Question

In: Statistics and Probability

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

Solutions

Expert Solution

1)

arrivals/time period = λ= 0.5
served/time period= μ= 4
utilization factor =      ρ                                        = λ/μ             = 0.125~12.5%

2)

average number of customers in system L = λ/(μ-λ)= 0.142857

3)

average number of customers in queue Lq = λ2/(μ(μ-λ))= 0.017857

4)

average time spend in system W                   = 1/(μ-λ)= 0.285714

5)

average time spend in queue Wq                   = λ/(μ(μ-λ))= 0.035714

6)

Probability all servers are busy=0.125

7)

Probability an arriving customer has to wait =0.125


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