A machine produces three different parts, two of which require two separate oper
ID: 3350981 • Letter: A
Question
A machine produces three different parts, two of which require two separate operations on the same machine. The batch size for all parts is 100; assume a setup is required for each batch processed. The data is given below:
The machine works 8 hours per day for 300 days per year, and has been observed to be unavailable due to maintenance, calibration or other causes 20% of its working time.
a) Compute the fraction of its uptime the machine will be busy processing the annual average workload described in the table above.
b) Assume batches arrive at the machine following a Poisson process. Compute the rate of the Poisson process from the data above, considering only the time the machine is operational
c) Compute the probability that a randomly selected batch will require one of the five operations above.
d) Compute the mean and variance of the processing time of a randomly selected batch at this machine.
e) Compute the probability of exactly one batch of any operation arriving during an 8 hour shift.
f) Compute the probability of exactly four batches of Part 3 arriving during a 40 hour period. (Hint: The arrival of batches of a given operation will follow a Poisson process; what will the rate be?)
Avg. Batches/yr Part Operation Setup time/batch (mins) Processing time/ part (mins) 80 ) 100 100 50 80 2 1.4 160 3 2Explanation / Answer
machine works =8*300=2400hrs
unavailble hours= 20*2400 =480
total number of available hours = 1920
for part 1
time in setting up =80*60= 4800 min=4800/60= 80 hours
processing time for 1 part =1 min
processing time for 80 baches= 80*100*1 =8000 min= 8000/60 =133.33 hrs.
setup time for operation 2 on part 1= 80*20 = 1600 min= 26.66 hrs.
processing time for operation 2 on part 1 = 80*100*2 =16000min = 266.66 hrs
total time for part 1 =80+ 133.33+26.66+266.66 = 506.66
for part 2
setup up time for operation 1= 100*80=8000min = 133.33 hrs
processing time for operation1 = 80*1*100 = 133.33 hrs
setup time for operation 2= 80*100/60=133.33 hrs.
processing time for operation 2= 80*1.5*100 = 200 hrs
total time for part 2 = 133.33+133.33+133.33+200 = 599.99
for part 3
setup time =160*50= 133.33 hrs
processing time= 160*2*100=533.33
total time for part 3 = 666.66
total time=506.66+599.99+666.66 =1773.22
time consumed in processing process of all the parts = 133.33+266.66+133.33+200+533.33 =1266.66
total fraction =1266.66/1773.22 = 0.714
b) total no of batches=80=80+160 =320
P(k events in interval, arrival of 320 baches in 1920 hrs.) =1
P=e- *k /k!
320/1920 =
e- *k /k!
0.16*320!=2.718- *320
2.718- *320-0.16*320!=0 this can be solved as quadratic equation
thus =2
c) 5/320==0.0156
d) mean= 133.33+266.66+133.33+200+533.33/3 =422.21
e) variance = (133.33+266.66-422.21)2+ (133.33+200-422.21)2 +(533.33-422.21)2 /3
493.28+7899.65+12348.4/3
20412.48
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