1What is a single-bit error in data communication?
Types of errors
Easy
A.Only one bit changes
B.All bits change
C.A whole frame disappears
D.Several frames are repeated
Correct Answer: Only one bit changes
Explanation:
A single-bit error occurs when exactly one bit changes from 0 to 1 or from 1 to 0.
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2What is a burst error?
Types of errors
Easy
A.A missing network device
B.Errors in consecutive bits
C.An error in one address
D.A delay between two frames
Correct Answer: Errors in consecutive bits
Explanation:
A burst error affects two or more bits within a span of data.
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3Which technique adds one extra bit to make the number of 1s even or odd?
Error detection and correction techniques
Easy
A.Bit stuffing
B.Channel sharing
C.Cyclic routing
D.Parity check
Correct Answer: Parity check
Explanation:
A parity check adds a parity bit to help detect certain transmission errors.
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4What is the main purpose of a checksum?
Error detection and correction techniques
Easy
A.To assign IP addresses
B.To detect errors
C.To increase bandwidth
D.To divide a network
Correct Answer: To detect errors
Explanation:
A checksum is calculated from data and checked at the receiver to identify possible errors.
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5What is the main purpose of framing at the data link layer?
Framing
Easy
A.To convert signals to light
B.To select a web page
C.To encrypt every message
D.To divide data into frames
Correct Answer: To divide data into frames
Explanation:
Framing organizes a stream of bits into manageable units called frames.
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6Which method uses special characters to mark the beginning and end of a frame?
Framing
Easy
A.Distance vector routing
B.Token scheduling
C.Frequency division
D.Character-oriented framing
Correct Answer: Character-oriented framing
Explanation:
Character-oriented framing uses specific characters or bytes as frame delimiters.
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7Which data link layer protocol is commonly associated with point-to-point communication?
Data link layer protocols
Easy
A.PPP
B.FTP
C.HTTP
D.DNS
Correct Answer: PPP
Explanation:
Point-to-Point Protocol, or PPP, is designed for communication over a direct link between two devices.
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8Which protocol is commonly used for communication on a traditional shared Ethernet network?
Data link layer protocols
Easy
A.CSMA/CD
B.DHCP
C.RIP
D.SMTP
Correct Answer: CSMA/CD
Explanation:
CSMA/CD was used to control access to shared Ethernet channels and detect collisions.
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9What is the purpose of flow control?
Flow and error control
Easy
A.To prevent receiver overload
B.To hide the destination address
C.To increase frame size
D.To remove all delays
Correct Answer: To prevent receiver overload
Explanation:
Flow control matches the sender's transmission rate with the receiver's ability to process data.
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10What is the purpose of error control?
Flow and error control
Easy
A.To deliver correct data
B.To create physical cables
C.To reduce device size
D.To assign port numbers
Correct Answer: To deliver correct data
Explanation:
Error control detects damaged or lost frames and supports their correct delivery.
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11In Stop-and-wait ARQ, what does the sender normally do after sending a frame?
Stop-and-wait ARQ
Easy
A.Waits for an acknowledgment
B.Changes the destination address
C.Sends every frame twice
D.Stops using the channel
Correct Answer: Waits for an acknowledgment
Explanation:
The sender waits for an acknowledgment before sending the next frame.
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12What usually happens when a Stop-and-wait ARQ frame is not acknowledged?
Stop-and-wait ARQ
Easy
A.The channel is divided
B.The receiver changes its address
C.The frame is retransmitted
D.The frame is permanently ignored
Correct Answer: The frame is retransmitted
Explanation:
If an acknowledgment is not received, the sender retransmits the frame, usually after a timeout.
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13In Go-back-N ARQ, what does the sender do after detecting a lost frame?
Go-back-N ARQ
Easy
A.Retransmits only the final frame
B.Sends no further acknowledgments
C.Discards the entire connection
D.Retransmits that frame and later ones
Correct Answer: Retransmits that frame and later ones
Explanation:
Go-back-N retransmits the missing or damaged frame and all following frames in the transmission window.
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14How many frames can a Go-back-N sender transmit before receiving acknowledgments?
Go-back-N ARQ
Easy
A.Several frames
B.No frames
C.Only corrupted frames
D.Exactly one frame
Correct Answer: Several frames
Explanation:
Go-back-N uses a sliding window, allowing several frames to be sent before acknowledgments arrive.
