Question 1
Determine the maximum length of the cable (in km) for transmitting data at a rate of 500 Mbps in an Ethernet LAN with frames of size 10,000 bits. Assume the signal speed in the cable to be 2,00,000 km/s.
A
1
B
2
C
2.5
D
5
Data Link Layer    GATE CS 2013    
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Question 1 Explanation: 
Data should be transmitted at the rate of 500 Mbps.
Transmission Time >= 2*Propagation Time
=> 10000/(500*1000000) <= 2*length/200000
=> lenght = 2km (max)
so, answer will be: (B) 2km 
Question 2
Let G(x) be the generator polynomial used for CRC checking. What is the condition that should be satisfied by G(x) to detect odd number of bits in error?
A
G(x) contains more than two terms
B
G(x) does not divide 1+x^k, for any k not exceeding the frame length
C
1+x is a factor of G(x)
D
G(x) has an odd number of terms.
Data Link Layer    GATE-CS-2009    
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Question 2 Explanation: 
Odd number of bit errors can be detected if G(x) contains (x+1) as a factor. See this for proof.
Question 3
Frames of 1000 bits are sent over a 10^6 bps duplex link between two hosts. The propagation time is 25ms. Frames are to be transmitted into this link to maximally pack them in transit (within the link). What is the minimum number of bits (i) that will be required to represent the sequence numbers distinctly? Assume that no time gap needs to be given between transmission of two frames.
A
i = 2
B
i = 3
C
i = 4
D
i = 5
Data Link Layer    GATE-CS-2009    
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Question 3 Explanation: 
Transmission delay for 1 frame = 1000/(10^6) = 1 ms Propagation time = 25 ms The sender can atmost transfer 25 frames before the first frame reaches the destination. The number of bits needed for representing 25 different frames = 5
Question 4
Consider the data of previous question. Suppose that the sliding window protocol is used with the sender window size of 2^i where is the number of bits identified in the previous question and acknowledgments are always piggybacked. After sending 2^i frames, what is the minimum time the sender will have to wait before starting transmission of the next frame? (Identify the closest choice ignoring the frame processing time.)
A
16ms
B
18ms
C
20ms
D
22ms
Data Link Layer    GATE-CS-2009    
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Question 4 Explanation: 
Size of sliding window = 2^5 = 32 Transmission time for a frame = 1ms Total time taken for 32 frames = 32ms The sender cannot receive acknoledgement before round trip time which is 50ms After sending 32 frames, the minimum time the sender will have to wait before starting transmission of the next frame = 50 – 32 = 18
Question 5
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
Data Link Layer    GATE-CS-2007    
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Question 5 Explanation: 
In Manchester encoding, the bitrate is half of the baud rate.
Question 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
(1-p)(n-1)
B
np(1-p)(n-1)
C
p(1-p)(n-1)
D
1-(1-p)(n-1)
Data Link Layer    GATE-CS-2007    
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Question 6 Explanation: 
 

P(X) = Probability that station X attempts to transmit = P P (-X) = Probability that station X does not transmit = 1-P Required is: Probability that only one station transmits = y

Y = (A1, -A2, -A3...... -An) + (-A1, A2, A3......-An) + (-A1, -A2, A3.....-An) + ........... + (-A1, -A2, -A3......An) = (p*(1-p)*(1-p)*...... (1-p) + (1-p)*p*(1-p)........(1-p) + ............. = p*(1-p)^(n-1) + p*(1-p)^n-1 + .................................................... + p*(1-p)^(n-1) = n*p*(1-p)^(n-1)

This solution is contributed Anil Saikrishna Devarasetty .

