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So @hokomoko here's your long awaited (and long) answer ;):
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So @hokomoko here's your long awaited (and long) answer ;):
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[quote]Easy to see that when E = 0 we get Output = M.[/quote]I have chosen the [url=http://zero-k.info/Forum/Thread/17596?page=5#139075]log system[/url] so that this is true for any a, c. I just use m for what you call M and M for what you call Output.
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[quote]Easy to see that when E = 0 we get Output = M.[/quote]I have chosen the [url=http://zero-k.info/Forum/Thread/17596?page=5#139075]log system[/url] so that this is true for any a, c. I just use m for what you call M and M for what you call Output.
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[quote]You can change log base to tweak rates, although base 10 gives you ~30% for first solar, 47% for 2nd and so on which is pretty cool.[/quote]A change of log base is equivalent to a change of a. Generally changes of a and c should be considered where possible.
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[quote]You can change log base to tweak rates, although base 10 gives you ~30% for first solar, 47% for 2nd and so on which is pretty cool.[/quote]A change of log base is equivalent to a change of a. Generally changes of a and c should be considered where possible.
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[quote]you're looking at E/M wrong. A grid can be a black box, you don't care about which mex the energy goes to. As stated before, 10m mex and 5 2m mex are equivalent (if already connected).[/quote]What I said wasn't wrong, too. Your system can be regarded as overdriving every single mex in a grid with E/m, which is technically s=infinity, because E is the whole grid energy. It is equivalent to regard a grid as a single mex that is overdriven the same way.[spoiler]Generally you have to distribute energies in a grid and between grids. In the systems so far every mex "demanded" his ideal energy and grid connection was a limitaion to that. However in @hokomoko's system a grid is a bit like a single mex and the distribution between grids is the crucial point.[/spoiler]
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[quote]you're looking at E/M wrong. A grid can be a black box, you don't care about which mex the energy goes to. As stated before, 10m mex and 5 2m mex are equivalent (if already connected).[/quote]What I said wasn't wrong, too. Your system can be regarded as overdriving every single mex in a grid with E/m, which is technically s=infinity, because E is the whole grid energy. It is equivalent to regard a grid as a single mex that is overdriven the same way.[spoiler]Generally you have to distribute energies in a grid and between grids. In the systems so far every mex "demanded" his ideal energy and grid connection was a limitaion to that. However in @hokomoko's system a grid is a bit like a single mex and the distribution between grids is the crucial point.[/spoiler]
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[quote]I'm 99% sure it's always better to combine grids but I've only proveמ it for simpler equations, feel free to find issues.[/quote]Proving grid monotony here seems difficult. But a system's grid monotony must be proven according to the [url=http://zero-k.info/Forum/Thread/11335#114411]system construction instruction[/url]. I found easier formulations, have done it for special cases and have proven some related things that are partially also visible in @TheEloIsALie's [url=http://tube.geogebra.org/m/fn46b5zW]graph[/url] (like that connection of grids with same E and m changes nothing). My easier formulations also make the gird connection monotony proof independant of a and c. Thus it's no problem that @TheEloIsALie used ln instead of log_10. I will respect anyone who states a correct proof of @hokomoko's system's grid monotony! Let me know if you want my intermediate results. It would be nice to generalize it for other gammas.
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[quote]I'm 99% sure it's always better to combine grids but I've only proveמ it for simpler equations, feel free to find issues.[/quote]Proving grid monotony here seems difficult. But a system's grid monotony must be proven according to the [url=http://zero-k.info/Forum/Thread/11335#114411]system construction instruction[/url]. I found easier formulations, have done it for special cases and have proven some related things that are partially also visible in @TheEloIsALie's [url=http://tube.geogebra.org/m/fn46b5zW]graph[/url] (like that connection of grids with same E and m changes nothing). My easier formulations also make the gird connection monotony proof independant of a and c. Thus it's no problem that @TheEloIsALie used ln instead of log_10. I will respect anyone who states a correct proof of @hokomoko's system's grid monotony! Let me know if you want my intermediate results. It would be nice to generalize it for other gammas.
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@TheEloIsALie this reminds me a bit of the principle of comparative advantages..
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@TheEloIsALie this reminds me a bit of the principle of comparative advantages..
