Ethernet
Question 1 
In an Ethernet local area network, which one of the following statements is TRUE?
A station stops to sense the channel once it starts transmitting a frame.  
The purpose of the jamming signal is to pad the frames that are smaller than the minimum frame size.  
A station continues to transmit the packet even after the collision is detected.  
The exponential backoff mechanism reduces the probability of collision on retransmissions. 
Question 1 Explanation:
An Ethernet is the most popularly and widely used LAN network for data transmission.
It is a protocol of data link layer and it tells how the data can be formatted to transmit and how to place the data on network for transmission.
Now considering the Ethernet protocol we will discuss all the options one by one
I. In Ethernet the station does not required, stops to sense the channel prior frame transmission.
II. A Jamming signal used to inform all the other devices or stations about collision that has occurred, so that further data transmission is stopped.
III. No, stations sends jamming signal if collusion is detected
IV. To reduce the probability of collision on retransmissions an exponential back off mechanism is used. This is only True.
Now considering the Ethernet protocol we will discuss all the options one by one
I. In Ethernet the station does not required, stops to sense the channel prior frame transmission.
II. A Jamming signal used to inform all the other devices or stations about collision that has occurred, so that further data transmission is stopped.
III. No, stations sends jamming signal if collusion is detected
IV. To reduce the probability of collision on retransmissions an exponential back off mechanism is used. This is only True.
Question 2 
The minimum frame size required for a CSMA/CD based computer network running at 1 Gbps on a 200 m cable with a link speed of 2 × 10^{8 }m/s is
125 bytes  
250 bytes  
500 bytes  
None of these 
Question 2 Explanation:
For CSMA/CD protocol to work, the transmission time of the frame must be more than minimum value. This minimum value is the RTT.
So,
So,
Question 3 
In Ethernet when Manchester encoding is used, the bit rate is:
Half the baud rate.
 
Twice the baud rate.
 
Same as the baud rate.
 
None of the above.

Question 3 Explanation:
Bit rate is half the baud rate in Manchester encoding as bits are transferred only during a positive transition of the clock.
Question 4 
Suppose the round trip propagation delay for a 10 Mbps Ethernet having 48bit jamming signal is 46.4 μs. The minimum frame size is:
94  
416  
464  
512 
Question 4 Explanation:
Given RTT = 46.4 μs, B.w. = 10 Mbps
Round trip propagation delay is RTT = 2*T_{p}
Minimum frame size of Ethernet can be found by using formula T_{t} = 2*T_{p}
Let L is minimum frame size. Then L / 10Mbps = 46.4 μs
L=464 Kbits
It has nothing to do with jamming signal.
Round trip propagation delay is RTT = 2*T_{p}
Minimum frame size of Ethernet can be found by using formula T_{t} = 2*T_{p}
Let L is minimum frame size. Then L / 10Mbps = 46.4 μs
L=464 Kbits
It has nothing to do with jamming signal.
Question 5 
A and B are the only two stations on an Ethernet. Each has a steady queue of frames to send. Both A and B attempt to transmit a frame, collide, and A wins the first backoff race. At the end of this successful transmission by A, both A and B attempt to transmit and collide. The probability that A wins the second backoff race is
0.5  
0.625  
0.75  
1.0 
Question 5 Explanation:
A has 5 chances to win out of 8 combinations.
The probability that A wins the second backoff race = 5/8 = 0.625
More explanation in the video.
The probability that A wins the second backoff race = 5/8 = 0.625
More explanation in the video.
Question 6 
A 2km long broadcast LAN has 10^{7} bps bandwidth and uses CSMA/CD. The signal travels along the wire at 2×10^{8} m/s. What is the minimum packet size that can be used on this network?
50 bytes
 
100 bytes
 
200 bytes
 
None of the above

Question 6 Explanation:
Minimum packet size for a CSMA/CD LAN is the frame which cover whole RTT(round trip time). i.e. T_{t}=2T_{p}
d= 2 km = 2 x 10^{3} m, v = 2 x 10^{8} m/s, B= 10^{7}
T_{p}= d / v = 2 x 10^{3} /(2 x 10^{8} ) seconds= 10^{5} seconds
Let L bits be minimum size of frame, then T_{t}=t L / B = L / 10^{7} seconds
Now, T_{t}=2T_{p}
L/10^{7} = 2 x 10^{5} = 200 bits = (200 / 8) bytes = 25 bytes
d= 2 km = 2 x 10^{3} m, v = 2 x 10^{8} m/s, B= 10^{7}
T_{p}= d / v = 2 x 10^{3} /(2 x 10^{8} ) seconds= 10^{5} seconds
Let L bits be minimum size of frame, then T_{t}=t L / B = L / 10^{7} seconds
Now, T_{t}=2T_{p}
L/10^{7} = 2 x 10^{5} = 200 bits = (200 / 8) bytes = 25 bytes
There are 6 questions to complete.