Cycle time for duplicated machines

[helathira]

Hi,

I have been faced with a challenge at work. I have a testing machine which has an operator to load and unload. After the load the machine operates automatically. Because the machine test time is high and we are unable to meet the Target Cycle time, we have made a machine with multiple cavities. In which case my operator will just load into multiple cavities but each cavity will operate immediately after loading, irrelevant of what is happening in the other cavities. This how the cycle time calculation is done.

Using 1 cavity
Loading - 3sec
Testing - 200sec
Unloading - 3sec
Giving CT per piece if 206sec

Using 12 cavity machine
Loading 3 sec
Testing : 200/12 = 16.7sec
Unloading : 3sec
Giving cycle time per piece of 22.7sec

But my team feel that the Loading and unloading time should also be divided by 12:
Using 12 cavity machine
Loading 3/12 = 0.25sec
Testing : 200/12 = 16.7sec
Unloading : 3/12 = 0.25sec
Giving cycle time per piece of 17.2sec

Below is a pictorial explanation of the situation.

Could you kindly advice please.

 

[IRstuff]

You guys can't do this thought experiment? Even with a spreadsheet?

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[EdStainless]

Thanks IR, I had started on that.
One slight modification, This will change if you have 66 parts in a row. At that point the operator will need to stagger load/unload activity and it will effectively double the load time for each part after that.
The real question is can one operator really load these every 3 sec, or will there be some additional time moving from one cavity to the next?
The test time for each part does not change, it is just a throughput issue.
As long as there is a supply of parts they will come off at the reload speed.

= = = = = = = = = = = = = = = = = = = =
P.E. Metallurgy

 

[IRstuff]

With a 12-cavity machine, they're stuck after 12; they have to wait 167 seconds before the first one finishes and they can unload, so if the machine is used continuously, it'll be (n*206+33)/(n*12), where n is the total number of dozens tested. As n goes to infinity, the average time is 206/12 = 17.2 seconds. Of course, this assumes the operator keeps up.

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[EdStainless]

Thanks IR, I presumed that we had multiple machines, bad assumption.


= = = = = = = = = = = = = = = = = = = =
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[IRstuff]

If they were really scaling, then sure, there would be multiple machines, but the operator couldn't handle more than 4 additional machines, since he only ever has 167 seconds of slack time -- 4*33 = 132 seconds. That would be an absurdly grueling work shift, reminiscent of the worker drones in the movie "Metropolis" One would hope that it would get automated,

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[EdStainless]

Even with a 12 bay machine I would think that having a loading magazine and then automating the insertions would be the way to go.

I made my wife watch Metropolis, let's just say that she isn't the fan that I am.

= = = = = = = = = = = = = = = = = = = =
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[IRstuff]

Agreed, there's not much fun in a 3-second operation for a human operator.

Yea, my wife's not much of a fan either, nor the rest of my family, although they simply think there's plenty to watch without resorting to a silent, B/W, movie.

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[helathira]

Thank you for the detailed explanation. From your calculation shown the below formulation would be suitable...

To calculate cycle time for a continuous production
Loading : 3 sec/12 = .25sec
Testing : 200sec/12 = 16.67sec
Unloading : 3/12 = 0.25sec
Cycle time per part : 17.2sec

But I assume that the operators time would be 6sec for every part.

Appreciate your support.

 

[IRstuff]

Yes, the starting 3 seconds of the 2nd is pipelined immediately after the 3 second load for the 1st unit, and does not add to the cycle time, per se, only the last 3 seconds sticks out

t=3 #1 loaded
t=6 #2 loaded
t=9 #3 loaded
t=203 #1 cycle complete
t=206 #1 unloaded
t=206 #2 cycle complete
t=209 #2 unloaded
t=209 #3 cycle complete
t=212 #3 unloaded

total time is 206 for #1, 6 additional for #2 and #3, as shown in my first timing chart

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