Roff D.A. Modeling Evolution: An Introduction to Numerical Methods
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Table 6.4
Calculating the expected distribution of patch choice for the foraging model at time
t = 2
Probability of x =
P(x, t)
State,
z
Proportion,
P(z, t – 1)
Best
Patch 3 4 5 6 7 8 9 10
P
10
x¼4
Pðx; tÞPð
z; t 1Þ
Predicted
proportiona
a
4 0.1
3 0.412 0 0 0 0 0.588 0 0
0.059
0.062
...................................................................................................................................................
.................................................................................................
5 0.1
3 0.020 0.392 0 0 0 0
0.588 0 0.098
0.104
...................................................................................................................................................
.................................................................................................
6 0.2
3 0.020 0 0.392 0 0 0 0 0.588
0.196
0.207
...................................................................................................................................................
.................................................................................................
7 0.3
3 0.020 0 0 0.392 0 0 0
0.588 0.294
0.311
...................................................................................................................................................
.................................................................................................
8 0.2
2 0.004 0 0 0 0.5976 0 0
0.3984 0.199
0.211
...................................................................................................................................................
.................................................................................................
9 0.1
2 0.004 0 0 0 0 0.5976 0
0.3984 0.100
0.105
...................................................................................................................................................
.................................................................................................
0.946
1.000
Note: The column
x =3 is ignored because animals in this category do not survive.
a
Predicted proportion = value in preceding cell divided
by the sum of the preceding column (0.946), that is, 0.07 =
0.059/0.946.
The predicted proportion in patch 3 is 0.062 + 0.104
+ 0.207 + 0.311 = 0.684.
The predicted proportion in patch 2 is 0.211 + 0.105 =
0.316.
where I is the lower integer value and x
I
is the values of x for index value I. The new
value of x is x x
I
¼ q. The value of F(x, t) with respect to the new origin is simply
½Fðx
Iþ1
; tÞFðx
I
; tÞq ð6:12Þ
And to translate back to the original scale we add F(x
I
, t) to give
Fðx; tÞ½Fðx
Iþ1
; tÞFðx
I
; tÞq þ Fðx
I
; tÞ¼qFðx
Iþ1
; tÞþð1 qÞFðx
I
; tÞð6:13Þ
Examples of this are given in Scenarios 3 and 5. The above formula is problematic
when x
I
¼ Xmax, because in this case x
I þ1
is greater than Xmax and will cause a
“subscript too large” error, even though q
x
is zero, and hence this term does not
contribute. To resolve this, insert a test such as x
Iþ 1
¼ min(x
Iþ 1
, Xmax).
The interpolation formula can be extended to two state variables, x and y:
Table 6.5 Example of a non-integer state distribution and its conversion to an integer index
State interval Midpoint = State Index Converting state to index
0.1−0.3 0.2 1 1+(0.2−0.2)/0.2=1
...................................................................................................................................................
0.3−0.5 0.4 2 1+(0.4−0.2)/0.2=2
...................................................................................................................................................
0.5−0.7 0.6 3 1+(0.6−0.2)/0.2=3
...................................................................................................................................................
0.7−0.9 0.8 4 1+(0.8−0.2)/0.2=4
...................................................................................................................................................
0.9−1.1 1.0 5 1+(1.0−0.2)/0.2=5
q = x–x
I
0
0
x
I + 1
−x
I
x
I + 1
x
I
x
Slope =
F(x
I + 1
, t) − F(x
I
, t)
F(x
I + 1
, t) − F(x
I
, t)
F(x
I + 1
, t)
F(x
I
, t)
F(x, t)
F(x, t) − F(x
I
, t)
Figure 6.2 Linear interpolation to account for non‐integer values in the state variable.
The value of F(x, t, T) is approximated by a straight line drawn through the two
enclosing integer values of the state variable.
