1Which elementary data link protocol assumes that the receiver can process incoming data at an infinite rate?
Elementary Datalink Protocols
Easy
A.Sliding Window Protocol
B.Stop-and-Wait Protocol
C.Unrestricted Simplex Protocol
D.Simplex Stop-and-Wait Protocol
Correct Answer: Unrestricted Simplex Protocol
Explanation:
The Unrestricted Simplex Protocol (also known as the Utopian protocol) assumes ideal conditions, including a receiver that can process data infinitely fast and a channel that does not lose frames.
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2In the Stop-and-Wait protocol, what must the sender do after transmitting a data frame?
Elementary Datalink Protocols
Easy
A.Wait for an acknowledgment (ACK) from the receiver
B.Terminate the connection
C.Immediately send the next frame
D.Wait for a predetermined time and send the next frame automatically
Correct Answer: Wait for an acknowledgment (ACK) from the receiver
Explanation:
In the Stop-and-Wait protocol, the sender transmits one frame and then stops to wait for an acknowledgment before sending the next one.
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3What is the primary difference between error detection and error correction?
Error Detection and Correction
Easy
A.Error detection fixes the error, while correction only identifies it
B.Error detection only identifies that an error occurred, while correction also figures out where it is and fixes it
C.Both are exactly the same concept
D.Error correction requires less redundancy than error detection
Correct Answer: Error detection only identifies that an error occurred, while correction also figures out where it is and fixes it
Explanation:
Error detection simply alerts the system that data was corrupted in transit, whereas error correction identifies the specific corrupted bits and restores their original values.
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4Which of the following describes a 'burst error' in data communication?
Error Detection and Correction
Easy
A.Two or more bits in a data unit have changed
B.Only a single bit is changed from 0 to 1 or 1 to 0
C.Frames arrive out of order
D.The entire frame is lost in transit
Correct Answer: Two or more bits in a data unit have changed
Explanation:
A burst error means that two or more bits in the data unit have changed from 1 to 0 or from 0 to 1, usually caused by sustained interference.
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5What does CRC stand for in computer networks?
Error Detection and Correction- Hamming code, CRC, Parity, Checksum
Easy
A.Computer Routing Code
B.Channel Redundancy Check
C.Cyclic Repeater Control
D.Cyclic Redundancy Check
Correct Answer: Cyclic Redundancy Check
Explanation:
CRC stands for Cyclic Redundancy Check, which is a powerful and widely used error-detecting code based on polynomial division.
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6If a system uses 'even parity', what happens if the data bits have an odd number of 1s?
Error Detection and Correction- Hamming code, CRC, Parity, Checksum
Easy
A.The parity bit is set to 1
B.An error is automatically corrected
C.The data is discarded
D.The parity bit is set to 0
Correct Answer: The parity bit is set to 1
Explanation:
In even parity, the total number of 1s (data bits plus parity bit) must be even. If the data bits have an odd number of 1s, the parity bit is set to 1 to make the total even.
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7The Hamming code is primarily known for its ability to perform which function?
Error Detection and Correction- Hamming code, CRC, Parity, Checksum
Easy
A.Compressing data
B.Single-bit error correction
C.Routing packets efficiently
D.Encrypting passwords
Correct Answer: Single-bit error correction
Explanation:
Hamming codes are a family of linear error-correcting codes that can detect up to two-bit errors or correct single-bit errors.
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8Which error detection method involves dividing the data into segments, adding them together using 1's complement arithmetic, and sending the complement of the sum?
Error Detection and Correction- Hamming code, CRC, Parity, Checksum
Easy
A.Hamming Code
B.CRC
C.Checksum
D.Parity Check
Correct Answer: Checksum
Explanation:
Checksum is an error detection technique where a sum of data segments is calculated (often using 1's complement arithmetic) and sent along with the data.
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9What does MAC stand for in the context of the Data Link Layer?
MAC SUBLAYER
Easy
A.Media Access Control
B.Machine Authentication Code
C.Multiple Access Carrier
D.Memory Access Code
Correct Answer: Media Access Control
Explanation:
MAC stands for Media Access Control. It is a sublayer of the Data Link Layer responsible for controlling how devices in a network gain access to data and permission to transmit it.
