Collective decisions in ants when foraging under crowded conditions

Abstract

In this paper we examine the effect of crowding on the selection of a path in the mass-recruiting ant Lasius niger. In our experiment, ants had to go from their nest to a food source by crossing a diamond-shaped bridge, giving the choice between two paths. Two types of bridges were used: the first had two branches of equal length but different width while the second had two branches of different length and width. Experiments at high traffic volume always ended up with the selection of the wider branch, even if it was longer. This result shows that overcrowding on the narrow branch plays an essential role in the mechanism underlying the choice of route in ants. A mathematical model was developed to evaluate the importance of two mechanisms that could account for this result. The first is based on the difference in travel duration between the two paths. The second is based on the repulsive interactions between workers making head-on encounters. The model shows that travel duration per se is not sufficient to explain path choice. Rather, it is the interplay between trail following behaviour and repulsive interactions that allows ants to choose the path that minimizes their travel time. When choosing a path ants thus prefer to trade time against energy. Our results demonstrate that any environmental constraint that alters the dynamics of trail recruitment can lead to the emergence of adaptive foraging decisions without any explicit coding of information by the foragers at the individual level.

Appendix

The concentration of the pheromone Cij on branch i (i=1, 2) immediately behind each choice point j (j=1, 2) changes with time t according to the equation:

$${dC_{{ij}} } \mathord{\left/ {\vphantom {{dC_{{ij}} } {dt}}} \right. \kern-\nulldelimiterspace} {dt} = q\Phi _{{ij}} {\left( t \right)} + q\Phi _{{ij\prime }} {\left( {t - \tau } \right)} - \nu C_{{ij}} {\left( t \right)}\quad {\text{with}}\quad j\prime = 3 - j$$

(2)

, where Φ _i1(t) represents the flow of foragers from the nest to the food source choosing branch i behind the choice point 1, Φ _i2(t) the opposite flow on branch i behind the other choice point, j′=3−j=2, τ the average time required for an ant to get from one choice point to the other, q the quantity of pheromone laid on the trail per forager, and νC _ij the dispersion rate of the pheromone (ν=1/mean life time of the pheromone).

At low density (without collision between ants) we have

$$\Phi _{{ij}} {\left( t \right)} = \phi _{j} {\left( t \right)}F_{{ij}} {\left( t \right)}$$

(3)

where φ ₁ is the outbound flow of foragers traveling from the nest to the food source and φ ₂ the opposite, nestbound flow. The function F _ij describes the relative attractiveness of the trail on branch i at each choice point j (Goss et al. 1989; Beckers et al. 1992a).

$$F_{{ij}} = \frac{{{\text{ }}{\left( {k + C_{{ij}} } \right)}^{2} }} {{{\left( {k + C_{{1{\text{ }}j}} } \right)}^{2} + {\left( {k + C_{{2{\text{ }}j}} } \right)}^{2} }} = 1 - F_{{i\prime j}}$$

(4)

Equation 3 describes the flow dynamics without interactions between ants. At high flow volumes, ants making an induced U-turn on a branch do not continue back to the nest or the food source when reaching the choice point but almost always turn to the other branch. The consequence of an induced U-turn is then essentially equivalent to a pushing behavior in which an ant that has just engaged on a branch after a choice point is pushed to the other branch by an ant coming from the opposite direction (Dussutour et al. 2004). Taking into account the probability to make an induced U-turn, the flow of ants arriving at choice point j and choosing branch i can then be expressed by the following formula:

$$\Phi _{{ij}} {\left( t \right)} = \phi _{j} {\left( t \right)}F_{{ij}} {\left( t \right)}{\left[ {1 - \gamma _{i} \Phi _{{ij\prime }} {\left( {t - \tau } \right)}} \right]} + \phi _{j} {\left( {\text{t}} \right)}F_{{i\prime j}} {\left( {\text{t}} \right)}\gamma _{{i\prime }} \Phi _{{i\prime j\prime }} {\left( {t - \tau } \right)}$$

(5)

with j′=3−j and i′=3−i; and with the proportionality factor γ _i quantifying the frequency of induced U-turns.

The first term on the right-hand side of Eq. 5 represents the flow of ants engaged on branch $i\,{\left[ {\phi _{{\underline{j} }} {\left( t \right)}F_{{ij}} {\left( t \right)}} \right]}$, diminished by the number of ants pushed toward the other branch i′=3−i by the ants arriving from the opposite direction ${\left[ {\Phi _{{ij\prime }} {\left( {t - \tau } \right)}} \right]}$. The second term represents the flow of ants that were engaged on branch i′ and were pushed toward branch i.

If γ ₁=γ ₂=0, Eq. 5 agrees with Eq. 3 for the case of low traffic volume.

