The column
for Q_Sample provides answers to the questions originally
posed. How many machines on the average will be waiting for
repair? This is given by the mean number in the queue or
3.51. The mean number in service (or actually being repaired)
is 2.50, so the total number in the system is 6.01. This
is an interesting number because it represents the average
inventory due to the maintenance operation.
How much time will a machine spend in
the repair facility? This is given by the mean time in the
system as 1.20 hours. The time dimensions are the same as
those assumed for the arrival and service rates. The system
time is broken into time in the queue (0.70 hours) and time
in service (0.50) hours. These numbers are interesting because
they describe the cycle time for the manufacturing process.
How often will the workers be idle?
The efficiency result indicates that on the average the repair
workers are busy 83.3% of the time. The state probabilities
indicate that all three workers are idle simultaneously 4.5%
of the time (P(0)), all three are busy 70% of the time (1
- P(0) - P(1) - P(2)), and the system is never full since
in this case there is no limit to the length of the queue.
The columns for Q_2 and Q_3 show the
results for systems with a finite queue and a finite population
respectively. For the finite population case, Q_3, we have
entered an arrival rate for an individual of the population
as 0.625. When the system is empty, every member of the population
arrives at this rate so the total arrival rate is 8*0.625
= 5. Thus, the third system is comparable to the other two,
which also have arrival rates of 5. |