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10The Data Link Layer is generally divided into two sublayers. One is the MAC sublayer; what is the other?
MAC SUBLAYER
Easy
A.Logical Link Control (LLC) Sublayer
B.Network Control Sublayer
C.Physical Access Sublayer
D.Transport Control Sublayer
Correct Answer: Logical Link Control (LLC) Sublayer
Explanation:
The IEEE 802 standard divides the Data Link Layer into the Logical Link Control (LLC) sublayer and the Media Access Control (MAC) sublayer.
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11What is the basic rule for a station transmitting data in Pure ALOHA?
Multiple Access Protocols- ALOHA, CSMA and CSMA/CD
Easy
A.Listen to the channel before transmitting
B.Transmit data whenever a frame is ready
C.Ask a primary station for permission
D.Wait for a token to be received
Correct Answer: Transmit data whenever a frame is ready
Explanation:
In Pure ALOHA, there is no channel sensing. A station simply transmits its data whenever it has a frame ready to send.
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12How does Slotted ALOHA improve upon Pure ALOHA?
Multiple Access Protocols- ALOHA, CSMA and CSMA/CD
Easy
A.By forcing stations to listen before talking
B.By using a token ring topology
C.By eliminating collisions entirely
D.By dividing time into discrete intervals (slots) and only allowing transmission at the beginning of a slot
Correct Answer: By dividing time into discrete intervals (slots) and only allowing transmission at the beginning of a slot
Explanation:
Slotted ALOHA improves efficiency by dividing time into slots. Stations are only allowed to begin transmitting at the start of a slot, which reduces the vulnerable time for collisions.
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13What does the 'CD' in CSMA/CD stand for?
Multiple Access Protocols- ALOHA, CSMA and CSMA/CD
Easy
A.Collision Detection
B.Control Data
C.Carrier Detection
D.Channel Division
Correct Answer: Collision Detection
Explanation:
CD stands for Collision Detection. In CSMA/CD, if a node detects a collision while transmitting, it stops and sends a jam signal.
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14What does a station do first in the CSMA protocol?
Multiple Access Protocols- ALOHA, CSMA and CSMA/CD
Easy
A.Sends a request-to-send (RTS) frame
B.Senses the carrier (listens to the channel) to see if it is idle
C.Waits for a token
D.Transmits immediately
Correct Answer: Senses the carrier (listens to the channel) to see if it is idle
Explanation:
CSMA (Carrier Sense Multiple Access) requires a station to 'listen before it talks'—meaning it senses the channel to check for ongoing transmissions before sending its own data.
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15Which of the following is a key characteristic of random access protocols?
Random Access
Easy
A.Stations transmit data based on a strict time schedule
B.One station acts as a master and controls the others
C.No station is superior to another, and any station can transmit if it follows the protocol rules
D.A token is passed sequentially between nodes
Correct Answer: No station is superior to another, and any station can transmit if it follows the protocol rules
Explanation:
In random access (or contention) protocols, all stations are peers. No station controls another, and they compete for the shared medium.
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16What is the consequence of two stations transmitting at the same time in a random access network?
Random Access
Easy
A.The router automatically separates the signals
B.The data speed doubles
C.A collision occurs, corrupting the frames
D.The network permanently shuts down
Correct Answer: A collision occurs, corrupting the frames
Explanation:
When two or more stations transmit simultaneously on a shared medium in a random access protocol, their signals interfere with each other, resulting in a collision and corrupted data.
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17Which of the following is an example of a controlled access protocol?
Controlled access
Easy
A.Pure ALOHA
B.Token Passing
C.Slotted ALOHA
D.CSMA/CD
Correct Answer: Token Passing
Explanation:
Token passing is a controlled access method where a special frame called a 'token' circulates around the network, and a station can only transmit if it holds the token.
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18In the polling access method, what is the role of the primary station?
Controlled access
Easy
A.It strictly acts as a backup for the network
B.It controls the link and dictates which secondary device can transmit
C.It generates random tokens
D.It deliberately creates collisions to test network strength
Correct Answer: It controls the link and dictates which secondary device can transmit
Explanation:
In polling, one device acts as a primary station and controls the data link. It polls the secondary stations to grant them permission to transmit.
