Question

Will runs a dog grooming service, and has asked you to analyze the current 'line situation'. He has gathered data and calculated that he can groom a dog in about 10 minutes, based on a negative exponential distribution. He also noted that about 3 dogs arrive to his shop per hour, during all operating hours, based on a Poisson distribution.

How long, on average, will a dog wait for grooming? ["10 minutes", "1 minute", "16.67 minutes", "20 minutes"]      

In total, how long will a dog spend in the shop waiting and being groomed? ["26.67 minutes", "20 minutes", "16.67 minutes", "10 minutes"]      

Will's shop is fairly small, and he only has space for 3 dogs to be in the shop before it gets crowded. What is the probability that there will be more than 3 dogs at any time? ["3.125%", "6.25%", "1.5625%", "12.5%"]      

Will wants to reduce the wait time, and is considering two options. The first would be to hire a helper, who would assist Will, and reduce the grooming time from 10 minutes to 8 minutes. (NOTE: The helper only helps, and does not work alone.)

What would the wait time be with a helper? ["3.33 minutes", "1.33 minutes", "13.33 minutes", "5.33 minutes"]      

The second option is to rent an Auto-Groomer, a new machine that would allow Will to groom a dog in a constant 9 minutes.

What would the wait time be with Auto-Groomer? ["3.68 minutes", "4.67 minutes", "3.33 minutes", "5.33 minutes"]      

Assuming both options cost the same amount and all else is equal, which should Will choose to deliver the shortest wait time? ["Choose the helper", "Choose either; wait times are equal", "Choose the Auto-Groomer"]      

Answer 1

Average rate of arrivals per hour , = 1/4= 3 Dog(s) per hour 0.3333 hrs = 20 mins Time between arrival of Dog(s)(A) = Average rate of service per hour t Time taken to serve one Dog(s)(B) = = 1/B = 8 Dog(s) per hour 0.1333 hrs = 8 mins 0.4 40.00% Probability that the system is occupied, Utilization rate,p, -— — 17.5 Average number of Dog(s) in the system per hour = Ls - 4- 0.6664 Dog(s) Average number of Dog(s) in the queue per hour = L = 0.2665 Dog(s) 44-2) = Average waiting time of Dog(s) in the system = 0.2221 hrs = 13.3277 mins U-2 Average waiting time of Dog(s) in the queue, for Auto groomer option= W == 0.0888 hrs 5.3297 mins u(1-2)

For auto groomer option

1/A=3 Dog(s) per hour Average rate of arrivals per hour ,2 Time between arrival of Dog(s)(A) Average rate of service per hour t Time taken to serve one Dog(s)(B) = = = = 0.3333 hrs = 20 mins 1/8 = 7 Dog(s) per hour 0.15 hrs = 9 mins .00 2 Probability that the system is occupied, Utilization rate,p, -- 3 — 0.45 45.00% 6.67 Average number of Dog(s) in the system per hour = Ls - u- 0.8182 Dog(s) Average number of Dog(s) in the queue per hour = L = 0.3682 Dog(s) 4(1-2) = Average waiting time of Dog(s) in the system = 0.2727 hrs 16.3635 mins Average waiting time of Dog(s) in the queue, for Auto groomer option = Wa=- 0.1227 hrs 7.3635 mins u(1-2)

As it is clear, the costs being the same for both the options we need to go with the Helper option because of the waiting time in the queue in case of a helper than the other option. This will also provide the opportunity for serving more dogs than the option 2