Unit 3
Data Link Layer: Introduction, Framing, Error Detection and Correction, Flow control (Elementary Data Link Protocols, Sliding Window protocols), Medium Access Control and Local Area Networks: Channel allocation, Multiple access protocols, LAN standards, Link layer switches & bridges (learning bridge and spanning tree algorithms).
Q41 – 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?
G(x) contains more than two terms
G(x) does not divide 1 + xk, for any not exceeding the frame length
1 + x is a factor of G(x)
G(x) has an odd number of terms.
Ans – (3)
Explanation –
If 1+x (which is also written as x+1) is a factor of the generator polynomial G(x), then G(x) can detect all single-bit errors, which implies it can detect any odd number of bit errors. This is because 1+x corresponds to the pattern that flips the parity of the bit sequence, making it capable of detecting all odd bit errors.
Q42 – Frames of 1000 bits are sent over a 106 bps duplex link between two hosts. The propagation time 25 ms. Frames are to be transmitted into this link to maximally pack them in transit (within the link). What is the minimum number of bits (l) that will be required to represent the sequence numbers distinctly? Assume that no time gap needs to be given between transmission of two frames.
I = 2
I = 3
I = 4
I = 5
Ans – (4)
Explanation –
To determine the minimum number of bits required to represent the sequence numbers distinctly for maximally packing frames in transit on a 106 bps duplex link with a propagation time of 25 milliseconds, we need to calculate the bandwidth-delay product. This product gives us the maximum number of bits that can be in transit at any given time on the link.
First, calculate the bandwidth-delay product:
Bandwidth = 106 bps
Propagation time = 25 ms = 0.025 seconds
Bandwidth-Delay Product = Bandwidth × Propagation time Bandwidth-Delay Product = 25,000 bits
Next, determine the number of frames that can be in transit. Each frame is 1000 bits long:
Number of frames in transit = Bandwidth-Delay Product / Frame size Number =25,000 bits / 1,000 bits/frame = 25 frame
This means there can be up to 25 frames in transit at any time. To represent the sequence numbers for these frames distinctly, we need to be able to distinguish at least 25 different sequence numbers.
Therefore, we need 2l ≥ 25
So, I = 5
Q43 – In the carrier sense network if the prevailing condition is a ‘channel busy’, then which of the following are correct?
If the technique used is non-persistent then it results in randomised wait and sense.
If the technique used is 1-persistent then the channel is continuously sensed.
If the technique used is p-persistent then randomized retransmission is done.
If the method used is non-persistent then continuous sensing results.
(Williams Timothy, McGraw Hill Education)
Ans – (1, 2)
Explanation –
Non-persistent: The station waits for a random period before re-sensing the channel, reducing the probability of collisions.
1-persistent: The station continuously senses the channel and transmits immediately once the channel is free, which can lead to higher chances of collisions.
Q44 – What is the maximum efficiency of pure aloha at G = 1/2?
1.89
17.99
18.999
18.4
Ans – (4)
Explanation –
Efficiency (η) = G * e^(-2G)
where G is the offered load (average number of packet transmissions per unit time per slot).
For G = 1/2
Efficiency (η) = (1/2) * e^(-2 * (1/2))
= (1/2) * e^(-1)
≈ (1/2) * 0.3679
≈ 0.1839
Therefore, the maximum efficiency of pure ALOHA at G = 1/2 is approximately 18.39%, which when rounded to two decimal places is 18.4%.
Q45 – What is the maximum efficiency of slotted aloha at G = 1?
36.8
35.8
35.5
37.8
Ans – (1)
Explanation –
Efficiency (η) = G * e^(-G)
where G is the offered load (average number of packet transmissions per unit time per slot).
For G = 1,
Efficiency (η) = 1 * e^(-1)
= e^(-1)
Using a calculator, e^(-1) is approximately 0.3679.
Therefore, the maximum efficiency of slotted ALOHA at G = 1 is approximately 36.79%, which when rounded to one decimal place is 36.8%.
Q46 – Consider a building a CSMA/CD network running at 10 Mbps over a cable with no repeaters. If the signal speed is in the cable is 106 km/sec and minimum frame size is 1500 bytes than what is cable length?
600 km
1200 km
12 km
120 km
Ans – (1)
Explanation –
Network speed: 10 Mbps
Signal speed in cable: 106 km/sec
Minimum frame size: 1500 bytes = 1500 x 8 = 12000 bits
Transmission Time Tt = 12000 / 10000000 = 1.2 ms
Tt = 2 x Distance / speed
Distance = (1.2 x 10-3 x 106) / 2 = 600 km
Q47 – The …………….. layer is responsible for resolving access to the shared media or resources.
