Re: Cost of natural selection revisited

Graham Jones wrote:
"Andrew Lang" <fireforeg@xxxxxxxxx> wrote in message
news:gpbcq9$1qhu$1@xxxxxxxxxxxxxxxxxxxxxx

g++.exe -Wall -pedantic -Wextra -Wno-unused-parameter Individual.cpp
main.cpp Options.cpp RandomGenerator.cpp Runner.cpp Species.cpp -o
nunney.exe

There were a few trivial compilation errors with these strict options. It
seems to run OK, but I haven't done anything except watch the output go by.

Thanks; I'll go fix those up.


2) Nunney compares his results to the Haldane figure of 1 substitution
per 300 generations. This is not correct; 300 is obtained from a figure
of 30K deaths divided by 0.1 selection intensity. The figure of 30 is
just a mean to account for occasional high values (according to
Haldane's paper), and the minimum given (which is 10) would be a better
match for this case, where we have gone looking for the fastest
survivable speed. Similarly Nunney's maximum selection intensity,
log(R/2), is 1.6, although that's with a zero population. Instrumenting
the code shows a typical value of 0.6 (about 3.6 children per female),
so the "Haldane reference line" should be about 16 generations per
substitution. That roughly matches what actually happens at a mutation
rate of 1 per generation per trait. If we adjust the graph that way, we
see that low mutation rates (< 1 per generation) result in substitution
speeds that are slower than the expected limit, and high rates (> 1 per
generation) result in speeds in excess of the limit.
[...]

I am not clear about your definitions here. It could be that most
females are producing .5 offspring, a few fit ones are producing 3 and the
selection intensity would be log(3/.5) = log(6). It seems this could be
larger than what you call 'Nunney's maximum selection intensity' when R=10.

All females have the same number of offspring (except to the extent that
number is non-integer, so if fecundity is 2.33, most have 2 children but
every third female has 3). Not all children survive, but this is based
on genetic fitness and is the point of the exercise.

Selection intensity is the log of (children born) / (children that
survive). For a species to maintain its numbers (as the simulated ones
do) two children must survive per female on average, in the long term.
The genes of females that only have 0.5 surviving children will die out.
Those that have 3 surviving children will become the norm and population
pressure will reduce the number to 2.


How are you implementing 'typically the fittest individuals mate with each
other'?

Sort all the males by fitness; sort all the females by fitness. Match
them one-one all the way down. There will be some leftovers of one sex
or the other; these fail to find mates. Just like real life.

Andrew

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