Mixing
How to formulate a queueing model where riders and drivers arrive randomly and are matched at every specific...
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I am going to develop a queueing model in which riders and drivers arrive with inter-arrival time exponentially distributed.
All the riders and drivers arriving in the system will wait for some amount of time until being matched.
The matching process will pair one rider with one driver and takes place every $Delta t$ unit of time (i.e., $Delta t$, $2Delta t$, $3Delta t$, $cdots$). Whichever side outnumbers the other, its exceeding portion will remain in the queue for the next round of matching.
The service follows the first come first served principle, and how they are matched in particular is not in the scope of this problem and will not affect the queue modelling.
I tried to formulate it as a double-ended queue, where state indicates the exceeding number in the system.
However, this formulation didn't incorporate the factor $Delta t$ in it, it is thereby not in a batch service fashion. I have no clue how I can formulate this $Delta t$ (somewhat like a buffer) into this model.
probability-theory stochastic-processes markov-chains markov-process queueing-theory
asked Feb 2 at 3:07
Guoyang QinGuoyang Qin
1327
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add a comment |
$begingroup$
I am going to develop a queueing model in which riders and drivers arrive with inter-arrival time exponentially distributed.
All the riders and drivers arriving in the system will wait for some amount of time until being matched.
The matching process will pair one rider with one driver and takes place every $Delta t$ unit of time (i.e., $Delta t$, $2Delta t$, $3Delta t$, $cdots$). Whichever side outnumbers the other, its exceeding portion will remain in the queue for the next round of matching.
The service follows the first come first served principle, and how they are matched in particular is not in the scope of this problem and will not affect the queue modelling.
I tried to formulate it as a double-ended queue, where state indicates the exceeding number in the system.
However, this formulation didn't incorporate the factor $Delta t$ in it, it is thereby not in a batch service fashion. I have no clue how I can formulate this $Delta t$ (somewhat like a buffer) into this model.
probability-theory stochastic-processes markov-chains markov-process queueing-theory
asked Feb 2 at 3:07
Guoyang QinGuoyang Qin
1327
$endgroup$
add a comment |
$begingroup$
I am going to develop a queueing model in which riders and drivers arrive with inter-arrival time exponentially distributed.
All the riders and drivers arriving in the system will wait for some amount of time until being matched.
The matching process will pair one rider with one driver and takes place every $Delta t$ unit of time (i.e., $Delta t$, $2Delta t$, $3Delta t$, $cdots$). Whichever side outnumbers the other, its exceeding portion will remain in the queue for the next round of matching.
The service follows the first come first served principle, and how they are matched in particular is not in the scope of this problem and will not affect the queue modelling.
I tried to formulate it as a double-ended queue, where state indicates the exceeding number in the system.
However, this formulation didn't incorporate the factor $Delta t$ in it, it is thereby not in a batch service fashion. I have no clue how I can formulate this $Delta t$ (somewhat like a buffer) into this model.
probability-theory stochastic-processes markov-chains markov-process queueing-theory
asked Feb 2 at 3:07
Guoyang QinGuoyang Qin
1327
$endgroup$
I am going to develop a queueing model in which riders and drivers arrive with inter-arrival time exponentially distributed.
All the riders and drivers arriving in the system will wait for some amount of time until being matched.
The matching process will pair one rider with one driver and takes place every $Delta t$ unit of time (i.e., $Delta t$, $2Delta t$, $3Delta t$, $cdots$). Whichever side outnumbers the other, its exceeding portion will remain in the queue for the next round of matching.
The service follows the first come first served principle, and how they are matched in particular is not in the scope of this problem and will not affect the queue modelling.
I tried to formulate it as a double-ended queue, where state indicates the exceeding number in the system.
However, this formulation didn't incorporate the factor $Delta t$ in it, it is thereby not in a batch service fashion. I have no clue how I can formulate this $Delta t$ (somewhat like a buffer) into this model.
probability-theory stochastic-processes markov-chains markov-process queueing-theory
probability-theory stochastic-processes markov-chains markov-process queueing-theory
asked Feb 2 at 3:07
Guoyang QinGuoyang Qin
1327
asked Feb 2 at 3:07
Guoyang QinGuoyang Qin
1327
asked Feb 2 at 3:07
Guoyang QinGuoyang Qin
1327
asked Feb 2 at 3:07
Guoyang QinGuoyang Qin
1327
asked Feb 2 at 3:07
Guoyang QinGuoyang Qin
1327
1327
add a comment |
add a comment |
1 Answer
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It seems that you are describing a discrete-time double ended batch system. You can view $Delta$ as a time slot and divide it into two stages:
1. All matches are made so one side of the queue goes down to zero and the other goes to the remainder. (e.g. (10,5) goes to (5,0) at this stage).
