# On average, how much time does it take to transmit a file on a noisy network channel?

Let's say I want to deliver a file of $2Kb$ over a network channel from A to B, where A can transmit in a transfer rate of R bits-per-second (let's ignore other parameters like propagation time, for the sake of argument).
And let's say that the channel is quite noisy: when you deliver a bit in that channel, it will be flipped (or "corrupted", if you will) with probability $p$.
I want to calculate two things:

1. The probability to successfully deliver the file from A to B.
2. The expected time (time expectancy? not sure what exactly the English term is...) to deliver the file. Or to put it simply: on average, how much time does it take...

the first one is not too hard; the probability to successfully deliver one bit from A to B is $1-p$, so successfully delivering the file will be $(1-p)^{\#bits\_in\_file}=(1-p)^{2048}$.

I'm not sure about 2 though.
I know the time to send a file from A to B is $\frac{\#bits\_in\_file}{transfer\_rate}=\frac{2Kb}{R}$.

The answers to 2 say: $(delivering\_time)\times \frac{1}{prob\_of\_success}=(\frac{2Kb}{R})\frac{1}{(1-p)^{2048}}$.

My question is: why is it the answer? How do you even calculate such thing? I know how to calculate an expected value given a random variable, but I can't see it here.

• By linearity and independence, the expected time for the file transfer is just $\frac{1}{1-p}$ times the time required in a perfect channel (no corruption). – Jack D'Aurizio Dec 13 '14 at 21:29

This is because we are transmitting the file with a geometric distribution on each variable, i.e. we can model transmitting one bit with a geometric distribution $N \sim \text{Geo}(1-p)$, as we keep transmitting the bit until it is transmitted succcessfully.

Thus we have intuitively that the mean expected number of bits that need to be transferred is given by:

$$\nu= \sum_{n=1}^{2048}\mathbb{E}[N]=\frac{2048}{1-p}$$

Where we use the known fact that if $X \sim \text{Geo}(p)$, then $\mathbb{E}[X]=\frac{1}{p}$. And thus the expected total time would be given by:

$$\tau = \frac{\nu}{R}=\frac{2048}{R}\cdot\frac{1}{1-p}$$

If we look at the problem using the Negative Binomial Distribution, we have that the entire file can be expressed using the random variable $N \sim \text{NB}(2048,p)$, we therefore have that:

$$\nu = \mathbb{E}[N] + 2048 = \frac{2048\:p}{1-p} + 2048 = \frac{2048p + 2048 - 2048p}{1-p}=\frac{2048}{1-p}$$

Where we have to add $2048$ to the expected value of $N$ as the Negative Binomial Distribution is effectively measuring the number of bits that we have to resend. And thus we get the same as before:

$$\tau = \frac{\nu}{R}=\frac{2048}{R}\cdot\frac{1}{1-p}$$

• Ohh! geometric distribution. didn't see that... But... shouldn't it be Negative binomial distribution then? Because we "roll a dice with probability $p$ until $k$ successes", aren't we? – so.very.tired Dec 13 '14 at 21:41