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A road transport company has one reservation clerk on duty at a time. He handles information of bus schedules and make reservation. Customer arrive at a rate of 8 per hour and the clerk can service 12 customers on an average per hour.
The probability of at least 2 customers in the queue would be ______.
- a)0.25
- b)0.20
- c)0.30
- d)0.35
Correct answer is option 'C'. Can you explain this answer?
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A road transport company has one reservation clerk on duty at a time. ...
We are given λ = 8 customers per hour
and μ = 12 customer per hour
Probability of at least two customer in the Queue
= Probability at least three customer in the system
=1-(P0+P1+P2)
Most Upvoted Answer
A road transport company has one reservation clerk on duty at a time. ...
Solution:
Given:
Arrival rate, λ = 8 customers per hour
Service rate, µ = 12 customers per hour
Number of servers, m = 1
To find: Probability of at least 2 customers in the queue, i.e. P(n ≥ 2)
We know that the system is an M/M/1 queue, where M represents Poisson distribution for arrivals and exponential distribution for service.
The traffic intensity, ρ = λ/µ = (8/12) = 0.67
As ρ < 1,="" the="" system="" is="" stable="" and="" queue="" />
Using Little's formula, we can find the expected number of customers in the system:
Ls = λ * Ws
where Ws is the expected time spent by a customer in the system.
Ws = Wq + 1/µ
where Wq is the expected time spent by a customer in the queue.
To find Wq, we can use the formula:
Wq = ρ^2/(1 - ρ) * (1/µ)
Plugging in the values, we get:
Wq = (0.67)^2 / (1 - 0.67) * (1/12) = 0.089 hours
Ws = Wq + 1/µ = 0.089 + 1/12 = 0.172 hours
Ls = λ * Ws = 8 * 0.172 = 1.376
Now, using the formula for P(n ≥ 2) in an M/M/1 queue with one server:
P(n ≥ 2) = (ρ^2 / (1 - ρ)) * ((m + 1)^2 / m^3) = (0.67^2 / (1 - 0.67)) * ((1 + 1)^2 / 1^3) = 0.3
Therefore, the probability of at least 2 customers in the queue is 0.30, which is option (c).
Given:
Arrival rate, λ = 8 customers per hour
Service rate, µ = 12 customers per hour
Number of servers, m = 1
To find: Probability of at least 2 customers in the queue, i.e. P(n ≥ 2)
We know that the system is an M/M/1 queue, where M represents Poisson distribution for arrivals and exponential distribution for service.
The traffic intensity, ρ = λ/µ = (8/12) = 0.67
As ρ < 1,="" the="" system="" is="" stable="" and="" queue="" />
Using Little's formula, we can find the expected number of customers in the system:
Ls = λ * Ws
where Ws is the expected time spent by a customer in the system.
Ws = Wq + 1/µ
where Wq is the expected time spent by a customer in the queue.
To find Wq, we can use the formula:
Wq = ρ^2/(1 - ρ) * (1/µ)
Plugging in the values, we get:
Wq = (0.67)^2 / (1 - 0.67) * (1/12) = 0.089 hours
Ws = Wq + 1/µ = 0.089 + 1/12 = 0.172 hours
Ls = λ * Ws = 8 * 0.172 = 1.376
Now, using the formula for P(n ≥ 2) in an M/M/1 queue with one server:
P(n ≥ 2) = (ρ^2 / (1 - ρ)) * ((m + 1)^2 / m^3) = (0.67^2 / (1 - 0.67)) * ((1 + 1)^2 / 1^3) = 0.3
Therefore, the probability of at least 2 customers in the queue is 0.30, which is option (c).
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A road transport company has one reservation clerk on duty at a time. He handles information of bus schedules and make reservation. Customer arrive at a rate of 8 per hour and the clerk can service 12 customers on an average per hour.The probability of at least 2 customers in the queue would be ______.a)0.25b)0.20c)0.30d)0.35Correct answer is option 'C'. Can you explain this answer?
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