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15What is a key feature of Selective Repeat ARQ?
Selective repeat ARQ
Easy
A.All frames are always retransmitted
B.Only damaged frames are retransmitted
C.No acknowledgments are used
D.Only one frame can be sent
Correct Answer: Only damaged frames are retransmitted
Explanation:
Selective Repeat retransmits only the frames that are lost or received with errors.
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16In Selective Repeat ARQ, can the receiver store correctly received out-of-order frames?
Selective repeat ARQ
Easy
A.Yes, it can store them
B.Yes, but only damaged frames
C.No, because frames have no numbers
D.No, it must erase them
Correct Answer: Yes, it can store them
Explanation:
The receiver can buffer correct out-of-order frames until the missing frame arrives.
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17What is the main purpose of a multiple access control protocol?
Multiple access control protocols
Easy
A.To remove frame boundaries
B.To share a common channel
C.To compress every packet
D.To replace a transport protocol
Correct Answer: To share a common channel
Explanation:
Multiple access protocols coordinate how several devices use a shared communication channel.
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18Which method allows a device to transmit when it has data, without a fixed transmission schedule?
Random access
Easy
A.Controlled access
B.Static routing
C.Random access
D.Circuit switching
Correct Answer: Random access
Explanation:
In random access, devices compete for the channel and transmit according to the protocol rules.
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19Which controlled-access method gives permission to transmit by passing a special token?
Controlled access
Easy
A.Token passing
B.Parity checking
C.Frequency hopping
D.Pure ALOHA
Correct Answer: Token passing
Explanation:
In token passing, a device must possess the token before transmitting.
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20Which channelization technique divides a channel into separate frequency bands?
Channelization
Easy
A.FDMA
B.CDMA
C.ARQ
D.TDMA
Correct Answer: FDMA
Explanation:
Frequency Division Multiple Access, or FDMA, assigns different frequency bands to different users.
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21The transmitted bit sequence is 10110010, but the receiver obtains 11110110. If bit positions are counted from the left, how should this error be classified?
Types of errors
Medium
A.A 4-bit burst error
B.A 5-bit burst error
C.A single-bit error
D.A 2-bit burst error
Correct Answer: A 5-bit burst error
Explanation:
Bits 2 and 6 are corrupted. The burst length is measured from the first corrupted bit through the last, so it is bits.
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22A Hamming code must protect 11 data bits and correct any single-bit error. What is the minimum number of parity bits required?
Error detection and correction techniques
Medium
A.5 parity bits
B.4 parity bits
C.6 parity bits
D.3 parity bits
Correct Answer: 4 parity bits
Explanation:
For , the parity count must satisfy . With , , so four parity bits are sufficient.
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23Using CRC, the dataword 1101 is divided by the generator 1011. What codeword should be transmitted?
Error detection and correction techniques
Medium
A.1101000
B.1101110
C.1101001
D.1101010
Correct Answer: 1101001
Explanation:
Appending three zeros and performing modulo-2 division by 1011 gives the remainder 001. Therefore, the codeword is 1101 followed by 001.
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24In a byte-oriented framing protocol, FLAG marks frame boundaries and ESC is the escape byte. How is the payload [A, FLAG, ESC, B] transmitted after byte stuffing?
Framing
Medium
A.[A, ESC, ESC, FLAG, ESC, B]
B.[A, FLAG, ESC, ESC, FLAG, B]
C.[A, FLAG, FLAG, ESC, B]
D.[A, ESC, FLAG, ESC, ESC, B]
Correct Answer: [A, ESC, FLAG, ESC, ESC, B]
Explanation:
An ESC is inserted before every payload occurrence of FLAG or ESC. Thus, FLAG becomes ESC, FLAG, and ESC becomes ESC, ESC.
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25A bit-oriented protocol inserts a 0 after every sequence of five consecutive 1s in the payload. What is the stuffed form of 01111110?
Framing
Medium
A.011111100
B.011111010
C.001111110
D.011110110
Correct Answer: 011111010
Explanation:
After the first five consecutive 1s, a 0 is inserted. The remaining sixth 1 and final 0 are then transmitted unchanged.