DYNAMIC PROGRAMMING 359
Fðx
I
; y
J
; tÞq
x
q
y
Fðx
Iþ1
; y
Jþ1
; tÞþq
x
ð1 q
y
ÞFðx
Iþ1
; y
J
; tÞþ
ð1 q
x
Þq
y
Fðx
I
; y
Jþ1
; tÞþð1 q
x
Þð1 q
y
ÞFðx
I
; y
J
; tÞ
ð6:14Þ
where q
x
¼ x x
I
and q
y
¼ y y
I
.
We now consider five scenarios showing how the basic patch-foraging model
can be modified to fit a range of biological situations.
6.2 Summary of scenarios
Scenario 1: In this scenario the fitness measure is changed from survival to the
accumulated value of some state variable, such as energy store, at the end of the
time period.
Scenario 2: The basic scenario considered in Chapter 1 is based on foraging among
patches. In the second scenario I show how “patches” can be considered simply as
“options” and hence how the basic scenario can be readily expanded.
Scenario 3: In the previous scenarios there were unique choices at each step,
the state variable could be directly related to its position in the relevant
vector and took integer values. In this scenario the problem of equivalent
choices is considered and the concepts of indexing and interpolation are illustrated.
Scenario 4: The state variable in all the above scenarios increased over time (e.g.,
accumulation of resources). In many cases fitness may be equated to the number
of eggs laid over the time period, in which case the state variable decreases over
time. This circumstance is illustrated using host choice in parasitoid wasps.
Scenario 5: This scenario considers the problem of solving for two state variables,
in this case egg size and number.
6.3 Scenario 1: A different terminal fitness
In the patch-foraging model examined above it was assumed that fitness was equal
to survival at the terminal period. In many cases fitness will depend not simply
upon survival but also upon the accumulation of resources, which might be used
to survive the following period or invested in offspring.
6.3.1 General assumptions
Assumptions 1–5 are the same as previously given.
6. Fitness is a function of the state variable at the end of some time interval (e.g., in
a small terrestrial vertebrate this may be the summer period, the end of the season
being the time at which hibernation is entered, or it could be the reproductive
episode).
360 MODELING EVOLUTION
6.3.2 Mathematical assumptions
Assumptions 1–5 are the same as previously given.
6. Fitness is equal to the expected value of the state variable at time T.
6.3.3 Outcome chart and expected lifetime fitness function
This is exactly the same as previously given.
6.3.4 Calculating the decision matrix
At time T we assume that the animal is alive and hence its fitness according to
assumption 6 above is equal to the value of the state variable. Thus the only change
we have to make to the coding in R is to adjust the terminal fitness function from
F.vectors[Xmin:Xmax,2] <-1
to
F.vectors[Xmin:Xmax,2] <- seq(from¼Xmin, to¼Xmax)
or in MATLAB
F_vectors[Xmin:Xmax,2] ¼ 1
to
F_vectors[Xmin:Xmax,2] = Xmin:Xmax
that is, the terminal fitness is now equal to the state variable, X. This principle is quite
general and enables the previous coding to be used for any number of terminal fitness
definitions. The effect of changing the fitness definition in this instance is to change
the decisions for animals late in the season and already of high value (Table 6.6).
6.4 Scenario 2: To forage or not to forage when patches
become options
This example illustrates how the basic patch-foraging model can be modified to fit
a scenario in which “patches” are equated to options. The scenario is based
on Kokko (2007), who presents MATLAB coding to analyze the model: For the
interested reader I suggest comparing the two approaches, simply to see how
the same result can be obtained by visually very different coding. The general
scenario is that of a bird that must gather resources during a given period, say
a winter day, to survive the following period, say the winter night.
6.4.1 General assumptions
1. During the day the bird can decide to forage or remain sedentary during any
given period.
2. A non-foraging bird risks losing condition.
DYNAMIC PROGRAMMING 361
3. A foraging bird faces a predation risk that is a function of its condition, well-fed
birds being more susceptible.
4. Fitness is a function of condition at the end of the period (day).
6.4.2 Mathematical assumptions
1. The condition of a bird ranges in unit increments from 1 to 7.
2. A bird in condition 1 is dead.
3. A non-foraging bird losing one unit of condition with some fixed probability,
Ploss.