See question 3 of http://www.geeksforgeeks.org/computer-networks-set-9/
Question 7
In a token ring network the transmission speed is 10^7 bps and the propagation speed is 200 metres/micro second. The 1-bit delay in this network is equivalent to:
A
500 metres of cable.
B
200 metres of cable.
C
20 metres of cable.
D
50 metres of cable.
Data Link Layer    GATE-CS-2007    
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Question 7 Explanation: 
Question 8
The message 11001001 is to be transmitted using the CRC polynomial x^3 + 1 to protect it from errors. The message that should be transmitted is:
A
11001001000
B
11001001011
C
11001010
D
110010010011
Data Link Layer    GATE-CS-2007    
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Question 8 Explanation: 
Question 9
The distance between two stations M and N is L kilometers. All frames are K bits long. The propagation delay per kilometer is t seconds. Let R bits/second be the channel capacity. Assuming that processing delay is negligible, the minimum number of bits for the sequence number field in a frame for maximum utilization, when the sliding window protocolis used, is: GATECS200770
A
A
B
B
C
C
D
D
Data Link Layer    GATE-CS-2007    
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Question 9 Explanation: 
Question 10
Consider a token ring network with a length of 2 km having 10 stations including a monitoring station. The propagation speed of the signal is 2 × 108 m/s and the token transmission time is ignored. If each station is allowed to hold the token for 2 μsec, the minimum time for which the monitoring station should wait (in μsec)before assuming that the token is lost is _______.
A
28 to 30
B
20 to 22
C
0 to 2
D
31 to 33
Data Link Layer    GATE-CS-2014-(Set-1)    
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Question 10 Explanation: 
Length = 2 km
Propagation Speed v = 2*10^8 m/s
Token Holding Time = 2 micro sec

Waiting time
= length/speed + (#stations - 1)*(token holding time) to 
  length/speed + (#stations)*(token holding time)
= 28 to 30
Question 11
Consider a selective repeat sliding window protocol that uses a frame size of 1 KB to send data on a 1.5 Mbps link with a one-way latency of 50 msec. To achieve a link utilization of 60%, the minimum number of bits required to represent the sequence number field is ________.
A
3
B
4
C
5
D
6
Data Link Layer    GATE-CS-2014-(Set-1)    
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Question 11 Explanation: 
Transmission delay = Frame Size/bandwidth
                   = (1*8*10^3)/(1.5 * 10^6)=5.33ms
Propagation delay = 50ms
Efficiency = Window Size/(1+2a) = .6

a = Propagation delay/Transmission delay
So, window size = 11.856(approx)
min sequence number = 2*window size = 23.712
bits required in Min sequence number = log2(23.712)
Answer is 4.56

Ceil(4.56) = 5 
Question 12
In the diagram shown below L1 is an Ethernet LAN and L2 is a Token-Ring LAN. An IP packet originates from sender S and traverses to R, as shown. The link within each ISP, and across two ISPs, are all point to point optical links. The initial value of TTL is 32. The maximum possible value of TTL field when R receives the datagram is GATECS2014Q25
A
25
B
24
C
26
D
28
Data Link Layer    GATE-CS-2014-(Set-2)    
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Question 12 Explanation: 
TTL(Time to Live) - It is a mechanism that limits the lifespan of a packet in a computer network. It is implemented with the help of a counter or timestamp which is set by the sender of the packet and embedded in the packet itself. It represents the maximum lifetime of a packet in the network. When a packet routes through a network, each router checks the current value of its TTL, if the TTL value is not zero then only the router accepts the packet, and decrements its value by 1. This process takes place at every router. If some router founds the TTL value of the incoming packet to be 0, then it simply discards/destroys the packet ( because the lifetime of the packet is over, and hence the packet is not eligible to be in the network ). One of the main purposes of setting the TTL value and doing all this process is to ensure that there is no undelivered packet in the network which is circulating indefinitely, and to avoid the problem of duplicate delivery of the same packet, which may arise in the case of network congestion. Now, routing decisions are taken place at the network layer. hence we have to see in the above question that when the packet is going through the network layer. LANs work at Data Link Layer only, hence the packet doesn't reach network layer in LANs. So in the question above, Except LANs, at all other points routing decisions have to be taken. Hence TTL value will be checked and manipulated at those points/routers. At Receiver end also, the packet has to go through the network layer so as to reach to application layer, hence Receiver will also check and decrement the TTL value. So there are 6 routers in the above diagram. Initially TTL value was 32, so at the Receiver it will become 32 - 6 = 26. TTL (1)    
Question 13
Consider the store and forward packet switched network given below. Assume that the bandwidth of each link is 106 bytes / sec. A user on host A sends a file of size 103 bytes to host B through routers R1 and R2 in three different ways. In the first case a single packet containing the complete file is transmitted from A to B. In the second case, the file is split into 10 equal parts, and these packets are transmitted from A to B. In the third case, the file is split into 20 equal parts and these packets are sent from A to B. Each packet contains 100 bytes of header information along with the user data. Consider only transmission time and ignore processing, queuing and propagation delays. Also assume that there are no errors during transmission. Let T1, T2 and T3 be the times taken to transmit the file in the first, second and third case respectively. Which one of the following is CORRECT? GATECS2014Q26
A
T1 < T2 < T3
B
T1 > T2 > T3
C
T2 = T3, T3 < T1
D
T1 = T3, T3 > T2
Data Link Layer    GATE-CS-2014-(Set-2)    
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Question 13 Explanation: 
The important thing to note here is in first case, the whole packet is being transmitted, so no piplelining of packet happens. In second and third case, we have advantage of pipelining (While packet 'i' is being transmitted from R1 to R2, packet 'i-1' is being transmitted from A to R1 at the same time). Following are complete calculations.
File Size = 1000 bytes
Header Size = 100 bytes
Transmission Speed of all  links = 10^6 bytes/sec