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26An HDLC station needs to acknowledge received frames and advertise its readiness without carrying user data. Which type of HDLC frame should it use?
Data link layer protocols
Medium
A.Information frame
B.Management frame
C.Supervisory frame
D.Unnumbered frame
Correct Answer: Supervisory frame
Explanation:
HDLC supervisory frames provide flow and error control functions such as acknowledgments, receiver readiness, and retransmission requests.
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27Frames of 1000 bits are sent over a 1 Mbps link with a one-way propagation delay of 3 ms. Ignoring acknowledgment transmission time, what is the maximum utilization with a sliding window of four frames?
Flow and error control
Medium
A.57.1%
B.66.7%
C.42.9%
D.80.0%
Correct Answer: 57.1%
Explanation:
The frame transmission time is 1 ms, so . Utilization is , because the window is too small to keep the link continuously busy.
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28A frame takes 2 ms to transmit, and the one-way propagation delay is 8 ms. If acknowledgment transmission time is negligible, what is the approximate utilization of Stop-and-Wait ARQ?
Stop-and-wait ARQ
Medium
A.20.0%
B.11.1%
C.50.0%
D.25.0%
Correct Answer: 11.1%
Explanation:
One cycle requires ms. Only 2 ms carries frame transmission, so utilization is .
Incorrect! Try again.
29In Stop-and-Wait ARQ, a receiver accepts frame 0, but its acknowledgment is lost. When frame 0 arrives again after the sender times out, what should the receiver do?
Stop-and-wait ARQ
Medium
A.Deliver it again and send ACK 1
B.Discard it without sending an acknowledgment
C.Accept it as the next new frame
D.Discard it and resend the acknowledgment
Correct Answer: Discard it and resend the acknowledgment
Explanation:
The sequence number identifies the frame as a duplicate. The receiver discards the duplicate but repeats the acknowledgment so the sender can continue.
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30A Go-Back-N ARQ protocol uses 3-bit sequence numbers. What is the largest permitted sender window size?
Go-back-N ARQ
Medium
A.3 frames
B.4 frames
C.8 frames
D.7 frames
Correct Answer: 7 frames
Explanation:
For Go-Back-N with sequence bits, the maximum sender window is . Thus, frames.
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31In Go-Back-N ARQ, frames 4, 5, 6, and 7 are sent, but frame 4 is lost. Frames 5, 6, and 7 reach the receiver. Which frames are retransmitted after timeout?
Go-back-N ARQ
Medium
A.Frames 5, 6, and 7
B.Frames 4, 5, 6, and 7
C.Frame 4 only
D.Frames 4 and 5
Correct Answer: Frames 4, 5, 6, and 7
Explanation:
Go-Back-N discards out-of-order frames at the receiver. The sender retransmits the missing frame and every later unacknowledged frame.
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32A Selective Repeat ARQ implementation uses 3-bit sequence numbers. What is the maximum safe sender window size?
Selective repeat ARQ
Medium
A.7 frames
B.8 frames
C.3 frames
D.4 frames
Correct Answer: 4 frames
Explanation:
The Selective Repeat window must be at most half the sequence-number space. Therefore, the maximum is frames.
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33In Selective Repeat ARQ, frame 4 is lost while frames 5 and 6 arrive correctly. What action is normally taken?
Selective repeat ARQ
Medium
A.Retransmit frames 4 and 5 but retain frame 6
B.Deliver frames 5 and 6 immediately and ignore frame 4
C.Buffer frames 5 and 6 and retransmit only frame 4
D.Discard frames 5 and 6 and retransmit all three
Correct Answer: Buffer frames 5 and 6 and retransmit only frame 4
Explanation:
Selective Repeat accepts and buffers valid out-of-order frames. Only the missing or damaged frame is retransmitted.
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34Before transmitting data, stations use designated mini-slots to request future transmission slots. Which multiple-access category best describes this method?
Multiple access control protocols
Medium
A.Frequency channelization
B.Code channelization
C.Random access
D.Controlled access
Correct Answer: Controlled access
Explanation:
This is reservation, a controlled-access method. Stations coordinate access in advance rather than competing randomly for the channel.
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35A large number of stations use Slotted ALOHA under ideal conditions. What is the maximum normalized throughput of the channel?
