TCP establishes a connection and provides reliable, ordered data delivery.
Incorrect! Try again.
5What is multiplexing at the transport layer?
Transport Layer Services
Easy
A.Combining data from multiple applications
B.Assigning addresses to multiple routers
C.Converting packets into electrical signals
D.Selecting routes between multiple networks
Correct Answer: Combining data from multiple applications
Explanation:
Multiplexing allows data from several application processes to share the transport and network layers.
Incorrect! Try again.
6What does throughput measure in a computer network?
Performance issues
Easy
A.The size of a network address
B.The count of transport protocols
C.The rate of successful data delivery
D.The number of connected routers
Correct Answer: The rate of successful data delivery
Explanation:
Throughput is the amount of data successfully delivered per unit of time.
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7What is network latency?
Performance issues
Easy
A.The number of open ports
B.The size of a TCP header
C.The rate of creating addresses
D.The delay in delivering data
Correct Answer: The delay in delivering data
Explanation:
Latency is the time required for data to travel from a source to a destination.
Incorrect! Try again.
8Which condition commonly causes packet loss and increased delay?
Performance issues
Easy
A.Port numbering
B.Process multiplexing
C.Network congestion
D.Header compression
Correct Answer: Network congestion
Explanation:
Congestion occurs when network demand exceeds available capacity, causing queues and packet loss.
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9Which TCP header field is used to place received data in the correct order?
TCP header format
Easy
A.Window size
B.Sequence number
C.Header checksum
D.Destination port
Correct Answer: Sequence number
Explanation:
TCP sequence numbers allow the receiver to arrange data in its original order.
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10What is the minimum size of a TCP header without options?
TCP header format
Easy
A.20 bytes
B.16 bytes
C.32 bytes
D.8 bytes
Correct Answer: 20 bytes
Explanation:
A TCP header has a minimum length of 20 bytes when no optional fields are included.
Incorrect! Try again.
11Which TCP header field helps detect corrupted segments?
TCP header format
Easy
A.Source port
B.Sequence number
C.Window size
D.Checksum
Correct Answer: Checksum
Explanation:
The checksum is used to detect errors in the TCP header and data.
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12How many main steps are used in the TCP connection-establishment handshake?
TCP handshaking operation
Easy
A.Five steps
B.Three steps
C.Four steps
D.Two steps
Correct Answer: Three steps
Explanation:
TCP establishes a connection using the three-way handshake: SYN, SYN-ACK, and ACK.
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13Which TCP flag is sent first when a client starts a connection?
TCP handshaking operation
Easy
A.FIN
B.SYN
C.RST
D.ACK
Correct Answer: SYN
Explanation:
The client begins the TCP handshake by sending a segment with the SYN flag set.
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14What does a server normally send in the second step of the TCP three-way handshake?
TCP handshaking operation
Easy
A.RST-ACK
B.FIN
C.SYN-ACK
D.PSH
Correct Answer: SYN-ACK
Explanation:
The server acknowledges the client's SYN and sends its own SYN using a SYN-ACK segment.
Incorrect! Try again.
15What is the fixed size of a UDP header?
UDP header format
Easy
A.8 bytes
B.4 bytes
C.20 bytes
D.16 bytes
Correct Answer: 8 bytes
Explanation:
A UDP header contains four 16-bit fields, giving it a total size of 8 bytes.
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16Which field is present in a UDP header?
UDP header format
Easy
A.Sequence number
B.Congestion window
C.Acknowledgment number
D.Destination port
Correct Answer: Destination port
Explanation:
The UDP header contains source port, destination port, length, and checksum fields.
Incorrect! Try again.
17In network-assisted congestion control, which devices provide congestion information?
Network-assisted Congestion Control Algorithms
Easy
A.File servers
B.Network routers
C.Domain names
D.Application users
Correct Answer: Network routers
Explanation:
Routers can detect congestion and provide feedback to hosts so that sending rates can be adjusted.
Incorrect! Try again.
18What is the purpose of a choke packet?
Network-assisted Congestion Control Algorithms
Easy
A.To verify a domain name
B.To report network congestion
C.To assign a port number
D.To establish a TCP connection
Correct Answer: To report network congestion
Explanation:
A choke packet informs a sender that congestion has occurred and that its sending rate should be reduced.
Incorrect! Try again.
