Computer Networking Midterm Exam Questions and Detailed
Solutions Latest Update 2026/2027 | Verified Questions and
Answers, Complete Examination - 160 Questions
Comprehensive examination on Computer Networking Midterm Exam Questions and Detailed Solutions Latest
Update 2026/2027 | Verified Questions and Answers, Complete Examination. It contains 160 multiple-choice
questions, each with four distractors and a fully worked rationale that explains why the keyed answer is correct.
Questions are organized into clearly labelled sections that mirror the major content areas of the course. Targeted
learning outcomes include: Demonstrate mastery of core concepts. Every item has been reviewed for clinical
accuracy, current guidelines, and clarity so that students can study with confidence and self-correct as they work
through the bank. Use it as a high-yield review immediately before the exam, or as a structured practice tool
during the unit - the rationales double as concise teaching notes. The recommended writing time is 3 hours, with a
passing score of 70%. Aligned with Aligned with US university standards. standards and reflects the question style
commonly seen on accredited program examinations. Students consistently achieving above the cut score on this
bank have historically gone on to earn A+ on the corresponding course exam. Read every stem carefully -
distractors are written to look plausible, and the best answer is sometimes the one that addresses the patient's
most immediate physiological or safety need. Where multiple options appear correct, prioritize airway, breathing,
circulation, safety, and Maslow's hierarchy before psychosocial interventions. Treat each rationale as a
mini-lecture: don't just confirm the right letter, study the reasoning so you can transfer it to similar items on the
Section 1: General (Questions 1-160)
1 In a network using TCP Reno, the congestion window is 20
segments and the slow start threshold is 32 segments. A triple
duplicate ACK is received. After the subsequent congestion
avoidance phase, the window reaches 24 segments and then a
timeout occurs. What is the congestion window size immediately
after the timeout?
A) 1 segment
B) 2 segments
C) 12 segments
D) 24 segments
Answer: A
Rationale: In TCP Reno, a triple duplicate ACK triggers fast retransmit
and fast recovery, halving the window to 10 and setting ssthresh to 10.
After recovery, congestion avoidance continues until a timeout occurs,
which resets the window to 1 segment and ssthresh to half of the
current window (12). Thus, immediately after timeout, cwnd is 1.
,2 A packet is sent from host A to host B through routers R1 and R2.
The TTL is initially 64. R1 takes 5 ms to process the packet, and the
link A-R1 has a propagation delay of 2 ms. R2 takes 8 ms, and link
R1-R2 has 3 ms delay. Link R2-B has 4 ms delay. If the packet
arrives at B with TTL=62, what is the total time spent in routers?
A) 13 ms
B) 15 ms
C) 17 ms
D) 22 ms
Answer: A
Rationale: The TTL decrement indicates the packet passed through
two routers (R1 and R2), since TTL goes from 64 to 62. The total
processing time in routers is the sum of processing delays: 5 ms + 8
ms = 13 ms. Propagation delays are not processing time.
3 A network administrator is designing a subnet plan for a /23
network. They need to accommodate the following number of hosts
per subnet: 300, 120, 60, 30, and 2 (for point-to-point links). Using
VLSM, what is the minimum number of subnets required, and what
is the largest subnet mask that can be used for the 300-host subnet?
A) 5 subnets, /23
B) 5 subnets, /22
C) 6 subnets, /23
D) 6 subnets, /22
Answer: A
Rationale: VLSM allows creating subnets of different sizes. The
300-host subnet requires at least 9 host bits (2^9 - 2 = 510), so a /23
mask (32-9 = 23) is sufficient. The other subnets can be carved from
the remaining address space. The minimum number of subnets is 5,
one for each requirement. A /22 would be too large for the 300-host
subnet, and 6 subnets are unnecessary.
,4 In a BGP-speaking router, two paths to the same destination are
available: Path A has local preference 150, AS path length 3, and
MED 100. Path B has local preference 200, AS path length 5, and
MED 50. Which path will be selected, and what is the first
tiebreaker that determines this?
A) Path A, because of higher local preference
B) Path B, because of higher local preference
C) Path A, because of shorter AS path
D) Path B, because of lower MED
Answer: B
Rationale: BGP best path selection first compares local preference;
higher is preferred. Path B has local preference 200 vs Path A's 150,
so Path B is selected. AS path length and MED are lower-priority
tiebreakers, but they are not considered because local preference
differs.
