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A single server queuing system with a Poisson arrival rate and exponential service time has an average arrival rate of 6 customers per hour and an average service rate of 8 customers per hour. The average length of time customers will spend in the system is: a. 0.375 minutes b. 30 minutes c. 0.5 minutes d. 22.5 minutes

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ANSWER AND EXPLAINATION:
To find the average length of time customers will spend in the system in a single server queuing system with a Poisson arrival rate and exponential service time, we can use Little's Law.

Little's Law states that the average number of customers in a stable queuing system is equal to the average arrival rate multiplied by the average time spent in the system. Mathematically, it can be expressed as:

L = λ * W

where:

L is the average number of customers in the system,

λ is the average arrival rate, and

W is the average time spent in the system.

In this case, the average arrival rate (λ) is given as 6 customers per hour, and the average service rate (μ) is given as 8 customers per hour.

Since the system is stable, the arrival rate (λ) must be less than the service rate (μ), ensuring that the system does not become overwhelmed with customers.

To calculate the average time spent in the system (W), we can use the following formula:

W = 1 / (μ - λ)

Substituting the given values:

W = 1 / (8 - 6)

W = 1 / 2

W = 0.5 hours

Now, to convert the time to minutes:

W = 0.5 hours * 60 minutes/hour

W = 30 minutes

Therefore, the average length of time customers will spend in the system is 30 minutes.

The correct answer is B. 30 minutes.

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