Suppose we have a single server in a shop and customers arrive in the shop with...

Suppose we have a single server in a shop and customers arrive in the shop with a Poisson arrival distribution at a mean rate of λ=0.5 customers per minute. The interarrival time have an exponential distribution with the average inter-arrival time being 2 minutes. The server has an exponential service time distribution with a mean service rate of 4 customers per minute. Calculate: 1. Overall system utilization 2. Number of customers in the system 3. Number of customers in the queue 4. Average time customers spend in the system 5. Average time customer spends in the queue 6. Probability all servers are busy 7. Probability an arriving customer has to wait

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Expert Solution

arrivals/time period =			λ=	0.5
served/time period=			μ=	4

utilization factor = ρ =

λ/μ =

0.125~12.5%

average number of customers in system L =

λ/(μ-λ)=

0.142857

average number of customers in queue L_q =

λ²/(μ(μ-λ))=

0.017857

average time spend in system W =

1/(μ-λ)=

0.285714

average time spend in queue W_q =

λ/(μ(μ-λ))=

0.035714

Probability all servers are busy=0.125

Probability an arriving customer has to wait =0.125

orchestra answered 1 year ago

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