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Routing setup scenario-REM

Oct 10, 2016 at 01:58 PM

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There are operations # 10 and 20.

Once the component is processed in opn# 10 there is a "cooling time* " of say 1 hr for each component before it could start in opn# 20.

Say Machine time = 1 min (opn #10 and # 20)

So each component that is processed in opn# 10,could start in opn# 20 only after "1 Hr".

OPN# 10.

Start Time: 8.00 am

Finish Time : 8.01 am ( 8.00 am + 1 min)

OPN# 20

Start Time : 9.01 am (start 1 hr* later (8.01 + 1 hr))

Finish Time : : 9.02 am (9.01 am + 1 min)

If the qty = 10 nos.

Each component should start in the opn# 20 only "1" hr after it is processed in opn# 10.. (The 2nd component would start & finish at 8.01 & 8.02 am (opn#10) and start & finish at 9.02am & 9.03 am (opn# 20) and so on..).

1) How do we set this up in routings- If opn# 10 and 20 are in same routing.?

2) If operations are in 2 different routing ,where component B is processed 1st in R2 and then header material A is processed in R1..?

R2- Routing for component B .(only 1 operation in routing)

R1- Routing for Header material A that uses component B (Only 1 operation in routing)


Extra Note:

In the above example both opn# 10 and 20 had same machine time = 1 min.

*However , If the machine time of opn# 20 = 5 mins, the 2nd component after "Cooling time" will be scheduled to start based on the availability of OPn# 20 (work center) (i.e) start only at 9.06 am.even if it ready at 9.02 am.

I am thinking of overlapping operation..in routing.

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2 Answers

Ritesh Dube Oct 12, 2016 at 02:02 PM
1

S S,

I read your operation timings as 1 hr and 1 minute instead of 1 minute as after every operation you have cooling period , it means for that 1 hour machine is not available and its still being used by the first component processed earlier.

You can use operation & sub operation to define this bifurcation of execution and cooling.

check and reply.

Thanks

Ritesh

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Solomon Macwan Oct 12, 2016 at 08:57 AM
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Hi SS

Try std move time in routing if one routing.

if different routing than use std queue time.

for overlapping-

lets say you start second operation once 5 pcs of the first operation completed. in that case you can use min send ahead qty.

or lets say after starting first operation, second operation time should start after 10 min. than in that case you maintain overlap time.

hope this is clear.

regards,

Solomon

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