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19What is the IEEE standard number for traditional Ethernet?
Ethernet protocol
Easy
A.IEEE 802.5
B.IEEE 802.3
C.IEEE 802.11
D.IEEE 802.15
Correct Answer: IEEE 802.3
Explanation:
The IEEE 802.3 standard defines the rules for traditional Ethernet networks.
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20How long is a standard Ethernet MAC address?
Ethernet protocol
Easy
A.48 bits
B.64 bits
C.32 bits
D.128 bits
Correct Answer: 48 bits
Explanation:
A standard Ethernet MAC address is 48 bits (or 6 bytes) long, typically represented as 12 hexadecimal digits.
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21If a data word has bits, what is the minimum number of redundant bits required for a Hamming code to correct single-bit errors?
Error Detection and Correction- Hamming code
Medium
A.6
B.3
C.4
D.5
Correct Answer: 4
Explanation:
The Hamming code requirement is . Substituting , we need . If , (False). If , (True). Therefore, 4 redundant bits are needed.
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22To guarantee the correction of up to errors in a data frame, the minimum Hamming distance of the block code must be at least:
Error Detection and Correction
Medium
A.
B.
C.
D.
Correct Answer:
Explanation:
To detect errors, a minimum Hamming distance of is required. To correct errors, the distance must be large enough that the erroneous codeword is still closer to the original valid codeword than any other, which requires a minimum distance of .
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23If a Cyclic Redundancy Check (CRC) uses a generator polynomial , how many bits will the CRC remainder appended to the data be?
CRC
Medium
A.4 bits
B.3 bits
C.5 bits
D.2 bits
Correct Answer: 3 bits
Explanation:
The degree of the generator polynomial is 3. The length of the CRC remainder is always equal to the degree of the generator polynomial, which means 3 bits will be appended to the data message.
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24In CRC, if the received message polynomial is perfectly divisible by the generator polynomial using modulo-2 arithmetic, what can be concluded?
CRC
Medium
A.The message definitely has no errors.
B.The message has an odd number of errors.
C.The message has burst errors equal to the degree of .
D.No errors are detected, though undetected errors are theoretically possible.
Correct Answer: No errors are detected, though undetected errors are theoretically possible.
Explanation:
A remainder of zero indicates that no errors were detected by the CRC. However, it does not guarantee absolute perfection, as the error polynomial could theoretically be a multiple of the generator polynomial.
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25Which of the following statements is true regarding a Two-Dimensional Parity Check?
Parity, Checksum
Medium
A.It adds redundant bits equal to the size of the data block.
B.It can detect any burst error of any length.
C.It only detects odd numbers of errors.
D.It can detect all 1-bit, 2-bit, and 3-bit errors, and correct single-bit errors.
Correct Answer: It can detect all 1-bit, 2-bit, and 3-bit errors, and correct single-bit errors.
Explanation:
Two-dimensional parity organizes data in a matrix with row and column parities. It can correct any single-bit error (by finding the intersection of the failed row and column) and can detect all 1, 2, and 3-bit errors.
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26A sender calculates the checksum for data words $1010$ and $0101$ using 4-bit one's complement arithmetic. What is the value of the checksum transmitted?
Parity, Checksum
Medium
A.$0000$
B.$0101$
C.$1010$
D.$1111$
Correct Answer: $0000$
Explanation:
The correct option follows directly from the given concept and definitions.
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27In a Stop-and-Wait protocol, if the propagation delay is and the transmission time is , the maximum link utilization (efficiency) is given by which formula? (Assume )
Elementary Datalink Protocols
Medium
A.
B.
C.
D.
Correct Answer:
Explanation:
In Stop-and-Wait, the total time to send one frame and receive an ACK is . The useful time is . Therefore, efficiency . Dividing numerator and denominator by , we get .
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28In a Go-Back-N ARQ protocol using bits for sequence numbers, what is the maximum permissible sender window size?
Elementary Datalink Protocols
Medium
A.
B.
C.
D.