Ist Case: 
 Transmission time for one link 
                = packetsize/bandwidth
                = (1000 + 100)/10^6
                = 1100 micros
 Total time = 3*1100
            = 3300 microsec.

Second case: 
  Transmission time for one link and one part
                   = (100 + 100)/10^6
                   =  200 microsec

  [Note the pipe-lining in packets.  While
  packet 'i' is being transmitted from R1 to R2,
  packet 'i-1' is being transmitted from A to R1
  at the same time]
  Total time = 3*200 + 9*200
             = 2400 micro sec

Third Case: 
  Transmission time for one link and one part
                                 = (50+100)/10^6 
                                 = 150microsec
  Total time = 3*150+19*150
             = 3300 microsec 
Question 14
A bit-stuffing based framing protocol uses an 8-bit delimiter pattern of 01111110. If the output bit-string after stuffing is 01111100101, then the input bit-string is
A
0111110100
B
0111110101
C
0111111101
D
0111111111
Data Link Layer    GATE-CS-2014-(Set-3)    
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Question 14 Explanation: 
Bit Stuffing is used to create framing.
8-bit delimiter pattern is 01111110. 

The output bit-string after stuffing is 01111100101.

The above highlighted bit is stuffed bit.  
So input bit-string must be 0111110101. 
Refer https://www.youtube.com/watch?v=FpyoALtvUTo&index=16&list=PLkHsKoi6eZnzJl1qTzmvBwTxrSJW4D2Jj for more details.
Question 15
Station A uses 32 byte packets to transmit messages to Station B using a sliding window protocol. The round trip delay between A and B is 80 milliseconds and the bottleneck bandwidth on the path between A and B is 128 kbps. What is the optimal window size that A should use?
A
20
B
40
C
160
D
320
Data Link Layer    GATE-CS-2006    
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Question 15 Explanation: 
Round Trip propagation delay = 80ms 
Frame size = 32*8 bits
Bandwidth = 128kbps
Transmission Time = 32*8/(128) ms = 2 ms

Let n be the window size.

UtiliZation = n/(1+2a) where a 
            = Propagation time / 
                transmission time
            = n/(1+80/2)