4. A foraging bird gains one unit of condition with some fixed probability, Pgain.
5. A foraging bird suffers a mortality probability from predators at a rate that is a
linear function of condition:
Pmortality ¼ a þ bx ð6:15Þ
Table 6.6 Decision matrices for foraging model with different fitness criteria (survival vs. expected
state value, the fitness criteria considered in Scenario 1)
State (Shown along bottom row)
Time [,1] [,2] [,3] [,4] [,5] [,6] [,7]
[1,] 3333221
...................................................................................................................................................
[2,] 3333221
...................................................................................................................................................
[3,] 3333221
...................................................................................................................................................
[4,] 3333221
...................................................................................................................................................
[5,] 3333221
...................................................................................................................................................
[6,] 3333221
...................................................................................................................................................
[7,] 3333221
...................................................................................................................................................
[8,] 3333221
...................................................................................................................................................
[9,] 3333221
...................................................................................................................................................
[10,] 3333221
...................................................................................................................................................
[11,] 3333221
...................................................................................................................................................
[12,] 3333221
...................................................................................................................................................
[13,] 3333222
...................................................................................................................................................
[14,] 3333221
...................................................................................................................................................
[15,] 333321,21
...................................................................................................................................................
[16,] 33331,21,21,2
...................................................................................................................................................
[17,] 3 3 3 1,3 1,2 1,2 1,2
...................................................................................................................................................
[18,] 3 3 1,3 1,3 1,3 1,3 1,2
...................................................................................................................................................
[19,] 3 1,3 1,3 1,3 1,3 1,3 1,3
...................................................................................................................................................
[20,] 45678910
Note: Where decisions differ, the second fitness criterion is given second.
362 MODELING EVOLUTION
where mortality at X ¼ 2 is 0 and Pmax at x ¼ Xmax, the constants a and b being
determined accordingly.
6. A non-foraging bird is not susceptible to predators.
7. Fitness is the expected state value (i.e., condition) at the end of the day.
6.4.3 Outcome chart and expected lifetime fitness function
Although, at first glance, this model may not appear to resemble the patch-
foraging model it is actually mathematically equivalent. First, we note that a
“patch” in this scenario is simply the decision to “not forage” (patch 1) or to
“forage” (patch 2). Second, the benefits in the previous foraging model are repre-
sented here by the loss or gain condition. Thus we can write the model using the
same outline as in the patch-foraging model.
The outcome chart can thus be written as (“—” indicates that this condition is not
applicable).
The expected lifetime fitness function is the same as the patch-foraging model
minus the Cost component.
6.4.4 Calculating the decision matrix
The changes to the program are shown in bold font. The most significant change
is the addition of the state-dependent mortality function. To code this function we
construct a matrix of two rows and Xmax columns. Row 1 contains the probability
of being depredated if not foraging, which is zero for all values of x, coded by
Pnoforage <- rep(0,Xmax)
Row 2 contains the probability of being depredated if foraging. It is zero in column
1, because that corresponds to dead animals. According to assumption 5 it is zero
in column 2, the minimum condition, and Pmax in column 7. To construct this
function we make use of the seq function:
Patch type Benefit Pbenefit Pmortality Cost
1=“Not forage” −1 Ploss 0 Not applicable
...................................................................................................................................................
2=“Forage” 1 Pgain a + bx Not applicable
Forage x >Xcritical Survives x
Yes — Yes min(x + Benefit
i
, Xmax)
...................................................................................................................................................
Yes — No Set at Xcritical
...................................................................................................................................................
No Yes — x + Benefit
i
...................................................................................................................................................
No No — Set at Xcritical
DYNAMIC PROGRAMMING 363
Pforage <- c(0,seq(from¼Pmin, to¼Pmax, length¼Xmax-1))
We then bind the two vectors together to form the predation matrix:
Pmortality <- rbind(Pnoforage,Pforage)
The probability of mortality is then passed to the fitness function as Pmortality
[i,X].
