Answer:
(a) 75% (b) 25% (c) 2.25 customers (d) 12 minutes (e) 0.25 (f)0.237
Step-by-step explanation:
Solution
Given that:
The Arrival rate at Poisson distribution = 15 per hour = λ
The Service rate at exponential distribution = 20 per hour = μ
(a) System utilization = λ/μ = 15 / 20 = 0.75 = 75%
(b) The Probability of zero requests in server = 1 - λ/μ = 1 - 0.75 = 0.25
or
The percentage of time server will be idle = 25%
(c)The expected number of customers waiting to be served = Average number of customers in line = λ^2/μ (μ-λ ) = 225 /20(20-15) = 45 /20 = 2.25
Therefore, it is expected that on an average, 2.25 customers are waiting in line to get served.
(d)The average time customers will spend in system = 1/(μ-λ )
=1/(20 - 15) = 1/5 hours = 12 minutes
(e)The probability of zero customer in system = 1 -λ/μ = 1 - 0.75 = 0.25
(f) The probability of more than 4 customers in the system = (λ/μ)^ 4+1
= (15/20)5 = 0.237