At the stationary state the conditions are defined by: ${dC_{{ij}} } \mathord{\left/ {\vphantom {{dC_{{ij}} } {dt}}} \right. \kern-\nulldelimiterspace} {dt} = 0,\;F_{{ij}} {\left( {t - \tau } \right)} = F_{{ij}} {\left( t \right)} = F_{{ij}} ,\;\Phi _{{ij}} {\left( {t - \tau } \right)} = \Phi _{{ij}} {\left( t \right)} = \Phi _{{ij}}$, and $\phi _{j} {\left( {t - \tau } \right)} = \phi _{j} {\left( t \right)} = \phi _{{j\prime }} {\left( t \right)} = \phi$ (as the nestbound flow and the outbound flow should be equal). This implies for Eq. 2

$${dC_{{ij}} } \mathord{\left/ {\vphantom {{dC_{{ij}} } {dt}}} \right. \kern-\nulldelimiterspace} {dt} = 0 = q{\left( {\Phi _{{ij}} + \Phi _{{ij\prime }} } \right)} - \nu C_{{ij}}$$

(6)

At equilibrium we get $\Phi _{{ij}} = \Phi _{{ij\prime }} = \Phi _{i} ,\,\Phi _{{i\prime j}} = \Phi _{{i\prime j\prime }} = \Phi _{{i\prime }} \;{\text{and}}\;F_{{ij}} = F_{{ij\prime }} = F_{{i\prime }}$.

From Eq. 5 we get:

$$\Phi _{i} + \Phi _{{i\prime }} = {\left( {\phi F_{i} {\left[ {1 - \gamma _{i} \Phi _{i} } \right]} + \phi F_{{i\prime }} \gamma _{{i\prime }} \Phi _{{i\prime }} } \right)} + {\left( {\phi F_{{i\prime }} {\left[ {1 - \gamma _{{i\prime }} \Phi _{{i\prime }} } \right]} + \phi F_{i} \gamma _{i} \Phi _{i} } \right)}$$

(7)

which gives

$$\Phi _{i} + \Phi _{{i\prime }} = \phi$$

(8)

Equation 6 can then be written as 2q(Φ _i )=νC _i and $C_{1} + C_{2} = \frac{{2q\phi }} {\nu } = A$, taking into account Eq. 8, Φ _i can be rewritten as

$$ \begin{array}{*{20}c} {\Phi _{i} = \phi F_{i} {\left[ {1 - \gamma _{i} \Phi _{i} } \right]} + \phi F_{{i\prime }} \gamma _{{i\prime }} {\left( {\phi - \Phi _{i} } \right)}} \\ {\Phi _{i} = \frac{{\phi F_{i} + \phi ^{2} F_{{i\prime }} \gamma _{{i\prime }} }} {{1 + \phi F_{i} \gamma _{i} + \phi F_{{i\prime }} \gamma _{{i\prime }} }}} \\ \end{array} $$

If we take γ ₁=0 for the wide branch and γ ₂>0 for the narrow branch, we obtain:

$$\Phi _{1} = \frac{{\phi F_{1} + \phi ^{2} F_{2} \gamma _{2} }} {{1 + \phi F_{2} \gamma _{2} }}\;{\text{and}}\;\Phi _{2} = \frac{{\phi F_{2} }} {{1 + \phi F_{2} \gamma _{2} }}$$

The concentration of the pheromone on each branch immediately after either choice point of the bridge is proportional to the flow of ants passing on the branch

$$\frac{{C_{1} }} {{C_{2} }} = \frac{{\Phi _{1} }} {{\Phi _{2} }} = \frac{{\phi F_{1} + \phi ^{2} F_{2} \gamma _{2} }} {{\phi F_{2} }} = \frac{{F_{1} + \phi F_{2} \gamma _{2} }} {{F_{2} }}$$

taking γ=γ ₂ and considering the choice function (Eq. 4), we get

$$C_{1} {\left( {k + C_{2} } \right)}^{2} - C_{2} {\left( {k + C_{1} } \right)}^{2} - \gamma \phi C_{2} {\left( {k + C_{2} } \right)}^{2} = 0$$

or

$$k^{2} {\left( {C_{1} - C_{2} } \right)} + C_{1} C_{2} {\left( {C_{2} - C_{1} } \right)} - \gamma \phi C_{2} {\left( {k + C_{2} } \right)}^{2} = 0$$

(9)

Substituting A=C ₁+C ₂ in Eq. 9, we obtain

$$k^{2} {\left( {A - 2C_{2} } \right)} + {\left( {A - C_{2} } \right)}C_{2} {\left( {2C_{2} - A} \right)} - \gamma \phi C_{2} {\left( {k + C_{2} } \right)}^{2} = 0$$

(10)

which can be rewritten as

$$aC^{3}_{2} + bC^{2}_{2} + cC_{2} + d = 0$$

(11)

with

$$ \begin{array}{*{20}l} {{a{\text{ }} = \gamma \phi + 2;} \hfill} \\ {{b{\text{ }} = 2\gamma \phi k - 3A;} \hfill} \\ {{c{\text{ }} = \gamma \phi k^{2} + 2k^{2} + A^{2} ;} \hfill} \\ {{d{\text{ }} = - k^{2} A} \hfill} \\ \end{array} $$

Equation 11 is solved numerically giving us three stationary solutions. However, we are only interested by positive real solutions. Depending on the value of the parameters, Eq. 11 may have one positive solution (always stable) or three positive real solutions (two stable and one unstable).