Mixing
Queuing Theory with Poisson Distribution
Suppose customers arrive in a one-server queue according to a Poisson distribution with rate lambda=1 (in hours). Suppose that the service times equal 1/4 hour, 1/2 hour, or one hour each with probability 1/3.
(a) Assume that the queue is empty and a customer arrives. What is the expected amount of time until that customer leaves?
(b) Assume that the queue is empty and a customer arrives. What is the expected amount of time until the queue is empty again?
(c) At a large time t what is the probability that there are no customers in the queue?
I'm trying to do couple of practice problems involving queuing before my exam and I am really confused, I would really appreciate it if someone can show me how to do this problem.
Thanks
probability statistics statistical-inference queueing-theory
asked Mar 3 '14 at 0:31
user130344user130344
425
add a comment |
Suppose customers arrive in a one-server queue according to a Poisson distribution with rate lambda=1 (in hours). Suppose that the service times equal 1/4 hour, 1/2 hour, or one hour each with probability 1/3.
(a) Assume that the queue is empty and a customer arrives. What is the expected amount of time until that customer leaves?
(b) Assume that the queue is empty and a customer arrives. What is the expected amount of time until the queue is empty again?
(c) At a large time t what is the probability that there are no customers in the queue?
I'm trying to do couple of practice problems involving queuing before my exam and I am really confused, I would really appreciate it if someone can show me how to do this problem.
Thanks
probability statistics statistical-inference queueing-theory
asked Mar 3 '14 at 0:31
user130344user130344
425
add a comment |
Suppose customers arrive in a one-server queue according to a Poisson distribution with rate lambda=1 (in hours). Suppose that the service times equal 1/4 hour, 1/2 hour, or one hour each with probability 1/3.
(a) Assume that the queue is empty and a customer arrives. What is the expected amount of time until that customer leaves?
(b) Assume that the queue is empty and a customer arrives. What is the expected amount of time until the queue is empty again?
(c) At a large time t what is the probability that there are no customers in the queue?
I'm trying to do couple of practice problems involving queuing before my exam and I am really confused, I would really appreciate it if someone can show me how to do this problem.
Thanks
probability statistics statistical-inference queueing-theory
asked Mar 3 '14 at 0:31
user130344user130344
425
Suppose customers arrive in a one-server queue according to a Poisson distribution with rate lambda=1 (in hours). Suppose that the service times equal 1/4 hour, 1/2 hour, or one hour each with probability 1/3.
(a) Assume that the queue is empty and a customer arrives. What is the expected amount of time until that customer leaves?
(b) Assume that the queue is empty and a customer arrives. What is the expected amount of time until the queue is empty again?
(c) At a large time t what is the probability that there are no customers in the queue?
I'm trying to do couple of practice problems involving queuing before my exam and I am really confused, I would really appreciate it if someone can show me how to do this problem.
Thanks
probability statistics statistical-inference queueing-theory
probability statistics statistical-inference queueing-theory
asked Mar 3 '14 at 0:31
user130344user130344
425
asked Mar 3 '14 at 0:31
user130344user130344
425
asked Mar 3 '14 at 0:31
user130344user130344
425
asked Mar 3 '14 at 0:31
user130344user130344
425
asked Mar 3 '14 at 0:31
user130344user130344
425
425
add a comment |
add a comment |
1 Answer
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I actually figured it out myself and decided to post the answer in case someone needed it.
(a) It is the mean service time: μ = (1/4 + 1/2 + 1)/3 = 7/12.
(b) This is μ/(1 − λμ) = μ/(1 − μ) = 7/5 hours.
(c) This is 1/(1 + 7/5) = 5/12.
answered Mar 3 '14 at 1:33
user130344user130344
425
We usually use $mu$ to refer to the service rate, not the mean service duration.
– Brian Tung
Jul 17 '18 at 20:18
add a comment |
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1 Answer
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1 Answer
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I actually figured it out myself and decided to post the answer in case someone needed it.
(a) It is the mean service time: μ = (1/4 + 1/2 + 1)/3 = 7/12.
(b) This is μ/(1 − λμ) = μ/(1 − μ) = 7/5 hours.
(c) This is 1/(1 + 7/5) = 5/12.
answered Mar 3 '14 at 1:33
user130344user130344
425
We usually use $mu$ to refer to the service rate, not the mean service duration.
– Brian Tung
Jul 17 '18 at 20:18
add a comment |
I actually figured it out myself and decided to post the answer in case someone needed it.
(a) It is the mean service time: μ = (1/4 + 1/2 + 1)/3 = 7/12.
(b) This is μ/(1 − λμ) = μ/(1 − μ) = 7/5 hours.
(c) This is 1/(1 + 7/5) = 5/12.
answered Mar 3 '14 at 1:33
user130344user130344
425
We usually use $mu$ to refer to the service rate, not the mean service duration.
– Brian Tung
Jul 17 '18 at 20:18
add a comment |
I actually figured it out myself and decided to post the answer in case someone needed it.
(a) It is the mean service time: μ = (1/4 + 1/2 + 1)/3 = 7/12.
(b) This is μ/(1 − λμ) = μ/(1 − μ) = 7/5 hours.
(c) This is 1/(1 + 7/5) = 5/12.
answered Mar 3 '14 at 1:33
user130344user130344
425
I actually figured it out myself and decided to post the answer in case someone needed it.
(a) It is the mean service time: μ = (1/4 + 1/2 + 1)/3 = 7/12.
(b) This is μ/(1 − λμ) = μ/(1 − μ) = 7/5 hours.
(c) This is 1/(1 + 7/5) = 5/12.
answered Mar 3 '14 at 1:33
user130344user130344
425
answered Mar 3 '14 at 1:33
user130344user130344
425
answered Mar 3 '14 at 1:33
user130344user130344
425
answered Mar 3 '14 at 1:33
user130344user130344
425
425
We usually use $mu$ to refer to the service rate, not the mean service duration.
– Brian Tung
Jul 17 '18 at 20:18
add a comment |
We usually use $mu$ to refer to the service rate, not the mean service duration.
– Brian Tung
Jul 17 '18 at 20:18
We usually use $mu$ to refer to the service rate, not the mean service duration.
– Brian Tung
Jul 17 '18 at 20:18
We usually use $mu$ to refer to the service rate, not the mean service duration.
– Brian Tung
Jul 17 '18 at 20:18
add a comment |
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