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When ants work in pulses

Recruitment only pays once enough workers are active at the same moment. A small colony that spreads its foragers evenly over time can sit below that minimum forever. The same colony, spending the same total effort in synchronized bursts, can cross it. The four views below let you test when that holds and when it does not.

Models from Bonabeau, Theraulaz and Deneubourg (1998), BioSystems 45, 195–211, with a fourth view on the idea of Bonabeau and Cogne (1996), From Animals to Animats 4, 537–544. Default parameters are those used for Tetramorium impurum.

active foragers, pulsing colonyactive foragers, steady colonycritical numberat the source, pulsing colonyat the source, steady colony
Both colonies spend the same worker time. The steady crew stays below the critical number, so recruitment never takes hold and almost nobody ends up at the source. The pulsing crew spends part of every cycle above it (shaded), recruitment builds during those windows, and foragers accumulate at the source with a short delay. Computed from Eqs. 5 and 6 with the default parameters (n = 30, T = 1196 min).

Below a critical number, recruitment cannot sustain itself

With N active foragers, the number X exploiting a source follows dX/dt = (aX + p)(N − X) − bX (Eq. 6). Recruitment gains scale with aN, losses with b. When p = 0 the source is exploited only if N > Nc = b/a. Chance discoveries (p > 0) soften the transition, and the useful threshold becomes the inflection point Ni = (b − p)/a.

Foragers at the source at steady state

X∞ (Eq. 7)p = 0 limitNiyour N

Time needed to establish exploitation

τe in minutes (Eq. 8), log scalep = 0 limit

Two things follow. Above the threshold every extra active worker adds almost one forager to the source, while below it extra workers add almost nothing. And the time to build up a recruitment peaks near the threshold, so a colony that only just crosses it needs to stay above it for a long time.

Same total effort, different timing

Two colonies each put n/2 foragers out on average. The steady colony keeps N = n/2 at all times. The pulsing colony follows N(t) = (n/2)(1 + cos 2πt/T) (Eq. 5), swinging between 0 and n. Compare how many foragers each manages to keep at the source.

Four cycles

N(t), active foragersat the source, pulsing colonyat the source, steady colonyNi

Are the conditions met?

    Where pulsing pays: colony size against period

    Effect of the period at your colony size

    mean, pulsingpeak, pulsingsteady colonyyour T

    The map shades the extra foragers at the source, on average, that pulsing buys. The advantage is confined to a band: the peak n must exceed Ni, yet the mean n/2 should not already be far above it, and the period has to be long compared with τe. Fast oscillations average out and the colony behaves as if steady. In large colonies the steady strategy already works and the gain shrinks relative to what is exploited anyway. Click the map to jump to a point.

    No external clock is needed

    Here each worker is an automaton that is inactive, active, or foraging (Eqs. 9–13). An unmet need D (hunger, say) rises at rate D0 and each forager reduces it by α per time unit. Need and already active nestmates wake inactive workers, so activity builds on itself, foraging overshoots, need collapses and the colony rests. The comparison colony keeps a fixed number of workers permanently active with the same rules.

    Colony with self-organized bursts

    active (N₂ + N₃)foragers (N₃)

    Colony with a fixed active crew

    active, held constantforagers (N₃)

    Unmet need D over time

    bursting colonyfixed crewresponse threshold Dc

    Both colonies end up with about the same average number of foragers, because D keeps rising until foraging matches D0/α. What differs is the level of need the colony must tolerate to get there. The fixed crew recruits poorly and only delivers once need is far above threshold. The bursting colony concentrates its workers, recruits well, and keeps need lower. Use "Match effort" to give the fixed crew the same average number of active workers as the bursting colony.

    Implementation note. With Eq. 10 read literally as aA(N₃ + pA) and pA = 0.03, almost no worker ever finds the source in my simulation and neither colony forages. Reading the spontaneous term as additive, (aAN₃ + pA), gives bursts about every 55 time units, a mean of about 2 foragers, and a fixed crew of 16 with mean need near 300, close to Figs. 8 to 11. This view uses the additive reading. Averages are taken over 3000 time units; the charts show the first 500.

    Troughs let a colony change its mind

    A colony that is well above threshold exploits efficiently but can be trapped: an established trail captures the recruits, so a richer source that appears later is ignored. If activity dips below the threshold between bursts, the old trail fades and the next burst starts a fresh competition that the better source tends to win. The colony gets efficiency during the peak and flexibility during the trough.

    Minimal two-source extension of Eq. 6 in Bonabeau and Cogne (1996): dXi/dt = (N − X₁ − X₂)[(a₁X₁ + a₂X₂) wi + p/2] − bXi, where recruits pick trail i with weight wi = (k + aiXi)² / Σj(k + ajXj)². If the 1996 equations differ, the numbers here will differ too.

    Pulsing colony

    N(t)on the original sourceon the richer sourceNc for the original sourcericher source appears

    Steady colony with the same average effort

    Non the original sourceon the richer source

    Are the conditions met?

      Try reducing the depth of the oscillation until the trough stays above Nc: the pulsing colony then stays locked on the original source like the steady one. Try lowering trail selectivity (larger k): trapping weakens and the steady colony can switch on its own, so oscillations matter less.