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.