Reduction of Foraging Work and Cooperative Breeding

Abstract

Using simple stochastic models, we discuss how cooperative breeders, especially wasps and bees, can improve their productivity by reducing foraging work. In a harsh environment, where foraging is the main cause of mortality, such breeders achieve greater productivity by reducing their foraging effort below full capacity, and they may thrive by adopting cooperative breeding. This could prevent the population extinction of cooperative breeders under conditions where a population of lone breeders cannot be maintained.

Appendix

1.1 E[H] for Cooperative Breeders

Let H _i be the number of dispersers reared in the interval (i, i + 1). The matured brood from the initial maturation cycle stay in the nest, so H ₀ = 0. Thus,

$$H = \sum\limits_{i = 1}^{\infty } H_{i} .$$

(7)

Assume the single cooperative breeder who founded the nest succeeded in rearing the first brood, and let H′ be the total number of dispersers by conditioning to this case. Because the first matured brood stay in the nest, there exist two adults at time 1. Indeed, until the termination of the nest, two cooperative breeders always exist at the start of each maturation cycle (see Fig. 3). In the maturation cycle (1, 2), depending on the survival of the adults, there are three possibilities: (a) two adults survive and the matured brood disperse at time 2, (b) one of the two adults dies and the brood stay in the nest, and (c) both adults die and the nest is terminated. As the death of the adults is a Poisson process with intensity μ ₂, we have

$$H^{\prime} = \left( {\begin{array}{*{20}l} {1 + H_{2} + H_{3} + \ldots } \hfill & {{\text{with probability }}e^{{ - m_{2} }} ,} \hfill \\ {0 + H_{2} + H_{3} + \ldots } \hfill & {{\text{with probability }}m_{2} e^{{ - m_{2} }} ,} \hfill \\ 0 \hfill & {{\text{with probability }}1 - e^{{ - m_{2} }} - m_{2} e^{{ - m_{2} }} .} \hfill \\ \end{array} } \right.$$

(8)

After time 2, dispersal from a nest of two cooperative breeders will be repeated stochastically in the same manner, and H ₂ + H ₃ + … is stochastically equivalent to $H^{\prime}$. Thus, taking the expectation on both sides of (8), we have

$$E[H^{\prime}] = e^{{ - \mu_{2} }} (1 + E[H^{\prime}]) + \mu_{2} e^{{ - \mu_{2} }} E[H^{\prime}].$$

(9)

Rearranging this, we have

$$E[H|{\text{the nest survives up to time 1}}] = E[H^{\prime}] = \frac{{e^{{ - \mu_{2} }} }}{{1 - e^{{ - \mu_{2} }} - \mu_{2} e^{{ - \mu_{2} }} }}.$$

(10)

Because the death of the founding single breeder is exponentially distributed with the rate μ ₁, we have

$$\begin{aligned} E[H] & = e^{{ - \mu_{1} }} E[H| {\text{the}}\, {\text{nest}} \, {\text{survives}}\, {\text{up}}\,{\text{to}}\,{\text{ time}} 1] \\ & = \frac{{e^{{ - \mu_{1} }} }}{{e^{{\mu_{2} }} - 1 - \mu_{2} }}. \\ \end{aligned}$$

(11)