19Which TCP congestion-control phase begins with a small congestion window?
TCP Congestion Control
Easy
A.Connection closing
B.Slow start
C.Fast recovery
D.Flow control
Correct Answer: Slow start
Explanation:
TCP slow start begins with a small congestion window and increases it as acknowledgments arrive.
Incorrect! Try again.
20What does TCP usually do when a retransmission timeout indicates congestion?
TCP Congestion Control
Easy
A.Increases the segment header
B.Reduces the congestion window
C.Changes the destination port
D.Disables acknowledgments
Correct Answer: Reduces the congestion window
Explanation:
TCP treats a timeout as a sign of congestion and reduces its sending rate by shrinking the congestion window.
Incorrect! Try again.
21A web browser sends data to a web server on another host. Which statement correctly describes the responsibilities of the network and transport layers?
Relationship between Transport and Network Layer
Medium
A.The network layer provides host-to-host delivery, while the transport layer provides process-to-process delivery
B.The network layer provides process-to-process delivery, while the transport layer selects routes
C.The network layer manages application ports, while the transport layer assigns IP addresses
D.The network layer guarantees reliability, while the transport layer forwards packets between routers
Correct Answer: The network layer provides host-to-host delivery, while the transport layer provides process-to-process delivery
Explanation:
IP delivers packets between hosts, whereas transport protocols use port numbers to deliver data to the appropriate application process.
Incorrect! Try again.
22Several IP packets carrying one TCP stream arrive at the destination out of order. Which layer is primarily responsible for restoring the original byte order?
Relationship between Transport and Network Layer
Medium
A.The transport layer at the receiving host
B.The application layer at the sending host
C.The network layer at the final router
D.The data-link layer at the receiving host
Correct Answer: The transport layer at the receiving host
Explanation:
IP may deliver packets out of order. TCP uses sequence numbers and buffering at the receiving host to reconstruct the original byte stream.
Incorrect! Try again.
23A TCP server uses destination port 443 and simultaneously communicates with several clients. How does TCP distinguish these connections?
Transport Layer Services
Medium
A.By using only the destination port number
B.By using the source IP, source port, destination IP, and destination port
C.By using the source and destination MAC addresses
D.By using only the source IP address
Correct Answer: By using the source IP, source port, destination IP, and destination port
Explanation:
A TCP connection is identified by its four-tuple: source IP address, source port, destination IP address, and destination port.
Incorrect! Try again.
24A live voice application prefers low delay and can tolerate occasional packet loss. Which transport service is most appropriate?
Transport Layer Services
Medium
A.TCP, because it preserves message boundaries automatically
B.UDP, because it avoids connection setup and retransmission delays
C.TCP, because it retransmits every missing segment immediately
D.UDP, because it guarantees ordered delivery without acknowledgments
Correct Answer: UDP, because it avoids connection setup and retransmission delays
Explanation:
UDP has low protocol overhead and does not delay later data while retransmitting lost packets, making it suitable for delay-sensitive traffic.
Incorrect! Try again.
25A path has a bandwidth of Mbps and a round-trip time of ms. What minimum amount of unacknowledged data is needed to approximately fill the path?
Performance issues
Medium
A. bytes
B. bytes
C. bytes
D. bytes
Correct Answer: bytes
Explanation:
The bandwidth-delay product is bits, which equals bytes.
Incorrect! Try again.
26A TCP sender is limited to a -byte window, and the RTT is ms. Ignoring overhead and loss, what is the approximate maximum throughput?
Performance issues
Medium
A. bytes/s
B. bytes/s
C. bytes/s
D. bytes/s
Correct Answer: bytes/s
Explanation:
Window-limited throughput is approximately bytes per second.
Incorrect! Try again.
27A TCP segment has a Data Offset field value of . How many bytes long is its TCP header?
TCP header format
Medium
A. bytes
B. bytes
C. bytes
D. bytes
Correct Answer: bytes
Explanation:
The Data Offset gives the header length in 32-bit words. Thus, bytes.
Incorrect! Try again.
28A TCP receiver has correctly received bytes numbered through . Assuming no gaps exist, what acknowledgment number should it send?
TCP header format
Medium
A.
B.
C.
D.
Correct Answer:
Explanation:
TCP acknowledgments are cumulative and identify the next byte expected. After byte , the receiver expects byte .
