Deriving the process of successfully consumed requests from the process of request-producers and the process of request-consumers

The title is not very straightforward I understand, but you will soon realize it was not so simple to describe in few words this problem.

The problem

Consider a system consisting of:

• A process of incoming requests.
• A process of request checking. Or, better, a process of request consumers.

So, basically, in this system, some requests (or calls) are performed by requestors and they are consumed by consumers. No need to detail further. It is just a simple producer/consumer model adaptable to many existing conditions.

Model is homogeneous

There is an important condition. Consider the model being homogeneous. Which means we focus on time intervals other than time instants. So I will always consider $\Delta t = t_2 - t_1, t_1,t_2 \in \mathbb{R}, t_1 < t_2$ as the quantity indicating the time interval.

Successful request consumption

Consumers cannot consume all requests. It depends on instants where consumers check for request and instants where requests are made. Let us say that if a request arrives in a certain instant $t_r$ and a consuming-check is performed, instant $t_c$, in that same time too ($t_r = t_c$), then the probability that the request is consumed is 1. The probability for a request to be successfully processed/consumed depends on $t_c - t_r$; the higher this interval, the lower the probability.

Model so far

I modeled the process of arrivals (requests) as a homogeneous Poisson process: $\Lambda(k,\Delta t)$. The probability that in $\Delta t$ there will be $k$ incoming requests is:

$$p_\lambda = \frac{e^{- \lambda \Delta t} \cdot (\lambda \Delta t)}{k!}$$

Where $\lambda > 0$ is the number of arrivals in $\Delta t$ experienced previously (the typical rate of arrivals).

The process of requests consumption is modeled similarly using Poisson: $M(k,\Delta t)$. The probability that in $\Delta t$ there will be $k$ request-check is:

$$p_\mu = \frac{e^{- \mu \Delta t} \cdot (\mu \Delta t)}{k!}$$

Where $\mu > 0$ is the number of request-check in $\Delta t$ experienced previously (the typical rate of request-check).

My question

I want to get the probability, say $p_\rho$, that in $\Delta t$ there will be $k$ successfully processed requests in $\Delta t$, considering Poisson-$\lambda$-distributed request arrivals (in $\Delta t$) and Poisson-$\mu$-distributed request-checks (always in $\Delta t$).

Just one more thing

I need a solution in order to find $p_\rho$ that depends on $k$ and $\Delta t$: $p_\rho(k,\Delta t)$.

Thankyou

I got a feeling that...

It is just a guessing but, is it something related to process convolution or similar?

I know that this is a problem can can easily be related to queueing models. But I tried those and my problem is that M/M/1 and M/M/m queues are modeled using death-birth time-continuos and time-discrete Markov chains. When I apply those models I can get an ergodic model and obtain steady state probabilities but those will not depend by $\Delta t$. Of course if you have some ideas how to get this done using Markov chains, thankyou.

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Check out queueing models (this is a modified M/M/1 queue, if my rusty notation isn't off), in particular, how the equations are set up and solved. – vonbrand Feb 17 '13 at 14:24
I had a feeling... But I cannot figure out how to adapt a death-birth model to this scenario. Btw, gonna check it out, thank you very much! – Andry Feb 17 '13 at 14:29
"The probability for a request to be successfully processed/consumed depends on $t_c - t_r$; the higher this interval, the lower the probability." It seems you wrote up a two-page question and didn't include the most important piece of information. How are we supposed to determine $p_\rho$ without knowing the probability for a request to be successfully processed? – joriki Feb 18 '13 at 9:50
@joriki: It is one of the things I am asking as well... – Andry Feb 18 '13 at 10:26
I don't understand -- how is this determined by any of what you wrote? – joriki Feb 18 '13 at 10:28