Speedy Oil provides a single-server automobile oil change and lubrication servic
ID: 339735 • Letter: S
Question
Speedy Oil provides a single-server automobile oil change and lubrication service. Customers provide an arrival rate of 3 cars per hour. The service rate is 4 cars per hour. Assume that arrivals follow a Poisson probability distribution and that service times follow an exponential probability distribution.
What is the average number of cars in the system? If required, round your answer to two decimal places
L =
What is the average time that a car waits for the oil and lubrication service to begin? If required, round your answer to two decimal places.
Wq = hours
What is the average time a car spends in the system? If required, round your answer to two decimal places.
W = hours
What is the probability that an arrival has to wait for service? If required, round your answer to two decimal places.
Pw =
Explanation / Answer
Following data are provided :
Arrival rate of cars = a = 3 / hour
Service rate of cars = s = 4 / hour
Average number of cars in the system
= a^2/S x ( S – a ) + a/s
= 3x3/4x(4-3) + ¾
= 9/4 + ¾
= 12/4
= 3
AVERAGE NUMBEROF CARS IN THE SYSTEM , L = 3
Average time the car waits for oil and lubricating service to begin
= a/ s x ( S – a ) hour
= 3/ 4 x ( 4 – 3 ) hour
= ¾ hour (0.75 hour )
AVERAGE TIME THE CAR WAITS FOR OIL AND LUBRICATING SERVICE TO BEGIN = 0.75 HOUR
Average time a car spends in the system
= a/s x ( s – a) + 1/S hour
= ¾ x ( 4 – 3 ) + ¼ hour
= ¾ + ¼ hour
= 1 hour
AVERAGE TIME A CAR SPENDS IN THE SYSTEM = 1 HOUR
Probability that there will be ZERO waiting on arrival = Po = 1 – a/s = 1 – ¾ = 0.25
Therefore ,
Probability that an arrival has to wait for service
= 1 – Probability that there will be ZERO waiting on arrival
= 1 – 0.25
= 0.75
PROBABILITY THAT AN ARRIVAL HAS TO WAIT FOR SERVICE, PW = 0.75
AVERAGE NUMBEROF CARS IN THE SYSTEM , L = 3
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