Mathematical Model of Courtship versus Survival

Introduction
In natural world, organisms strive to survive long enough to be able to pass on their genes to the next generation. Thus biologically speaking, every individual has two primary tasks - survival and procreation. Since natural selection would work against those individuals who do not strategize in such a manner, so it is pertinent for us to discuss only those who do. But often individuals are in dilemma when faced with having to choose between gathering food and nesting over searching and impressing potential mates.

Put differently, what strategy is best suited for resource or time allocation between the tasks of foraging versus courting? To identify the optimal approach in tackling this dilemma is the motivation behind this article...

It is also worth mentioning that we are not, in any way, suggesting that individuals in the real world actively strategize in any manner. Quoting Dobzhansky "nothing in biology makes sense except in the light of evolution", we point out that this analysis only helps us to discover the strategy of those individuals that are best adapted for survival given a particular set of circumstances.

Methods

Here we construct a simple mathematical model to elucidate the optimal strategy in time allocation between foraging and courtship.
Without loss of generality, the life span of the individual under discussion is assumed to be only two years. This is convenient as it leads to only one unknown variable – the time spent in courtship during the first year denoted 'm1' and is normalized to a value that lies between 0 and 1 where '0' implies all the time is spent in foraging and '1' implies that all the time is spent in courtship. It is to be noted that the mating effort in second year 'm2' is not a variable as the individual will go all out in favour of courtship as it does not expect to survive to year three i.e. 'm2' = 1 so we can denote variable 'm1' as just 'm'.

Since time spent in courtship denoted as 'm' is inversely related to the chances of survival to the following year denoted as 's', so the relation between these variables is as follows:  s=1-m^2

To capture the effect of all parameters on survival chance that are beyond the control of the individual i.e. irrespective of the choices made, we introduce a variable 'Smax' that captures the maximum survival chance from year one to year two, and thus its value lies between 0 to 1. Table 1 captures the list of variables used in this model.

Table 1 List of variables used in the model

Variable	What the variable denotes	Valid range
m	Mating effort (m1 – first year, m2 – second year)	0 to 1
W	Fitness of an individual	>=0
Smax	Maximum survival chance from year 1 to year 2	0 to 1
f1	Number of potential mates available in year 1	>= 0
f2	Number of potential mates available in year 2	>= 0
m*	Optimal mating effort for a given Smax	0 to 1

A strategy is said to be optimal if it maximizes the fitness of the individual denoted as 'W', measured as the number of potential mates the individual can successfully court over its entire life time, in our case it is two years. Thus, by definition, the fitness equation is given below, where 'f1' and 'f2' denote the number of potential mates available to court in the first and second year respectively. 

W = m_1 * f_1 + S_max * s * m_2 * f_2

Substituting for 'm1', 'm2', and ‘s’ we reduce the fitness equation to: 

W = m * f_1 + S_max * (1-m^2 ) * f_2

Results

To understand the relationship between the different variables used in the model, we use the fitness model and plot some results. Figure 1 presents an example plot of fitness ‘W’ as a function of mating effort ‘m’ for Smax = 1, f1=10, f2=10. We can note that the maximum fitness occurs when mating effort is 0.5 for this configuration. This result is as expected since the number of potential mates across the years is the same and maximum survival chance is 1 so it does not skew the preference towards either year.

Figure 1 Fitness (W) as a function of mating effort (m)

To find the optimum mating effort which maximizes the fitness of an individual, we adopt an analytical method. In this approach, we find the value of optimal mating effort denoted 'm*' at which a maxima occurs in the fitness equation, using the first derivative of ‘W’ with respect to ‘m’. 

dW/dm = f_1 + S_max * (-2m) * f_2

Since the slope of curve i.e. first derivative is zero at either maxima or minima or at saddle points, so we equate dW/dm to zero to find the corresponding value of mating effort m*. 

dW/dm = 0 = f_1- 2∙ S_max ∙ m* ∙ f_2

Thus we obtain the optimal value of mating effort as m^*=f_1/(2∙S_max∙f_2 ) subject to a maximum value of 1 to ensure that m* is within valid range.

m* = min(f_1/(2 ∙ S_max ∙ f_2), 1)

As expected, the optimal mating effort m* in first year is proportional to the number of mates in first year but inversely to those in second year. It is also inversely proportional to the maximum survival advantage across years which is also intuitive as m* corresponds to mating effort in first year.

Further, to verify whether it corresponds to a point of maxima and not minima, we evaluate the sign of the second derivative of ‘W’ with respect to ‘m’ to see if is negative for all valid values of variables.

(d^2 W)/dm^2 = -2 * S_max * f_2 ≤0

Since we observe that the second derivate is always negative, we can conclude that m* corresponds to a point of maximum fitness. 