Correct Answer:
Explanation:
In Go-Back-N, the maximum sender window size is . If the window size were , the sender could not distinguish between an acknowledgment for a new frame and a duplicate acknowledgment for an old frame.
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29For a Selective Repeat ARQ protocol using 4 bits for sequence numbering, what is the optimal window size for both the sender and receiver?
Elementary Datalink Protocols
Medium
A.15
B.16
C.7
D.8
Correct Answer: 8
Explanation:
In Selective Repeat ARQ, the window size for both sender and receiver must be at most to prevent overlapping sequence numbers between old and new windows. For , the max window size is .
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30Which of the following best describes the primary responsibility of the Medium Access Control (MAC) sublayer?
MAC SUBLAYER
Medium
A.Providing mechanisms to control access to the shared transmission medium.
B.Encrypting data frames before they are placed on the physical medium.
C.Resolving logical IP addresses to physical addresses.
D.Routing packets across different networks.
Correct Answer: Providing mechanisms to control access to the shared transmission medium.
Explanation:
The MAC sublayer is responsible for coordinating access to a shared medium, resolving collisions, and managing physical addressing in broadcast networks.
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31What is the maximum throughput (efficiency) of Pure ALOHA and Slotted ALOHA, respectively?
Multiple Access Protocols- ALOHA, CSMA and CSMA/CD
Medium
A.36.8% and 18.4%
B.18.4% and 50.0%
C.50.0% and 81.6%
D.18.4% and 36.8%
Correct Answer: 18.4% and 36.8%
Explanation:
The maximum efficiency of Pure ALOHA occurs at yielding . For Slotted ALOHA, the vulnerable time is halved, peaking at yielding .
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32In Pure ALOHA, if the frame transmission time is , what is the vulnerable time during which a collision can occur?
Multiple Access Protocols- ALOHA, CSMA and CSMA/CD
Medium
A.
B.
C.
D.
Correct Answer:
Explanation:
In Pure ALOHA, a frame can collide with another frame that starts transmitting up to time before it or up to time after it starts. Thus, the vulnerable time window is .
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33In a CSMA/CD network, a station must transmit a frame of at least a minimum length () to ensure it can detect a collision before it finishes transmitting. If the network bandwidth is bps and the one-way propagation delay is , what is the formula for ?
Multiple Access Protocols- ALOHA, CSMA and CSMA/CD
Medium
A.
B.
C.
D.
Correct Answer:
Explanation:
To detect a collision, the transmission time must be at least twice the propagation delay (). Since , we get . Therefore, .
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34Why is CSMA/CD generally not utilized in wireless networks (like Wi-Fi)?
Multiple Access Protocols- ALOHA, CSMA and CSMA/CD
Medium
A.Wireless networks do not experience collisions.
B.It cannot solve the hidden terminal problem, and collision detection is difficult due to signal attenuation.
C.CSMA/CD requires a token to operate, which cannot be passed wirelessly.
D.CSMA/CD forces half-duplex operation, making wireless networks too slow.
Correct Answer: It cannot solve the hidden terminal problem, and collision detection is difficult due to signal attenuation.
Explanation:
In wireless networks, stations cannot easily listen and transmit at the same time (detect collisions) due to rapid signal attenuation. Additionally, CSMA/CD fails to address the hidden terminal problem, making CSMA/CA the preferred choice.
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35In the standard Ethernet Binary Exponential Backoff algorithm, after collisions (where ), a station chooses a random delay time multiplied by the slot time. The random number is chosen from the range:
Random Access
Medium
A.$0$ to
B.$0$ to
C.$1$ to
D.$0$ to
Correct Answer: $0$ to
Explanation:
The Binary Exponential Backoff algorithm dictates that after collisions, a station waits for a random number of slot times chosen uniformly from the interval .
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36Which of the following controlled access methods requires one specific node to act as a primary node that authorizes secondary nodes to transmit data?
Controlled access
Medium
A.Reservation
B.Token Passing
C.CSMA/CA
D.Polling
Correct Answer: Polling
Explanation:
In the Polling method, one device acts as a primary controller and the others are secondary. The primary node polls the secondary nodes to determine if they have data to send, effectively controlling access to the medium.