Incorrect! Try again.
29A TCP sender has a congestion window of bytes and receives an advertised receive window of bytes. If bytes are already unacknowledged, how much additional data may it send?
TCP header format
Medium
A. bytes
B. bytes
C. bytes
D. bytes
Correct Answer: bytes
Explanation:
The usable sending window is limited to the smaller of the congestion and receive windows, which is bytes. With bytes outstanding, more bytes may be sent.
Incorrect! Try again.
30A TCP client sends a SYN segment with sequence number . What acknowledgment number should appear in the server's SYN-ACK?
TCP handshaking operation
Medium
A.
B.
C.
D.
Correct Answer:
Explanation:
A SYN consumes one sequence number, so the server acknowledges the client's SYN by sending acknowledgment number .
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31A client sends SYN with sequence number , and the server responds with SYN-ACK using sequence number . Which values should the client's final ACK contain?
TCP handshaking operation
Medium
A.Sequence number and acknowledgment number
B.Sequence number and acknowledgment number
C.Sequence number and acknowledgment number
D.Sequence number and acknowledgment number
Correct Answer: Sequence number and acknowledgment number
Explanation:
Each SYN consumes one sequence number. The client therefore continues with sequence number and acknowledges the server's SYN with .
Incorrect! Try again.
32The final ACK of a TCP three-way handshake is lost. The server later retransmits its SYN-ACK. What should the client normally do?
TCP handshaking operation
Medium
A.Retransmit an ACK for the duplicate SYN-ACK
B.Send a reset and immediately close the connection
C.Wait for the server to transmit application data
D.Start a new connection using another destination port
Correct Answer: Retransmit an ACK for the duplicate SYN-ACK
Explanation:
The client recognizes the retransmitted SYN-ACK and sends the final ACK again, allowing the server to complete connection establishment.
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33The UDP Length field contains the value . How many bytes of application data are present in the datagram?
UDP header format
Medium
A. bytes
B. bytes
C. bytes
D. bytes
Correct Answer: bytes
Explanation:
The UDP Length field includes the 8-byte UDP header. Therefore, the payload length is bytes.
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34Why does the UDP checksum calculation include an IP pseudoheader containing source and destination IP addresses?
UDP header format
Medium
A.To calculate the route followed by the datagram
B.To provide reliable retransmission of lost datagrams
C.To replace the IPv4 header checksum at each router
D.To detect delivery to an incorrect network-layer endpoint
Correct Answer: To detect delivery to an incorrect network-layer endpoint
Explanation:
The pseudoheader lets the UDP checksum detect corruption involving important IP-layer information, including the source and destination addresses.
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35A router experiences incipient congestion and sets an ECN mark instead of dropping a packet. What is the expected end-to-end response?
Network-assisted Congestion Control Algorithms
Medium
A.The receiver retransmits the marked packet using a different route
B.The router increases the receiver's advertised transport window
C.The router reserves bandwidth for all later packets in the flow
D.The receiver reports the mark, and the sender reduces its sending rate
Correct Answer: The receiver reports the mark, and the sender reduces its sending rate
Explanation:
With ECN, routers signal congestion by marking packets. The receiver echoes this information to the sender, which responds by reducing its congestion window.
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36In a choke-packet congestion-control scheme, what occurs when a router detects serious congestion?
Network-assisted Congestion Control Algorithms
Medium
A.The router acknowledges all queued packets on behalf of the receiver
B.The router converts the transport connection from TCP to UDP
C.The router enlarges the receiver's window to drain queued packets
D.The router sends control feedback requesting the source to reduce traffic
Correct Answer: The router sends control feedback requesting the source to reduce traffic
Explanation:
A choke packet is explicit feedback generated by a congested router and sent toward the traffic source to request a lower transmission rate.
Incorrect! Try again.
37A TCP sender begins slow start with a congestion window of MSS. Assuming no loss and no threshold is reached, what will the congestion window be after two complete RTTs?
TCP Congestion Control
Medium
A. MSS
B. MSS
C. MSS
D. MSS
Correct Answer: MSS
Explanation:
During slow start, the congestion window approximately doubles each RTT: MSS.
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38A TCP connection is in congestion avoidance with a congestion window of MSS. After one RTT in which all segments are acknowledged, what is the approximate new window?
