Cisco CCNP Enterprise 350-401 ENCOR QoS Configuration Interpretation Practice Test

 

Topic 03 interprets QoS configurations for Cisco ENCOR v1.2. These original practice scenarios state the relevant IOS XE MQC support, units, directions, congestion conditions and evidence limits. Select one answer unless Choose TWO. or Choose THREE. is stated. For broader preparation, visit the Cisco 350-401 ENCOR Exam Dumps page. Every option includes a scenario-specific explanation.

Question 1

An IOS XE router uses this supported MQC class:
class-map match-all PREMIUM
match dscp ef
match access-group name APPROVED
Packet P is marked EF but is not selected by APPROVED. Packet Q is marked AF41 and is selected by APPROVED. No other user class matches either packet. What changes if only match-all is replaced by match-any?

1. Neither packet matches before the change; only Q matches afterward.
2. Neither packet matches before the change; both match afterward.
3. Neither packet matches before the change; only P matches afterward.
4. Only P matches before the change; both match afterward.
5. Only Q matches before the change; both match afterward.

Correct Answer: B

 

Correct Answer

Answer B is correct because each packet satisfies only one predicate. The original AND rejects both packets, while the replacement OR admits each packet through its successful predicate.

Incorrect Answers

Answer A is incorrect because the ACL is not a mandatory gate under match-any. P satisfies the EF predicate and therefore also matches the revised class.

Answer C is incorrect because match-any can succeed through either predicate. Q is selected by APPROVED even though its marking is AF41, so it also matches afterward.

Answer D is incorrect because this treats the first DSCP predicate as sufficient under match-all. P fails the ACL predicate and therefore does not match before the change.

Answer E is incorrect because this gives ACL selection precedence over the other predicate. Q fails the required EF match while match-all is in use.

 

Question 2

A supported MQC class VOICE uses only match access-group name CALLS. The ACL contains permit udp 10.30.0.0 0.0.0.255 any range 16384 32767, with no other permit entries. Five first-fragment IPv4 packets traverse this policy; their fields appear below. Which packet is selected by VOICE? No additional DSCP criterion is configured.

10. UDP, source 10.30.0.8:20000, destination 192.0.2.10:40000, DSCP EF.
11. UDP, source 10.30.1.8:40000, destination 192.0.2.10:20000, DSCP EF.
12. UDP, source 10.30.0.8:40000, destination 192.0.2.10:20000, DSCP 0.
13. UDP, source 192.0.2.10:20000, destination 10.30.0.8:40000, DSCP 0.
14. TCP, source 10.30.0.8:40000, destination 192.0.2.10:20000, DSCP EF.

Correct Answer: C

 

Correct Answer

Answer C is correct because the source lies in the permitted /24 and the destination UDP port lies in the stated range. The source port and DSCP are not restricted by this entry.

Incorrect Answers

Answer A is incorrect because the permitted range is applied to the destination port. An in-range source port does not compensate for destination port 40000.

Answer B is incorrect because the destination port is suitable, but the source belongs to 10.30.1.0/24 rather than the permitted 10.30.0.0/24. EF marking does not repair that mismatch.

Answer D is incorrect because this is a reverse-direction tuple. The ACL constrains the source subnet and destination port, so reversing a qualifying conversation does not preserve the match.

Answer E is incorrect because the addresses and destination port fit, but this ACL entry selects UDP. A similarly numbered TCP port does not meet that protocol condition.

 

Question 3

A supported ingress MQC policy contains:
class BUSINESS
set dscp af31
class class-default
set dscp default
A test packet enters marked EF and is positively matched to BUSINESS. There is no subsequent marking action on this router. An analyst expects a capture immediately after this policy to show EF because the original marking had higher priority. Which interpretation is correct?

1. The capture should show AF31 because BUSINESS applies an unconditional set action.
2. The capture should show EF unless BUSINESS has congestion and invokes remarking.
3. The capture should show EF because set dscp supplies a value only for unmarked packets.
4. The capture should show default because the later default class overwrites BUSINESS.
5. The capture should show EF while only internal QoS-group metadata becomes AF31.

