Marcia operates an expresso stand where customers arrive according to a poisson

Question

Marcia operates an expresso stand where customers arrive according to a poisson process with a mean interarrival time of 1 minute and 40 sec. Her service times have an exponential distribution with a mean of 1 minute and 40 seconds. What is the average waiting time in the queue? Why? Marcia hears that one of the principles of queueing theory is that reducing service time variability lowers the average waiting time. So she may buy a machine that would change the exponential service times to deterministic times with the same mean of 1 minute and 40 seconds. If she boys it, what would be the average number of customers in the system? Why? Marcia didn't buy the machine but she is still worried about the lines and delays, so she works harder and tells the cashier to help her where there is just one customer at the expresso stand. If there is more than One customer, that is if there is a then the cashier does not help her and the mean is 1 minute and 13 seconds. When the cashier helps her, the service time is still exponentially distributed but the mean is reduced to 1 minute. With this change, what fraction of the time is the expresso stand empty of customers?

Explanation / Answer

N follows poissson process with mean inter arrival time of 1minute 40 sec
so the inter arrival time follows exp(1 minute 40 secs)
and inter service time also follows exp(1 minute 40 secs)

here we have r = mu/lambda = 1
The average time spent waiting is 1/(???)?1/?= r/(???) = infinite

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