For maximum utilization: n = 41 
which is close to option (B)
Source : Question 1 of http://www.geeksforgeeks.org/computer-networks-set-11/
Question 16
Consider the diagram shown below where a number of LANs are connected by (transparent) bridges. In order to avoid packets looping through circuits in the graph, the bridges organize themselves in a spanning tree. First, the root bridge is identified as the bridge with the least serial number. Next, the root sends out (one or more) data units to enable the setting up of the spanning tree of shortest paths from the root bridge to each bridge. Each bridge identifies a port (the root port) through which it will forward frames to the root bridge. Port conflicts are always resolved in favour of the port with the lower index value. When there is a possibility of multiple bridges forwarding to the same LAN (but not through the root port), ties are broken as follows: bridges closest to the root get preference and between such bridges, the one with the lowest serial number is preferred. cs200682 For the given connection of LANs by bridges, which one of the following choices represents the depth first traversal of the spanning tree of bridges? [2 marks]
A
B1, B5, B3, B4, B2
B
B1, B3, B5, B2, B4
C
B1, B5, B2, B3, B4
D
B1, B3, B4, B5, B2
Data Link Layer    GATE-CS-2006    
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Question 16 Explanation: 
Spanning tree is
                B1
                /  \
              /      \
             B5      B3
                     /  \
                   /     \
                  B4     B2 
Note that B4 and B2 are connected through B3 (Not B5) because B3 has lower serial number than B5. One DFS traversal of tree is B1 B5 B3 B4 B2 Hence Option A is the Answer
Question 17
Consider the data given in above question. Consider the correct spanning tree for the previous question. Let host H1 send out a broadcast ping packet. Which of the following options represents the correct forwarding table on B3?[2 marks] (A)
 Hosts  Port
 H1, H2, H3, H4  3
 H5, H6, H9, H10  1
 H7, H8, H11, H12  2
(B)
 Hosts  Port
 H1, H2  4
 H3, H4  3
 H5, H6  1
 H7, H8, H9, H10,H11,H12  2
(C)
 Hosts  Port
H3, H4  3
H5, H6, H9, H10  1
 H1, H2  4
 H7, H8, H11, H12  2
(D)
 Hosts  Port
 H1, H2, H3, H4  3
 H5, H7, H9, H10  1
 H7, H8, H11, H12  4
A
A
B
B
C
C
D
D
Data Link Layer    GATE-CS-2006    
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Question 18
Suppose the round trip propagation delay for a 10 Mbps Ethernet having 48-bit jamming signal is 46.4 ms. The minimum frame size is
A
94
B
416
C
464
D
512
Data Link Layer    GATE-CS-2005    
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Question 18 Explanation: 
Question 19
How many 8-bit characters can be transmitted per second over a 9600 baud serial communication link using asynchronous mode of transmission with one start bit, eight data bits, two stop bits, and one parity bit ?
A
600
B
800
C
876
D
1200
Data Link Layer    GATE-CS-2004    
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Question 19 Explanation: 
  Background required - Physical Layer in OSI Stack In serial Communications, information is transferred in or out one bit at a time. The baud rate specifies how fast data is sent over a serial line. It’s usually expressed in units of bits-per-second (bps). Each block (usually a byte) of data transmitted is actually sent in a packet or frame of bits. Frames are created by appending synchronization and parity bits to our data. frame
"9600 baud" means that the serial port is capable of transferring a 
maximum of 9600 bits per second.

Total Data To send = 1 bit(start) + 8 bits(char size) + 1 bit(Parity) + 2 bits(Stop) 

= 12 bits.

Number of 8-bit characters that can be transmitted per second  = 9600/12 = 800.
This explanation is contributed by Pranjul Ahuja.
Question 20
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
Data Link Layer    GATE-CS-2004    
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Question 20 Explanation: 
This is basically the question related to unfairness of exponential back-off algorithm called 'capture effect'. You can find more info about it here: http://intronetworks.cs.luc.edu/current/html/ethernet.html#capture-effect The solution to the above problem goes like this: At every attempt to transit a frame, both A and B chooses value of 'k' randomly. Based on the value of 'k', back-off time is calculated as a multiple of 'k'. The station or node having the smaller back-off time gets to send the frames eariler. 1st attempt: Value of 'k' would be k=0 or k=1 (0 <= k <= 2^n-1; where n=nth attempt). Since A won the first race, A must have chosen k=0 and B must have chosen k=1 (A wins here with probability 0.25). As A won, A will again choose k=0 or k=1 for its 2nd frame, but B will choose k=0,1,2 or 3 as B failed to send its first frame in the first attempt. 2nd attempt: Let kA= value of k chosen by A and kB = value of k chosen by B. We will use notation (kA,kB) to show the possible values. Now the sample space for the 2nd attempt is (kA,kB) = (0,0),(0,1),(0,2),(0,3),(1,0),(1,1),(1,2) or (1,3) i.e. 8 possible outcomes. For A to win, kA should be less than kB (kA < kB). Thus, our event space is (kA, kB) = (0,1),(0,2),(0,3),(1,2),(1,3) i.e. 5 possible outcomes. Thus the probability that A wins the 2nd back-off race = 5/8 = 0.625
Question 21
A 2 km long broadcast LAN has 107 bps bandwidth and uses CSMA/CD. The signal travels along the wire at 2 × 108 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 these
Data Link Layer    GATE-CS-2003    
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Question 21 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 worst-case 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(2000/2×108) 

Therefore to find minimum size of the packet, 
RTT = Length of packet / Bandwidth
Length of packet = RTT x Bandwith
                         = 2(2000/2×108) x 107 = 200bits = 25bytes

