SSA performance bug regarding order of reactions

[leon9812]

If the last reaction defined in a model file is comparatively fast, then the simulation is significantly slower compared to when a slower reaction is last.

Example: consider a circular model with all equal initial populations and rate constants, but with two additional reactions with higher rate constants

S0 -> S1 (slow)
S1 -> S2 (slow)
...
Sn -> S0 (slow)

S0 -> S1 (fast)
S1 -> S0 (fast)

For such a model with 500 species and the fast reactions being defined last (last.bngl), I get the following results:

TOTAL STEPS: 2525732
CPU TIME: simulate 11.76 s.

If one slow reaction is defined after both fast reactions at the end of the file (second_to_last.bngl), then these are my results:

TOTAL STEPS: 2491560
CPU TIME: simulate 3.15 s.

[wshlavacek]

I notice that S0->S1 is defined twice in your example. Does the last instance simply replace the first instance? I don’t know the answer - check the NET file to see if the reaction is there only once or twice. I’m assuming a NET file can be generated. If only the last instance of S0->S1 is kept, that would explain why simulation is slower when the fast rule appears last. Try this instead of two rules: S0->S1 kfast+kslow Sent from my iPhoneOn Jun 28, 2023, at 12:59 AM, Leon Herrmann @.***> wrote: If the last reaction defined in a model file is comparatively fast, then the simulation is significantly slower compared to when a slower reaction is last. Example: consider a circular model with all equal initial populations and rate constants, but with two additional reactions with higher rate constants S0 -> S1 (slow) S1 -> S2 (slow) ... Sn -> S0 (slow)

S0 -> S1 (fast) S1 -> S0 (fast)

For such a model with 500 species and the fast reactions being defined last (last.bngl), I get the following results: TOTAL STEPS: 2525732 CPU TIME: simulate 11.76 s.

If one slow reaction is defined after both fast reactions at the end of the file (second_to_last.bngl), then these are my results: TOTAL STEPS: 2491560 CPU TIME: simulate 3.15 s.

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[wshlavacek]

I notice that S0->S1 is defined twice in your example. Does the last instance simply replace the first instance? I don’t know the answer - check the NET file to see if the reaction is there only once or twice. I’m assuming a NET file can be generated. If only the last instance of S0->S1 is kept, that would explain why simulation is slower when the fast rule appears last. Try this instead of two rules: S0->S1 kfast+kslow Sent from my iPhoneOn Jun 28, 2023, at 12:59 AM, Leon Herrmann @.***> wrote: If the last reaction defined in a model file is comparatively fast, then the simulation is significantly slower compared to when a slower reaction is last. Example: consider a circular model with all equal initial populations and rate constants, but with two additional reactions with higher rate constants S0 -> S1 (slow) S1 -> S2 (slow) ... Sn -> S0 (slow)

S0 -> S1 (fast) S1 -> S0 (fast)

For such a model with 500 species and the fast reactions being defined last (last.bngl), I get the following results: TOTAL STEPS: 2525732 CPU TIME: simulate 11.76 s.

If one slow reaction is defined after both fast reactions at the end of the file (second_to_last.bngl), then these are my results: TOTAL STEPS: 2491560 CPU TIME: simulate 3.15 s.

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[leon9812]

Both generated NET files contain all 502 reactions. The example also works if I remove the slow "S0 -> S1" reaction.

[jrfaeder]

Sorry for the delay in getting to this. We will have a look and get back to you next week.