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CS6250 COMPUTER NETWORKS ACTUAL
FINAL EXAM PREP 2026 ALL QUESTIONS
AND CORRECT DETAILED ANSWERS WITH
RATIONALES ALREADY A GRADED WITH
EXPERT FEEDBACK |CURRENTLY
TESTING |NEW AND REVISED
1. Which layer of the OSI model is responsible for end-to-end error
recovery and flow control?
A. Network layer
B. Data link layer
C. Transport layer
D. Session layer
*Rationale: The transport layer (Layer 4) provides logical
communication between processes, including reliable data transfer,
congestion control, and flow control. The network layer handles routing,
and the data link layer handles node-to-node error detection.*
2. In the TCP/IP model, which layer corresponds to the OSI presentation
and session layers?
A. Transport layer
B. Network layer
C. Application layer
D. Link layer
Rationale: The TCP/IP application layer encompasses the functions of
the OSI application, presentation, and session layers, including data
encoding, encryption, and session management.
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3. A sender transmits a packet with a TTL (Time to Live) value of 32.
The packet traverses 25 routers before reaching the destination. What is
the TTL value when the packet arrives?
A. 32
B. 25
C. 7
D. 0
*Rationale: Each router decrements the TTL by 1. Starting at 32, after
25 hops, TTL = 32 – 25 = 7. If TTL reaches 0, the packet is discarded.*
4. Which HTTP method is idempotent and typically used to retrieve a
resource without side effects?
A. POST
B. GET
C. PUT
D. DELETE
Rationale: GET is idempotent and safe – multiple identical requests
have the same effect as a single request and do not change server state.
POST is not idempotent.
5. A TCP sender experiences a timeout. Which congestion control
mechanism is invoked?
A. Slow start
B. Fast retransmit
C. Congestion avoidance (after slow start reset) – but the immediate
action: the sender sets cwnd to 1 MSS and enters slow start.
*Clarify: Upon timeout, TCP sets ssthresh = cwnd/2, cwnd = 1, and
enters slow start.*
A. Slow start is correct. The correct answer is Slow start (since timeout
resets cwnd to 1). But many exams list both. The best answer: A. Slow
start because that is the phase entered after timeout. However, the
question may expect: Timeout causes cwnd=1 and slow start. I'll
answer A. Slow start.
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*Rationale: On a timeout, TCP Reno sets ssthresh = cwnd/2, cwnd = 1
MSS, and enters slow start to probe for available bandwidth.*
6. The TCP three-way handshake involves which sequence of flags?
A. SYN, SYN-ACK, ACK
B. SYN, SYN-ACK, ACK
C. SYN, ACK, FIN
D. SYN, SYN, ACK
Rationale: The connection establishment uses: client sends SYN, server
responds with SYN-ACK, client sends ACK.
7. In the Distance Vector routing algorithm, the problem of “count to
infinity” is often mitigated by:
A. Poison reverse
B. Poison reverse or split horizon
C. Link state advertisements
D. Dijkstra’s algorithm
Rationale: Split horizon and poison reverse are techniques to avoid
routing loops and the count-to-infinity problem in distance vector
protocols like RIP.
8. Which of the following is a loop-free interior gateway protocol that
uses link-state advertisements and Dijkstra’s algorithm?
A. RIP
B. OSPF
C. BGP
D. EIGRP
Rationale: OSPF (Open Shortest Path First) is a link-state routing
protocol that floods LSAs and computes shortest paths using Dijkstra’s
algorithm.
9. A network uses CIDR notation 192.168.10.0/25. How many usable
host addresses are in this subnet?
A. 128
B. 126
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C. 254
D. 64
*Rationale: /25 leaves 7 bits for hosts (2^7 = 128 addresses). Subtract
network and broadcast addresses (2) → 126 usable hosts.*
10. What is the primary purpose of the ARP (Address Resolution
Protocol)?
A. Resolve domain names to IP addresses
B. Resolve IP addresses to MAC addresses
C. Resolve MAC addresses to IP addresses
D. Assign IP addresses dynamically
Rationale: ARP maps an IPv4 address to a physical (MAC) address on a
local network segment.
11. An HTTP/1.1 persistent connection allows:
A. Multiple requests and responses over a single TCP connection
B. Multiple requests and responses over a single TCP connection
C. Only one request per connection
D. Parallel connections without overhead
*Rationale: HTTP/1.1 persistent connections (keep-alive) reuse the same
TCP connection for multiple HTTP requests/responses, reducing
latency.*
12. Which of the following is NOT a characteristic of User Datagram
Protocol (UDP)?
A. Connectionless
B. No congestion control
C. Reliable ordered delivery
D. Low overhead
Rationale: UDP provides no reliability, ordering, or congestion control;
it is a simple datagram service.
13. A TCP flow has a congestion window (cwnd) of 12 MSS and a
receiver window (rwnd) of 8 MSS. How many bytes can the sender
transmit before waiting for an ACK?
CS6250 COMPUTER NETWORKS ACTUAL
FINAL EXAM PREP 2026 ALL QUESTIONS
AND CORRECT DETAILED ANSWERS WITH
RATIONALES ALREADY A GRADED WITH
EXPERT FEEDBACK |CURRENTLY
TESTING |NEW AND REVISED
1. Which layer of the OSI model is responsible for end-to-end error
recovery and flow control?
A. Network layer
B. Data link layer
C. Transport layer
D. Session layer
*Rationale: The transport layer (Layer 4) provides logical
communication between processes, including reliable data transfer,
congestion control, and flow control. The network layer handles routing,
and the data link layer handles node-to-node error detection.*
2. In the TCP/IP model, which layer corresponds to the OSI presentation
and session layers?
A. Transport layer
B. Network layer
C. Application layer
D. Link layer
Rationale: The TCP/IP application layer encompasses the functions of
the OSI application, presentation, and session layers, including data
encoding, encryption, and session management.
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3. A sender transmits a packet with a TTL (Time to Live) value of 32.
The packet traverses 25 routers before reaching the destination. What is
the TTL value when the packet arrives?
A. 32
B. 25
C. 7
D. 0
*Rationale: Each router decrements the TTL by 1. Starting at 32, after
25 hops, TTL = 32 – 25 = 7. If TTL reaches 0, the packet is discarded.*
4. Which HTTP method is idempotent and typically used to retrieve a
resource without side effects?
A. POST
B. GET
C. PUT
D. DELETE
Rationale: GET is idempotent and safe – multiple identical requests
have the same effect as a single request and do not change server state.
POST is not idempotent.
5. A TCP sender experiences a timeout. Which congestion control
mechanism is invoked?
A. Slow start
B. Fast retransmit
C. Congestion avoidance (after slow start reset) – but the immediate
action: the sender sets cwnd to 1 MSS and enters slow start.
*Clarify: Upon timeout, TCP sets ssthresh = cwnd/2, cwnd = 1, and
enters slow start.*
A. Slow start is correct. The correct answer is Slow start (since timeout
resets cwnd to 1). But many exams list both. The best answer: A. Slow
start because that is the phase entered after timeout. However, the
question may expect: Timeout causes cwnd=1 and slow start. I'll
answer A. Slow start.
,3|P a g e
*Rationale: On a timeout, TCP Reno sets ssthresh = cwnd/2, cwnd = 1
MSS, and enters slow start to probe for available bandwidth.*
6. The TCP three-way handshake involves which sequence of flags?
A. SYN, SYN-ACK, ACK
B. SYN, SYN-ACK, ACK
C. SYN, ACK, FIN
D. SYN, SYN, ACK
Rationale: The connection establishment uses: client sends SYN, server
responds with SYN-ACK, client sends ACK.
7. In the Distance Vector routing algorithm, the problem of “count to
infinity” is often mitigated by:
A. Poison reverse
B. Poison reverse or split horizon
C. Link state advertisements
D. Dijkstra’s algorithm
Rationale: Split horizon and poison reverse are techniques to avoid
routing loops and the count-to-infinity problem in distance vector
protocols like RIP.
8. Which of the following is a loop-free interior gateway protocol that
uses link-state advertisements and Dijkstra’s algorithm?
A. RIP
B. OSPF
C. BGP
D. EIGRP
Rationale: OSPF (Open Shortest Path First) is a link-state routing
protocol that floods LSAs and computes shortest paths using Dijkstra’s
algorithm.
9. A network uses CIDR notation 192.168.10.0/25. How many usable
host addresses are in this subnet?
A. 128
B. 126
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C. 254
D. 64
*Rationale: /25 leaves 7 bits for hosts (2^7 = 128 addresses). Subtract
network and broadcast addresses (2) → 126 usable hosts.*
10. What is the primary purpose of the ARP (Address Resolution
Protocol)?
A. Resolve domain names to IP addresses
B. Resolve IP addresses to MAC addresses
C. Resolve MAC addresses to IP addresses
D. Assign IP addresses dynamically
Rationale: ARP maps an IPv4 address to a physical (MAC) address on a
local network segment.
11. An HTTP/1.1 persistent connection allows:
A. Multiple requests and responses over a single TCP connection
B. Multiple requests and responses over a single TCP connection
C. Only one request per connection
D. Parallel connections without overhead
*Rationale: HTTP/1.1 persistent connections (keep-alive) reuse the same
TCP connection for multiple HTTP requests/responses, reducing
latency.*
12. Which of the following is NOT a characteristic of User Datagram
Protocol (UDP)?
A. Connectionless
B. No congestion control
C. Reliable ordered delivery
D. Low overhead
Rationale: UDP provides no reliability, ordering, or congestion control;
it is a simple datagram service.
13. A TCP flow has a congestion window (cwnd) of 12 MSS and a
receiver window (rwnd) of 8 MSS. How many bytes can the sender
transmit before waiting for an ACK?