Therefore, minimum size of the packet = 25bytes 
Source: http://www.btechonline.org/2012/12/gate-computer-networks-ethernet.html
Question 22
Host A is sending data to host B over a full duplex link. A and B are using the sliding window protocol for flow control. The send and receive window sizes are 5 packets each. Data packets (sent only from A to B) are all 1000 bytes long and the transmission time for such a packet is 50 µs. Acknowledgement packets (sent only from B to A) are very small and require negligible transmission time. The propagation delay over the link is 200 us. What is the maximum achievable throughput in this communication?
A
7.69 × 106 bytes per second
B
11.11 × 106 bytes per second
C
12.33 × 106 bytes per second
D
15.00 × 106 bytes per second
Data Link Layer    GATE-CS-2003    
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Question 22 Explanation: 
Network throughput ≈ Window size / roundtrip time

Roundtrip time = 2 × Packet delivery time + processing delay
= ransmission delay+2*propagation delay 
=50microsec+2*200microsec=450microsec

Now Throughput = ((5*1000*bytes)/450microsec) = 11.1111 * 106 bytes per second
Question 23
In serial data transmission, every byte of data is padded with a ‘0’ in the beginning and one or two ‘I’ s at the end of byte because
A
Receiver is to be synchronized for byte reception
B
Receiver recovers lost ‘0’ and ‘1’s from these padded bits
C
Padded bits are useful in parity computation
D
None of these
Data Link Layer    GATE-CS-2002    
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Question 23 Explanation: 
In serial communication, the sender & receiver needs to be synchronized with each other.
'0' is added in the beginning of data as start bit and '1' is added at the end as a stop bit. The start signal tells the receiver about the arrival of data and the stop signal resets the receiver's state for the arrival of a new data.
 
Thus, in serial communication receiver is to be synchronized for byte reception.
 
Please comment below if you find anything wrong in the above post.
Question 24
Consider a LAN with four nodes S1, S2, S3 and S4. Time is divided into fixed-size slots, and a node can begin its transmission only at the beginning of a slot. A collision is said to have occurred if more than one node transmit in the same slot. The probabilities of generation of a frame in a time slot by S1, S2, S3 and S4 are 0.1, 0.2, 0.3 and 0.4, respectively. The probability of sending a frame in the first slot without any collision by any of these four stations is _________.
A
0.462
B
0.711
C
0.5
D
0.652
Data Link Layer    GATE-CS-2015 (Set 1)    
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Question 24 Explanation: 
The probability of sending a frame in the first slot 
without any collision by any of these four stations is
sum of following 4 probabilities

Probability that S1 sends a frame and no one else does + 
Probability thatS2 sends a frame and no one else does +
Probability thatS3 sends a frame and no one else does +
Probability thatS4 sends a frame and no one else does

= 0.1 * (1 - 0.2) * (1 - 0.3) *(1 - 0.4) + 
  (1 -0.1) * 0.2 * (1 - 0.3) *(1 - 0.4) + 
  (1 -0.1) * (1 - 0.2) * 0.3 *(1 - 0.4) + 
  (1 -0.1) * (1 - 0.2) * (1 - 0.3) * 0.4

= 0.4404 
Question 25
Suppose that the stop-and-wait protocol is used on a link with a bit rate of 64 kilobits per second and 20 milliseconds propagation delay. Assume that the transmission time for the acknowledgment and the processing time at nodes are negligible. Then the minimum frame size in bytes to achieve a link utilization of at least 50% is _________.
A
160
B
320
C
640
D
220
Data Link Layer    GATE-CS-2015 (Set 1)    
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Question 25 Explanation: 
Transmission or Link speed = 64 kb per sec
Propagation Delay = 20 milisec

Since stop and wait is used, a packet is sent only
when previous one is acknowledged.

Let x be size of packet, transmission time = x / 64 milisec

Since utilization is at least 50%, minimum possible total time
for one packet is twice of transmission delay, which means 
x/64 * 2 = x/32

