Variability Basics - Causes of Variability
8 important questions on Variability Basics - Causes of Variability
True or false?
A: A nonpreemptive outage typically occurs between jobs, rather than during them, because we have some control as to exactly when they will inevitably occur.
B: A preemptive outage cannot occur between jobs.
A: True
B: False (i.e. a power outage could randomly occur during a setup that is in between jobs, but also during jobs)
Consider two (single machine) production stations in series (Penny Fab One). The first station is subject to random failures. It takes 0,5 hour to repair the machine and the number of failures are distributed with an average of 5 per day. Assume a production of 16 hours per day. What is the availability of station 1?
It takes 0,5 hour to repair the machine at station 1. So, MTTR = 0,5 hours
The time per day that the machine is up and running equals = 16 - (5 * 0,5) = 13,5 hours
Consequently, the MTTF equals: 13,5/5 = 2,7 hours
As a result, the avialability of the machine at station 1 equals: A = MTTF/ (MTTF + MTTR) = 0,5/ (0,5 + 2,7) =0.16
Explain the difference between Nonpreemptive Outages and Preemptive Outages.
Nonpreemtive Outages: represent downtimes that will inevitably occur but for which we have some control as to exactly when.
Preemptive Outages: Might be caused by catastrophic failure of a machine or when the machine becomes radically out of adjustment, forces a stoppage whether or not the current job is completed.
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Does describe a complete relationship between effective mean process time and natural mean process time?
Not fully, since there still relatively small detractors not considered in this relationship (i.e. operator unavailability). This means that could be underestimated when using this formula.
Suppose a machine is just repaired and starts working again at 1 PM. The machine works fine and is turned off at 5 PM. The next day it is turned on again at 8 AM and it breaks down at 1 PM. Is the time to failure 24 hours, or 9 hours?
9 hours, MTTF (or 'TTF' here, since we only have 1 failure) does not include time when a machine is off. (Can a machine fail if it is not doing anything?)
Does it matter for the SCV of effective mean process times if the availability is 80% or 20% (considering all other things being equal)? Does it matter in any other way?
1: No. Consider formula 8.6. 0,2 (1-0,2) = 0,8 (1-0,8)
2: Of course. An availability of 0,2 will have considerably less throughput for the same system as an availability of 0.2.
What is the CV of process times for a system that has exponentially distributed process times and no downtimes?
No downtime means 100% availability ->
Exponentially distributed process times -> CV = 1
Consider formulas 8.7, 8.8 & 8.9 for variability from nonpreemptive outages. Now consider a system that is totally equivalent, but also suffers from variability from rework. Obviously from 8.4.4 you should use the same formulas, but different values. Which value changes in these formulas?
Ns decreases (in comparison to the first situation there will be less parts produced in between setups).
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