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This is question #1:
1) During busy times, 60 potential customers per hour arrive at the booth (assume a Poisson distribution). A booth worker takes 3.75 minutes, on average, to meet the information needs of a potential customer (assume an exponential distribution).
This is question #2:
2) During slack times, 10 potential customers per hour arrive at the booth (assume a Poisson distribution). A booth worker takes 3.75 minutes, on average, to meet the information needs of a potential customer (assume an exponential distribution).
Expert Answer
1) Arrival rate, λ = 60 customers per hour
Service rate, μ = (1/3.75)*60 = 16 customers per hour
Arrival rate is more than the service rate, therefore multiple booth workers are required.
Minimum number of booth workers required = 60/16 = 3.75 ~ 4 workers (rounded-up to integer, because worker cannot be in fraction)
This is M/M/s queue model with s=4
With 4 workers, Average waiting time in line, Wq = 0.22 hour = 13 minutes (approx)
Calculation of waiting time is following
2) Arrival rate, λ = 10 customers per hour
Service rate, μ = (1/3.75)*60 = 16 customers per hour
Arrival rate is less than the service rate, therefore 1 worker is enough.
This is M/M/1
Average waiting time in line, Wq = λ/(μ*(μ-λ)) = 10/(16*(16-10)) = 0.1042 hour = 6.25 minutes (approx)