Physical
Mac sub layer
Network
Transport
Ans – (2)
Explanation –
The layer responsible for resolving access to shared media or resources is the Media Access Control (MAC) sublayer, which is a part of the Data Link layer.
In the OSI model, the Data Link layer consists of two sublayers
Logical Link Control (LLC) Sublayer – The LLC sublayer is responsible for providing an interface between the higher-layer protocols and the lower-layer media access control methods. It ensures reliable data transfer across the physical link and handles error correction and flow control.
Media Access Control (MAC) Sublayer – The MAC sublayer is responsible for controlling access to the physical network medium. It determines how devices on the same network gain access to the medium and manage the transmission of data frames. This sublayer ensures that data is transmitted reliably across the physical network segment.
Q48 – If the bit string 0111101111101111110 is subjected to bit stuffing for the flag string 01111110, the output string is
011110111110011111010
01111011111011111100
01111011111011111010
0111101111101111110
(Timothy Williams, McGraw Hill Education)
Ans – (1)
Explanation –
In bit stuffing, the flag sequence is used to detect the start and end of frames. When the flag sequence appears in the data, a special bit is inserted after a certain number of consecutive 1s to ensure that the flag sequence does not occur within the data.
Let’s go through the given bit string and apply bit stuffing according to the flag sequence “01111110”:
Original bit string: 0111101111101111110
After bit stuffing, we insert a 0 after every sequence of five consecutive 1s – 0111101111100111111010
Q49 – 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.
1
2
2.5
5
Ans – (2)
Explanation –
Transmission delay = Data rate / Frame size = 10,000 / 500 x 106
Transmission delay = 0.00002 sec
Next, we need to calculate the maximum propagation delay. For Ethernet, the maximum propagation delay should be such that the frame can traverse the cable and return (round-trip time) within the duration of the transmission delay.
Propagation delay = 2*Cable length / Signal speed
0.00002 = 2*Cable length / 2,00,000
Cable length = 2 km
Q50 – At a transmission rate of 5 Mbps and propagation speed of 200 m/µsec, to how many meters of cable is the 1-bit display in a token ring interface equivalent?
40 m
30 m
20 m
10 m
Ans – (1)
Explanation –
Transmission rate = 5 Mbps
Propagation Speed = 200 m/µsec
Time to travel 1 bit = 1 / 5*106 = 0.2 µsec
Distance = Propagation Speed * time to travel 1 bit
Distance = 200 m/µsec * 0.2 µsec = 40 m
Q51 – The purpose of ethernet frame preamble is to provide
A synchronization of the receiver with the signal
Information that identifies the type of data in the frame.
A mechanism where by Ethernet’s CSMA/CD can determine if a collision has occurred
The destination MAC address at the beginning of the frame
Ans – (1)
Explanation –
The preamble in an Ethernet frame consists of a sequence of bits (typically 7 bytes of alternating 1s and 0s) that helps the receiver to synchronize with the incoming signal and prepare for the reception of the actual data that follows. This ensures that the receiver is ready to interpret the frame correctly.
Q52 – Assume that, in a stop and wait ARQ system, the bandwidth of the line is 1 mbps, and 1 bit takes 20 ms to make a round trip. What is the bandwidth delay product utilization percentage of the link if we send 1000 bits?
1%
5%
10%
50%
Ans – (2)
Explanation –
In a stop and wait ARQ system, we can calculate the bandwidth-delay product utilization percentage by considering both the bandwidth of the line and the round-trip time (RTT). The bandwidth of the line is given as 1 megabit per second (1 mbps), which is equivalent to 1 × 106 bits per second. The round-trip time is 20 milliseconds (ms), or 0.02 seconds.
First, we determine the Bandwidth-Delay Product (BDP), which is the product of the bandwidth and the RTT. The calculation is as follows
BDP = 1 × 106 bits/second × 0.02 seconds = 20,000 bits
This means that a maximum of 20,000 bits can be in transit in the network at any given time. Next, we calculate the utilization of the link, which is the ratio of the frame size to the BDP. In this scenario, the frame size is 1000 bits. Thus, the utilization is
U = 1000 bits / 20,000 bits = 0.05
So, utilization is 5%.
Q53 – If we run N simultaneous Stop-n-wait ARQ processors parallelly over the transmission channel, then it is equal to
Go back N protocol
SR protocol
Stop-n-wait protocol
None
Ans – (2)
Explanation –
If we run 𝑁 simultaneous Stop-and-Wait ARQ processors in parallel over the transmission channel, it effectively mimics the behaviour of the Selective Repeat (SR) protocol. This is because, like the SR protocol, each of the 𝑁 Stop-and-Wait processors can independently manage its own acknowledgment and retransmission process.