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37In a Token Passing network, what happens if the station holding the token currently has no data to send?
Controlled access
Medium
A.It holds the token until a timeout occurs.
B.It destroys the token and generates a new one.
C.It immediately passes the token to its logical successor in the ring.
D.It sends a dummy frame to keep the channel active.
Correct Answer: It immediately passes the token to its logical successor in the ring.
Explanation:
In Token Passing, the token circulates sequentially. If a station has the token but no data to transmit, it simply passes the token to the next station in the logical ring without delay.
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38An Ethernet MAC address is a unique identifier assigned to a network interface controller. What is the length of an Ethernet MAC address?
Ethernet protocol
Medium
A.32 bits
B.48 bits
C.128 bits
D.64 bits
Correct Answer: 48 bits
Explanation:
Standard Ethernet MAC addresses are 48 bits (6 bytes) long, typically represented as six groups of two hexadecimal digits.
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39In an IEEE 802.3 Ethernet frame, what is the purpose of the 7-byte Preamble followed by the 1-byte Start Frame Delimiter (SFD)?
Ethernet protocol
Medium
A.To identify the MAC address of the destination.
B.To provide error correction bits for the frame payload.
C.To indicate the type of network layer protocol encapsulated.
D.To synchronize the clocks of the receiving stations and mark the start of the frame.
Correct Answer: To synchronize the clocks of the receiving stations and mark the start of the frame.
Explanation:
The Preamble consists of alternating 1s and 0s that allow the receiver's clock to synchronize with the incoming data. The SFD (10101011) signals the exact beginning of the actual frame.
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40What is the minimum and maximum size of an Ethernet frame payload (data field) in standard IEEE 802.3?
Ethernet protocol
Medium
A.46 bytes and 1500 bytes
B.32 bytes and 1500 bytes
C.64 bytes and 1518 bytes
D.48 bytes and 1024 bytes
Correct Answer: 46 bytes and 1500 bytes
Explanation:
In a standard IEEE 802.3 frame, the data payload must be a minimum of 46 bytes (to ensure the frame meets the 64-byte minimum length requirement) and a maximum of 1500 bytes.
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41In a sliding window protocol, if the sequence numbers are represented using bits, what is the maximum permissible sender window size for Go-Back-N and Selective Repeat protocols, respectively, to ensure correct operation under worst-case delay?
Elementary Datalink Protocols
Hard
A. and
B. and
C. and
D. and
Correct Answer: and
Explanation:
For Go-Back-N, the sender window size can be at most to distinguish between a new packet and a retransmission. For Selective Repeat, the window size must be at most to prevent the new window from overlapping with the old window.
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42If a code has a minimum Hamming distance of , what is the maximum number of errors it can detect () and the maximum number of errors it can correct ()?
Error Detection and Correction
Hard
A.,
B.,
C.,
D.,
Correct Answer: ,
Explanation:
A code with minimum Hamming distance can detect up to errors. To correct errors, the distance must be at least , yielding .
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43To transmit data bits along with redundant bits using a Hamming code such that all single-bit errors are correctable, which of the following inequalities must strictly hold?
Error Detection and Correction- Hamming code
Hard
A.
B.
C.
D.
Correct Answer:
Explanation:
The redundant bits can form states. One state indicates no error, and the remaining states must cover single-bit errors in all data bits and parity bits. Hence, .
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44Given the generator polynomial and the message polynomial , what is the transmitted codeword polynomial in systematic form?
Error Detection and Correction- CRC
Hard
A.
B.
C.
D.
Correct Answer:
Explanation:
Multiply by to get . Dividing this by over GF(2) yields a remainder of . Adding the remainder gives the codeword .
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45Consider a Two-Dimensional Parity Check scheme applied to an matrix of data bits. Which of the following error patterns is GUARANTEED to go undetected?
Error Detection and Correction- Parity
Hard
A.4 errors forming the corners of a rectangle in the matrix
B.Any 3 errors in the same row
C.3 errors forming an L-shape
D.Any 2 errors in the same column
Correct Answer: 4 errors forming the corners of a rectangle in the matrix
Explanation:
In 2D parity, 4 errors located at flip exactly two bits in rows and two bits in columns . The parity of these rows and columns remains unchanged, so the error goes undetected.