2. New arrivals occur on either side according to a Poisson distribution with rates $Delta lambda$ and $Delta mu$. (e.g. from (5,0) you may go to (6,2) if one new passenger and two new drivers arrive).
You can write this process down as a discrete-time Markov Chain. Note that if $Deltato 0$ then this system is basically a standard double ended queue.
Also take care that if both sides of the queue are unbounded then this process will not have a steady-state distribution as it will either be transient or null-recurrent (if the rates on both sides are equal).
answered Feb 7 at 10:42
QQQQQQ
63155
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1 Answer
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1 Answer
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$begingroup$
It seems that you are describing a discrete-time double ended batch system. You can view $Delta$ as a time slot and divide it into two stages:
1. All matches are made so one side of the queue goes down to zero and the other goes to the remainder. (e.g. (10,5) goes to (5,0) at this stage).
2. New arrivals occur on either side according to a Poisson distribution with rates $Delta lambda$ and $Delta mu$. (e.g. from (5,0) you may go to (6,2) if one new passenger and two new drivers arrive).
You can write this process down as a discrete-time Markov Chain. Note that if $Deltato 0$ then this system is basically a standard double ended queue.
Also take care that if both sides of the queue are unbounded then this process will not have a steady-state distribution as it will either be transient or null-recurrent (if the rates on both sides are equal).
answered Feb 7 at 10:42
QQQQQQ
63155
$endgroup$
add a comment |
$begingroup$
It seems that you are describing a discrete-time double ended batch system. You can view $Delta$ as a time slot and divide it into two stages:
1. All matches are made so one side of the queue goes down to zero and the other goes to the remainder. (e.g. (10,5) goes to (5,0) at this stage).
2. New arrivals occur on either side according to a Poisson distribution with rates $Delta lambda$ and $Delta mu$. (e.g. from (5,0) you may go to (6,2) if one new passenger and two new drivers arrive).
You can write this process down as a discrete-time Markov Chain. Note that if $Deltato 0$ then this system is basically a standard double ended queue.
Also take care that if both sides of the queue are unbounded then this process will not have a steady-state distribution as it will either be transient or null-recurrent (if the rates on both sides are equal).
answered Feb 7 at 10:42
QQQQQQ
63155
$endgroup$
add a comment |
$begingroup$
It seems that you are describing a discrete-time double ended batch system. You can view $Delta$ as a time slot and divide it into two stages:
1. All matches are made so one side of the queue goes down to zero and the other goes to the remainder. (e.g. (10,5) goes to (5,0) at this stage).
2. New arrivals occur on either side according to a Poisson distribution with rates $Delta lambda$ and $Delta mu$. (e.g. from (5,0) you may go to (6,2) if one new passenger and two new drivers arrive).
You can write this process down as a discrete-time Markov Chain. Note that if $Deltato 0$ then this system is basically a standard double ended queue.
Also take care that if both sides of the queue are unbounded then this process will not have a steady-state distribution as it will either be transient or null-recurrent (if the rates on both sides are equal).
answered Feb 7 at 10:42
QQQQQQ
63155
$endgroup$
It seems that you are describing a discrete-time double ended batch system. You can view $Delta$ as a time slot and divide it into two stages:
1. All matches are made so one side of the queue goes down to zero and the other goes to the remainder. (e.g. (10,5) goes to (5,0) at this stage).
2. New arrivals occur on either side according to a Poisson distribution with rates $Delta lambda$ and $Delta mu$. (e.g. from (5,0) you may go to (6,2) if one new passenger and two new drivers arrive).
You can write this process down as a discrete-time Markov Chain. Note that if $Deltato 0$ then this system is basically a standard double ended queue.
Also take care that if both sides of the queue are unbounded then this process will not have a steady-state distribution as it will either be transient or null-recurrent (if the rates on both sides are equal).
answered Feb 7 at 10:42
QQQQQQ
63155
answered Feb 7 at 10:42
QQQQQQ
63155
answered Feb 7 at 10:42
QQQQQQ
63155
answered Feb 7 at 10:42
QQQQQQ
63155
63155
add a comment |
add a comment |
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