TCP Congestion Control
Medium
A. MSS
B. MSS
C. MSS
D. MSS
Correct Answer: MSS
Explanation:
Congestion avoidance uses additive increase, raising the congestion window by approximately one MSS per RTT.
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39Using classic TCP behavior, a retransmission timeout occurs when the congestion window is MSS. What values are normally assigned to ssthresh and cwnd?
TCP Congestion Control
Medium
A.ssthresh = MSS and cwnd = MSS
B.ssthresh = MSS and cwnd = MSS
C.ssthresh = MSS and cwnd = MSS
D.ssthresh = MSS and cwnd = MSS
Correct Answer: ssthresh = MSS and cwnd = MSS
Explanation:
After a timeout, TCP normally halves the previous congestion window to set ssthresh and resets cwnd to one MSS for slow start.
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40TCP Reno receives a third duplicate ACK when cwnd is MSS. Immediately after entering fast recovery, what are the usual values of ssthresh and cwnd?
TCP Congestion Control
Medium
A.ssthresh = MSS and cwnd = MSS
B.ssthresh = MSS and cwnd = MSS
C.ssthresh = MSS and cwnd = MSS
D.ssthresh = MSS and cwnd = MSS
Correct Answer: ssthresh = MSS and cwnd = MSS
Explanation:
TCP Reno sets ssthresh to half of MSS and temporarily sets cwnd to ssthresh + 3 MSS for the three duplicate ACKs.
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41A TCP server receives two IP packets with identical destination IP addresses and destination ports. Their source ports are also identical, but their source IP addresses differ. Assuming both packets belong to established connections, how does the transport layer demultiplex them?
Relationship between Transport and Network Layer
Hard
A.It uses the IP identification and destination port fields
B.It uses only the destination port and protocol number
C.It uses the TCP sequence and acknowledgment numbers
D.It uses the source and destination socket-address pairs
Correct Answer: It uses the source and destination socket-address pairs
Explanation:
TCP identifies a connection by the four-tuple of source IP, source port, destination IP, and destination port. Different source IP addresses therefore identify different connections.
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42A transport protocol runs over a network layer that may lose, duplicate, reorder, or indefinitely delay packets. Which service cannot be guaranteed solely by adding acknowledgments, sequence numbers, and retransmissions at the transport layer?
Transport Layer Services
Hard
A.In-order delivery when communication eventually succeeds
B.Duplicate suppression using transport sequence numbers
C.Corruption detection using an end-to-end checksum
D.Delivery within a fixed finite deadline under all conditions
Correct Answer: Delivery within a fixed finite deadline under all conditions
Explanation:
Reliability mechanisms can recover when communication remains possible, but they cannot guarantee a finite deadline if the network can delay packets indefinitely or remain partitioned.
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43A bulk TCP flow uses a Mb/s path with an RTT of ms and an MSS of bytes. Its congestion window is at least the bandwidth-delay product, but its receive window is fixed at MSS. Ignoring headers and loss, what is its maximum throughput?
Performance issues
Hard
A. Mb/s
B. Mb/s
C. Mb/s
D. Mb/s
Correct Answer: Mb/s
Explanation:
The receive window limits one RTT of data to bits. Thus throughput is Mb/s.
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44An interactive TCP application performs two small writes. After transmitting the first segment, the sender's Nagle algorithm buffers the second write because unacknowledged data exists. The receiver uses delayed ACKs and waits briefly for another segment. What is the most likely performance effect?
Performance issues
Hard
A.A latency spike lasting approximately one delayed-ACK interval
B.A throughput increase caused by larger advertised receive windows
C.A congestion collapse caused by repeated fast retransmissions
D.A connection reset caused by an invalid acknowledgment number
Correct Answer: A latency spike lasting approximately one delayed-ACK interval
Explanation:
Nagle waits for an ACK before sending more small data, while delayed ACK waits for more data or a timer. Their interaction can introduce a delay approximately equal to the delayed-ACK timer.
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45An IPv4 packet has total length bytes and IHL . Its enclosed TCP segment has Data Offset . Assuming no lower-layer padding, how many bytes of application data does the packet contain?
TCP header format
Hard
A. bytes
B. bytes
C. bytes
D. bytes
Correct Answer: bytes
Explanation:
The IPv4 header is bytes and the TCP header is bytes. The application payload is bytes.