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46If the 16-bit Internet checksum is used, and a bit sequence of 16 zeros is altered to 16 ones during transmission while another 16-bit block changes from 16 ones to 16 zeros, what is the effect on the receiver's checksum validation?
Error Detection and Correction- Checksum
Hard
A.The error is never detected.
B.The checksum computation loops infinitely.
C.The error is detected with probability 0.5.
D.The error is always detected.
Correct Answer: The error is never detected.
Explanation:
The Internet checksum uses 1's complement arithmetic. Swapping a word of all 0s with a word of all 1s (and vice versa) does not change the 1's complement sum of the data. Thus, the error will go undetected.
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47Which of the following best describes the fundamental trade-off addressed by the Medium Access Control (MAC) sublayer in a broadcast network?
MAC SUBLAYER
Hard
A.Minimizing propagation delay vs. Maximizing bandwidth
B.Minimizing collision probability vs. Maximizing channel utilization
C.Minimizing bit error rate vs. Maximizing frame size
D.Maximizing routing efficiency vs. Minimizing overhead
Correct Answer: Minimizing collision probability vs. Maximizing channel utilization
Explanation:
The MAC sublayer coordinates access to a shared medium. Protocols must balance the need to keep stations from transmitting simultaneously (causing collisions) with the goal of keeping the channel busy transmitting useful data (high utilization).
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48In Pure ALOHA, if the frame transmission time is , what is the vulnerable time, and at what offered load (in frames per ) is the maximum throughput achieved?
Multiple Access Protocols- ALOHA
Hard
A.Vulnerable time = ; Max throughput at
B.Vulnerable time = ; Max throughput at
C.Vulnerable time = ; Max throughput at
D.Vulnerable time = ; Max throughput at
Correct Answer: Vulnerable time = ; Max throughput at
Explanation:
In Pure ALOHA, a collision occurs if another frame is generated anywhere within before or after the start of transmission, making the vulnerable time . Throughput is , which is maximized when the derivative with respect to is zero, yielding .
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49In a CSMA/CD network operating at 1 Gbps with a maximum cable length of 200 meters and signal propagation speed of m/s, what is the minimum frame size required to ensure collision detection?
Multiple Access Protocols- CSMA and CSMA/CD
Hard
A.2500 bits
B.250 bits
C.200 bits
D.2000 bits
Correct Answer: 2000 bits
Explanation:
Round-trip propagation time . To detect collisions, transmission time . Thus, , meaning minimum frame length bits.
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50In the Binary Exponential Backoff algorithm used after collisions in Ethernet, the sender chooses a random wait time from the interval . If , what is the probability that two colliding stations choose the same backoff time?
Random Access
Hard
A.
B.
C.
D.
Correct Answer:
Explanation:
After collisions, the interval is , which has 32 slots. The probability that both stations pick the exact same slot is .
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51In a token ring network with stations, assuming the token propagation time around the ring is and frame transmission time is , what is the maximum channel utilization if the protocol uses Single-Token operation (release after reception)?
Controlled access
Hard
A.
B.
C.
D.
Correct Answer:
Explanation:
In Single-Token operation, a station transmits one frame of length , waits for it to travel around the ring (), and then releases the token. The cycle time per frame is , making utilization .
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52Which of the following fields in the classic IEEE 802.3 Ethernet frame is responsible for clock synchronization, and what is its specific bit pattern?
Ethernet protocol
Hard
A.SFD: 1 byte of 11111111
B.Preamble: 7 bytes of 10101010
C.Preamble: 8 bytes of 10101010
D.SFD: 1 byte of 10101011
Correct Answer: Preamble: 7 bytes of 10101010
Explanation:
The Preamble consists of 7 bytes of alternating 1s and 0s (10101010), allowing the receiver's clock to synchronize with the sender's. It is followed by the 1-byte Start Frame Delimiter (SFD) which is 10101011.
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53A CRC generator polynomial is defined as , where is a primitive polynomial of degree . Which set of errors is this guaranteed to detect?
