Dr. Meyers, a California vet, is running a rabies vaccination clinic for dogs at
ID: 410440 • Letter: D
Question
Dr. Meyers, a California vet, is running a rabies vaccination clinic for dogs at the local grade school. Meyers can "shoot" a dog every 2 minutes. It is estimated that the dogs will arrive independently and randomly throughout the day at a rate of one dog every 6 minutes according to a Poisson distribution. Also assume that Tarun's shooting times are negative exponentially distributed.
Compute the following:
a) The probability that Meyers is idle.
b) The proportion of the time that Meyers is busy.
c) The average number of dogs being vaccinated and waiting to be vaccinated.
d) The average number of dogs waiting to be vaccinated.
e) The average time a dog waits before getting vaccinated.
f) The average amount of time a dog spends waiting in line and being vaccinated.
Explanation / Answer
The clinic represents single line-single server queuing model.
Operating Characteristics
Scenario 1: Single operator
Inter-arrival rate
1 dog every 6 minutes
Arrival rate (Customers/hour)
= (1/6) x 60 = 10 dogs per hour
Service (vaccination) time per dog (minutes/customer)
2.0 minutes/dog
Service Rate (calls/hour)
µ= (60 mins/hour)/(2 mins./dog)
µ= 30 customers per hour
A
Probability that Meyers is idle
Po = 1 - /µ
Po = 1 - /µ = 1 – (10/30) = 0.6667
B
Probability of time that Meyers is busy
= /µ
= /µ = 10/30 = 0.3333
C
Average number of dogs in the system (being vaccinated and waiting to be vaccinated)
Ls = /(µ - )
Ls = 10/(30 – 10)]
Ls= 0.5
Average number of dogs in the system (being vaccinated and waiting to be vaccinated) is 0.5 dogs per hour
D
Average number of dogs in the line (waiting to be vaccinated)
Lq = Ls – /µ
Lq = 0.5 – 0.33
Lq = 0.1667
Average number of dogs in the line (waiting to be vaccinated) is 0.1667 dogs per hour
E
Average waiting time in line (waiting to be vaccinated)
Wq = /[µ(µ - )]
Wq = 10/[30(30 – 10)]
Wq = 0.0167 hour
Wq = 1 minutes
Average waiting time in line (waiting to be vaccinated) is 1 minutes per dog
F
Average waiting time in the system (being vaccinated and waiting to be vaccinated)
Ws = /[µ(µ - )] + 1/µ
Ws= Wq+ 1/µ
Ws = 0.0167 + 1/30
Ws = 0.05 hour
Ws = 3 minutes
Average waiting time in the system (being vaccinated and waiting to be vaccinated) is 3 minutes per dog
Operating Characteristics
Scenario 1: Single operator
Inter-arrival rate
1 dog every 6 minutes
Arrival rate (Customers/hour)
= (1/6) x 60 = 10 dogs per hour
Service (vaccination) time per dog (minutes/customer)
2.0 minutes/dog
Service Rate (calls/hour)
µ= (60 mins/hour)/(2 mins./dog)
µ= 30 customers per hour
A
Probability that Meyers is idle
Po = 1 - /µ
Po = 1 - /µ = 1 – (10/30) = 0.6667
B
Probability of time that Meyers is busy
= /µ
= /µ = 10/30 = 0.3333
C
Average number of dogs in the system (being vaccinated and waiting to be vaccinated)
Ls = /(µ - )
Ls = 10/(30 – 10)]
Ls= 0.5
Average number of dogs in the system (being vaccinated and waiting to be vaccinated) is 0.5 dogs per hour
D
Average number of dogs in the line (waiting to be vaccinated)
Lq = Ls – /µ
Lq = 0.5 – 0.33
Lq = 0.1667
Average number of dogs in the line (waiting to be vaccinated) is 0.1667 dogs per hour
E
Average waiting time in line (waiting to be vaccinated)
Wq = /[µ(µ - )]
Wq = 10/[30(30 – 10)]
Wq = 0.0167 hour
Wq = 1 minutes
Average waiting time in line (waiting to be vaccinated) is 1 minutes per dog
F
Average waiting time in the system (being vaccinated and waiting to be vaccinated)
Ws = /[µ(µ - )] + 1/µ
Ws= Wq+ 1/µ
Ws = 0.0167 + 1/30
Ws = 0.05 hour
Ws = 3 minutes
Average waiting time in the system (being vaccinated and waiting to be vaccinated) is 3 minutes per dog
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