In: Statistics and Probability
A manufacturing company employs a maintenance crew to repair its machines as needed. Management now wants a simulation study done to analyse what the size of the crew should be, where the crew sizes are 2, 3, and 4. The time required by the crew to repair a machine has a uniform distribution over the interval from 0 and twice the mean, where the mean depends on the crew size. The mean is 4 hours with two crew members, 3 hours for three crew members, and 2 hours with four crew members. The time between breakdowns of some machine has an exponential distribution with a mean of 5 hours. When a machine breaks down and so requires repair, management wants its average waiting time before repair begins to be no more than 3 hours. Management also wants the crew size to be no larger than necessary to achieve this.
(a) Consider the case of a crew of size 2. That means that the
corresponding repair time has the uniform distribution between 0
and 8 hours.
Assume that the following numbers represent times between
breakdowns (in minutes):
100; 120; 90; 230; 20; 60; 280; 220; 110; 20; 270; 220; 140; 90;
50; 260; 240
Assume also that the following numbers represent repair times (in
minutes):
80; 170; 130; 370; 300; 120; 420; 80; 190; 250; 120; 180; 310; 460;
30; 170:
Simulate this system by hand for 20 hours. (That is stop the
simulation process when the simulation time is >= 20 hours or
1200 minutes.) Indicate the number of machines that have completed
their delay in the queue and compute the time-average number of
machines in the repair queue and other relevant performance measure
of this queuing system.
Solution
For ease in presentation, the following codes are used in the following Simulation Table:
BD: Break-down; IB: Inter-breakdown Time; BT: Break-down Time;
RT: Repair Time; RS: Repair Start Time; RE: Repair End Time;
WN: Number of machines waiting for repair; WT: Waiting Time of machines;
ST: Cumulative Simulation Time; IT: Idle Time of Repair Crew.
All times are in minutes
If BT of a particular BD# > RE of the preceding BD#, the BD machine goes for repair straight without ant wait. but, the repair crew is idle for duration of (BT of a particular BD# - RE of the preceding BD# implying there mus be an entry under IT.
If BT of a particular BD# < RE of the preceding BD#, the BD machine has to wait up to the RE of the preceding BD# implying there must be an entry under WN and WT.
BD# |
IB |
BT |
RT |
RS |
RE |
WN |
WT |
ST |
IT |
Remarks |
1 |
100 |
100 |
80 |
100 |
180 |
180 |
100 |
Till the first breakdown occurs at 100 minutes, the repair crew is idle |
||
2 |
120 |
220 |
170 |
220 |
390 |
390 |
40 |
IT: 180 to 220 |
||
3 |
90 |
310 |
130 |
390 |
520 |
1 |
80 |
520 |
WT: 310 to 390 |
|
4 |
230 |
540 |
370 |
540 |
910 |
910 |
20 |
IT: 520 to 540 |
||
5 |
20 |
560 |
300 |
910 |
1210 |
1 |
350 |
1210 |
WT: 560 to 910 |
Since Cumulative Simulation Time has crossed, simulation is stopped as directed in the question.
Out of 5 breakdowns, 2 machines waited and hence average number of machines waiting
= 2/5 = 0.4 Answer 1
Total waiting time is 430 for 5 breakdowns and hence average waiting time
= 83 minutes Answer 2
Total waiting time is 160 over a total of 1210 and hence idle time percentage
= 13.22% Answer 3
DONE