Other changes consist of ignoring the parameter Cost (we could have deleted it
from the program but as it is not used there is no harm in leaving it in the
program) and resetting the initial values. Here we set Ploss ¼ 0.4 and Pgain ¼ 0.8
using the vector Pbenefit:
Pbenefit <– c(0.4,0.8) # Probability of “Benefit”
Benefit <–c(1,1) # “Benefit”
Finally, as in Scenario 1, we set the terminal fitness to be equal to the expected
state value:
F.vectors[Xmin:Xmax,2] <- seq(from ¼ Xmin, to ¼ Xmax)
The program is thus
R CODE:
rm(list¼ls()) # Remove all objects from memory
# Function to calculate fitness when organism is in state X
FITNESS <- function(X, Xcritical, Xmax, Xmin, Cost, Benefit,
Pbenefit, Pmortality, F.vectors)
{
# Benefit gained
# State in patch if “Benefit” gained. Note that “Benefit” can be -1
X.Food <-Xþ Benefit # Note that Cost is omitted
# If X.Food greater than Xmax then X.Food must be set to Xmax
X.Food <- min(X.Food, Xmax)
# If X.Food less than or equal to Xcritical then set to Xcritical
X.Food <- max(X.Food, Xcritical)
# State in patch if forager does not gain “Benefit”
X.NoFood <- X # Note that Cost is omitted
# If X.NoFood is less than Xcritical set X.NoFood to Xcritical
X.NoFood <- max(X.NoFood, Xcritical)
# Now gather terms together
Term1 <- Pbenefit*F.vectors[X.Food,2] # If benefit gained
Term2 <- (1-Pbenefit)*F.vectors[X.NoFood,2] # If no benefit
gained
W <- (1 - Pmortality)*(Term1 þ Term2) # Fitness in patch
return(W) # Return Fitness
} # End of function
# ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ...
# Function to iterate over patches
OVER.PATCHES <
- function(X, F.vectors, Xcritical,Xmax, Xmin,
Npatch,
Cost, Benefit, Pbenefit, Pmortality)
364 MODELING EVOLUTION
{
RHS <- matrix(0,Npatch,1) # Set matrix for Right Hand Side of equn
for (i in 1: Npatch) # Cycle over patches
{
# Call Fitness function. Note that mortality a function of patch and X
RHS[i] <- FITNESS(X, Xcritical, Xmax, Xmin, Cost, Benefit[i],
Pbenefit[i], Pmortality[i,X], F.vectors)
} # End of i loop
# Now find optimal patch Best row is in Best[1]
Best <- order(RHS, na.last¼TRUE, decreasing¼TRUE)
F.vectors[X,1] <- RHS[Best[1]]
Best.Patch <- Best[1]
# Concatenate F(x,t,T) and the optimal patch number
Temp <- c(F.vectors[X,1], Best.Patch)
# Add Temp to bottom of F.vectors and rename to Temp
Temp <- rbind(F.vectors, Temp)
return (Temp)
} # End of function
# ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ....
# Function to iterate over states of X
OVER.STATES <- function(F.vectors, Xcritical, Xmax, Xmin, Npatch,
Cost, Benefit, Pbenefit, Pmortality)
{
These lines are the same as in the patch-foraging model:
