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COE 768 Midterm Solution-v1

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Q1. For each of the following questions, circle T (true) or F (false). (10 marks) T/F Stop and wait protocol can be implemented in the application layer. T/F HDLC uses bit stuffing to prevent occurrence of flag inside the frame. T/F The longer the frames or the shorter the propagation time, the higher link efficiency can be achieved in CSMA/CD. T/F In CSMA/CD, the amount of time that it takes to detect a collision is never greater than the propagation delay. T/F When using CSMA/CD technique, if the medium is busy, the station continues to listen until the channel is sensed idle and then transmits immediately. Q2. Multiple choices (10 marks) 1. If a network uses CSMA/CD, we require transmission delay Tf 2Tc (propagation delay) because: (a.) Tc is the round-trip time from a source to the destination and back again. Therefore, it must be at least twice the one way propagation delay. (b.) Otherwise, the signal would degrade too much along the wire making it difficult to detect collisions. (c.) The sender needs to determine that a frame encountered a collision before it finishes transmitting the frame. (d.) In any network (regardless of whether we use CSMA/CD or not) the transmission time of a frame is a function of both the bit rate, and speed of propagation along the wire. 2. After the kth consecut

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Q1. For each of the following questions, circle T (true) or F (false). (10 marks)

T/F Stop and wait protocol can be implemented in the application layer.
T/F HDLC uses bit stuffing to prevent occurrence of flag 01111110 inside the frame.
T/F The longer the frames or the shorter the propagation time, the higher link efficiency can
be achieved in CSMA/CD.
T/F In CSMA/CD, the amount of time that it takes to detect a collision is never greater than
the propagation delay.
T/F When using CSMA/CD technique, if the medium is busy, the station continues to listen
until the channel is sensed idle and then transmits immediately.

Q2. Multiple choices (10 marks)
1. If a network uses CSMA/CD, we require transmission delay Tf >2Tc (propagation delay)
because:
(a.) Tc is the round-trip time from a source to the destination and back again. Therefore, it must
be at least twice the one way propagation delay.
(b.) Otherwise, the signal would degrade too much along the wire making it difficult to detect
collisions.
(c.) The sender needs to determine that a frame encountered a collision before it finishes
transmitting the frame.
(d.) In any network (regardless of whether we use CSMA/CD or not) the transmission time of

a frame is a function of both the bit rate, and speed of propagation along the wire.
2. After the kth consecutive collision, each colliding station waits for a random time chosen
from the interval-
(a) (0 to 2k) * 2Tc
(b) (0 to 2k-1) * 2Tc
(c) (0 to 2k-1) * Maximum Propagation delay
(d) (0 to 2k-1) * Maximum Propagation delay

3. The Cyclic Redundancy Check (CRC) used on frames of Ethernet provides:
(a) very powerful error detection
(b) very powerful error correction
(c) very powerful error detection and error correction
(d) none of them
4. Routers and switches are different because:
(a) routers operate at the network layer, while switches operate at the data link layer
(b) switches operate at the network layer, while routers operate at the data link layer
(c) routers are for wired networks, while switches are for wireless networks
(d) switches are for wired networks, while routers are for wireless networks




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, Q3 (5 marks)
If the TCP server supports n simultaneous connections, each from a different client host, how
many sockets would the TCP server need?

Solution: n+1

Q4 (5 marks)
The following bit stream is to be inserted into an HDLC frame that uses bit stuffing. Show the
resulting frame including the beginning and ending flag (01111110)
11111101 01111101 11111011 11000001 01111110

Solution:




Q5. (10 marks)
A bit stream 10011101 is transmitted using the standard CRC method. The generator
polynomial is x3 + 1.
Show the actual bit string transmitted.

Suppose that the third bit from the left is inverted during transmission. Show that this error is
detected at the receiver’s end.




2

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