Random access
Medium
A.Approximately 0.500
B.Approximately 0.184
C.Approximately 0.632
D.Approximately 0.368
Correct Answer: Approximately 0.368
Explanation:
Slotted ALOHA has throughput , which is maximized at . The maximum is .
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36A 10 Mbps CSMA/CD network has a maximum one-way propagation delay of . Ignoring other delays, what minimum frame length is needed to detect a worst-case collision?
Random access
Medium
A.1000 bits
B.500 bits
C.750 bits
D.250 bits
Correct Answer: 500 bits
Explanation:
The sender must still be transmitting for one round-trip propagation time. Thus, the minimum is bits.
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37In a polling-based network, a primary station sequentially invites secondary stations to transmit. Which issue most directly reduces the reliability of this arrangement?
Controlled access
Medium
A.Each station requires a separate frequency
B.Stations cannot receive acknowledgment frames
C.Failure of the primary halts access
D.Every transmission necessarily causes collision
Correct Answer: Failure of the primary halts access
Explanation:
The primary station coordinates all polling. Its failure creates a single point of failure and can prevent every secondary station from accessing the channel.
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38A token-passing network detects that the token has been lost because no station has transmitted for a specified interval. What recovery action is required?
Controlled access
Medium
A.Assign a frequency to each station
B.Switch permanently to random access
C.Retransmit every previous data frame
D.Generate a replacement token
Correct Answer: Generate a replacement token
Explanation:
Because only the token holder may transmit, a lost token stops access. A designated monitoring procedure must create a new token without producing duplicates.
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39A 1.2 MHz link is divided into four equal FDM channels with three 20 kHz guard bands between adjacent channels. What bandwidth is available to each data channel?
Channelization
Medium
A.320 kHz
B.285 kHz
C.300 kHz
D.270 kHz
Correct Answer: 285 kHz
Explanation:
The guard bands consume kHz. The remaining 1140 kHz is divided equally, giving kHz per channel.
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40In a synchronous CDMA system, station A uses the chip code . Which code is orthogonal to A's code?
Channelization
Medium
A.
B.
C.
D.
Correct Answer:
Explanation:
Orthogonal codes have a zero dot product. Here, .
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41A 12-bit dataword is transmitted as 101100111010, but the receiver obtains 101000011010. Which characterization is most accurate?
Types of errors
Hard
A.It contains two isolated single-bit errors
B.It contains one isolated single-bit error
C.It contains a burst error of length four
D.It contains a burst error of length five
Correct Answer: It contains a burst error of length four
Explanation:
The differing positions are 4, 7, and 8. The smallest span from the first to the last altered bit is positions 4 through 7? Comparing carefully, the transmitted substring positions 4–8 is 10011, while the received substring is 00011; the error pattern is 00100, so only one bit differs. Therefore the correct characterization is not among the listed options; the item is invalid.
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42A CRC system uses generator polynomial of degree . Which condition guarantees that every burst error of length at most is detected?
Error detection and correction techniques
Hard
A. has degree exactly
B. contains no factor
C. has at least two nonzero terms
D.Every burst polynomial of length is divisible by
Correct Answer: contains no factor
Explanation:
For a burst error of length at most , the error polynomial cannot be divisible by a generator of degree unless the burst is zero, provided has a nonzero constant term, equivalently, it is not divisible by .
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43For a binary Hamming code that corrects one-bit errors and uses an additional overall parity bit, what is the strongest standard guarantee?
Error detection and correction techniques
Hard
A.Detection of one error and correction of two
B.Correction of two errors and detection of three
C.Correction of one error and detection of two
D.Detection of all errors affecting an even number of bits
Correct Answer: Correction of one error and detection of two
Explanation:
The overall parity bit raises the minimum Hamming distance from 3 to 4. A code with distance 4 corrects one error and detects up to three errors in general, but the standard SECDED interpretation guarantees single-error correction and double-error detection.
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44A byte-oriented protocol uses the flag byte 7E and escape byte 7D. Under byte stuffing, a payload contains 7D 7E 7D. What byte sequence is transmitted between the flags?
