Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 356 / 2 / e

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- Solution e

Solution

An exact Gillespie algorithm for one particle is:

Set the current well $k$ and time $t = 0$ .
If $k$ is absorbing, stop.
Set $a = k_{+} + k_{-}$ and draw $Δ t = - lo g U_{1} / a$ .
Move to $k + 1$ if $U_{2} < k_{+} / a$ , and otherwise to $k - 1$ .
Set $t \leftarrow t + Δ t$ and repeat, stopping if the jump crosses an absorbing end.

For

N ≪ K

, simulate particle identities independently and maintain a priority queue of their next event times. For

N ≫ K

, store occupation numbers

n_{k}

and use aggregate event rates

n_{k} k_{+}

and

n_{k} k_{-}

for each well; one population-level Gillespie event then decrements one

n_{k}

and increments its neighbor. This replaces work proportional to particle number by work proportional to the number of occupied wells.

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