An upper-layer packet is split into 10 frames, each of which has an 80 percent chance of arriving undamaged.

Q: An upper-layer packet is split into 10 frames, each of which has an 80 percent chance of arriving undamaged. If no error control is done by the data link protocol, how many times must the message be sent on average to get the entire thing through?

Solution:

First notice that no error checking is done by the data link protocol. What this means is that if an error occurs in a frame this information will not be known until all 10 frames have been transmitted and reassembled(the transport layer is doing some error control presumably, and will detect if there was an error in 1 or more frames and if so, request re-transmission). So in the case of this problem, we can't just repeatedly send the first frame until it gets through without error, then do the same for the second,etc. Instead, we must send all 10 frames. If they all get through without error we are done. Otherwise, we must send all 10again.
Because each frame has a probability of 0.8 of getting through correctly and errors can be considered independent events, the probability of the whole message getting through correctly on any given attempt is which is about 0.107 (or 10.7%). Call this value p.
Now notice that the probability of the message requiring i transmissions is .The reason is that if it takes exactly i transmissions, the first i-1 attempts must have failed(this happens with probability since the probability of a failed transmission is (1- probability of successful transmission) = (1-p). After these failed attempts,the ith attempt must succeed. This happens with probability p.
To compute the expected number of transmissions required(i.e.the average transmissions required) we weight all the possible transmissions required with their respective probabilities and then sum over all these possibilities. So the expected number of transmissions, E, is:

Notice that the last sum starts at i=0. This is fine since at i=0 =0, so we are just adding 0, something that doesn't change the value of the expression.
To evaluate this sum, we again use the trick of differentiating a series for which we already know the sum to get a form that is useful in the problem at hand.
It was proven in class that
Taking the derivative of both sides shows that .
Now, we just set r=1-p to get .Notice that we must have r<1 for the above equations to hold. But since r=1-p and ,r < 1 as needed.
So the answer is 1/0.107 or about 9.3 transmissions.
Now, we do the same problem, but take the probability of error-free transmission for a frame to be 0.99. In this case which is about 0.904 (quite a bit higher than the first part). The work we did in the first part still holds, so al lwe need to do is calculate which in this case is about 1.1.