A mechanic services 5 drilling machines for a steel plate manufacturer. Machines
ID: 344266 • Letter: A
Question
A mechanic services 5 drilling machines for a steel plate manufacturer. Machines break down on an average of once every 6 working days, and breakdowns tend to follow a Poisson distribution. The mechanic can handle an average of one repair job per day. Repairs times follow an exponential distribution.
1. What is the average waiting time in the system?
2. What is the average waiting time in the queue?
3. On average, how many machines are waiting to be repaired?
4. On average, how many machines are in working order?
5. On average, how many machines are being repaired?
Explanation / Answer
Given are following details :
Machine breakdown rate = a = 1/6 = 0.1666 per working day
Machine service rate = s = 1 per day
2. Average waiting time in the queue
= = a/S x ( s – a )
= 0.1666 / 1 x ( 1 – 0.1666)
= 0.1666 / 1 x 0.8334
= 0.1999 days ( 0.20 rounded to two decimal places)
AVERAGE WAITING TIME IN THE QUEUE = 0.20 DAYS
= Average waiting time in the queue + 1/s
= 0.20 + 1/1
= 1.20 days
AVERAGE WAITING TIME IN THE SYSTEM = 1.20 DAYS
3.Number of machines waiting to be repaired
= a^2 / S x ( s – a )
= 0.1666 x 0.1666 / 1 x ( 1 – 0.1666)
= 0.1666 x 0.1666/ 0.8334
= 0.033
AVERAGE NUMBER OF MACHINES WAITING TO BE REPAIRED = 0.033
5. Average number of machines being repaired = a/s = 0.1666/1 = 0.1666
4. Average number of machines in working order = 1 – a/s = 1 – 0.1666 = 0.8334
AVERAGE WAITING TIME IN THE QUEUE = 0.20 DAYS
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