5 A TCP connection is in established state. The receive window
advertised by the receiver is 0. The sender has data to send. What
action does the sender take, and why?
A) It sends a segment with one byte of data to probe the window
B) It stops sending and waits for a window update
C) It sends a segment with zero bytes but with the ACK flag set
D) It sends a segment with the urgent pointer set to force delivery
Answer: A
Rationale: When the receive window is zero, the sender must stop
sending data. To avoid deadlock, it periodically sends a segment with
one byte of data (a window probe) to elicit an ACK that may contain a
new window size. This is called the persist timer mechanism.
6 A network uses a distance-vector routing protocol. Router A
receives an update from neighbor B that advertises a route to
network X with metric 5. Router A's current route to X has metric 7
via C. How does A update its routing table?
, A) It replaces the route with B's route if B's metric is lower than the
current metric
B) It replaces the route only if B's metric is lower than the current
metric and B is the next hop
C) It always replaces the route with the new route if the metric is
lower
D) It ignores the update because the metric is not lower than the
current metric
Answer: A
Rationale: In distance-vector protocols, a router updates its table to the
route with the lowest metric, regardless of the next hop. Since B's
metric (5) is lower than the current (7), A will replace the route with
B's route. If the metric were higher, it would ignore (unless it came
from the same next hop).
7 A network engineer is troubleshooting a VLAN trunk between two
switches. The trunk is configured as 802.1Q. Hosts on VLAN 10
can communicate, but hosts on VLAN 20 cannot. What is the most
likely cause?
A) The native VLAN is set to 20 on one switch
B) VLAN 20 is not allowed on the trunk
C) The trunk is configured in access mode
D) The encapsulation on the trunk is ISL
Answer: B
Rationale: If VLAN 10 works but VLAN 20 does not, the trunk is
likely not carrying VLAN 20 traffic. The most common cause is that
VLAN 20 is not in the allowed list on the trunk. Native VLAN
mismatch (A) would cause issues on the native VLAN, not a specific
non-native VLAN. Access mode would prevent all VLANs. ISL vs
802.1Q mismatch would affect all VLANs.
Solutions Latest Update 2026/2027 | Verified Questions and
Answers, Complete Examination - 160 Questions
Comprehensive examination on Computer Networking Midterm Exam Questions and Detailed Solutions Latest
Update 2026/2027 | Verified Questions and Answers, Complete Examination. It contains 160 multiple-choice
questions, each with four distractors and a fully worked rationale that explains why the keyed answer is correct.
Questions are organized into clearly labelled sections that mirror the major content areas of the course. Targeted
learning outcomes include: Demonstrate mastery of core concepts. Every item has been reviewed for clinical
accuracy, current guidelines, and clarity so that students can study with confidence and self-correct as they work
through the bank. Use it as a high-yield review immediately before the exam, or as a structured practice tool
during the unit - the rationales double as concise teaching notes. The recommended writing time is 3 hours, with a
passing score of 70%. Aligned with Aligned with US university standards. standards and reflects the question style
commonly seen on accredited program examinations. Students consistently achieving above the cut score on this
bank have historically gone on to earn A+ on the corresponding course exam. Read every stem carefully -
distractors are written to look plausible, and the best answer is sometimes the one that addresses the patient's
most immediate physiological or safety need. Where multiple options appear correct, prioritize airway, breathing,
circulation, safety, and Maslow's hierarchy before psychosocial interventions. Treat each rationale as a
mini-lecture: don't just confirm the right letter, study the reasoning so you can transfer it to similar items on the
Section 1: General (Questions 1-160)
1 In a network using TCP Reno, the congestion window is 20
segments and the slow start threshold is 32 segments. A triple
duplicate ACK is received. After the subsequent congestion
avoidance phase, the window reaches 24 segments and then a
timeout occurs. What is the congestion window size immediately
after the timeout?
A) 1 segment
B) 2 segments
C) 12 segments
D) 24 segments
Answer: A
Rationale: In TCP Reno, a triple duplicate ACK triggers fast retransmit
and fast recovery, halving the window to 10 and setting ssthresh to 10.
After recovery, congestion avoidance continues until a timeout occurs,
which resets the window to 1 segment and ssthresh to half of the
current window (12). Thus, immediately after timeout, cwnd is 1.
