Previous GATE questions with solutions on Computer Networks (Ethernet)  CS/IT
GATE2005
1. In a network of LANs connected by bridges, packets are sent from one LAN to another through intermediate bridges. Since more than one path may exist between two LANs, packets may have to be routed through multiple bridges. Why is the spanning tree algorithm used for bridgerouting?
GATE2005
1. In a network of LANs connected by bridges, packets are sent from one LAN to another through intermediate bridges. Since more than one path may exist between two LANs, packets may have to be routed through multiple bridges. Why is the spanning tree algorithm used for bridgerouting?
(a) For shortest path routing between LANs
(b) For avoiding loops in the routing paths
(c) For fault tolerance
(d) For minimizing collisions
Ans: Option (b)
Explanation:
The Spanning Tree Protocol is used by OSI Data Link Layer devices to create a spanning tree using the existing links (as the source graph) in order to avoid avoiding loops in the routing paths.
Note: To avoid infinite looping, Network Layer uses TTL (Time To Live) field.
GATE2007
2. In Ethernet when Manchester encoding is used, the bit rate is:
(a) Half the baud rate
(b) Twice the baud rate
(c) Same as the baud rate
(d) None of the above
Ans: Option (a)
Explanation:
Manchester coding has the advantage of enabling data to be transmitted without the need for an extra clocking signal.
GATE2003
3. A 2 km 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?
Hence A has 5 chances to win out of 8 combinations.
(a) Half the baud rate
(b) Twice the baud rate
(c) Same as the baud rate
(d) None of the above
Ans: Option (a)
Explanation:
Manchester coding is the process by which digital information in a binary bit stream is converted into electrical signals for transmission. It uses a twostate transition of line voltage to represent one bit of information. In other words, two baud (voltage changes) are used for one bit (piece of information).
GATE2003
3. A 2 km 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?
(a) 50 bytes (b) 100 bytes (c) 200 bytes (d) None of the above
Ans: Option (d)
Explanation:
In CSMA/CD, the transmitting node is listening for collisions while it transmits it's frame. Once it has finished transmitting the final bit without hearing a collision, it assumes that the transmission was successful. In this worstcase collision scenario, the time that it takes for a Node to detect that its frame has been collided with is twice the propagation delay. Hence to confirm that the collision has not occurred the condition for the minimum size of the packet is:
RTT = Transmission Time
Transmission Time = Length of packet / Bandwidth
RTT = 2 (d/v) = 2(2000/2×10^{8})
Therefore to find minimum size of the packet,
RTT = Length of packet / Bandwidth
Length of packet = RTT x Bandwith
= 2(2000/2×10^{8}) x 10^{7} = 200bits = 25bytes
Therefore, minimum size of the packet = 25bytes
GATE2004
4. A and B are the only two stations on an Ethernet. Each has a steady queue of
4. 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
(a) 0.5 (b) 0.625 (c) 0.75 (d) 1.0
Ans: Option (b)
Explanation:
In Ethernet networks, the Exponential backoff algorithm is commonly used to schedule retransmissions after collisions. This algorithm gives waiting time for the stations that are involved in collision.
Waiting time for station = k x 51ms
K is randomly chosen from 0 to 2^{n}1 where n= no of collisions a station is involved.
51ms is a generic RTT for a standard Ethernet.
Station A and Station both try to access a link at the same time. Since they detect a collision, A waits for a random time between 0 and 1 time units and so does B. It's given that A wins the first backoff race. Hence A has no need to wait and begins to use the link and B waits for 1 x 51ms (k=1 is the number selected by B according to the algorithm). At the end of this successful transmission by A, both A and B attempts to transmit, and collide. A will once again choose a random backoff time between 0 and 1, but B will choose a backoff time between 0 and 3 – because this is his second time colliding in a row.
Value of K selected by A

Value of K selected by B

Winner

0

0

X

0

1

A

0

2

A

0

3

A

1

0

B

1

1

X

1

2

A

1

3

A

Therefore that A wins the second backoff race = 5/8 = 0.625
Gate2005
5. Suppose the round trip propagation delay for a 10 Mbps Ethernet having 48bit jamming signal is 46.4 ms. The minimum frame size is:
5. Suppose the round trip propagation delay for a 10 Mbps Ethernet having 48bit jamming signal is 46.4 ms. The minimum frame size is:
(a) 94 (b) 416 (c) 464 (d) 512
Ans: Option (c)
Explanation:
As explained in question no 3, condition for the minimum size of the packet is:
RTT = Transmission Time
RTT = Length of packet / Bandwidth
Minimum frame size = RTT x Bandwidth
= 46.4ms x 10Mbps = 464bits.
GATE2007
6. There are n stations in a slotted LAN. Each station attempts to transmit with a probability p in each time slot. What is the probability that ONLY one station transmits in a given time slot?
6. There are n stations in a slotted LAN. Each station attempts to transmit with a probability p in each time slot. What is the probability that ONLY one station transmits in a given time slot?
(a) (1p)^{n1 }(b) np(1p)^{n1 }(c) p(1p)^{n1 }(d) 1(1p)^{n1}
Ans: Option (b)
Explanation:
Given,
n= total number of stations
p= probability of a station to transmit a packet
1p = probability of a station not to transmit a packet
The probability that a particular station transmits and no body else transmits = p(1p)^{n1}
The probability that any station can transmit among n stations = n p(1p)^{n1}
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