Correct Answer: A

 

Correct Answer

Answer A is correct because the selected action replaces the original DSCP with AF31. No trust exception or conditional preservation action appears in the supplied configuration.

Incorrect Answers

Answer B is incorrect because the supplied set action has no congestion condition. Scheduling pressure does not determine whether this matching packet is remarked.

Answer C is incorrect because a set action is not an implicit fill-in for DSCP 0. A separate classifier or conditional action would be needed to preserve already marked packets.

Answer D is incorrect because class-default is used for packets that fail the explicit classes; it is not a second marking stage for this already selected BUSINESS packet.

Answer E is incorrect because set dscp modifies the IP marking. An internal QoS-group assignment is a different action and does not describe this command.

 

Question 4

A supported MQC policy lists GOLD before GENERAL, then class-default. GOLD selects EF packets from an approved subnet and applies set dscp af41. GENERAL selects all EF packets and applies set dscp af21. The default class applies set dscp default. One packet arrives marked EF from the approved subnet. What happens within this single policy?

41. GOLD handles the packet, which leaves the policy marked AF41.
42. GOLD remarks the packet, then class-default resets it because AF41 no longer matches either class.
43. The packet stays EF because its two matching classes specify different markings.
44. GENERAL handles the packet, which leaves the policy marked AF21.
45. GOLD and GENERAL both handle the packet, leaving it marked AF21.

Correct Answer: A

 

Correct Answer

Answer A is correct because both explicit classifiers initially match, but the first matching class determines the action. The policy does not then apply GENERAL or class-default to this packet.

Incorrect Answers

Answer B is incorrect because a set action does not restart class selection against the newly written DSCP in this same policy. The packet remains handled by its selected class.

Answer C is incorrect because overlapping matches are resolved by class order; they do not create an implicit preserve-marking result when actions differ.

Answer D is incorrect because this would follow if GENERAL preceded GOLD. The supplied order puts the more specific GOLD match first.

Answer E is incorrect because the packet does not collect actions from every class it initially matches. The first match selects GOLD and ends class selection for this policy.

 

Question 5

At an enterprise trust boundary, a supported ingress policy classifies voice by validated source and application criteria, sets that class to EF, and explicitly sets class-default to DSCP default. No policing or dropping action is present. Captures identify two flows: a file transfer that fails the voice criteria but arrives marked EF, and approved voice that meets the criteria but arrives marked default. Which TWO outcomes follow? Choose TWO.

1. The transfer retains EF because default-class remarking applies only to packets initially marked default.
2. The approved voice remains default because the voice class requires the endpoint to set EF first.
3. The approved voice leaves this policy marked EF.
4. The unmatched file transfer leaves this policy marked default.
5. The transfer is discarded because failure of the voice classifier denies forwarding.

Correct Answers: C, D

 

Correct Answers

Answer C is correct because the voice classifier is satisfied independently of the incoming DSCP, and the selected set action establishes EF at the trusted ingress boundary.

Answer D is correct because its original EF mark is not a classifier criterion here. Because it fails the validated voice criteria, the explicit default-class remarking action applies.

Incorrect Answers

Answer A is incorrect because membership in class-default depends on failing explicit classes, not on arriving with DSCP 0. The nonvoice transfer is therefore eligible for the supplied remarking action.

Answer B is incorrect because the stem specifies validated source and application criteria, with no incoming-EF requirement. Requiring EF would add a condition absent from the classifier.

Answer E is incorrect because a classification miss directs this packet to class-default; it is not an interface ACL denial. The only default-class action stated is remarking.

 

Question 6

On a congested IOS XE output, a correctly classified voice stream offers 800 kb/s within its tested burst allowance. Two supported policies are proposed: A uses bandwidth 1000; B uses priority 1000. Other classes are backlogged. Both numeric values use kb/s. Which comparison correctly identifies the scheduling consequence?