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46A NAT changes both the source IPv4 address and source TCP port of a segment without modifying its payload. Why must the TCP checksum normally be updated?
TCP header format
Hard
A.Only the source address participates in the TCP checksum calculation
B.Only the source port participates in the TCP checksum calculation
C.Neither value participates unless TCP options are present
D.Both modified values participate in the TCP checksum calculation
Correct Answer: Both modified values participate in the TCP checksum calculation
Explanation:
The source port is in the TCP header, while the source IP address is in the pseudoheader included in the TCP checksum. Changing either affects the checksum.
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47A client chooses ISN and a server chooses ISN . The client's third handshake segment acknowledges the server's SYN and carries bytes of data. What sequence and acknowledgment numbers does that segment contain, and what acknowledgment should the server return after receiving all its data?
TCP handshaking operation
Hard
A.Client ; server acknowledgment
B.Client ; server acknowledgment
C.Client ; server acknowledgment
D.Client ; server acknowledgment
Correct Answer: Client ; server acknowledgment
Explanation:
Each SYN consumes one sequence number. The client data begins at , acknowledges , and occupies sequence numbers through , so the next expected number is .
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48Two TCP endpoints simultaneously perform active opens and send SYN segments to each other. Under standard TCP state transitions, what happens when each endpoint receives the peer's SYN while in SYN-SENT?
TCP handshaking operation
Hard
A.Each sends SYN+ACK and enters SYN-RECEIVED
B.Each sends only ACK and enters ESTABLISHED directly
C.Each retransmits SYN and remains permanently in SYN-SENT
D.Each sends a reset and returns immediately to CLOSED
Correct Answer: Each sends SYN+ACK and enters SYN-RECEIVED
Explanation:
TCP supports simultaneous open. Each endpoint acknowledges the peer's SYN with SYN+ACK, enters SYN-RECEIVED, and reaches ESTABLISHED after the corresponding ACK.
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49A server enables SYN cookies during a SYN flood. Which trade-off follows directly from avoiding per-connection state between receipt of SYN and receipt of the final ACK?
TCP handshaking operation
Hard
A.The client must retransmit every data segment after handshaking
B.Some negotiated options must be encoded with limited precision
C.The server cannot validate the final acknowledgment number
D.The server must disable checksums for half-open connections
Correct Answer: Some negotiated options must be encoded with limited precision
Explanation:
A SYN cookie encodes connection information into the server's initial sequence number. The limited encoding space restricts how much option state, such as MSS or window scaling, can be preserved.
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50What is the maximum UDP application payload that can fit in a conventional IPv4 datagram with a -byte IPv4 header and no IPv4 options, assuming the IPv4 total-length limit is bytes?
UDP header format
Hard
A. bytes
B. bytes
C. bytes
D. bytes
Correct Answer: bytes
Explanation:
The maximum IPv4 payload is bytes. Removing the -byte UDP header leaves bytes of application data.
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51A UDP datagram has a UDP Length value of bytes, so its header and data contain an odd number of octets. How is the Internet checksum computed?
UDP header format
Hard
A.A zero octet is transmitted and included in the UDP Length field
B.The final data octet is omitted from the checksum calculation
C.A zero octet is appended for calculation but is not transmitted
D.The final data octet is duplicated before checksum calculation
Correct Answer: A zero octet is appended for calculation but is not transmitted
Explanation:
The Internet checksum operates on 16-bit words. An odd final octet is paired with a zero octet for computation, but this padding is neither transmitted nor counted in UDP Length.
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52A router experiences incipient congestion and receives an ECN-capable TCP packet. Which sequence correctly describes standard ECN feedback and reaction?
Network-assisted Congestion Control Algorithms
Hard
A.Router drops the ACK; receiver marks CE; sender disables congestion control
B.Router marks CE; receiver echoes ECE; sender reduces its window and sends CWR
C.Router marks CWR; receiver clears CE; sender retransmits without reducing its window
D.Router marks ECE; receiver sends CWR; sender increases its congestion window
Correct Answer: Router marks CE; receiver echoes ECE; sender reduces its window and sends CWR
Explanation:
A congested router marks the IP ECN field as CE. The TCP receiver reports this with ECE, and the sender responds by reducing its congestion window and signaling CWR.