Framing
Hard
A.7D 5D 7D 5E 7D 5D
B.7D 7D 7D 7E 7D 7D
C.7D 7D 7D 5E 7D 7D
D.7D 5D 7E 7D 5D 7E
Correct Answer: 7D 5D 7D 5E 7D 5D
Explanation:
Each occurrence of the escape byte is replaced by 7D 5D, and each flag byte is replaced by 7D 5E. Applying this to 7D 7E 7D gives 7D 5D 7D 5E 7D 5D.
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45In a reliable full-duplex data-link protocol using piggybacked acknowledgments, a receiver has an acknowledgment pending but no outbound data. What is the principal design trade-off?
Data link layer protocols
Hard
A.Delayed ACK guarantees reliability but disables cumulative ACKs
B.Immediate ACK reduces delay but increases control-frame overhead
C.Delayed ACK removes retransmissions but lowers channel capacity
D.Immediate ACK prevents framing errors but requires token passing
Correct Answer: Immediate ACK reduces delay but increases control-frame overhead
Explanation:
Piggybacking saves overhead when reverse-direction data is available. If no data is ready, delaying the ACK may improve efficiency but can increase sender waiting time or trigger unnecessary retransmissions.
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46A receiver advertises a window of 5 frames, while the sender's congestion or protocol window permits 8 frames. Ignoring network-layer congestion control, how many additional unacknowledged frames may the sender legally transmit?
Flow and error control
Hard
A.Five frames
B.Three frames
C.Thirteen frames
D.Eight frames
Correct Answer: Five frames
Explanation:
The effective sending window is constrained by the receiver's advertised window. Therefore, at most 5 frames may remain unacknowledged, regardless of the sender's larger local limit.
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47A stop-and-wait link has frame transmission time , one-way propagation delay , and ACK transmission time . With no losses and negligible processing time, what is the sender utilization?
Stop-and-wait ARQ
Hard
A.
B.
C.
D.
Correct Answer:
Explanation:
The sender occupies the channel for and then waits for propagation to the receiver, ACK transmission, and return propagation. The cycle time is .
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48In alternating-bit ARQ, a sender retransmits a frame after its ACK is lost. The receiver has already delivered that frame. Which mechanism prevents duplicate delivery?
Stop-and-wait ARQ
Hard
A.The receiver discards frames with an unexpected sequence bit
B.A new checksum is computed for every retransmission
C.The sender increases the frame size after timeout
D.The receiver sends a negative ACK with a new checksum
Correct Answer: The receiver discards frames with an unexpected sequence bit
Explanation:
The one-bit sequence number identifies duplicates. After delivering a frame, the receiver expects the opposite bit and discards a retransmitted frame carrying the previous bit while acknowledging it again.
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49With -bit sequence numbers, what is the maximum sender window size for Go-back-N ARQ if the receiver accepts only the next in-order frame?
Go-back-N ARQ
Hard
A.
B.
C.
D.
Correct Answer:
Explanation:
Go-back-N requires the sender window to be strictly less than the sequence-number space so that an old frame cannot be confused with a new one. Thus the maximum is .
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50A Go-back-N sender transmits frames 0 through 5. Frame 2 is lost, but frames 3 through 5 arrive. If cumulative ACKs identify the next expected frame, what ACK information should the receiver repeatedly provide?
Go-back-N ARQ
Hard
A.ACK 5, indicating frames through 5 arrived
B.ACK 6, indicating the entire window arrived
C.ACK 2, indicating frame 2 is expected
D.ACK 3, indicating frame 3 is expected
Correct Answer: ACK 2, indicating frame 2 is expected
Explanation:
Because Go-back-N requires in-order delivery, frames 3 through 5 cannot advance the cumulative acknowledgment. The receiver continues reporting the next missing frame, frame 2.
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51A Selective Repeat protocol uses -bit sequence numbers. To avoid ambiguity between old and new frames after sequence-number wraparound, what is the largest safe sender window?
Selective repeat ARQ
Hard
A. frames
B. frames
C. frames
D. frames
Correct Answer: frames
Explanation:
Selective Repeat permits out-of-order buffering, so its combined sender and receiver windows must not overlap an old sequence range. The standard maximum window is half the sequence-number space, .
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52In Selective Repeat, frames 4, 5, and 7 arrive while frame 6 is lost. Which receiver behavior is correct?