Q54 – P1 – logical token bus is contention protocol
P2 – CSMA (Carrier Sense Multiple Access) is contention less protocol
Which of the above will true?
P1 only
P2 only
Both P1 and P2
None
Ans – (4)
Explanation –
A logical token bus (as specified in IEEE 802.4) uses a token-passing mechanism to control access to the shared medium. This means that a token is passed around the network, and only the device holding the token can transmit data. This prevents collisions and makes it a contention-free protocol, not a contention protocol.
CSMA (such as used in Ethernet) is a contention-based protocol. Devices sense the carrier to check if the medium is free before transmitting. If multiple devices transmit simultaneously, collisions can occur, and a backoff algorithm is used to resolve these collisions.
Q55 – A channel with latency of 50ms and bandwidth of 45Mbps can hold
280 KB of data
2.25 x 106 bits
1.11 x 10-9 bits
0.9 x 109 bits
Ans – (2)
Explanation –
50 ms can hold = 45 x 106 bps x 50 x 10-3 = 2.25 x 106
Q56 – In Ethernet CSMA/CD, what is the special bit sequence transmitted by media access management collision handling?
Jamming Signal
Preamble
Post amble
All of these
Ans – (2)
Explanation –
In Ethernet CSMA/CD (Carrier Sense Multiple Access with Collision Detection), when a collision is detected, the collision handling mechanism transmits a special bit sequence known as the “Jamming Signal” to ensure that all stations on the network are aware that a collision has occurred and that they should refrain from transmitting for a certain period of time.
Q57 – Consider an error free 64 kbps satellite channel used to send 512 bytes data frames in one direction, with very short ACK coming back the other way, what is the maximum throughput for window size of one. Assume propagation time to be 270 ms.
63.99 kbps
8.62 kbps
6.67 kbps
Data insufficient
Ans – (3)
Explanation –
Data rate (R) = 64 kbps
Frame size (F) = 512 bytes = 512 * 8 bits = 4096 bits
Propagation time (Tp) = 270 ms
ACK time is very short (can be considered negligible)
Transmission time = Tt = F/R = 4096 bits / 64 × 103 bits per second = 64 ms
RTT = Tt + 2×Tp = 64 ms+540 ms = 604 ms
Throughput = F/RTT = 4096 / 604 = 6.785 kbps is nearly similar to 6.67 kbps.
Q58 – A group of 2n – 1 routers are interconnected in a centralized binary tree, with a router at each tree node. Router I communicate with router J by sounding a message to the root of the tree. The root then sends the message back down to J. Derive an approximate expression for the mean number of hops per message for large N, assuming that all router pairs are equally likely
2N – 4
(N – 4)/2
N – 2
N – 4
Ans – (1)
Explanation –
Binary tree with 2n-1 routers.
Each node (router) at the Nth level of the tree takes n-1 hops to reach the root, and this level contains half of all routers. (n-1)*1/2 hops
The nodes at the N-1 level will take n-2 hops to reach the root, and this level contains half of the remaining routers. (n-2)*1/4 hops
The nodes at the N-2 level will require n-3 hops to reach the root, and this level contains half of the remaining routers. (n-3)*1/8 hops
In all, the number of routers to the root is (n-1).*1/2+ (n-2)*1/4+ (n-3)*1/8+ …… = n-2
Minimum number of hops = 2(n-2) = 2n – 4
Q59 – A 3000 km long T1 trunk (data rate for which is 1.544 Mbps) is used to transmit a 64-byte frame. Assume that the propagation speed is 6 µsec/km, and stop-n-wait protocol is being used. What is the transmission time and channel utilization.
0.332 ms, 0.81%
0.232 ms, 0.64%
0.232 ms, 0.91%
0.332 ms, 0.91%
Ans – (4)
Explanation –
Transmission time = Tt = Frame Size / Data Rate
Converting the frame size to bits (1 byte = 8 bits)
Frame Size = 64 × 8 = 512 bits
Data rate = 1.544 Mbps
Tt = 0.332 ms
Propagation time = Tp = Distance x Propagation Speed
Tp = 3000 x 6 x 10-6 seconds
Propagation time = 0.018
Channel Utilization U = Tt / (Tt + 2xTp)
U = 0.00913
Convert into percentage, U = 0.91%
Q60 – Adaptive or dynamic directory used in packet routing changes
Within each user session
With each user session
At system generation time only
Both (1) and (2)
(Timothy Williams, McGraw Hill Education)
Ans – (1)
Explanation –
In an adaptive or dynamic directory used in packet routing, the changes occur within each user session. This means that the routing decisions or directory updates are made dynamically during the course of each individual user session based on factors such as network conditions, traffic load, and available routes. This adaptive approach allows for more efficient and flexible routing, optimizing the network performance for each session’s specific requirements.
