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Travka [436]
3 years ago
14

Customers enter the waiting line to pay for food as they leave a cafeteria on a first-come, first-served basis. The arrival rate

follows a Poisson distribution, while service times follow an exponential distribution. If the average number of arrivals is four per minute and the average service rate of a single server is seven per minute, what proportion of the time is the serverbusy?A) 0.43B) 0.57C) 0.75D) 0.25E) 0.33
Mathematics
1 answer:
Elan Coil [88]3 years ago
3 0

Answer:

The answer is B) 0.57.

Step-by-step explanation:

In this problem we have to apply queueing theory.

It is a single server queueing problem.

The arrival rate is  \lambda=4 and the service rate is \mu=7.

The proportion of time that the server is busy is now as the "server utilization"and can be calculated as:

p=\frac{\lambda}{c\mu} =\frac{4}{1*7}=0.57

where c is the number of server (in this case, one server).

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4 0
3 years ago
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valina [46]

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hope it helps:)
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