpful when you start with only partially damaged factory.
<snip>

quote:

ORIGINAL: kmitahj


For your perl script there is one minor change you may consider:

quote:

# A factory generates research if it was fully repaired at the start of the day.
$p = ($maxsize+1)/30;

Reason for that "+1" is that production formula is (or at least was last time I've checked): (RepairedSize+RAND1(30))/30 which result on average in one "bonus" production point every 30 days. Minor fix and relevant only to small facilities (maxsize<30) which are suboptimal anyway.

Another - and really challenging task I think - may be taking into account in your script that number of days till availability ($days) isn't really a fixed number but random variable. Every day there is a nonzero probability p that total of 100 r/d pts is reached is a day when $days variable is reduced by 30 (well, actually by length of calendar month which introduces minor variability) with probability p or stay unchanged with probability (1-p). Such "quantum" leaps potentially influence further calculations of distribution as well as limit number of days over which research points can be cumulated. This of course does not matter that much when we have only one factory ($facilities=1) beacuse in this case whole distribution must be already condensed on the last (maxsize) column but when having many factories some of them will repair and start generating r/d pts earlier and will influence probabilities of repair process for slower factories.

quote:

ORIGINAL: n01487477

Would you mind if I borrowed/adapted this formula for later iterations of Tracker - not 1.9 as it is about due but 2.0(as an estimate of repairs to R&D factories) ?

quote:

ORIGINAL: n01487477

Would you mind if I borrowed/adapted this formula for later iterations of Tracker - not 1.9 as it is about due but 2.0(as an estimate of repairs to R&D factories) ?

quote:

ORIGINAL: Kitakami

If it is so, then it makes MUCH sense to set a few factories to research the late fighters... they will come online slowly, so the economic impact should not be felt as much, but 37% of many months is a LOT of research time... a couple of airframes could be accelerated quite a bit. I guess the Frank is a given, but as I see things, there are enough R&D factories to research another late (maybe IJN?) fighter.

Just my 2 centavos.

quote:

ORIGINAL: alimentary

The approach I took on the latter challenge was to model the possibility of an
advance in availability date (and the resulting increase in factory repair rate)
by taking the expected value of the research to date _from all of the "other"
factories_. When that exceeds 100 points, availability moves up and repair
rates increase to match.

If you are estimating for 1 factory, this correction has no effect.
If you are estimating for 30 factories, I _think_ that this results
in an unbiased estimator.

quote:

# First, figure the number of research points per factory per turn
# complete repairs.
$p = ($maxsize+1)/30;
if ( $p > 30 ) { <--here: $p > 1
$p = 1;
};

quote:

ORIGINAL: kmitahj

quote:

ORIGINAL: Kitakami

Btw, before I forget: I think there is a typo in the script in place marked below:

quote:

# First, figure the number of research points per factory per turn
# complete repairs.
$p = ($maxsize+1)/30;
if ( $p > 30 ) { <--here: $p > 1
$p = 1;
};

quote:

ORIGINAL: Mike Solli

quote:

ORIGINAL: alimentary

quote:

ORIGINAL: kmitahj

quote:

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 . 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.