In a factory, the machines breakdown on an average rate of 8 machines per hour.
ID: 3291976 • Letter: I
Question
Explanation / Answer
Solution
Given,
the rate of breakdown, = 8/hr
Work hours = 8 per day
Service rate, µ:
Crew 1: µ1 = 2/hr [30 minutes per repair => service rate = 60/30 = 2]
Crew 2: µ2 = 9/hr
Crew 1: µ3 = 10/hr [6 minutes per repair => service rate = 60/6 = 10]
Service cost per hour:
Crew 1: 50 AED [1 mechanic at 50 AED per mechanic]
Crew 2: 120 AED [3 mechanic at 40 AED per mechanic]
Crew 3: 160 AED [4 mechanic at 40 AED per mechanic]
Machine idle time cost per hour = 50 AED.
Back-up Theory
Approach
With each crew, the average total time spent in the system is computed employing M/M/1 Theory and multiplying it by idle time cost per hour, total cost of idle time is obtained.
Then, the total cost per hour = total cost of idle time + service cost for the crew. Decision criterion would be the minimum total cost.
Machine idle time = waiting time + repair time [= total time spent in the system, v, in Queueing Theory parlance]
For M/M/1, E(v) = {1/(µ - )}.
Now, to work out the solution,
Crew 1: Since average service rate of 2 per hour is much less than the average arrival (machine breakdown) rate, the queue will explode and hence the total cost would be huge.
Hence, Crew 1 is ruled out.
Crew 2: E(v) = {1/(µ - )} = 1/(9 - 8) = 1 hour and hence total cost = 50 + 120 = 170 AED.
Crew 3: E(v) = {1/(µ - )} = 1/(10 - 8) = ½ hour and hence total cost = 25 + 160 = 185 AED.
Since 170 AED is least, the optimum decision is: Crew 2 should be considered. ANSWER
[Note: Since all parameters are converted to per hour basis, the fact that company works 8 hours a day need not be taken into account.]
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.