Reasons why process instances can end quickly if there is without waiting state

Hi, camunda

I know that if there is without stopping at a wait state on the process instance, it is not written to the runtime database, which can reduce the decreases load.
(https://camunda.com/best-practices/_book/#_understanding_the_influencing_factors)
In the above case, is it possible to perform fast execution without the process of being recorded in the runtime database because instance information is cached in DeploymentCache?
(DeploymentCache: https://docs.camunda.org/manual/7.14/user-guide/process-engine/deployment-cache/)
Regards,
Choi

Hi @H_C,

if you don’t have any wait states, the final state of the process instance (with all variables) will be saved in the history once it is completed.

The deployment cache saves only process definitions, the instances are objects in the Java heap.

Hope this helps, Ingo

Hi, @Ingo_Richtsmeier,

Thank you for your answer.
In the case of a process instance without a wait state, where is the state and variable, i.e., process instance information during execution stored?
I want to know why I can keep all the information of the process instance without storing it in the DB.
Is it a ProcessDefinitionEntity?
I want to check with code, but I couldn’t find it properly.

Regards,
Choi

Hi @H_C,

I assume that the process instance lives as a ProcessInstanceEntity in the Java heap.

If you want to gain more insight, you can start the JUnit test from this template (https://github.com/camunda/camunda-engine-unittest) in a debugger.

Hope this helps, Ingo

Hi @Ingo_Richtsmeier,
Thank you for your reply.
I will check your advice.

Regards,
Choi