1. B shapes voice to 1000 kb/s by buffering traffic above that rate; A only schedules it.
2. B gives voice strict-priority service; A provides a bandwidth guarantee without that service.
3. A limits voice to 1000 kb/s at all loads, while B specifies only an unconstrained preference.
4. A gives strict-priority service, while B gives voice a weighted bandwidth share.
5. A and B give the same strict-priority service because both allocate more than the offered rate.

Correct Answer: B

 

Correct Answer

Answer B is correct because priority selects the low-latency scheduling behavior. The voice load is within the specified allocation and burst allowance, whereas bandwidth alone supplies a minimum class allocation.

Incorrect Answers

Answer A is incorrect because priority is not an output shaper. Its treatment of excess traffic under congestion differs from deliberately buffering traffic to smooth it to a shaping rate.

Answer C is incorrect because bandwidth is not a hard policer, and priority does not remove its configured allocation under congestion. Both parts misinterpret the roles of the rate values.

Answer D is incorrect because this reverses the command semantics. The priority command introduces the strict-priority behavior, while bandwidth defines the ordinary class guarantee.

Answer E is incorrect because sufficient bandwidth is not the same as priority scheduling. Other backlogged classes can compete with A under ordinary class-based scheduling.

 

Question 7

A supported output policy gives DATA bandwidth percent 30. The documented scheduling basis for this policy is 100 Mb/s. DATA has no rate limiter, and other queues are idle. The class is observed transmitting at 60 Mb/s. Which interpretation of the configured percentage is consistent with those facts?

1. It reserves 30 Mb/s for DATA and requires the remaining 70 Mb/s to stay idle when other classes are idle.
2. It gives DATA 30 percent of its own offered rate, making the allocation 18 Mb/s.
3. It caps DATA at 30 Mb/s, so the 60-Mb/s counter proves the policy was bypassed.
4. It guarantees DATA 60 Mb/s during future congestion because that rate was measured while queues were idle.
5. It supplies a 30-Mb/s minimum allocation under contention and permits use of spare capacity.

Correct Answer: E

 

Correct Answer

Answer E is correct because 30 percent of the stated 100-Mb/s basis is 30 Mb/s. A bandwidth guarantee is not a hard cap, so using otherwise idle capacity explains the 60-Mb/s observation.

Incorrect Answers

Answer A is incorrect because minimum allocations do not prohibit a class from using unused link capacity. That interpretation would discard the documented distinction between a guarantee and a cap.

Answer B is incorrect because the percentage uses the explicitly stated scheduling bandwidth, not the currently observed traffic rate. Multiplying 60 by 30 percent uses the wrong denominator.

Answer C is incorrect because a bandwidth allocation does not impose that cap. The measured rate can be valid with the policy active, so bypass cannot be deduced from this counter.

Answer D is incorrect because available spare capacity in one observation does not increase the configured contention guarantee. The supplied basis and percentage still establish 30 Mb/s.

 

Question 8

A supported policy has one explicit class matching EF. Its unmatched-traffic action is:
class class-default
police 5000000 10000 conform-action transmit exceed-action drop
Two long-running AF11 bulk flows each offer 4 Mb/s and match no explicit class. This is one aggregate policer instance, not a per-flow policer; ignore initial burst effects. Which interpretation is correct?

1. The AF11 flows use the explicit EF class because both marks indicate preferential service.
2. Only packets marked default enter this policer, so the AF11 flows bypass it.
3. The two flows share a 5-Mb/s shaping queue that delays their combined excess.
4. The two flows share the default-class 5-Mb/s policing rate.
5. Each flow receives a separate 5-Mb/s policing allowance because it is a different conversation.

Correct Answer: D

 

Correct Answer

Answer D is correct because both flows enter the same class and the same aggregate meter. Their nonzero DSCP does not exempt them, and the command does not allocate a separate rate to each flow.

Incorrect Answers

Answer A is incorrect because the explicit class matches the EF codepoint, not every marking associated with differentiated service. AF11 fails that exact match.

