Processors and Overall Efficiencies

Question

question

Karina GM2 asked Sep 19, '19 Joerg Vogel answered Oct 10, '19

Processors and Overall Efficiencies

Hello, I am trying to understand how FlexSim runs production, according to my theoretical output, I am supposed to be getting 46 pieces out of my system in 3600 seconds in a one piece flow (no queues). My math is: (3600/12)*(.80)=240pcs*(.83)=199pcs... and so on (multiplication of all machines' efficiency). Is there a way I can simulate my math and get a total output of 46 at the end? Am I putting the wrong inputs? I used "Breakdown" tabs with the OE being the UPtime and the Downtime column as the Downtime. I am getting an output of 195 pcs even though some of my processors run as low as 68%. Also, I ran a trial with queues to compensate the differences in OE of the machines as a decoupled system and those seem to work fine. 1launch-metrics-no-queues.fsm

CT (Sec)

OE
100%

Output
300pcs

DownTime

12

80%

240

20%

12

83%

199

17%

12

75%

149

25%

12

70%

105

30%

12

80%

84

20%

12

68%

57

32%

12

81%

46

19%

Software Version:

FlexSim 19.1.0

processors oe one piece flow

1launch-metrics-no-queues.fsm (27.4 KiB)

Answer 1 · 2019-09-19T13:15:19Z

Joerg Vogel answered Sep 19, '19

For the first row you have a down time of 3 seconds. A total available time of 15 seconds and a process time of 12 seconds. But we don't know, how the mean time before failure is distributed. Does the down time occur for each processed item? Can the down time be accumulated for several items? Does the down occurs as a single block of 720seconds in the run time of 3600 seconds? FlexSim expects time intervals between events and not probabilities. You can say the first failure is normal distributed with a mean value of 12 seconds and deviation of 1 seconds. The failure lasts for about 3 seconds normal distributed with a deviation of 0.5 seconds. But you have chosen a distribution by percentage. This is a probability distribution for randomly time intervals:

a time interval of 5 seconds has got a probability of 20%
a time interval of 7.34 seconds has got a probability of 10%
a time interval of 8.03 seconds has got a probability of 70%

The result of the distribution is that the first failure will happen in the interval of 5 to 8.03 seconds at times of 5.00, 7.34 or 8.03 seconds distributed by the given percentages, if you let run the model several times.

You see at the example the sum is up to 100% as you can read it in the function you have edited.

The first failure time can happen equal to the mean time before failure. It is an additional field to estimate the first failure in a life time, because you are able to repair the machine. Often you will dispose products if the first failure occures.

question