x/32 > x/64 + 2*20
x/64 > 40
x > 2560 bits = 320 bytes
Answer in GATE keys says 160 bytes, but the answer keys seem incorrect. See question 36 here.
Question 26
Consider a CSMA/CD network that transmits data at a rate of 100 Mbps (108 bits per second) over a 1 km (kilometer) cable with no repeaters. If the minimum frame size required for this network is 1250 bytes, what is the signal speed (km/sec) in the cable?
A
8000
B
10000
C
16000
D
20000
Data Link Layer    GATE-CS-2015 (Set 3)    
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Question 26 Explanation: 
Data should be transmitted at the rate of 100 Mbps.
Transmission Time >= 2*Propagation Time
=> 1250*8 / (100 * 10^6) <= 2*length/signal_speed
=> signal_speed  <= (2 * 10^3 * 100 * 10^6) / (1250 * 8)
                 <= 2 * 10 * (10 ^ 3) km/sec 
                 <= 20000
Refer http://pet.ece.iisc.ernet.in/course/E2223/Problems.pdf for more details.
Question 27
Which one of the following statements is FALSE?
A
Packet switching leads to better utilization of bandwidth resources than circuit switching.
B
Packet switching results in less variation in delay than circuit switching.
C
Packet switching requires more per packet processing than circuit switching
D
Packet switching can lead to reordering unlike in circuit switching
Data Link Layer    GATE-IT-2004    
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Question 27 Explanation: 
Packet-switched networks move data in separate, small packets -- based on the destination address in each packet. When received, packets are reassembled in the proper sequence to make up the message. Circuit-switched networks require dedicated point-to-point connections during calls.
  • Packet switching leads to better utilization of bandwidth resources than circuit switching //Because data move in packets therefore better use of resources
  • Packet switching results in less variation in delay than circuit switching. //Queueing Delay fluctuation is more in packet switching as  there is no dedicated path
  • Packet switching requires more per packet processing than circuit switching   //All the info needs to be in packet as well as packet reassembling has to be done at end therefore more processing
  • Packet switching can lead to reordering unlike in circuit switching //because reassembling is done at destination,packing reordering might occur
Therefore only B is false 
Question 28
Which of the following statements is TRUE about CSMA/CD  
A
IEEE 802.11 wireless LAN runs CSMA/CD protocol
B
Ethernet is not based on CSMA/CD protocol
C
CSMA/CD is not suitable for a high propagation delay network like satellite network
D
There is no contention in a CSMA/CD network
Data Link Layer    Gate IT 2005    
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Question 28 Explanation: 
CSMA/CD requires that sender should be transmitting at least till the first bit reaches the receiver, so it can detect collision if any. For Networks with high prorogation delay this time becomes too long hence the minimum packet size required becomes too big to be feasible.
Question 29
Which of the following statements is FALSE regarding a bridge?
A
Bridge is a layer 2 device
B
Bridge reduces collision domain
C
Bridge is used to connect two or more LAN segments
D
Bridge reduces broadcast domain
Data Link Layer    Gate IT 2005    
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Question 29 Explanation: 
  • Bridge is a layer 2 device -TRUE //A device used to connect two separate Ethernet networks into one extended Ethernet and Ethernet works on DATA LINK Layer
  • Bridge reduces collision domain-TRUE//A bridge is a two interfaces device that creates 2 collision domains, since it forwards the traffic it receives from one interface only to the interface where the destination layer 2 device (based on his mac address) is connected to.
  • Bridge is used to connect two or more LAN segments -TRUE // It is its sole purpose
  • Bridge reduces broadcast domain- FALSE //Bridge can reduce collision domain but can NOT reduce broadcast domain
Therefore, Answer is D    
Question 30
A network with CSMA/CD protocol in the MAC layer is running at 1 Gbps over a 1 km cable with no repeaters. The signal speed in the cable is 2 x 108 m/sec. The minimum frame size for this network should be  
A
10000 bits
B
10000 bytes
C
5000 bits
D
5000 bytes
Data Link Layer    Gate IT 2005    
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Question 30 Explanation: 
Frame Size S >= 2BL/P

Where,

Cable Length L = 1KM = 1000M 
Propogation Speed P = 2 x 10^8 m/sec
Bandwidth = 1 Gbps = 10^9 bps

See this for details of above formula.

S >= (2 *  10^9 * 1000) / (2 x 10^8)
  >= 10000 bits
Question 31
A channel has a bit rate of 4 kbps and one-way propagation delay of 20 ms. The channel uses stop and wait protocol. The transmission time of the acknowledgement frame is negligible. To get a channel efficiency of at least 50%, the minimum frame size should be  
A
80 bytes
B
80 bits
C
160 bytes
D
160 bits
Data Link Layer    Gate IT 2005    
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Question 31 Explanation: 

Bit rate = 4 kbps One-way propagation delay = 20 ms
Efficiency = Transmission time of packet/(Transmission time of packet + 2 * Propagation delay) 0.5 = x/(x + 2 * 20 * 10-3) x = 20 * 10-3 x = 40 * 10-3
Minimum frame size / Bit rate = 40 * 10-3 Therefore, Minimum frame size = 40 * 10-3 * 4 * 103 = 160 bits
 
Thus, option (D) is correct.
 