Answer B is incorrect because class-default contains traffic unmatched by explicit classes regardless of its DSCP. Both supplied flows therefore reach the configured meter.

Answer C is incorrect because police with exceed-action drop discards excess instead of buffering it for later eligibility. The configured rate does not make the mechanism a shaper.

Answer E is incorrect because the stem identifies a single aggregate meter. Per-flow allowances would require a different policy behavior not present in this configuration.

 

Question 9

A congested 10-Mb/s router output has VOICE priority 1000, BULK bandwidth 4000, and a backlogged default class. The commands use kb/s and this policy is supported. VOICE offers 0.8 Mb/s within its allowed burst profile; BULK is also backlogged. There are no additional class rate limiters. Which TWO interpretations are correct? Choose TWO.

1. VOICE receives an ordinary weighted share because its offered rate is below 1000 kb/s.
2. BULK has a 4-Mb/s minimum bandwidth allocation during this contention.
3. The unused 0.2 Mb/s of the voice allocation must remain idle instead of serving another queue.
4. Eligible voice is scheduled ahead of ordinary queued bulk traffic.
5. BULK is capped at 4 Mb/s even if the scheduler has spare capacity to assign.

Correct Answers: B, D

 

Correct Answers

Answer B is correct because bandwidth 4000 establishes the stated class guarantee in the supported policy. It does not place BULK in the voice priority queue.

Answer D is correct because priority provides LLQ service, and the stated voice load and bursts are within the priority allocation. This describes scheduling order, not an end-to-end delivery guarantee.

Incorrect Answers

Answer A is incorrect because being below the allocation does not change priority into ordinary weighted scheduling. Eligible packets retain the configured priority treatment.

Answer C is incorrect because an unused allocation does not require the output to waste available capacity. The priority setting protects eligible traffic without requiring idle reservations.

Answer E is incorrect because the bandwidth command provides a guarantee rather than a hard rate limiter. The stem explicitly excludes additional class rate limiters.

 

Question 10

A proposed scheduler gives REALTIME strict priority without a rate bound. Its classifier includes both voice and bulk video; design measurements show that admitted video can keep REALTIME backlogged at the full output rate. A lower-priority transactional queue also stays backlogged. The scheduler does not enforce a minimum service share for lower queues. Which conclusion should drive the design correction? These are explicit design assumptions, not claimed Cisco LLQ defaults.

1. A higher drop-precedence marking within the video packets is sufficient without any scheduler or drop-policy change.
2. Increasing the transactional queue limit is sufficient to preserve its sustained service.
3. The measured short voice bursts show that the lower queue will receive adequate service between calls.
4. Separating transactional traffic into several lower-priority queues ensures it obtains a service share.
5. Transactional traffic can be starved; the priority admission or service bound needs correction.

Correct Answer: E

 

Correct Answer

Answer E is correct because continuous full-rate demand in an unbounded strict-priority queue can consume every transmission opportunity. The lower queue has no enforced service share to interrupt that behavior.

Incorrect Answers

Answer A is incorrect because a marking has an effect only when a policy acts on it. The stated scheduler still admits the video into the continuously served priority queue.

Answer B is incorrect because a larger queue can store more waiting traffic, but it does not create a transmission opportunity while the unbounded priority queue remains continuously backlogged.

Answer C is incorrect because voice is not the only admitted load. The measured bulk video can keep REALTIME backlogged even when the voice component is temporarily quiet.

Answer D is incorrect because more lower-priority queues do not overcome the specified service order. None has an enforced minimum that can take service from the continuously backlogged priority class.

 

Question 11

A 1-Gb/s enterprise handoff has a fixed provider contract: 10 Mb/s with a small permitted burst. An immediate-drop output policer enforces it. Synchronized tests show that application bursts exceed the permitted burst but drain if smoothed to 10 Mb/s; the required buffer and resulting delay fit the application budget. The provider will not enlarge its burst allowance. Which change best preserves these bursts while meeting the contract?

1. Use output shaping with a provider-compliant rate and burst profile plus the validated buffer capacity.
2. Increase only the policer burst allowance until all application bursts are admitted immediately.
3. Change exceed-action to transmit while retaining the current rate and burst values.
4. Keep the current policer and add a deeper queue after its drop action.
5. Replace policing with bandwidth 10000 and leave the 1-Gb/s handoff otherwise unconstrained.

Correct Answer: A

 

Correct Answer

Answer A is correct because the tested buffer can absorb the temporary excess and shaping can release it within the contract. The explicit delay and capacity checks rule out an unbounded-buffer assumption.

Incorrect Answers

Answer B is incorrect because that could reduce local discards, but the stem says those bursts exceed the unchanged provider allowance. Admitting them immediately would move the loss risk to the provider.

Answer C is incorrect because that would forward packets identified as outside the contract rather than pace them. A conforming meter configuration alone does not enforce its envelope if excess is transmitted unchanged.

Answer D is incorrect because a later queue cannot buffer packets the policer has already discarded. The buffering must participate in the rate-control mechanism before avoidable loss occurs.

Answer E is incorrect because a bandwidth guarantee is not a 10-Mb/s output envelope. It can allow bursts into the provider faster than the contract permits.

 

Question 12

A supported ingress three-color policer uses conform-action transmit, exceed-action set-qos-transmit 4, and violate-action drop. Three otherwise identical packets arrive with DSCP AF31 and are reported as conform, exceed and violate respectively. No other action changes DSCP. Which THREE interpretations follow at this processing stage? Choose THREE.

1. The exceeding packet is remarked and then discarded because exceed is a nonconforming state.
2. The exceeding packet is transmitted with internal QoS group 4 and its AF31 DSCP preserved.
3. The violating packet is discarded at this policer.
4. The violating packet waits in the policer until tokens allow transmission.
5. The conforming packet is transmitted without a DSCP change by this policer.
6. The exceeding packet is transmitted with DSCP 4 because the action specifies the value 4.

Correct Answers: B, C, E

 

Correct Answers

Answer B is correct because set-qos-transmit changes internal QoS-group metadata and permits transmission. That action is distinct from writing a DSCP value in the IP header.

Answer C is correct because the reported violate state selects the explicitly configured drop action. This outcome is independent of the packet having arrived with an AF31 marking.

Answer E is correct because its action is transmit rather than a remarking action. With no other DSCP writer, the incoming AF31 value is preserved at this stage.

Incorrect Answers

Answer A is incorrect because the configured exceed action explicitly includes transmission. A nonconforming classification does not override the selected transmit action with an implicit drop.

Answer D is incorrect because the configured action discards that packet; a policer does not turn a reported drop outcome into deferred shaping service.

Answer F is incorrect because the number belongs to the QoS-group action, not to a DSCP action. Applying it to the IP header would confuse two different forms of marking.

 

Question 13

A supported single-bucket policer has a maximum bucket size of 12000 bytes and currently holds 10000 bytes of tokens. Packet charging includes exactly the stated packet length in bytes. A valid 8000-byte test packet arrives; transmit is the conform action and drop is the exceed action. Ignore replenishment during its evaluation. Which outcome follows?

1. The packet is transmitted and the token balance becomes 4000 bytes.
2. The packet is transmitted and the token balance becomes 2000 bytes.
3. The packet is dropped because its charge is 64000 against a balance of 10000.
4. The packet waits until the token balance reaches 12000 bytes before transmission.
5. The packet is transmitted and the token balance remains 10000 bytes.

Correct Answer: B

 

Correct Answer

Answer B is correct because the current 10000-byte balance covers the 8000-byte charge. The current balance, rather than the maximum bucket size, determines the remaining 2000 bytes.

Incorrect Answers

Answer A is incorrect because this subtracts the packet from the maximum bucket size instead of the current balance. The bucket was not full when this packet arrived.

Answer C is incorrect because that comparison converts the packet to bits while retaining a balance measured in bytes. The stem supplies consistent byte-based accounting.

Answer D is incorrect because there are already enough tokens for conformance, and this is a transmit/drop policer. Neither a full-bucket requirement nor a shaping wait is configured.

Answer E is incorrect because successful conformance consumes tokens for the admitted packet. Ignoring that deduction would permit subsequent packets to reuse the same allowance.

 

Question 14

A single aggregate policer is configured for 8,000,000 bits/s with conform-action transmit and exceed-action drop. During a 30-second steady test it receives 12 Mb/s. Initial burst credit is negligible, the bucket can accommodate the test packets, and no other losses occur. Use decimal units and ignore packet-boundary error. Which approximate byte-counter deltas match this test?

1. 3,750,000 bytes transmitted and 41,250,000 bytes dropped.
2. 45,000,000 bytes transmitted and no bytes dropped.
3. 30,000,000 bytes transmitted, no bytes dropped, and 15,000,000 bytes buffered.
4. 15,000,000 bytes transmitted and 30,000,000 bytes dropped.
5. 30,000,000 bytes transmitted and 15,000,000 bytes dropped.

Correct Answer: E

 

Correct Answer

Answer E is correct because the admitted rate is 8,000,000/8 = 1,000,000 bytes/s, giving 30,000,000 bytes in 30 seconds. The remaining 4 Mb/s contributes 15,000,000 dropped bytes.

Incorrect Answers

Answer A is incorrect because this divides the allowed bit rate by eight twice, effectively admitting only 1 Mb/s. A single conversion from bits to bytes is required for the displayed counters.

Answer B is incorrect because 45,000,000 bytes is the full offered load over the interval. Transmitting all of it would disregard the sustained 8-Mb/s meter and explicit excess-drop action.

Answer C is incorrect because those buffered bytes would require shaping behavior. The supplied policer discards excess instead of retaining it for later transmission.

Answer D is incorrect because the 4-Mb/s excess determines the smaller dropped amount, not the transmitted amount. This reverses the admitted rate and the excess rate.

 

Question 15

An engineer intends a sustained 8-Mb/s limit but configures police 8000 10000 conform-action transmit exceed-action drop. The supported command form takes rate in bits/s, followed by burst in bytes. A long test with adequate demand confirms about 8 kb/s of conforming traffic. The 10000-byte burst has already been approved and must remain unchanged. Which correction meets the requirement?

8000000. Leave the first value at 8000 and change the second value to 8000000.
8000001. Replace police with bandwidth 8000 while retaining the class match.
8000002. Change the first value to 1000000 and leave the second value at 10000.
8000003. Change the first value to 8000000 and leave the second value at 10000.
8000004. Change exceed-action to transmit and leave both numeric values unchanged.

Correct Answer: D

 

Correct Answer

Answer D is correct because 8 Mb/s requires 8,000,000 bits/s. The existing first value is 1000 times too small, while the approved byte-based burst is a separate parameter.

Incorrect Answers

Answer A is incorrect because enlarging the burst may admit more initial traffic but does not correct the sustained 8000-bit/s replenishment rate. It also changes the explicitly fixed burst allowance.

Answer B is incorrect because bandwidth uses a scheduling guarantee, not the required sustained policing ceiling. Even when its numeric unit is kb/s, it does not preserve the requested enforcement behavior.

Answer C is incorrect because 1,000,000 is the target rate expressed in bytes/s, but this field expects bits/s. Entering it would configure only 1 Mb/s.

Answer E is incorrect because that can raise delivered throughput by ceasing to discard excess, but it does not establish the required 8-Mb/s limit. Traffic outside the meter would be transmitted.

 

Question 16

A branch must meter all uploads together at one common WAN interface, including uploads arriving on three LAN interfaces. Routing confirms that every upload exits that WAN. A supported policing-only policy LIMIT is currently attached as service-policy input LIMIT on the WAN; upload counters do not increment. Its classifier has been verified against the upload packets. Which TWO statements follow? Choose TWO.

1. Attaching LIMIT on LAN output is the appropriate direction for traffic arriving from those LANs.
2. The current attachment evaluates traffic arriving from the WAN instead of these uploads.
3. Reversing source and destination fields in the WAN input classifier will catch these uploads.
4. Separate copies of LIMIT on each LAN ingress provide the same single aggregate ceiling.
5. An output attachment on this common WAN can meter the required combined upload traffic.

Correct Answers: B, E

 

Correct Answers

Answer B is correct because input is relative to the router interface. These uploads leave the WAN interface, so they do not traverse its input policy.

Answer E is correct because all three upload paths converge at that egress. The stated supported output policer can therefore apply one aggregate meter at the specified common point.

Incorrect Answers

Answer A is incorrect because uploads enter the router from the LANs rather than leave toward them. LAN output policies would evaluate the opposite traffic direction.

Answer C is incorrect because reversing a classifier does not make outbound packets traverse the interface input policy. The stem already confirms that the upload classifier itself is valid.

Answer D is incorrect because each attachment would have its own meter. Their individual ceilings do not enforce one combined ceiling across the three LAN paths.

 

Question 17

A supported hierarchical output policy attaches a parent shape average 10000000 to the WAN. Its child has VOICE priority 1000 and DATA bandwidth 3000. The parent covers all child traffic. Rates are bits/s for shape and kb/s for priority and bandwidth. During sustained contention, which interpretation correctly separates the functions?

1. The aggregate envelope is 4 Mb/s because the child allocations add to that value.
2. DATA receives a hard 3-Mb/s ceiling while the parent supplies an ordinary bandwidth guarantee.
3. VOICE and DATA each obtain a separate 10-Mb/s envelope from the parent.
4. VOICE can exceed the parent envelope because its child priority action takes precedence over shaping.
5. The parent provides a 10-Mb/s aggregate envelope; the child schedules classes within it.

Correct Answer: E

 

Correct Answer

Answer E is correct because the parent shapes the combined traffic, while the child differentiates priority and ordinary guaranteed service. These roles operate at different levels of the hierarchy.

Incorrect Answers

Answer A is incorrect because 1 Mb/s of priority allocation plus a 3-Mb/s guarantee does not rewrite the parent shape rate. Spare capacity can exist within the explicitly configured 10-Mb/s envelope.

Answer B is incorrect because this reverses the roles of bandwidth and shape. The child bandwidth value is a guarantee, whereas the parent shape controls the aggregate rate envelope.

Answer C is incorrect because the parent covers their combined traffic. Per-class independent envelopes would require separate shaping instances rather than this one aggregate parent.

Answer D is incorrect because priority selects service among the child classes. It does not detach the selected traffic from the parent shaper that covers the entire child policy.

 

Question 18

On one router, an ingress policy assigns QoS group 4 to an approved application. Verification confirms that the same packets retain group 4 at a later egress classifier; no other class precedes OTHER in this policy. OTHER uses match not qos-group 4. In an isolated test containing only this application, OTHER records zero matches. Which interpretation fits both the configuration and verification?

1. The zero count identifies loss of the group metadata between ingress and egress.
2. An earlier matching class consumed the test packets before OTHER was evaluated.
3. The zero count identifies a DSCP mismatch that should be repaired with match dscp 4.
4. OTHER intentionally excludes the test packets because their retained group is 4.
5. Replacing match not qos-group 4 with match qos-group 4 would preserve the intended exclusion.

Correct Answer: D

 

Correct Answer

Answer D is correct because the negative predicate admits QoS-group values other than 4. The verified group-4 traffic therefore should not increment this class.

Incorrect Answers

Answer A is incorrect because the stem directly verifies that group 4 is retained at the later classifier. Its exclusion, rather than a metadata-loss theory, explains the counter.

Answer B is incorrect because class order can matter, but the stem excludes an earlier class here. That competing explanation does not apply to this test.

Answer C is incorrect because this classifier examines QoS-group metadata, not the IP DSCP field. Changing fields would change the policy instead of explaining its current predicate.

