Optimisation strategies from Darwinian competition?

Second NetLogo ABM

In a more realistic situation, small random mutations will alter the fertility and enrichment at each time step (or generation). A question can then be asked: what will happen to fertility and enrichment over many generations?

We outline the model in the table below and run it in NetLogo (see link after the 2nd table). Assumptions behind the model are given in Appendix 4.

We start our 2nd ABM with all the organisms/replicators (agents or turtles in NetLogo language) at the same values of fertility and enrichment. Two normal-distribution random variables with mean zero and standard deviation that is set by a parameter, are added to the fertility and enrichment of each replicator/agent. The table below sets out the operations to calculate each successive generation.

ABM operations in orderOperation details
1: MutationAdd 2 random variables to fertility f and enrichment p to each agent
2: ReproductionCopy each agent f times
3: Calculate carrying-capacity locallyC=1/n\sum p where n is the number of agents locally
4: Kill excess agents locallyexcess agents are selected randomly
5: Move agents a short distanceTurn by a small random angle and move forward by 0.3 of a ‘patch’
6: Repeatrepeat

We note that the space the agents inhabit is 2-dimensional as that gives clearer results even though the simulation is slower to run. There are 2 sliders which set the initial conditions for all agents when the ‘Setup’ button is clicked. After selecting the amount of random zero mean noise to add to fertility and enrichment with another 2 sliders, the ‘Go’ button can be clicked.

It is suggested that the first 4 runs follow the table below, before the user explores further.

Run numberInitial f , pChange in f s.d.Change in p s.d.
10.2 , 200
20.2 , 20.0250
30.2 , 200.05
40.2 , 20.0250.05

The S.D. in the table headings is shorthand for standard deviation. This 2nd model can be run at page (This page is also a 5MB download). We note the expected result for each run.

  1. All the agents/replicators start at the same initial values of fertility and enrichment and stay there for all time. The average fertility graph and average ‘enrichment’ graph are both horizontal lines. The graph of the total number of agents bounces randomly around a constant value.
  2. In this run, just the fertility experiences mutations. The average enrichment graph is a horizontal line. The average fertility climbs upwards as agents with higher fertility out-compete those with lower fertility. The histogram of fertility spreads out broadly from a single peak as mutations (both positive and negative) accrue. Naturally the agents on the right of the peak out-compete the agents on the left of the peak, so the broad peak moves rightwards.
  3. In run 3, the fertility doesn’t experience mutations. The average enrichment climbs upwards as those with higher enrichment out-compete those with lower enrichment. Accordingly the population size increases upwards. The histogram of enrichment shows a broad peak moving rightwards to ever higher enrichment. The fertility graph is horizontal.
  4. In the final run, agents fertility and enrichment are mutated at each generation. Both the average fertility and average enrichment increase without limit. The histograms show both peaks moving rightwards.