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Hello Tim, Good job with this amazing peace of code. I was wondering If you could give me a hand with some doubts that i have.
I have to set up a maintenance task between two tasks, but how can I do that without knowing their order(tasks order)
transfer = dict()
for k, v in malha.items():
list = []
--if(k in hectos):
----for i in range(v['volume']):
----list.append(S.Task('t' + k + str(i+1), length=1))
----transfer[k] = o
--else:
----for i in range(v['volume']):
----list.append(S.Task('t' + k + str(i+1), length=1))
----transfer[k] = o
This is the result of code above : a dict where the values are a list of tasks.
{'STE': [tSTE1, tSTE2, tSTE3],
'BRZ': [tBRZ1, tBRZ2, tBRZ3],
'ANT': [tANT1, tANT2, tANT3]
'BOH': [tBOH1, tBOH2, tBOH3, tBOH4]
}
Tasks of the same value list must be scheduling in sequence until a maximum of 6.
When ending this pack of 6 and the next tasks is not in the same value list, it must put a break of one perid between this tasks.
There is only two resources to deal with this tasks
Resource 2 can schedule ( STE, BRZ, ANT, BOH)
resource 1 can schedule (ANT, BOH)
The text was updated successfully, but these errors were encountered:
Hello Tim, Good job with this amazing peace of code. I was wondering If you could give me a hand with some doubts that i have.
I have to set up a maintenance task between two tasks, but how can I do that without knowing their order(tasks order)
transfer = dict()
for k, v in malha.items():
list = []
--if(k in hectos):
----for i in range(v['volume']):
----list.append(S.Task('t' + k + str(i+1), length=1))
----transfer[k] = o
--else:
----for i in range(v['volume']):
----list.append(S.Task('t' + k + str(i+1), length=1))
----transfer[k] = o
This is the result of code above : a dict where the values are a list of tasks.
{'STE': [tSTE1, tSTE2, tSTE3],
'BRZ': [tBRZ1, tBRZ2, tBRZ3],
'ANT': [tANT1, tANT2, tANT3]
'BOH': [tBOH1, tBOH2, tBOH3, tBOH4]
}
The text was updated successfully, but these errors were encountered: