CS6250 Computer Networks Exam 2 Actual Exam
Newest Complete Questions And Correct Detailed
Answers| Already Graded A+
1. A network operator configures a BGP router to prefer routes with the shortest AS_PATH.
However, the router receives two paths to the same prefix: Path A with AS_PATH {100, 200, 300}
and Path B with AS_PATH {100, 400}. Both paths have equal local preference and are learned
from eBGP peers. The router selects Path B. Later, the operator applies an inbound route map on
the session for Path A that sets local preference to 200. After the change, which path will the router
prefer, and why?
A. Path B, because AS_PATH length is still the tie-breaker after local preference and the route map did not
affect Path B.
B. Path A, because the route map increased its local preference to 200, making it higher than Path B's default
local preference of 100.
C. Path B, because the route map is applied inbound, so it only affects routes received from that peer, but the
router will still prefer the shortest AS_PATH among paths with equal local preference.
D. Path A, because the route map also prepends an extra AS to Path B's AS_PATH, making it longer.
Answer: B
Rationale: BGP's decision process checks local preference first. By setting Path A's local preference to
200 (higher than Path B's default 100), Path A becomes preferred regardless of AS_PATH length.
AS_PATH length is only considered if local preference is equal. Options A and C incorrectly prioritize
AS_PATH; D is false because prepending is not mentioned.
2. In a data center using a Fat-Tree topology with k=4 (each switch has 4 ports), how many distinct
paths exist between two hosts in different pods that are both connected to edge switches in their
respective pods? Assume the Fat-Tree is built with k-port switches and uses k/2 core switches.
A. 2
B. 4
C. 8
D. 16
Answer: B
Rationale: In a k-ary Fat-Tree, each host in a pod connects to an edge switch. There are k/2 core
switches. A path from a host in one pod to a host in another pod goes: host -> edge -> aggregate ->
core -> aggregate -> edge -> host. Since there are k/2 = 2 core switches, and each core switch provides
one path, there are exactly 2 distinct paths. However, note that the question asks for paths between two
hosts in different pods, each connected to edge switches. The number of distinct paths equals the number
of core switches, which is k/2 = 2. Wait, but for k=4, k/2=2, but the correct answer is 4? Let me recalc:
Actually, there are (k/2)^2 = 4 core switches in a k-ary Fat-Tree? Standard Fat-Tree: k=4, each pod has
k/2=2 core switches? No, the number of core switches is (k/2)^2 = 4. Each core switch provides one path
between a given pair of edge switches. So for two specific hosts, there are (k/2)^2 = 4 distinct paths. Yes,
Page 1
,answer is 4.
3. Consider a TCP cubic flow with a current congestion window of 500 packets. The flow
experiences a packet loss due to a timeout. Immediately after the timeout, what is the new
congestion window size, assuming standard TCP behavior?
A. 1
B. 250
C. 500
D. 0
Answer: A
Rationale: In standard TCP (including Cubic), when a timeout occurs, the congestion window is reset to
1 MSS (slow-start threshold is set to half of the window before loss). So the new window is 1. Option B
would be the new ssthresh, not cwnd. Option C is the old window. Option D is not standard.
4. An SDN controller uses OpenFlow to manage switches. A packet arrives at a switch, and the
switch's flow table has no matching entry. The switch sends a Packet-In message to the controller.
The controller decides to install a flow entry that forwards packets based on destination MAC
address. Which fields must the controller include in the flow mod message to match on destination
MAC and forward out a specific port?
A. Match on eth_dst, set output port to the desired port, and set idle timeout.
B. Match on eth_dst and ip_dst, set output port, and set priority.
C. Match on eth_dst, set action to output to controller, and set hard timeout.
D. Match on in_port, set output port, and set buffer_id.
Answer: A
Rationale: To forward based on destination MAC, the match must include eth_dst. The action is output to
a port. Idle timeout is optional but typical. Option B adds unnecessary ip_dst match; Option C outputs to
controller, not forwarding; Option D matches on in_port, not destination.
5. A network employs MPLS traffic engineering. A router receives an MPLS packet with label
stack {100, 200} where label 100 is at the top. The router's LFIB shows that for incoming label 100,
the outgoing label is 300 and the next hop is R2. Which action does the router perform?
A. Swap label 100 with 300, push label 400, and forward to R2.
B. Pop the top label, then swap the new top label 200 with 300, and forward to R2.
C. Swap label 100 with 300, keep label 200 unchanged, and forward to R2.
D. Pop the entire label stack and forward the IP packet to R2.
Answer: C
Rationale: The LFIB entry for incoming label 100 specifies an outgoing label of 300. The router performs
a swap: it replaces label 100 with 300. The rest of the stack (label 200) remains untouched. The packet
is then forwarded to R2. Option A incorrectly adds a push; Option B incorrectly pops the top; Option D
pops everything, which is not indicated.
Page 2
,6. In a network using DCTCP, a sender observes that the ECN-marked fraction of packets in a
window is 0.3. The current congestion window is 100 packets. According to DCTCP's congestion
control algorithm, what will be the new congestion window after processing this window's
acknowledgments?
A. 85
B. 70
C. 50
D. 100
Answer: A
Rationale: DCTCP reduces the window by a factor of (1 - ±/2) where ± is the fraction of marked packets.
Here = 0.3, so new window = 100 * (1 - 0.3/2) = 100 * 0.85 = 85. Option B would be 100*(1-0.3)=70,
which is the reduction if using standard TCP with ECN (which halves window). Option C is half; D is no
reduction.
7. A BGP speaker in AS 100 receives a route for prefix 10.0.0.0/8 from two different peers: Peer A
(AS 200) with AS_PATH {200, 300} and Peer B (AS 400) with AS_PATH {400, 500, 600}. Both
routes have equal local preference and MED. Which route will the BGP speaker prefer, and what
is the reason?
A. Route from Peer A, because it has a shorter AS_PATH length.
B. Route from Peer B, because it has a longer AS_PATH, indicating more specific path.
C. Route from Peer A, because it has a lower origin code (IGP < EGP < incomplete).
D. Route from Peer B, because it is learned from a peer with higher IP address.
Answer: A
Rationale: BGP prefers the route with the shortest AS_PATH length when local preference, MED, and
other attributes are equal. AS_PATH length for Peer A is 2 (200,300) and for Peer B is 3 (400,500,600).
So Peer A's route is preferred. Option B is incorrect; longer AS_PATH is worse. Option C: origin code is
not given; Option D: IP address tie-breaker is only used if all else equal, but AS_PATH is considered
earlier.
8. A network uses BBR congestion control. The sender estimates the bandwidth (BW) as 10 Mbps
and the round-trip propagation delay (RTprop) as 20 ms. What is the BDP (Bandwidth-Delay
Product) in bytes?
A. 25000
B. 200000
C. 250000
D. 20000
Answer: A
Rationale: BDP = BW * RTprop. BW = 10 Mbps = 10 * 10^6 bits/sec = 1.25 * 10^6 bytes/sec. RTprop =
20 ms = 0.02 sec. BDP = 1.25e6 * 0.02 = 25000 bytes. Option B is if using bits (10e6*0.02=200000
bits, but question asks bytes). Option C is 10 times too high; D is off by factor.
Page 3
, 9. In a data center using a BCube topology, a server has a single network interface. How many
switches are directly connected to that server in a BCube(k, n) network?
A. k
B. n
C. k+1
D. n-1
Answer: C
Rationale: In BCube(k, n), each server has k+1 ports (one per level plus one for the server itself?
Actually, BCube is built recursively. A server in BCube(k, n) has k+1 network interfaces, each connected
to a different switch (one per level 0 to k). So the number of switches directly connected to a server is
k+1. Option A is k, which is off by one. Option B is n, which is the number of servers per switch? Not
correct.
10. A network experiences a BGP hijack where an attacker announces a more specific prefix (e.g.,
10.0.0.0/24) than the legitimate prefix (10.0.0.0/16). Which of the following mitigation techniques
would be most effective in preventing this attack?
A. Implementing RPKI (Resource Public Key Infrastructure) to validate origin AS.
B. Using BGP community to filter prefixes longer than /24.
C. Enforcing prefix filtering based on IRR (Internet Routing Registry) data.
D. Deploying BGP Flowspec to block traffic to the hijacked prefix.
Answer: A
Rationale: RPKI allows a network to verify that the origin AS is authorized to announce the prefix. A
more specific prefix hijack would be detected because the attacker's AS is not listed in the RPKI ROA for
that prefix. Option B is a blunt tool that may block legitimate longer prefixes. Option C relies on IRR
which can be inaccurate. Option D is reactive, not preventive.
11. In the context of Software-Defined Networking (SDN), which of the following best describes the
relationship between the control plane and data plane when using the OpenFlow protocol?
A. The control plane is centralized and communicates with switches via OpenFlow to install flow table entries,
while the data plane remains in the switches for packet forwarding.
B. The control plane is distributed across switches, and OpenFlow is used for inter-switch communication to
maintain routing tables.
C. OpenFlow enables the data plane to make forwarding decisions independently, with the control plane only
monitoring performance.
D. Both planes are merged into a single entity in each switch, with OpenFlow providing a standardized API for
external applications.
Answer: A
Rationale: In SDN, the control plane is logically centralized and separated from the data plane.
OpenFlow is the standard protocol that allows the controller to install flow entries in switches, which
then forward packets based on those entries. Option B incorrectly describes a distributed control plane;
option C reverses the roles; option D describes a non-SDN architecture.
Page 4
Newest Complete Questions And Correct Detailed
Answers| Already Graded A+
1. A network operator configures a BGP router to prefer routes with the shortest AS_PATH.
However, the router receives two paths to the same prefix: Path A with AS_PATH {100, 200, 300}
and Path B with AS_PATH {100, 400}. Both paths have equal local preference and are learned
from eBGP peers. The router selects Path B. Later, the operator applies an inbound route map on
the session for Path A that sets local preference to 200. After the change, which path will the router
prefer, and why?
A. Path B, because AS_PATH length is still the tie-breaker after local preference and the route map did not
affect Path B.
B. Path A, because the route map increased its local preference to 200, making it higher than Path B's default
local preference of 100.
C. Path B, because the route map is applied inbound, so it only affects routes received from that peer, but the
router will still prefer the shortest AS_PATH among paths with equal local preference.
D. Path A, because the route map also prepends an extra AS to Path B's AS_PATH, making it longer.
Answer: B
Rationale: BGP's decision process checks local preference first. By setting Path A's local preference to
200 (higher than Path B's default 100), Path A becomes preferred regardless of AS_PATH length.
AS_PATH length is only considered if local preference is equal. Options A and C incorrectly prioritize
AS_PATH; D is false because prepending is not mentioned.
2. In a data center using a Fat-Tree topology with k=4 (each switch has 4 ports), how many distinct
paths exist between two hosts in different pods that are both connected to edge switches in their
respective pods? Assume the Fat-Tree is built with k-port switches and uses k/2 core switches.
A. 2
B. 4
C. 8
D. 16
Answer: B
Rationale: In a k-ary Fat-Tree, each host in a pod connects to an edge switch. There are k/2 core
switches. A path from a host in one pod to a host in another pod goes: host -> edge -> aggregate ->
core -> aggregate -> edge -> host. Since there are k/2 = 2 core switches, and each core switch provides
one path, there are exactly 2 distinct paths. However, note that the question asks for paths between two
hosts in different pods, each connected to edge switches. The number of distinct paths equals the number
of core switches, which is k/2 = 2. Wait, but for k=4, k/2=2, but the correct answer is 4? Let me recalc:
Actually, there are (k/2)^2 = 4 core switches in a k-ary Fat-Tree? Standard Fat-Tree: k=4, each pod has
k/2=2 core switches? No, the number of core switches is (k/2)^2 = 4. Each core switch provides one path
between a given pair of edge switches. So for two specific hosts, there are (k/2)^2 = 4 distinct paths. Yes,
Page 1
,answer is 4.
3. Consider a TCP cubic flow with a current congestion window of 500 packets. The flow
experiences a packet loss due to a timeout. Immediately after the timeout, what is the new
congestion window size, assuming standard TCP behavior?
A. 1
B. 250
C. 500
D. 0
Answer: A
Rationale: In standard TCP (including Cubic), when a timeout occurs, the congestion window is reset to
1 MSS (slow-start threshold is set to half of the window before loss). So the new window is 1. Option B
would be the new ssthresh, not cwnd. Option C is the old window. Option D is not standard.
4. An SDN controller uses OpenFlow to manage switches. A packet arrives at a switch, and the
switch's flow table has no matching entry. The switch sends a Packet-In message to the controller.
The controller decides to install a flow entry that forwards packets based on destination MAC
address. Which fields must the controller include in the flow mod message to match on destination
MAC and forward out a specific port?
A. Match on eth_dst, set output port to the desired port, and set idle timeout.
B. Match on eth_dst and ip_dst, set output port, and set priority.
C. Match on eth_dst, set action to output to controller, and set hard timeout.
D. Match on in_port, set output port, and set buffer_id.
Answer: A
Rationale: To forward based on destination MAC, the match must include eth_dst. The action is output to
a port. Idle timeout is optional but typical. Option B adds unnecessary ip_dst match; Option C outputs to
controller, not forwarding; Option D matches on in_port, not destination.
5. A network employs MPLS traffic engineering. A router receives an MPLS packet with label
stack {100, 200} where label 100 is at the top. The router's LFIB shows that for incoming label 100,
the outgoing label is 300 and the next hop is R2. Which action does the router perform?
A. Swap label 100 with 300, push label 400, and forward to R2.
B. Pop the top label, then swap the new top label 200 with 300, and forward to R2.
C. Swap label 100 with 300, keep label 200 unchanged, and forward to R2.
D. Pop the entire label stack and forward the IP packet to R2.
Answer: C
Rationale: The LFIB entry for incoming label 100 specifies an outgoing label of 300. The router performs
a swap: it replaces label 100 with 300. The rest of the stack (label 200) remains untouched. The packet
is then forwarded to R2. Option A incorrectly adds a push; Option B incorrectly pops the top; Option D
pops everything, which is not indicated.
Page 2
,6. In a network using DCTCP, a sender observes that the ECN-marked fraction of packets in a
window is 0.3. The current congestion window is 100 packets. According to DCTCP's congestion
control algorithm, what will be the new congestion window after processing this window's
acknowledgments?
A. 85
B. 70
C. 50
D. 100
Answer: A
Rationale: DCTCP reduces the window by a factor of (1 - ±/2) where ± is the fraction of marked packets.
Here = 0.3, so new window = 100 * (1 - 0.3/2) = 100 * 0.85 = 85. Option B would be 100*(1-0.3)=70,
which is the reduction if using standard TCP with ECN (which halves window). Option C is half; D is no
reduction.
7. A BGP speaker in AS 100 receives a route for prefix 10.0.0.0/8 from two different peers: Peer A
(AS 200) with AS_PATH {200, 300} and Peer B (AS 400) with AS_PATH {400, 500, 600}. Both
routes have equal local preference and MED. Which route will the BGP speaker prefer, and what
is the reason?
A. Route from Peer A, because it has a shorter AS_PATH length.
B. Route from Peer B, because it has a longer AS_PATH, indicating more specific path.
C. Route from Peer A, because it has a lower origin code (IGP < EGP < incomplete).
D. Route from Peer B, because it is learned from a peer with higher IP address.
Answer: A
Rationale: BGP prefers the route with the shortest AS_PATH length when local preference, MED, and
other attributes are equal. AS_PATH length for Peer A is 2 (200,300) and for Peer B is 3 (400,500,600).
So Peer A's route is preferred. Option B is incorrect; longer AS_PATH is worse. Option C: origin code is
not given; Option D: IP address tie-breaker is only used if all else equal, but AS_PATH is considered
earlier.
8. A network uses BBR congestion control. The sender estimates the bandwidth (BW) as 10 Mbps
and the round-trip propagation delay (RTprop) as 20 ms. What is the BDP (Bandwidth-Delay
Product) in bytes?
A. 25000
B. 200000
C. 250000
D. 20000
Answer: A
Rationale: BDP = BW * RTprop. BW = 10 Mbps = 10 * 10^6 bits/sec = 1.25 * 10^6 bytes/sec. RTprop =
20 ms = 0.02 sec. BDP = 1.25e6 * 0.02 = 25000 bytes. Option B is if using bits (10e6*0.02=200000
bits, but question asks bytes). Option C is 10 times too high; D is off by factor.
Page 3
, 9. In a data center using a BCube topology, a server has a single network interface. How many
switches are directly connected to that server in a BCube(k, n) network?
A. k
B. n
C. k+1
D. n-1
Answer: C
Rationale: In BCube(k, n), each server has k+1 ports (one per level plus one for the server itself?
Actually, BCube is built recursively. A server in BCube(k, n) has k+1 network interfaces, each connected
to a different switch (one per level 0 to k). So the number of switches directly connected to a server is
k+1. Option A is k, which is off by one. Option B is n, which is the number of servers per switch? Not
correct.
10. A network experiences a BGP hijack where an attacker announces a more specific prefix (e.g.,
10.0.0.0/24) than the legitimate prefix (10.0.0.0/16). Which of the following mitigation techniques
would be most effective in preventing this attack?
A. Implementing RPKI (Resource Public Key Infrastructure) to validate origin AS.
B. Using BGP community to filter prefixes longer than /24.
C. Enforcing prefix filtering based on IRR (Internet Routing Registry) data.
D. Deploying BGP Flowspec to block traffic to the hijacked prefix.
Answer: A
Rationale: RPKI allows a network to verify that the origin AS is authorized to announce the prefix. A
more specific prefix hijack would be detected because the attacker's AS is not listed in the RPKI ROA for
that prefix. Option B is a blunt tool that may block legitimate longer prefixes. Option C relies on IRR
which can be inaccurate. Option D is reactive, not preventive.
11. In the context of Software-Defined Networking (SDN), which of the following best describes the
relationship between the control plane and data plane when using the OpenFlow protocol?
A. The control plane is centralized and communicates with switches via OpenFlow to install flow table entries,
while the data plane remains in the switches for packet forwarding.
B. The control plane is distributed across switches, and OpenFlow is used for inter-switch communication to
maintain routing tables.
C. OpenFlow enables the data plane to make forwarding decisions independently, with the control plane only
monitoring performance.
D. Both planes are merged into a single entity in each switch, with OpenFlow providing a standardized API for
external applications.
Answer: A
Rationale: In SDN, the control plane is logically centralized and separated from the data plane.
OpenFlow is the standard protocol that allows the controller to install flow entries in switches, which
then forward packets based on those entries. Option B incorrectly describes a distributed control plane;
option C reverses the roles; option D describes a non-SDN architecture.
Page 4
