[email protected] and the signal: evolution and game theory1/44
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Sex and the Signal: Evolution and Game Theory 2/44
Simple Foraging for Simple Foragers
Frank Thuijsman
joint work with
Bezalel Peleg, Mor Amitai, Avi Shmida
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Outline
Two approaches that explain certain
observations of foraging behavior
The Ideal Free Distribution
The Matching Law
…Risk Aversity
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The Ideal Free Distribution
Stephen Fretwell & Henry Lucas (1970):
Individual foragers will distribute themselves over various patches proportional to the amounts of resources available in each.
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The Ideal Free Distribution
Many foragers
For example: if patch A contains twice as much food as patch B, then there will be twice as many individuals foraging in patch A as in patch B.
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The Matching Law
Richard Herrnstein (1961):
The organism allocates its behavior over various activities in proportion to the value derived from each activity.
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The Matching Law
Single forager
For example: if the probability of finding food in patch A is twice as much as in patch B, then the foraging individual will visit patch A twice as often as patch B
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Simplified Model
?Yellow Blue
p qy b
Two patches
Nectar quantitiesNectar probabilities
One or more bees
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IFD and Simplified Model
Yellow Blue
y bnectar quantities:
nY nBnumbers of bees:
two patches:
IFD: nY / nB y / b
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Matching Law and Simplified Model
Yellow Blue
p qnectar probabilities:
nY nBvisits by one bee:
two patches:
Matching Law: nY / nB p / q
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How to choose where to go?
Alone …
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…or with others
How to choose where to go?
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No Communication !
How to choose where to go?
bzzz, bzzz, …
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How to choose where to go?
ε-sampling orfailures strategy!
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The Critical Level cl(t)
cl(t+1) = α·cl(t) + (1- α)·r(t)
0 < α < 1
r(t) reward at time t = 1, 2, 3, …
cl(1) = 0
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The ε-Sampling Strategy
Start by choosing a color at randomAt each following stage, with probability:
ε sample other color1 - ε stay at same color.
If sample “at least as good”,then stay at new color,otherwise returnimmediately.
ε > 0
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IFD, ε-Sampling, Assumptions
• reward at Y: 0 or 1 with average y/nY
reward at B: 0 or 1 with average b/nB
• no nectar accumulation
• ε very small: only one bee sampling
• At sampling cl is y/nY or b/nB
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ε-Sampling gives IFD
Proof:
Let P(nY, nB) = y·(1 + 1/2 + 1/3 + ··· + 1/nY) - b·(1 + 1/2 + 1/3 + ··· + 1/nB)
If bee moves from Y to B,
then we go from (nY, nB) to (nY - 1, nB + 1)
and
P(nY - 1, nB + 1) - P(nY, nB)
= b/(nB +1) - y/nY > 0
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ε-Sampling gives IFD
So P is increasing at each move,until it reaches a maximum
At maximum
b/(nB +1) < y/nY and y/(nY +1) < b/nB
Therefore
y/nY ≈ b/nB and so
y/b ≈ nY/nB
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ML, ε-Sampling, Assumptions
• Bernoulli flowers: reward 1 or 0
• with probability p and 1-p resp. at Y
• with probability q and 1-q resp. at B
• no nectar accumulation
• ε > 0 small
• at sampling cl is p or q
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ML, ε-Sampling, Movementsε
ε
1- ε
1- ε
1- p
1- q
qp
Y1
Y2
B2
B1
nY/nB = (p + qε)/ (q + pε) ≈ p/q
Markov chain
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The Failures Strategy A(r,s)
Start by choosing a color at random
Next:
Leave Y after r consecutive failures
Leave B after s consecutive failures
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ML, Failures, Assumptions
• Bernoulli flowers: reward 1 or 0
with probability p and 1-p resp. at Y
with probability q and 1-q resp. at B
• no nectar accumulation
• ε > 0 small
• “Failure” = “reward 0”
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ML and Failures Strategy A(3,2)
Now nY/nB = p/q if and only if
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ML and Failures Strategy A(r,s)
Generally: nY/nB = p/q if and only if
This equality holds for many pairs of reals (r, s)
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ML and Failures Strategy A(r,s)
If 0 < δ < p < q < 1 – δ, and M is such that (1 – δ)2 < 4δ (1 – δM), then there are 1 < r, s < Msuch that A(r,s) matches (p, q)
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ML and Failures Strategy A(fY,fB)
e.g. If 0 < 0.18 < p < q < 0.82, then there are 1 < r, s < 3such that A(r,s) matches (p, q)
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ML and Failures Strategy A(r,s)
If p < q < 1 – p, then there is x > 1such that A(x, x) matches (p, q)Proof: Ratio of visits Y to B for A(x, x) is
It is bigger than p/q for x = 1,while it goes to 0 as x goes to infinity
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IFD 1 and Failures Strategy A(r,s)
Assumptions:•Field of Bernoulli flowers: p on Y, q on B•Finite population of identical A(r,s) bees •Each individual matches (p,q)
Then IFD will appear
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IFD 2 and Failures Strategy A(r,s)
Assumptions:•continuum of A(r,s) bees•total nectar supplies y and b•“certain” critical levels at Y and B
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IFD 2 and Failures Strategy A(r,s)
If y > b and ys > br, then there exist probabilities p and q and related critical levels on Y and B such that
i.e. IFD will appear
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Attitude Towards Risk
?
1 3
2
22
2
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Attitude Towards Risk
Assuming normal distributions:
If the critical level is less than the mean, then any probability matching forager will favour higher variance
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Attitude Towards Risk
Assuming distributions like below:
If many flowers empty or very low nectar quantities, then any probability matching forager will favour higher variance
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Concluding Remarks• A(r,s) focussed on statics of stable situation; no dynamic procedure to
reach it• ε-sampling does not really depend on ε• ε-sampling requires staying in same color for long time• Field data support failures behavior
Simple Foraging?The Truth is in the Field
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?
F. Thuijsman, B. Peleg, M. Amitai, A. Shmida (1995): Automata, matching and foraging behaviour of bees. Journal of Theoretical Biology 175, 301-316.
Questions