Error Detection and Correction- CRC
Hard
A.Only single-bit errors and double-bit errors
B.All odd-numbered bit errors and all burst errors of length
C.All burst errors of length
D.All odd-numbered bit errors and all burst errors of length
Correct Answer: All odd-numbered bit errors and all burst errors of length
Explanation:
The factor ensures all odd numbers of bit errors are detected. The polynomial has degree , so it guarantees detection of all burst errors of length up to .
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54In the Stop-and-Wait ARQ protocol, what is the link utilization given a frame transmission time , propagation delay , and assuming acknowledgment transmission time is negligible?
Elementary Datalink Protocols
Hard
A.
B., where
C., where
D.
Correct Answer: , where
Explanation:
The total time for one frame cycle is (transmission + round-trip propagation). Utilization is the useful time divided by total time: . Dividing by yields where .
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55Why is CSMA/CD fundamentally unsuitable for wireless LANs (like 802.11), necessitating the use of CSMA/CA instead?
Multiple Access Protocols- CSMA and CSMA/CD
Hard
A.Wireless networks do not have a MAC sublayer.
B.Wireless signals propagate too slowly compared to wired networks.
C.Wireless stations cannot transmit and receive simultaneously on the same frequency to detect collisions.
D.CSMA/CA offers higher theoretical throughput by ignoring collisions.
Correct Answer: Wireless stations cannot transmit and receive simultaneously on the same frequency to detect collisions.
Explanation:
In wireless networks, the signal power degrades heavily over distance. A transmitting station's own signal overwhelms any incoming signal from a collision, making reliable collision detection hardware impractical. Furthermore, the hidden terminal problem makes CSMA/CD ineffective.
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56If an Ethernet frame is padded to reach the minimum frame size of 64 bytes, how does the receiving network layer know the actual length of the payload?
Ethernet protocol
Hard
A.The IEEE 802.3 Length field indicates the true payload size.
B.The receiver relies on the CRC to calculate payload size.
C.The padding is a sequence of alternating 1s and 0s that the receiver strips.
D.A special end-of-frame delimiter (EOF) is placed before the padding.
Correct Answer: The IEEE 802.3 Length field indicates the true payload size.
Explanation:
In the IEEE 802.3 standard, the 2-byte Length/Type field is used as a Length field (when ) to specify the exact number of bytes in the data field. This allows the receiver to discard the padding.
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57When constructing a Hamming code, parity bits are placed at bit positions that are powers of 2. For an 8-bit data sequence, at which positions (starting from 1) will the parity bits be placed, and what is the total codeword length?
The parity bits are placed at positions (1, 2, 4, 8). Since 4 parity bits are needed for 8 data bits (), the total codeword length is bits.
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58In a slotted ALOHA system with active nodes where each transmits with probability in a slot, the efficiency is maximized at . As , what does the maximum efficiency asymptotically approach?
Random Access
Hard
A.
B.
C.
D.
Correct Answer:
Explanation:
The probability of a successful transmission in slotted ALOHA is . Setting yields . As , this limit approaches .
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59In a Reservation protocol where time is divided into a reservation frame containing minislots followed by data frames, if the system is heavily loaded such that every minislot results in a reservation, what is the maximum efficiency? (Let minislot length be and data frame length be )
Controlled access
Hard
A.
B.
C.
D.
Correct Answer:
Explanation:
In one cycle, there are minislots taking time , which successfully reserve data frames taking time . The total useful time is out of a total cycle time of . Thus, efficiency is .
Incorrect! Try again.
60In a communication channel with a high error rate, why does the Selective Repeat protocol generally outperform Go-Back-N, despite having higher memory requirements?
Elementary Datalink Protocols
Hard
A.Selective Repeat does not require timers for frame retransmission.
C.Selective Repeat utilizes implicit acknowledgments via cumulative ACKs.
D.Selective Repeat avoids retransmitting frames that were successfully received after a lost frame.
Correct Answer: Selective Repeat avoids retransmitting frames that were successfully received after a lost frame.
Explanation:
Go-Back-N discards all out-of-order frames received after a lost frame, requiring retransmission of the entire window. Selective Repeat buffers out-of-order frames and only retransmits the specific lost frame, significantly saving bandwidth on high-error links.