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53In an explicit-rate congestion-control scheme, a packet starts with an allowed rate of Mb/s. Three switches successively determine allowable rates of , , and Mb/s and may only reduce the rate field. What rate should be returned to the source?
Network-assisted Congestion Control Algorithms
Hard
A. Mb/s
B. Mb/s
C. Mb/s
D. Mb/s
Correct Answer: Mb/s
Explanation:
Each switch reduces the explicit-rate field to the minimum of its current value and its local allowance. The path's bottleneck rate is therefore Mb/s.
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54A TCP Reno sender has MSS when it receives three duplicate ACKs. No SACK is used. What are the congestion window and slow-start threshold immediately after entering fast recovery?
TCP Congestion Control
Hard
A. MSS and MSS
B. MSS and MSS
C. MSS and MSS
D. MSS and MSS
Correct Answer: MSS and MSS
Explanation:
Reno sets to half the prior window, giving MSS, and inflates to MSS for the three packets represented by duplicate ACKs.
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55A TCP sender's congestion window is MSS, but only MSS are outstanding when a retransmission timeout occurs. Using the rule , what values should be selected?
TCP Congestion Control
Hard
A. MSS and MSS
B. MSS and MSS
C. MSS and MSS
D. MSS and MSS
Correct Answer: MSS and MSS
Explanation:
The threshold is based on the actual flight size, not merely the congestion window: MSS. After a timeout, the sender restarts with MSS.
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56During TCP NewReno fast recovery, an ACK advances the acknowledgment number but does not cover all data outstanding when fast recovery began. How should the sender interpret this partial ACK?
TCP Congestion Control
Hard
A.Exit fast recovery and immediately enter ordinary congestion avoidance
B.Ignore the ACK because it does not acknowledge the recovery point
C.Reset the congestion window to one MSS and invoke slow start
D.Retransmit the next missing segment and remain in fast recovery
Correct Answer: Retransmit the next missing segment and remain in fast recovery
Explanation:
NewReno treats a partial ACK as evidence of another loss in the same window. It retransmits the next unacknowledged segment and remains in fast recovery until the recovery point is acknowledged.
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57A receiver performs an ACK-division attack by acknowledging tiny portions of a segment with many separate ACKs. Which sender behavior most directly prevents excessive congestion-window growth during slow start?
TCP Congestion Control
Hard
A.Increase the window only after a retransmission timeout expires
B.Increase the window according to newly acknowledged bytes
C.Increase the window according to the advertised receive window
D.Increase the window once for every received ACK packet
Correct Answer: Increase the window according to newly acknowledged bytes
Explanation:
Appropriate Byte Counting ties congestion-window growth to the amount of newly acknowledged data rather than the number of ACK packets, preventing ACK splitting from creating artificial growth.
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58A sender has MSS, an advertised receive window of MSS, and MSS currently in flight. An ACK then cumulatively acknowledges MSS. If both windows are treated as unchanged for this calculation, how much new data may the sender immediately transmit?
TCP Congestion Control
Hard
A. MSS
B. MSS
C. MSS
D. MSS
Correct Answer: MSS
Explanation:
After the ACK, flight size becomes MSS. The usable sending limit is MSS, leaving room for MSS.
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59Two long-lived AIMD flows have congestion windows of MSS and MSS. Between shared congestion events, both increase by the same total amount; at each event, both halve their windows. What is the difference between their windows immediately after the third such decrease?
TCP Congestion Control
Hard
A. MSS
B. MSS
C. MSS
D. MSS
Correct Answer: MSS
Explanation:
Equal additive increases do not change the difference. Each multiplicative decrease halves it, so the initial difference of MSS becomes MSS.
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60Without TCP timestamps, a segment is retransmitted after timeout and an ACK later arrives that could acknowledge either the original or retransmitted copy. Under Karn's algorithm, how should the sender handle RTT estimation?
Performance issues
Hard
A.Average both possible RTT samples and reset the current RTO
B.Measure RTT from the retransmission and halve the current RTO
C.Exclude the ACK from RTT sampling and retain timeout backoff
D.Measure RTT from the original transmission and clear backoff
Correct Answer: Exclude the ACK from RTT sampling and retain timeout backoff
Explanation:
The ACK is ambiguous because the sender cannot determine which transmission it acknowledges. Karn's algorithm omits such RTT samples and preserves exponential timeout backoff until an unambiguous sample is obtained.
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