simple foraging for simple foragers
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Simple Foraging for Simple Foragers. Frank Thuijsman joint work with Bezalel Peleg, Mor Amitai, Avi Shmida. Outline. Outline. Two approaches that explain certain observations of foraging behavior The Ideal Free Distribution The Matching Law …Risk Aversity. - PowerPoint PPT PresentationTRANSCRIPT
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
Sex and the Signal: Evolution and Game Theory 4/44
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