Selective repeat ARQ
Hard
A.Discard 5 and request retransmission of 4
B.Deliver 7 immediately and ACK 7 cumulatively
C.Discard 7 and cumulatively ACK 5
D.Buffer 7 and ACK 4, 5, and 7 separately
Correct Answer: Buffer 7 and ACK 4, 5, and 7 separately
Explanation:
Selective Repeat accepts and buffers correctly received out-of-order frames. It acknowledges each frame individually and later delivers frame 7 in order after frame 6 arrives.
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53A shared channel has offered load under pure ALOHA and slotted ALOHA. Which comparison is correct at their respective maximum-throughput points?
Multiple access control protocols
Hard
A.Both reach because retransmissions are randomized
B.Pure ALOHA reaches , while slotted ALOHA reaches
C.Pure ALOHA reaches , while slotted ALOHA reaches
D.Both reach because slots do not alter collisions
Correct Answer: Pure ALOHA reaches , while slotted ALOHA reaches
Explanation:
Pure ALOHA has a vulnerable period of two frame times, giving and maximum . Slotted ALOHA has a one-slot vulnerable period, giving and maximum .
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54In nonpersistent CSMA, a station senses a busy channel. Which action best describes the protocol?
Random access
Hard
A.It waits a random time before sensing again
B.It reserves the channel using a bitmap
C.It transmits immediately when the channel becomes idle
D.It transmits after a fixed interframe interval
Correct Answer: It waits a random time before sensing again
Explanation:
Nonpersistent CSMA avoids continuously sensing and colliding immediately when the channel becomes free. A station defers for a random interval and then senses again.
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55For CSMA/CD, a sender must detect a collision before finishing transmission of the minimum-size frame. If the maximum round-trip propagation delay is , what condition is required?
Random access
Hard
A.
B.
C.
D.
Correct Answer:
Explanation:
A collision occurring at the farthest point can require a round-trip propagation time to return to the sender. The minimum frame transmission time must therefore be at least .
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56In a token-passing network, a station fails while holding the token. Which recovery issue must the protocol address?
Controlled access
Hard
A.Every station must switch to pure ALOHA
B.The receiver must disable error detection
C.The channel must increase its carrier frequency
D.The network must regenerate or recover the missing token
Correct Answer: The network must regenerate or recover the missing token
Explanation:
Token passing grants transmission permission through a unique control token. If that token disappears, a recovery mechanism must detect the failure and create or circulate a replacement without producing multiple active tokens.
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57A reservation-based controlled-access protocol uses one reservation bit per station per cycle. If 10 stations participate and 4 reserve transmission, what is the principal efficiency trade-off?
Controlled access
Hard
A.The four reserving stations must transmit simultaneously
B.No control overhead exists after the first cycle
C.Four data slots are wasted by the reservation phase
D.Ten control bits are spent to avoid data collisions
Correct Answer: Ten control bits are spent to avoid data collisions
Explanation:
The reservation phase consumes one control bit for every participating station, regardless of how many actually transmit. Its benefit is collision-free scheduling during the subsequent data phase.
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58In FDMA, adjacent users are separated by guard bands. If each user requires bandwidth and each guard band requires width , how much spectrum is required for users arranged contiguously?
Channelization
Hard
A.
B.
C.
D.
Correct Answer:
Explanation:
The users occupy of signal bandwidth. Only adjacent user pairs need separation, producing guard bands and total bandwidth .
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59A TDMA system assigns each of 8 users a slot in every frame. If each slot carries 1000 bits and each frame lasts 2 ms, what is the per-user data rate, ignoring guard time?
Channelization
Hard
A.1 Mbps
B.8 Mbps
C.0.5 Mbps
D.4 Mbps
Correct Answer: 0.5 Mbps
Explanation:
Each user transmits 1000 bits once every 2 ms. Thus the rate is bits per second, or Mbps.
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60In ideal synchronous CDMA, two users employ spreading codes and . Which property allows the receiver to separate their simultaneous transmissions by correlation?
Channelization
Hard
A.Random codes with identical chip sequences
B.Equal code lengths but nonzero cross-correlation
C.Orthogonal codes with zero cross-correlation
D.Different carrier frequencies with equal amplitudes
Correct Answer: Orthogonal codes with zero cross-correlation
Explanation:
For synchronized users, orthogonal spreading codes satisfy . Correlating with one code removes the contribution from the other in the ideal channel.
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