Re: poly: Re: ESS for HPLD

From: Hal Finney <hal@rain.org>
Date: Fri Dec 05 1997 - 12:31:37 PST

Trying to reproduce/understand Robin's algebra:

Robin Hanson, <hanson@econ.berkeley.edu>, writes:
> Assume (with Carl F.) that spores traveling between oases at a constant
> speed V have a 1/A chance of being destroyed per unit distance. Thus given
> N1 initial spores the number of spores surviving after distance X is
> N1*exp(X/A).

This should be N1/exp(X/A), right? The number of surviving spores should
decrease as we travel farther.

> Assume that spores deliver unit wealth to an oasis, and at an oasis wealth W
> grows a fractional rate of 1/G per unit time, up to a maximum Wmax. At any
> point growth can be stopped and N2 = W/C spores can be sent out, where C is
> the cost per spore. Thus after time T, N2 = exp(T/G)/C spores can be sent.

So you have W = exp(T/G) where T is the growth time (capped by Wmax).

> Finally, assume a delay D for spores to accelerate, decelerate, and set up
> shop.
> And assume only a fraction S of oases are discovered on arrival to be
> suitable.
>
> The maximum possible speed is where on average only one spore survives to
> grow in a new oasis per each previous colonized oasis. So assume N1 = N2.

I don't really follow this. Are you saying that it is wasteful to have
two spores hit one oasis?

I'm not sure what it means to assume that N1=N2. N1 is only used in one
equation, which estimates how many spores will be left after a given
distance if we start with N1. N2 tells how many spores can be sent out
after time T starting with unit wealth.

> A little algebra shows X = A*ln(WS/C) and T = D + g*ln(W) + X/V .

There must be some equations leading up to this. What is T? Is it the
time between generations, emission of spore to emission of daughter spore?
Instead of "g", you mean "G", right? V is velocity, so X/V is travel time
not counting accel/decel.

So you second equation says:

        Time between generations T
        equals =
        Time to accel, decel, set up shop D
        Plus time to grow + G * ln W
        Plus time to travel + X/V

The next to last term comes from W = exp(T/G) above.

Trying to derive your first equation, we start with N1 = N2. Parent
generation emits N1 spores. After distance X, N1/exp(X/A) survive.
Fraction S of oases are good, so SN1/exp(X/A) take root. These each
emit N2=W/C spores, so total in next generation is SN1(W/C)/exp(X/A).
We set this equal to N1(???) and get:

        SN1(W/C)/exp(X/A) = N1. Divide through by N1 and rearrange to get:

        (WS/C) = exp(X/A) or

        X = A*ln(WS/C) as you had it.

I'm not sure about my derivation here. Not clear that the total number
of spores should remain constant in each generation. This makes sense
though if the growth region is big enough that the edge is a plane.

Hal
Received on Fri Dec 5 12:46:47 1997

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