,2 A packet is sent from host A to host B through routers R1 and R2.
The TTL is initially 64. R1 takes 5 ms to process the packet, and the
link A-R1 has a propagation delay of 2 ms. R2 takes 8 ms, and link
R1-R2 has 3 ms delay. Link R2-B has 4 ms delay. If the packet
arrives at B with TTL=62, what is the total time spent in routers?
A) 13 ms
B) 15 ms
C) 17 ms
D) 22 ms
Answer: A
Rationale: The TTL decrement indicates the packet passed through
two routers (R1 and R2), since TTL goes from 64 to 62. The total
processing time in routers is the sum of processing delays: 5 ms + 8
ms = 13 ms. Propagation delays are not processing time.
3 A network administrator is designing a subnet plan for a /23
network. They need to accommodate the following number of hosts
per subnet: 300, 120, 60, 30, and 2 (for point-to-point links). Using
VLSM, what is the minimum number of subnets required, and what
is the largest subnet mask that can be used for the 300-host subnet?
A) 5 subnets, /23
B) 5 subnets, /22
C) 6 subnets, /23
D) 6 subnets, /22
Answer: A
Rationale: VLSM allows creating subnets of different sizes. The
300-host subnet requires at least 9 host bits (2^9 - 2 = 510), so a /23
mask (32-9 = 23) is sufficient. The other subnets can be carved from
the remaining address space. The minimum number of subnets is 5,
one for each requirement. A /22 would be too large for the 300-host
subnet, and 6 subnets are unnecessary.
,4 In a BGP-speaking router, two paths to the same destination are
available: Path A has local preference 150, AS path length 3, and
MED 100. Path B has local preference 200, AS path length 5, and
MED 50. Which path will be selected, and what is the first
tiebreaker that determines this?
A) Path A, because of higher local preference
B) Path B, because of higher local preference
C) Path A, because of shorter AS path
D) Path B, because of lower MED
Answer: B
Rationale: BGP best path selection first compares local preference;
higher is preferred. Path B has local preference 200 vs Path A's 150,
so Path B is selected. AS path length and MED are lower-priority
tiebreakers, but they are not considered because local preference
differs.
5 A TCP connection is in established state. The receive window
advertised by the receiver is 0. The sender has data to send. What
action does the sender take, and why?
A) It sends a segment with one byte of data to probe the window
B) It stops sending and waits for a window update
C) It sends a segment with zero bytes but with the ACK flag set
D) It sends a segment with the urgent pointer set to force delivery
Answer: A
Rationale: When the receive window is zero, the sender must stop
sending data. To avoid deadlock, it periodically sends a segment with
one byte of data (a window probe) to elicit an ACK that may contain a
new window size. This is called the persist timer mechanism.
6 A network uses a distance-vector routing protocol. Router A
receives an update from neighbor B that advertises a route to
network X with metric 5. Router A's current route to X has metric 7
via C. How does A update its routing table?
, A) It replaces the route with B's route if B's metric is lower than the
current metric
B) It replaces the route only if B's metric is lower than the current
metric and B is the next hop
C) It always replaces the route with the new route if the metric is
lower
D) It ignores the update because the metric is not lower than the
current metric
Answer: A
Rationale: In distance-vector protocols, a router updates its table to the
route with the lowest metric, regardless of the next hop. Since B's
metric (5) is lower than the current (7), A will replace the route with
B's route. If the metric were higher, it would ignore (unless it came
from the same next hop).
7 A network engineer is troubleshooting a VLAN trunk between two
switches. The trunk is configured as 802.1Q. Hosts on VLAN 10
can communicate, but hosts on VLAN 20 cannot. What is the most
likely cause?
A) The native VLAN is set to 20 on one switch
B) VLAN 20 is not allowed on the trunk
C) The trunk is configured in access mode
D) The encapsulation on the trunk is ISL
Answer: B
Rationale: If VLAN 10 works but VLAN 20 does not, the trunk is
likely not carrying VLAN 20 traffic. The most common cause is that
VLAN 20 is not in the allowed list on the trunk. Native VLAN
mismatch (A) would cause issues on the native VLAN, not a specific
non-native VLAN. Access mode would prevent all VLANs. ISL vs
802.1Q mismatch would affect all VLANs.
