Lets assume we have a router that transmits packets at 24,000 bytes/msec and that packet lengths are uniformly distributed between 100 and 1,500 bytes. Packets are incoming to the router at a rate of 1,200,000 packets/min. (M/M/1)
I am trying to calculate the following:
- Mean time it takes to send a packet through the line
- Std dev for the time is takes to transmit a packet
- Average latency
- Time a packet spends waiting in the queue
I know that I can calculate the mean transmission time by the following:
$$E[T_s] = ((100/24000) + (1500/24000)) / 2 = .0333 {bytes/msec}$$
However, how would I go about calculating the std dev without summing all 1,400 possible outcomes?
In order to find the latency, I know that the utilization is the following:
$$\rho = \lambda/\mu$$
But substituting $$\lambda$$ for 1,200,000 and $$\mu$$ for .0333 gives wacky results.
I know the latency is:
$$T_q = 1 / \mu(1-\rho)$$
But how would I calculate the time the packet spends in the queue?
Any help would be appreciated. Thanks.
The M/M/1 model assumes that job sizes are exponentially distributed with rate parameter $\mu$, in your setup the job sizes are uniformly distributed so you need to consider an M/G/1 queueing model.
What is the difference between 'Mean time it takes to send a packet through the line' and 'Average latency'? They sound like a description of the same phenomena to me. Use the result for waiting/response time in an M/G/1 queue, recalling that the mean can be found by differentiating the Laplace–Stieltjes transform using the formula $$\mathbb E [X] = -\left.\frac{\text{d} \{\mathcal{L}^*F\}(s)}{\text{d}s} \right|_{s=0}.$$
The times taken to send a packet (service times) are uniformly distributed, so the 'Std dev for the time is takes to transmit a packet' will be the standard deviation of the uniform distribution.
'Time a packet spends waiting in the queue' is the response time less the service time. You can use the waiting/response time result for this also, can you see how?