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RE: R&D Investment Returns

 
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RE: R&D Investment Returns - 1/31/2012 5:43:03 PM   
Alfred

 

Posts: 3710
Joined: 9/28/2006
Status: offline
The supply cost is not frontloaded. So the demand on the December 7th supply stocks in the Home islands is considerably less.

Alfred

(in reply to awaw)
Post #: 31
RE: R&D Investment Returns - 5/3/2012 10:38:24 PM   
alimentary

 

Posts: 44
Joined: 3/22/2010
Status: offline
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)
Post #: 32
RE: R&D Investment Returns - 5/3/2012 11:58:11 PM   
kmitahj

 

Posts: 95
Joined: 4/25/2011
Status: offline

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


(in reply to alimentary)
Post #: 33
RE: R&D Investment Returns - 5/4/2012 12:36:26 AM   
alimentary

 

Posts: 44
Joined: 3/22/2010
Status: offline
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.

(in reply to kmitahj)
Post #: 34
RE: R&D Investment Returns - 5/4/2012 2:55:31 AM   
Mike Solli


Posts: 14012
Joined: 10/18/2000
From: the flight deck of the Zuikaku
Status: offline

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

(in reply to alimentary)
Post #: 35
RE: R&D Investment Returns - 5/4/2012 9:51:33 AM   
BigBadWolf


Posts: 571
Joined: 8/8/2007
From: Serbia
Status: offline

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.







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

_____________________________


(in reply to Mike Solli)
Post #: 36
RE: R&D Investment Returns - 5/4/2012 4:37:55 PM   
kmitahj

 

Posts: 95
Joined: 4/25/2011
Status: offline

quote:

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

[...]

Q.E.D.


Thanks, I'm convinced now. Mathematics has spoken, the case is closed! .
It's refreshing to see somebody still able to use pencil and his brain instead of depending on the computer at every occcasion.

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.
More interesting from practical point of view may be taking into accout for research results estimation (as an option?) new (mis?)feature of of engine bonus which was discovered by Damian (n01487477).
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: 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


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.

As for AFB complains one thing is clear. Due to mentioned already in that thread new features like engine bonus and the trick (or rather cheat) of damage-free transition of r/d factories along aircraft upgrade path aircraft research is currently much, much, much easier in AE then it was in classic WitP. And contrary to OP you could achieve really decent research results in classic WitP if you knew what you were doing!

(in reply to alimentary)
Post #: 37
RE: R&D Investment Returns - 5/4/2012 8:22:37 PM   
Kitakami

 

Posts: 520
Joined: 5/3/2002
From: the bridge of the DNTK Kitakami
Status: offline
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>


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.

_____________________________

Tenno Heika Banzai!


(in reply to kmitahj)
Post #: 38
RE: R&D Investment Returns - 5/4/2012 10:51:17 PM   
alimentary

 

Posts: 44
Joined: 3/22/2010
Status: offline
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.



Both of your comments on the script are apt. I have incorporated corrections. The result is far from adequately tested but is provided here for your use. I have removed the earlier code version from the previous reply.

EDIT: Far from adequately tested was an understatement. Several errors corrected

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.

Attachment (1)

< Message edited by alimentary -- 5/6/2012 11:25:35 AM >

(in reply to kmitahj)
Post #: 39
RE: R&D Investment Returns - 5/5/2012 3:38:26 AM   
n01487477


Posts: 4717
Joined: 2/21/2006
Status: offline
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) ?

_____________________________

-Damian-
EconDoc
TrackerAE
Tutes&Java

(in reply to alimentary)
Post #: 40
RE: R&D Investment Returns - 5/5/2012 10:44:51 AM   
alimentary

 

Posts: 44
Joined: 3/22/2010
Status: offline

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) ?


Please feel free.

(in reply to n01487477)
Post #: 41
RE: R&D Investment Returns - 5/5/2012 12:51:51 PM   
Hanzberger


Posts: 771
Joined: 4/26/2006
From: SE Pennsylvania
Status: offline

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) ?


OH GOOD, this is why I LOVE Tracker~!!!! Let it do the work.

_____________________________


Japan AC wire chart here
http://www.matrixgames.com/forums/tm.asp?m=2769286&mpage=1&key=?

(in reply to n01487477)
Post #: 42
RE: R&D Investment Returns - 5/6/2012 1:13:40 AM   
kmitahj

 

Posts: 95
Joined: 4/25/2011
Status: offline

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.

Yes, I agree. That is what I used to aim at in old WitP: make a plan, setup research factories early and then... be patient. Very patient. At the beginning tempo of repair for late-war planes will be agonizingly slow which - as you said - has a good side effect of beeing light on supply usage.
This slowlness is btw clearly seen in the probability distribution table generated in alimentary script. Every entry $sizedist[D][S] in that table tells us precisely what is a probability that our factory will have size S after first D days. So nice... If applied mathematics can be cute then I dare to say that this script is really cute!

Just practical example from classic WitP age: if we setup Frank research early enough we have about 28 months till availability. Using above estimation: 28*0.37, rounded down to be on the safe side, leave us some 10 months for actual r/d pts generation. If we decide to invest in 5 Frank r/d factories of 30+ size we can expect 300 r/d pts every two months, that is we expect to get 3 months of advance every two months of work. Apply this twice and we use up whole 10 months: 4 months of r/d points generation resulting in 6 months of advance time - not bad I think. This assumes near ideal conditions of course. One thing which may make conditions less then ideal is supply stress in HI area (this may result in less then 10k supply in some research sites which negatively impact repair chances).
And of course there is another serious danger when researching very late-war airframes: allied player may get his 4E hordes in range of research facilities before they will be able to deliver. In classic WitP aggressive allied player using Iwojima or Sakhalin Gambits could start strategic bombing of HI somewhere in late 43...


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.

Thanks for taking the challenge. To be honest I can't say anything (at least anything consistent) pro or contra regarding correctnies of your approximation. I see it reduces nicely when $facilities=1 but that's it - beyond that my intuition simply refuses to cooperate with me
I also briefly comtemplated strict solution (that is one in which $days variable would have its own distribution table calculated synchronously with factory repairs dist-table [or rather tables]) but quickly came to conclusion that complexity of such approach is far beyond me.
Still I don't thing it is that importatnt as we are trying to catch here secondary if not tertiary effect relevant - if at all - only for larger number of factories. So it's mostly academic though interesting (imho) problem. From practical point of view the distribution generated for single factory is all we need I think. Beatuy of your script is that for single factory it gave us on the golden plate mathematicaly correct and complete solution which may be used as base for whatever prediction one would be interested in. For example one can easily (and precisely!) answer the question about expected supply cost of r/d factory repair after any given number of days. So, who knows, some day we may even see in Tracker a plot predicting day-after-day supply usage of r/d repairs for given initial setup of r/d factories.

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;
};




(in reply to Kitakami)
Post #: 43
RE: R&D Investment Returns - 5/6/2012 11:27:08 AM   
alimentary

 

Posts: 44
Joined: 3/22/2010
Status: offline
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;
};






Quite right. Corrected.

< Message edited by alimentary -- 5/6/2012 11:28:17 AM >

(in reply to kmitahj)
Post #: 44
RE: R&D Investment Returns - 5/6/2012 2:07:01 PM   
Hanzberger


Posts: 771
Joined: 4/26/2006
From: SE Pennsylvania
Status: offline

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.







I am with you Mike, and just when I thought I started to understand a few things....

_____________________________


Japan AC wire chart here
http://www.matrixgames.com/forums/tm.asp?m=2769286&mpage=1&key=?

(in reply to Mike Solli)
Post #: 45
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