Please comment below if you find anything wrong in the above post.
Question 32
In a TDM medium access control bus LAN, each station is assigned one time slot per cycle for transmission. Assume that the length of each time slot is the time to transmit 100 bits plus the end-to-end propagation delay. Assume a propagation speed of 2 x 108 m/sec. The length of the LAN is 1 km with a bandwidth of 10 Mbps. The maximum number of stations that can be allowed in the LAN so that the throughput of each station can be 2/3 Mbps is
A
3
B
5
C
10
D
20
Data Link Layer    Gate IT 2005    
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Question 32 Explanation: 
Tt = 10^(-5) seconds

Tp = 0.5*10^(-5) seconds

efficiency = [1/(1+a)]

where a = (Tp/Tt) = 0.5

therefore efficiency = 2/3

effective bandwidth = (2/3)*10Mbps

number of users = {(2/3)*10} / (2/3) = 10

therefore answer is (C)
Question 33
Consider the following message M = 1010001101. The cyclic redundancy check (CRC) for this message using the divisor polynomial x5 + x4 + x2 + 1 is :  
A
01110
B
01011
C
10101
D
10110
Data Link Layer    Gate IT 2005    
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Question 33 Explanation: 
M = 1010001101
Divisor polynomial: 1.x5 +1.x4+0.x3+1.x2+0.x2+1.x0   
Divisor polynomial bit= 110101
Bits to be appended to message= (divisor polynomial bits – 1) = 5
Append 5 zeros to message bits, modified message: 101000110100000
Now, divide and XOR the message with divisor polynomial bits. Make resultant reminder to 5 bit again and that is the CRC send along with the message. CRC This explanation has been contributed by Sandeep Pandey. Please visit the following links to learn more on CRC and its calulation: Wikipedia article: Cyclic Redundancy Check GeeksforGeeks article: Error Detection | Computer Networks
Question 34
A sender uses the Stop-and-Wait ARQ protocol for reliable transmission of frames. Frames are of size 1000 bytes and the transmission rate at the sender is 80 Kbps (1Kbps = 1000 bits/second). Size of an acknowledgement is 100 bytes and the transmission rate at the receiver is 8 Kbps. The one-way propagation delay is 100 milliseconds. Assuming no frame is lost, the sender throughput is __________ bytes/second.
A
2500
B
2000
C
1500
D
500
Data Link Layer    GATE-CS-2016 (Set 1)    
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Question 34 Explanation: 
Total time = Transmission-Time + 2* Propagation-Delay + Ack-Time.
Trans. time = (1000*8)/80*1000 = 0.1 sec
2*Prop-Delay = 2*100ms = 0.2 sec
Ack time = 100*8/8*1000 = 0.1 sec.
Total Time = 0.1 + 0.2 + 0.1 = 0.4 sec.
Throughput = ((L/B)/Total time) * B, 
L = data packet to be sent and
B = BW of sender.

