4
$\begingroup$

We have a fleet of taxis with $t$ taxis available. All taxis are identical in the sense that they have the same capacity for $p$ passengers and each taxi is dispatched only when its capacity is full.

Each taxi takes passengers from district A to district B and returns to district A without any passengers!

Passengers arrive individually at district A following a Poisson process with parameter $\lambda$ and the time it takes for each taxi to go from district A to B and return back to A is $T$.

Based on this information, I think we can view this fleet as a queuing system $M/D^p/t$ that is a batch processing queue with batch size $p$ with fixed service time that is $T$ and multiple severs that is $t$.

We are interested in computing the time between two consecutive dispatches. We did some calculations for different queues with different values of $\lambda$, $p$, $T$ and $t$ and it looks like the average time looks something like $\frac{\lambda T}{t}$ but we cannot show this rigorously.

Any suggestions would be appreciated.

$\endgroup$

1 Answer 1

3
$\begingroup$

Let $\tau$ represent the average time between dispatches.

I'm skeptical about $\frac{\lambda T}{t}$ as a possible value of $\tau$ because it fails dimensional analysis. $\tau$ is in units of time. $T$ is time, $\lambda$ is customers per unit time, and $t$ is customers, so $\frac{\lambda T}{t}$ is dimensionless.

Queuing theory is not my "jam", so I may be way off with the following. Assuming that you have enough taxis to keep up with demand, in steady state I would expect the time between dispatches to be (roughly?) the time required to get another $p$ customers into the system. The average time for $p$ arrivals is $\frac{p}{\lambda}$, so that would be my first guess for $\tau$. If you don't have enough taxis to reach steady state, then the taxis would be running nonstop. Each taxi would make an average of $\frac{1}{T}$ trips per unit time, so with $t$ taxis there would be an average of $\frac{t}{T}$ dispatches per unit time and thus an average of $\frac{T}{t}$ time between dispatches. Note that at full blast the system can carry $\frac{tp}{T}$ passengers per unit time, which means that steady state requires $\lambda<\frac{tp}{T}$ or equivalently $\frac{p}{\lambda}>\frac{T}{t}.$

$\endgroup$
3
  • $\begingroup$ Quick addendum: I ran a little discrete event simulation (in R) under conditions consistent with steady state (but without tossing the warm-up observations), and the mean time between dispatches matched my first guess pretty closely. $\endgroup$
    – prubin
    Feb 6 at 21:18
  • $\begingroup$ Thank you. This means that we can have a lower bound on the size of the fleet but out of curiosity, is there a way to have an upper bound on $t$? $\endgroup$
    – Vitamin Z
    Feb 14 at 15:22
  • $\begingroup$ There is nothing in the problem statement that limits the number of taxis. In practice, there would probably be a space constraint that would limit the number of idle taxis, and before you reached that there would be an economic limit. If you could assign costs to the time each taxi is idle and the time each passenger weights, you could try to minimize the steady-state expected cost per day (or hour). $\endgroup$
    – prubin
    Feb 14 at 19:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.