} # End of function
# ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ....
# MAIN PROGRAM
# Initialize parameters
Xmax <- 7 # Maximum value of X
Xcritical <-1 # Value of X at which death occurs
Xmin <- Xcriticalþ1 # Smallest value of X allowed
Cost <- 0.0 # Dummy not required but kept
# Probability of mortality if foraging
Pmin <-0
Pmax <- 0.01
# Create mortality function. Make Pmin at state 2
# Probability of mortality if not foraging
Pnoforage <- rep(0,Xmax)
# Foraging mortality
Pforage <- c(0,seq(from¼Pmin, to¼Pmax, length¼Xmax-1))
Pmortality <- rbind(Pnoforage,Pforage) # Mortality function
# Probability of foraging
Pbenefit <- c(0.4,0.8) # Probability of “Benefit”
Benefit <- c(-1,1) # “Benefit”
Npatch <
- 2 # Number of patches ¼ resting
or foraging
DYNAMIC PROGRAMMING 365
Horizon <- 6 # Number of time steps
# Set up matrix for fitnesses
# Column 1 is F(x,tþ1). Column 2 is F(x,t)
F.vectors <- matrix(0, Xmax,2)
F.vectors[Xmin:Xmax,2] <- seq(from¼Xmin, to¼Xmax) # Final wts
# Create matrices for output
FxtT <- matrix(0,Horizon,Xmax) # F(x,t,T)
Best.Patch <- matrix(0,Horizon,Xmax) # Best patch number
# Start iterations
Time <- Horizon # Initialize Time
while ( Time > 1)
{
Time <- Time -1 # Decrement Time by 1 unit
# Call OVER.STATES to get best values for this time step
Temp <- OVER.STATES(F.vectors, Xcritical, Xmax, Xmin,
Npatch, Cost, Benefit, Pbenefit, Pmortality)
# Extract F.vectors
TempF <- Temp[,1:2]
# Update F1
for ( J in Xmin: Xmax) { F.vectors[J,2] <- TempF[J,1]}
# Store results
Best.Patch[Time,] <- Temp[,4]
FxtT[Time,] <- Temp[,3]
} # End of Time loop
# Output information. For display add wts to last row of matrices
X <- seq(from¼1, to¼Xmax)
Best.Patch[Horizon,] <-X
FxtT[Horizon,] <-X
Best.Patch[,Xmin:Xmax] # Print Decision matrix
signif(FxtT[,Xmin:Xmax],3) # Print Fxt of Decision matrix: 3 sig
places
OUTPUT:
Table 6.7 Decision matrix for the daily foraging model (Scenario 3)
State (Shown along bottom row)
Time [,1] [,2] [,3] [,4] [,5] [,6]
[1,] 222211
........................................................................................................................
[2,] 222221
........................................................................................................................
[3,] 222221
........................................................................................................................
[4,] 222222
........................................................................................................................
[5,] 222222
........................................................................................................................
[6,] 234567
Note: Decision “1” means do not forage, whereas decision “2” means forage.
366 MODELING EVOLUTION
The decision matrix is shown in Table 6.7. Birds beginning in states 2–5 forage
throughout the day but birds in state 6 do not forage in the first period and birds in
state 7 do not forage for the first three periods.
MATLAB CODE: See Section 6.9.4.
6.5 Scenario 3: Testing for equivalent choices, indexing, and
interpolation
Thus far we have assumed that among the choices at each time step that there will
be one that is superior to all others. However, it is possible that two or more
choices may be equivalent in which case the programmer must decide if there is a
method to resolve the tie or if indeed the choice is simply made by “tossing a
coin.” The present scenario is taken from chapter 3 of Mangel and Clark (1988).
The object of the analysis is to determine the circumstances in which it would
profit animals to hunt in packs rather than singly. The specific case studied
by Mangel and Clark was that of lions hunting Thomson’s gazelle or zebra.
Parameters used here are for Thomson’s gazelle. The scenario can be placed
within the framework of the patch-foraging model by noting that patches can
be equated with pack (¼ pride in lions) size (i.e., a group of size 2 is a “patch” of
value 2). In addition to the question of equivalence of choices, this scenario
introduces two other new components: the transformation between index
value and state variable value; the use of interpolation to estimate fitness when
the index value of the state variable is not an integer.
6.5.1 General assumptions
1. The animal can hunt singly or in packs, changing pack size at each time
increment.
2. Each animal has a daily food requirement.
3. An animal whose gut contents fall below a critical value dies.
4. Food within the pack is shared equally.
5. Capture probability increases with the size of the pack (possibly to a limit).
6. Fitness is a function of the probability of surviving some specified time interval.
6.5.2 Mathematical assumptions
1. Pack (patch) size varies from 1 to N.
2. Each prey animal is of a fixed size, Y.
3. The daily food requirement per individual is the same, irrespective of pack size.
4. The probability of making a kill, p
i
, is a function of pack size, i.
DYNAMIC PROGRAMMING 367