Figure 2 Optimal mating effort (m*) as a function of Maximum Survival Chance (Smax)

Figure 2 captures the relationship between the optimal mating effort as a function of maximum survival chance for values of f1 =10 and f2 =10. One can observe that as Smax reduces from 1, m* increases from 0.5 until as Smax reaches 0.5 at which point m* attains 1. Since Smax skews in favour of first year as it reduces from 1, it is intuitive to expect optimum mating effort in year one to increase.

However, the effect of unequal availability of potential mates across years on the optimal mating effort as a function of Smax is also of interest to us. Figure 3 captures the relationship between the optimal mating effort (m*) as a function of both maximum survival chance and ratio of f1 to f2 varied from 10 to 0.1 as f2 is varied from 1 to 100 while f1 is fixed at 10. One can observe that when the number of potential mates is unequal across years then the critical Smax occurs at exactly 0.5 when f1 is same as f2 while before 0.5 if the f1 exceeds f2 and occurs after 0.5 if f2 exceeds f1. This behaviour of critical Smax captures the effect of unequal potential mates across the years.

m* = 1 =  f_1 / ( 2 ∙ CS_max ∙ f_2 )

We can evaluate the equation for critical maximal survival chance, denoted as CS_max and defined as the value of Smax at which optimal mating effort m* attains 1, by equating the optimal mating effort to 1 and obtaining an expression for CSmax as given below:

CS_max = f_1/ ( 2 ∙ f_2 )

Figure 3 Relationship between Optimal (m*) and Maximum survival chance (Smax) as a function of ratio of mates available in year 2 to year 1 presented  in a 3D plot

Discussion

In this brief analysis of a basic mathematical model of courtship versus foraging, we understood the relationship between the dependent variable viz. mating effort and independent variables viz. maximum survival chance, availability of potential mates across time. In particular, we were interested in the optimal value of mating effort given a particular "world" i.e. set of independent variables and its relation with those independent variables. The critical-maximum survival chance conveyed when an individual should go all out in favour of courtship during the first year without any concern towards making it to second year given the chances of survival are, in any case, very slim as conveyed by low value of critical Smax. We evaluated the expressions for optimal mating effort and critical Smax.

Further, to analyse individuals that have life expectancy more than 2 years, one can extrapolate this analysis using Bellman’s principle of optimality the essence of which is stated as "for an n-stage process to be optimal, all the remaining n-1 stages of the process resulting from the first process have to be optimal". Thus, the current analysis would be valid for the last two years of the lifespan of an individual and the analysis for preceding years would have to be evaluated in reverse.

The current attempt at analysing this "fundamental dilemma" of life viz. how to strike a balance between individual survival and species perpetuation, has given us a brief glimpse, at best, into the intricate complexities of life in the natural world, which we are just beginning to decipher.

R Source Code

• Fitness function

# func takes smax, f1, f2 as input and returns fitness in a array

fitness_search <- function(smax=1, f1=10, f2=10)

{

  m = seq(0,1,by=0.01)  # array to search for best mating effort

  w = (m * f1 + smax * (1-m^2) * f2)  # fitness equation

  return(w)  # return the fitness array found

}

• Optimal mating effort function

# Input: f1, f2; returns optimal m as a function of Smax

fitness_smax <- function(f1=10, f2=10)

{

  i=0 # index initialization

  optimal_m = 0 # variable creation

  # loop over all values of Smax while locating optimal mating effort

  for (smax in seq(0,1,by=0.01))

  {

    m = seq(0,1,by=0.01) # search array to locate the optimal value

    w = (m * f1 + smax * (1-m^2) * f2)    # fitness equation

    optimal_m[i] <- m[which.max(w)] # retrieve optimal m at current Smax

    i=i+1 # increment loop counter

  }

  return(optimal_m)  # return the optimal mating effort array

}

• Script for Three-Dimensional plot

f1 = 10 # number of potential mates in 1st year is fixed

f2 = 1:100   # number of potential mates in second year

opt_m = NULL # variable initialization

for (x in f2)   

{

# obtain optimal_m 2D plot for all values of f2=x at f1=10

opt_m = c(opt_m, fitness_smax(f1,x) ) # append for 3D plot

}

dim(opt_m) <- c(100, 100)     # update the dimension to form a matrix

persp(x=smax, xlab='Maximal survival chance Smax', y=f2, ylab='f2', z=opt_m, zlab='Optimal m', xlim=c(0,1), ylim=c(1,100), zlim=c(0,1))

title('Relation between optimal m and Smax as a function of f2')

mtext('f1=10, f2 = 1 to 100')