Throughput = L/Total Time
           = 1000/0.4
           = 2500 bytes/sec.
Question 35
In an Ethernet local area network, which one of the following statements is TRUE ?
A
A station stops to sense the channel once it starts transmitting a frame.
B
The purpose of the jamming signal is to pad the frames that are smaller than the minimum frame size.
C
A station continues to transmit the packet even after the collision is detected.
D
The exponential backoff mechanism reduces the probability of collision on retransmissions
Data Link Layer    GATE-CS-2016 (Set 2)    
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Question 35 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 (A) This option is false as in Ethernet the station is not required to stop to sense for the channel prior frame transmission. (B) A signal is jammed to inform all the other devices or stations about collision that has occurred so that further data transmission is stopped. Thus this option is also false (C) Once the collision has occurred the data transmission is stopped as the jam signal is sent. Thus this option is also incorrect. (D) To reduce the probability of collision on retransmissions an exponential back off mechanism is used. Thus, only this option is true. This solution is contributed by Namita Singh.
Question 36
A network has a data transmission bandwidth of 20 × 10 6 bits per second. It uses CSMA/CD in the MAC layer. The maximum signal propagation time from one node to another node is 40 microseconds. The minimum size of a frame in the network is _________ bytes.   Note : This question was asked as Numerical Answer Type.
A
200
B
250
C
400
D
1200
Data Link Layer    GATE-CS-2016 (Set 2)    
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Question 36 Explanation: 
For frame size to be minimum, its transmission time should be equal to twice of one way propagation delay. i.e, Tx = 2Tp suppose minimum frame size is L bits. Tx = L / B where B is the bandwidth  = 20 * 10^6 bits/sec and given Tp = 40 micro seconds as Tx = 2 Tp L / B = 80 micro seconds L = B * 80 micro seconds = 20 * 10^6 bits/sec * 80 micro seconds = 1600 bits Now as the answer has to be given in Bytes, hence, 1600 / 8 bytes = 200 bytes.   Thus, A is the correct answer.
Question 37
Consider a 128×10 3 bits/second satellite communication link with one way propagation delay of 150 milliseconds. Selective retransmission (repeat) protocol is used on this link to send data with a frame size of 1 kilobyte. Neglect the transmission time of acknowledgement. The minimum number of bits required for the sequence number field to achieve 100% utilization is ___________
A
2
B
4
C
6
D
8
Data Link Layer    GATE-CS-2016 (Set 2)    
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Question 37 Explanation: 
To achieve 100% efficiency, the number of frames that we should send = 1 + 2 * Tp / Tt where, Tp is propagation delay, and Tt is transmission delay. So, Number of frames sent = 1 + 4.687 = 5.687 (approx 6) As the protocol used is Selective repeat, Receiver window size should be equal to Sender window size. Then, Number of distinct sequence numbers required = 6 + 6 = 12 Number of bits required to represent 12 distinct numbers = 4 So, Answer is (B)
Question 38
How many bytes of data can be sent in 15 seconds over a serial link with baud rate of 9600 in asynchronous mode with odd parity and two stop bits in the frame?
A
10,000 bytes
B
12,000 bytes
C
15,000 bytes
D
27,000 bytes
Data Link Layer    Misc Topics in Computer Networks    Gate IT 2008    
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Question 38 Explanation: 
  Background required - Physical Layer in OSI Stack In serial Communications, information is transferred in or out one bit at a time. The baud rate specifies how fast data is sent over a serial line. It’s usually expressed in units of bits-per-second (bps). Each block (usually a byte) of data transmitted is actually sent in a packet or frame of bits. Frames are created by appending synchronization and parity bits to our data. frame
"9600 baud" means that the serial port is capable of transferring a 
maximum of 9600 bits per second.

1 sec--------> 9600 bits
15 sec------->9600*15 bits

Total Data To send in 1 frame = 1 bit(start) + 8 bits(char size) + 1 bit(Parity) + 2 bits(Stop) 

= 12 bits.

Number of 8-bit characters that can be transmitted per second  = (9600 * 15)/12 = 12000 bytes.
This explanation is contributed by Pranjul Ahuja.
Question 39
A 1Mbps satellite link connects two ground stations. The altitude of the satellite is 36,504 km and speed of the signal is 3 × 108 m/s. What should be the packet size for a channel utilization of 25% for a satellite link using go-back-127 sliding window protocol? Assume that the acknowledgment packets are negligible in size and that there are no errors during communication.
A
120 bytes
B
60 bytes
C
240 bytes
D
90 bytes
Data Link Layer    Gate IT 2008    
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Question 39 Explanation: 
2008_64_sol
Question 40
The minimum frame size required for a CSMA/CD based computer network running at 1 Gbps on a 200m cable with a link speed of 2 × 108m/s is
A
125 bytes
B
250 bytes
C
500 bytes
D
None of these
Data Link Layer    Misc    Gate IT 2008    
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Question 40 Explanation: 
2008_65_sol
Question 41
Data transmitted on a link uses the following 2D parity scheme for error detection: Each sequence of 28 bits is arranged in a 4×7 matrix (rows r0 through r3, and columns d7 through d1) and is padded with a column d0 and row r4 of parity bits computed using the Even parity scheme. Each bit of column d0 (respectively, row r4) gives the parity of the corresponding row (respectively, column). These 40 bits are transmitted over the data link.
2008_66
The table shows data received by a receiver and has n corrupted bits. What is the mini­mum possible value of n?
A
1
B
2
C
3
D
4
Data Link Layer    Gate IT 2008    
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Question 41 Explanation: 
(r1,d5)->1, (r4,d2)->0, (d4,d0)->1
=> n=3
There are 41 questions to complete.
 

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