Re: Sandbox Discussions
Posted: Fri Jan 30, 2015 3:46 am
Ok, so let's make some theoretical inroads on case 2b.
To start, I agree with DanielH that you are limited by daily consumption and by built-up averaging. For the situation with a fresh Vampire baby and an indeterminately aged Dragon, assuming the Dragon always leaves 100 days of "spare life":
n = Number of days = number of feedings
d(0) = Dragon lifespan on the first day = 2000*ElcenianYear ~= 1 Million
The Vampire feeds once, gaining a lifespan of 1 Million days, and then is drained of all but 101 (100+1 for the day already lived)
v(1) = 101, d(1) = 1,999,899
The next day, the Vampire feeds once again, gaining a lifespan which is the average of his old lifespan (101), weighted once (for his one previous feeding) and the Dragon's (weighted once). His lifespan becomes 1 Million again. Immediately afterwards the Dragon steals him down to 102, ending at 1,999,899 + 1,000,000 - 102
v(2) = 102, d(2) = 2,999,797
This sort of exchange can be modeled in perpetuity by
v(n) = 100 + n
d(n) = d(n-1) + ((n-1)*(100+n-1)+d(n-1))/n - 100 - n; We may wish to simplify this to:
d(n) = d(n-1) + ((n-1)*(n+99)+d(n-1))/n - 100 - n
d(n) = d(n-1) + ((n^2+98n-99)+d(n-1))/n - 100 - n
d(n) = d(n-1) + n+98-99/n + d(n-1)/n - 100 - n
d(n) = [(n+1)/n]*d(n-1) - 2 - 99/n
If we can persuade ourselves to discard the small rump terms:
d(n) ~= [(n+1)/n]*d(n-1)
Translating to derivatives (more small approximations here):
dd(n)/dn ~= d(n) - d(n-1) ~= d(n)/n
dd(n)/d(n) = dn/n
Integrating:
ln(d(n)) = ln(n) + c
d(n) = C*n
We see that we have linear(ish) growth. Taking a look at the first few terms of the series, it's fairly clear that C ~= d(0) ~= 1,000,000.
The end result is that if you start with a baby vampire and a dragon, and feed every day, such a project will gain roughly one dragon's worth of lifespan per day in perpetuity.
This means that one such pair can support roughly one million extra individuals, though obviously it would probably not be wise to stress the system that hard. Separating out the lifespan would decrease your efficiency by a lot, however you could just take one of your invested dragons and start with a new baby vampire, taking off with a new, much higher d(0) and reaping vastly increased gains. Practically, with a steady supply of baby vampires and a little vaguely clever maneuvering it's easy to bank up huge sums of lifespan and gain exponentially increasing amounts of it.
I'll look at 2a later, if anybody cares.
To start, I agree with DanielH that you are limited by daily consumption and by built-up averaging. For the situation with a fresh Vampire baby and an indeterminately aged Dragon, assuming the Dragon always leaves 100 days of "spare life":
n = Number of days = number of feedings
d(0) = Dragon lifespan on the first day = 2000*ElcenianYear ~= 1 Million
The Vampire feeds once, gaining a lifespan of 1 Million days, and then is drained of all but 101 (100+1 for the day already lived)
v(1) = 101, d(1) = 1,999,899
The next day, the Vampire feeds once again, gaining a lifespan which is the average of his old lifespan (101), weighted once (for his one previous feeding) and the Dragon's (weighted once). His lifespan becomes 1 Million again. Immediately afterwards the Dragon steals him down to 102, ending at 1,999,899 + 1,000,000 - 102
v(2) = 102, d(2) = 2,999,797
This sort of exchange can be modeled in perpetuity by
v(n) = 100 + n
d(n) = d(n-1) + ((n-1)*(100+n-1)+d(n-1))/n - 100 - n; We may wish to simplify this to:
d(n) = d(n-1) + ((n-1)*(n+99)+d(n-1))/n - 100 - n
d(n) = d(n-1) + ((n^2+98n-99)+d(n-1))/n - 100 - n
d(n) = d(n-1) + n+98-99/n + d(n-1)/n - 100 - n
d(n) = [(n+1)/n]*d(n-1) - 2 - 99/n
If we can persuade ourselves to discard the small rump terms:
d(n) ~= [(n+1)/n]*d(n-1)
Translating to derivatives (more small approximations here):
dd(n)/dn ~= d(n) - d(n-1) ~= d(n)/n
dd(n)/d(n) = dn/n
Integrating:
ln(d(n)) = ln(n) + c
d(n) = C*n
We see that we have linear(ish) growth. Taking a look at the first few terms of the series, it's fairly clear that C ~= d(0) ~= 1,000,000.
The end result is that if you start with a baby vampire and a dragon, and feed every day, such a project will gain roughly one dragon's worth of lifespan per day in perpetuity.
This means that one such pair can support roughly one million extra individuals, though obviously it would probably not be wise to stress the system that hard. Separating out the lifespan would decrease your efficiency by a lot, however you could just take one of your invested dragons and start with a new baby vampire, taking off with a new, much higher d(0) and reaping vastly increased gains. Practically, with a steady supply of baby vampires and a little vaguely clever maneuvering it's easy to bank up huge sums of lifespan and gain exponentially increasing amounts of it.
I'll look at 2a later, if anybody cares.