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CS-4333 Computer Network Exam 1 Questions with Complete Answers
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CS-4333 Computer Network Exam 1 Questions with Complete Answers
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CS-4333 Computer Network Exam 1Questions with Complete Answers
Exam 1 Focus (7 Question) - Answer-Topics include Ch.1.1-1.5 and Ch. 2.1-2.7
- Checksum
- CRC
Chapter 1.3.4 OSI Model - Answer-7. Application
6. Presentation
5. Session
4. Transport
3. Network
2. Data Link
1. Physical
Chapter 1.5 Latency (TEST) - Answer-Latency (second) = Propagation + Transmit +
Queue
Propagation = Distance/Speed Of Light
Ex: P = 2.5km / (3.0 x 10^8 m/s)
P = (2.5 x 10^3 m) / (3.0 x 10^8 m/s)
P = 0.833 x 10^-5 second
Transmit = Size(bits)/Bandwidth(bps)
Chapter 2.2 Encoding - Answer-non-return to zero (NRZ)
baseline wander
clock recovery
non-return to zero inverted (NRZI)
Manchester
4B/5B
non-return to zero (NRZ) - Answer-- The data value 1 onto the high signal and the data
value 0 onto the low signal.
- There are two fundamental problems caused by long strings of 1s or 0s
1. Baseline Wander problem
2. Clock Recovery problem
Baseline Wander problem - Answer-- Specifically, the receiver keeps an average of the
signal it has seen so far and then uses this average to distinguish between low and high
signals.
- Whenever the signal is significantly lower than this average, the receiver concludes
that it has just seen a 0; likewise, a signal that is significantly higher than the average is
interpreted to be a 1.
, The problem, of course, is that too many consecutive 1s or 0s cause this average to
change, making it more difficult to detect a significant change in the signal.
Clock Recovery problem - Answer-The clock recovery problem is that both the encoding
and decoding processes are driven by a clock—every clock cycle the sender transmits
a bit and the receiver recovers a bit.
The sender's and the receiver's clocks have to be precisely synchronized in order for
the receiver to recover the same bits the sender transmits.
If the receiver's clock is even slightly faster or slower than the sender's clock, then it
does not correctly decode the signal.
Non-Return to Zero Inverted (NRZI) - Answer-- A change at bit boundary means logical
1, no change means 0.
One approach that addresses this problem, called non-return to zero inverted (NRZI),
has the sender make a transition from the current signal to encode a 1 and stay at the
current signal to encode a 0.
This solves the problem of consecutive 1s, but obviously does nothing for consecutive
0s.
Manchester - Answer-The problem with the Manchester encoding scheme is that it
doubles the rate at which signal transitions are made on the link, which means that the
receiver has half the time to detect each pulse of the signal. The rate at which the signal
changes is called the link's baud rate. In the case of the Manchester encoding, the bit
rate is half the baud rate, so the encoding is considered only 50% efficient. Keep in
mind that if the receiver had been able to keep up with the faster baud rate required by
the Manchester encoding
4B/5B - Answer-- 4B/5B, attempts to address the inefficiency of the Manchester
encoding without suffering from the problem of having extended durations of high or low
signals.
- The idea of 4B/5B is to insert extra bits into the bit stream so as to break up long
sequences of 0s or 1s. Specifically, every 4 bits of actual data are encoded in a 5-bit
code that is then transmitted to the receiver; hence, the name 4B/5B.
- The 5-bit codes are selected in such a way that each one has no more than one
leading 0 and no more than two trailing 0s.
- Thus, when sent back-to-back, no pair of 5-bit codes results in more than three
consecutive 0s being transmitted.
- The resulting 5-bit codes are then transmitted using the NRZI encoding, which
explains why the code is only concerned about consecutive 0s—NRZI already solves
the problem of consecutive 1s.
- Note that the 4B/5B encoding results in 80% efficiency.
Chapter 2.3 Framing - Answer-Byte-Oriented Protocols (PPP)
Bit-Oriented Protocols (HDLC)
Clock-Based Framing (SONET)
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