Question

Hoping to improve customer service, Bob is considering hiring a second mechanic to work in the second bay. As before, on average customers arrive at a rate of 2 per hour and he can typically service a customer in 20 minutes on average. Assume interarrival times and service times are exponentially distributed. Again, Bob would like to understand how many customers will be waiting and how long they will have to wait, but also wants to keep an eye on how much free time mechanics will have. Should Bob hire the second mechanic? How will the performance of the queuing process be affected?

Answer 1

For one mechanic in the system:

Average rate of arrivals per hour ,2 = 1 1/A= 2 Customer(s) per hour Time between arrival of Customer(s)(A) = 0.5 hrs = 30 mins = = Average rate of service per hour t Time taken to serve one Customer(s)(B) 2 Probability that the system is occupied, Utilization rate,p, =— H 1/B = 3 Customer(s) per hour 0.3333 hrs = 19.998 mins 2 .00 — 0.6667 3 66.67% Average number of Customer(s) in the system per hour = Ls = 2 Customer(s) Average number of Customer(s) in the queue per hour = L = 1.3333 Customer(s) uu- 1) = Average waiting time of Customer(s) in the system = Ws= 1 hrs 60 mins uri = Average waiting time of Customer(s) in the queue 0.6667 hrs 40 mins Wa Luu - A) Probability that the system is empty, the idle time, P(O) 1 - Utilization rate 0.333 33.33%

Clearly as calculated above, the average number of customers waiting in queue = 1.33 customers

Probability or the percentage of the time the mechanic is idle = 33.33%

Now for two mechanics, the calculation is as below:

Given, Time between arrival of Customer(s) = 0.5 hrs = 30 mins Average rate of arrivals per hour, Time taken to serve one Customer(s) = = 2 Customer(s) per hour 0.33 hrs = 20 mins Average rate of service per hour rtl = 3 Customer(s) per hour For 2 servers, P = 0.333333 ,= = 2*11 33.33% 22. Utilization rate, = = 0.5 50.00% 1 1+p Average number of Customer(s) in the system per hour L= - 2p 1-02 0.75 Customer(s) Average number of Customer(s) in the queue per hour = 0.0833 Customer(s) Le = \W.= 2 LqW , 1-p2 W- 20 - 1 W = ū pa) u(1 - p2)" W111 Average waiting time of Customer(s) in the system = 0.375 hrs = 22.5 mins

Average waiting time of Customer(s) in the queue = W =W 1 pz --= ulu(1 - p2) 'hrs 0.045 hrs 2.7 mins Probability that the system is empty, P(0) = 1 - Utilization rate 0.50

In the two mechanics system, the number of customers waiting is decreased, but on the other hand the average utilisation of the workers has also reduced or in other words the idle time of the mechanics is increased.