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[quote]When deciding how to distribute the excess energy between grids, just try to even out E/M in all grids which didn't hit their maximum already.[/quote]Indeed the ideal distribution of grid energies (if maximum allows it) is E_i=m_i*E/(sum from k to n (m_k)), where E is total OD energy and m_i the total base metal rates of the i-th grid.[spoiler]But how did you know that? Even if you know that the inversion of the differentiation of gamma_log is affine linear, it is not that easy to see that the change of the dependancy to E_i/m_i allows a factorization of m_i so that the additive constant is cancelled out and affine linear becomes linear.[/spoiler]However I have proven that the ideal solution of any system with reasonable gamme [color=grey](continous differentiable with strictly monotonic decreasing differentiation)[/color] that depends in e_i/m_i is linear in m_i for any s: The condition for ideal distribution is that the following is a constant r in m_i forall i: r = m_i*∂gamma(e_i/m_i)/∂(e_i^s) = m_i/m_i*∂gamma(e_i/m_i)/∂((e_i/m_i)^s) = ∂gamma(e_i/m_i)/∂((e_i/m_i)^s) = ∂gamma(e_i/m_i)/∂(e_i/m_i)*(1/s)*(e_i/m_i)^(1/s-1) =: f(e_i/m_i). We can apply the inversion of f and get e_i = m_i * f^-1(r), where f^-1(r) is a constant in m_i.
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[quote]When deciding how to distribute the excess energy between grids, just try to even out E/M in all grids which didn't hit their maximum already.[/quote]Indeed the ideal distribution of grid energies (if maximum allows it) is E_i=m_i*E/(sum from k to n (m_k)), where E is total OD energy and m_i the total base metal rates of the i-th grid.[spoiler]But how did you know that? Even if you know that the inversion of the differentiation of gamma_log is affine linear, it is not that easy to see that the change of the dependancy to E_i/m_i allows a factorization of m_i so that the additive constant is cancelled out and affine linear becomes linear.[/spoiler]However I have proven that the ideal solution of any system with reasonable gamme [color=grey](continous differentiable with strictly monotonic decreasing differentiation)[/color] that depends in e_i/m_i is linear in m_i for any s: The condition for ideal distribution is that the following is a constant r in m_i forall i: r = m_i*∂gamma(e_i/m_i)/∂(e_i^s) = m_i/m_i*∂gamma(e_i/m_i)/∂((e_i/m_i)^s) = ∂gamma(e_i/m_i)/∂((e_i/m_i)^s) = ∂gamma(e_i/m_i)/∂(e_i/m_i)*(1/s)*(e_i/m_i)^(1/s-1) =: f(e_i/m_i). We can apply the inversion of f and get e_i = m_i * f^-1(r), where f^-1(r) is a constant in m_i.
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[quote]also,
since
it's
log
based,
eco
growth
should
be
linear.
[/quote]If
we
assume
that
eco
growth
is
proportional
to
exp(
gamma)
,
and
gamma=a*ln,
eco
growth
is
proportional
to
exp(
ln(
x^a)
)
=x^a.
In
@hokomoko's
system
a=1/ln(
10)
.
[spoiler]Actually
there
are
more
factors
to
consider:
Eco
growth
is
proportional
to
exp(
z*gamma)
=exp(
z+z*a*ln(
e_i/c+1)
)
=e^z*(
e_i/c+1)
^(
z*a)
,
where
z
determines
the
maximum
reinvestment
efficiency
in
the
game.
So
eco
growth
is
proportional
to
a
polynomial
of
degree
z*a.
[/spoiler]
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[quote]also,
since
it's
log
based,
eco
growth
should
be
linear.
[/quote]If
we
assume
that
eco
growth
is
proportional
to
exp(
gamma)
,
and
gamma=a*ln,
eco
growth
is
proportional
to
exp(
ln(
x^a)
)
=x^a.
In
@hokomoko's
system
a=1/ln(
10)
.
[spoiler]Actually
there
are
more
factors
to
consider:
Eco
growth
is
proportional
to
exp(
z*gamma(
x)
)
=exp(
z+z*a*ln(
x/c+1)
)
=e^z*(
x/c+1)
^(
z*a)
,
where
z
determines
the
maximum
reinvestment
efficiency
in
the
game.
So
eco
growth
is
proportional
to
a
polynomial
of
degree
z*a.
[/spoiler]
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Of course it will still be beneficial to build more energy generators next to big mexes, but only proportional to extraction rate, not more like currently. It is interesting to see that the principle of dependancy in e_i/m_i can not only be applied to any gamma, it also always yields extraction proportional energy distribution (as far as grid allows).
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14 |
Of course it will still be beneficial to build more energy generators next to big mexes, but only proportional to extraction rate, not more like currently. It is interesting to see that the principle of dependancy in e_i/m_i can not only be applied to any gamma, it also always yields extraction proportional energy distribution (as far as grid allows).
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make caretaker eco shared via whole team