Quiz 3.

Ms Kanigiluppai is grading quiz 3. I will not see it until Friday morning.
If you have questions, please come to my office. Friday 2:00 - 4:00.
Wed and Thu only by appointment. Send me Email.

I had a quick look. It seems to have been done much better. But several 
people had problem 2 wrong. (See below). If you are one of these and it 
was not a ``slip of the pen'' please come to my office.

1A. OSI = Open Systems Interconnection. (Also asked on Quiz 1).
1B. PCM = Pulse Code Modulation.        (Also asked on Quiz 2).

2. k bits gives 2^(k) levels. (two to the power k).
   I am very bothered by the fact that some people had this wrong.
   1 bit: two possible outcomes.  zero and one .
   2 bits: (0, 0), (1, 0), (0, 1), (1, 1)  four possible outcomes.
   etc.
   k bits: two to the power k possible outcomes.
   I had assumed every student who has had CIS114 knows this. 

Question 3 had been asked before on Quiz 2. If you still have trouble 
with this problem, please come to my office.

3A: Designers of the telephone system with digital transport wanted to
    represent frequences to somewhat over 3000 Hz reasonably accurately.
    They rounded this off to 1 - 4000 Hz. The Nyquist result says that
    if you want to represent frequencies up to H Hz you must sample 
    2H times per second. That is where the sampling rate of 8000 
    samples/sec comes from.
3B. 8000 samples/sec at 56,000 bits/sec gives 7 bits/sample.
    7 bits allows 2^(7) = 128 levels.
    A CIS (or COE, or IT) student should know the difference between 
    ``levels'' and ``bits''. 
    k bits allow 2^(k) levels.

Question 4 had been asked (with different numbers) in quiz 1 (problem 3)
and in quiz 2 (problem 3). If you still have trouble with this problem, 
please come to my office.

4.  S/N is 10 dB, hence 10*log(S/N) = 10 (log Base 10).
    log(S/N) = 1 (log base 10), S/N = 10.
    The frequency range is
    H = 10,000,001,000 - 10,000,000,000 =  1000 Hz. 
    By Shannon's result, the theoretical highest possible data rate is
    H*log(1 + S/N) (log base 2) = 1000*log(11) (log base 2) =
    1000*((ln(11))/(ln(2))) = 1000*2.39789/(.693147) = 3459.43 bits/sec. 

5. The drawing will appear shortly in this website.

Two possible problems with the ``no sequence number'' situation were 
discussed in class. You are OK if you got one of the two.

A. Suppose Data Frame n makes it allright to the destination, but the 
acknowledgement for that frame falls on the floor. There is a time-out 
for dataframe(n) in the source and the frame is retransmitted. Suppose 
the re-transmitted frame also makes it OK to the destination.

This means the destination twice receives dataframe(n), but because 
there are no sequence numbers the destination ``thinks'' it got frames 
n and (n+1). Result: Erroneous data.
(If Email: a couple of characters repeated).

B. Suppose Data Frame n makes it allright to the destination, and its 
acknowledgement makes it back, but late, after the time-out for frame n
in the source.
So dataframe n is re-transmitted and the receiver receives dataframe n
twice.

Same result.

The ``recipe'' for problems like question 5 in quiz 3, and question 4
in quiz 2, is easy:
Unless a protocol is really stupid, it will work fine as long as 
``all packets are delivered OK and in time''. For a more sophisticated
analysis: First consider the possibility a packet is lost. (dropped). 
If the protocol survives, consider the possibility an acknowledgement 
is lost. (dropped). If the protocol still survives, consider the 
possibility a packet is acknowledged OK, but the acknowledgement arrives 
after the packet has timed out and has been re-transmitted.

If the protocol still survives it may still be bad (!) but to show 
whether it is good or bad is too hard for a quiz. It might be asked 
as homework.