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Task:
2 operators are required to do a process, but not start the process until two operators are free. It is important to not seize one operator then wait for the other. Both have to be in a state of idle to begin the process, and if they are never both idle at the same time then the process won't get done.
Solution:
Control the output of the Queue of the Processor (ProcessorA). OnMessage check the situation and if two Operators are available the item can leave to the Processor. The delayed message is send at OnEntry of the Queue and OnRessourceAvailable of each Operator.
In other words the main logic is on QueueA where the output gets controlled (OnEntry, OnExit and OnMessage). Beside this everything else is Flexsim standard.
The other objects (B) are only in the model to also use the Operators.
Attention:
Do not delete or rename the object Logo_FlexsimD, because otherwise the model will not work correctly. Also do not rename, delete or re-rank other objects.