Answer E is incorrect because that replacement reverses the predicate and admits the application. It would increase matches by changing the intended policy rather than repairing an error.

 

Question 19

R1 correctly classifies a business flow and sets DSCP AF31. Captures show that it reaches R2 still marked AF31. R2 has one explicit output class, PREMIUM, matching only AF21; class-default performs no remarking. The business requirement is to place this flow in PREMIUM. Which TWO conclusions follow from the supplied configuration? Choose TWO.

1. The marking and classifier must be aligned through a deliberate change to the enterprise QoS contract.
2. Increasing PREMIUM’s bandwidth allocation will move this AF31 flow into that class.
3. AF31 should match AF21 because the higher AF class number implicitly includes lower AF classes.
4. Adding set dscp af21 inside PREMIUM will make this AF31 flow select PREMIUM.
5. The observed flow falls outside R2’s AF21 criterion and enters its default class.

Correct Answers: A, E

 

Correct Answers

Answer A is correct because either an approved upstream marking change or a downstream match change could align this flow. The evidence establishes their disagreement but does not decide which enterprise marking policy is intended.

Answer E is correct because AF31 and AF21 are different DSCP values. With no other explicit class, the received AF31 traffic fails PREMIUM and reaches class-default.

Incorrect Answers

Answer B is incorrect because the scheduling allocation acts on traffic already selected into PREMIUM. It does not modify the AF21 predicate that currently excludes the flow.

Answer C is incorrect because an exact DSCP criterion is not an ordered threshold across AF classes. A broader match must be configured explicitly if that is the desired policy.

Answer D is incorrect because the action is reached only after the class matches. A packet failing the AF21 criterion cannot use that class action to make its own initial match succeed.

 

Question 20

An output QoS policy was attached to GigabitEthernet0/0 when that interface carried uploads. A required routing change moves the uploads to GigabitEthernet0/1, as confirmed by forwarding checks and capture. QoS support on the new interface is confirmed, but the policy remains attached only to GigabitEthernet0/0. Its old counters stay flat. Which change addresses the demonstrated gap while retaining the new routing design?

1. Apply the policy in the input direction on GigabitEthernet0/1.
2. Clear the old interface counters and repeat the same upload test.
3. Broaden the old interface’s class map to match any upload marking.
4. Restore the old route so uploads again traverse GigabitEthernet0/0.
5. Apply the intended supported output policy on GigabitEthernet0/1.

Correct Answer: E

 

Correct Answer

Answer E is correct because the packets now leave that interface, so this places treatment on the verified path. The required route is preserved rather than changed merely to reach an obsolete attachment.

Incorrect Answers

Answer A is incorrect because these uploads leave that interface. An input attachment would inspect traffic received from the opposite direction.

Answer B is incorrect because clearing counters can improve observation, but it does not attach treatment to the current egress. The supplied forwarding and capture checks already establish that mismatch.

Answer C is incorrect because even a broader classifier cannot evaluate packets that never traverse its attachment. The verified egress change must be addressed first.

Answer D is incorrect because that could restore traversal of the old policy but conflicts with the explicit requirement to retain the new routing design. Moving the supported attachment addresses both constraints.

 

Question 21

During an isolated test, BUSINESS is the only flow in a supported output class. Across the same bounded interval, class matches increase by 20,000 packets and the class queue’s tail-drop counter increases by 2,000. The queue reaches its configured limit under sustained output congestion. No policer is attached to this class. Which conclusion is supported?

1. The 2,000 drops show that BUSINESS exceeds a configured policer rate.
2. A larger queue limit is proven to eliminate sustained loss without any throughput change.
3. The flow is selected into BUSINESS and experiences loss at that output queue.
4. The interface has a physical-error fault that should be corrected before the class policy is examined.
5. The classifier fails for 2,000 packets, which are therefore counted as BUSINESS tail drops.

Correct Answer: C

 

Correct Answer

Answer C is correct because correlated match and tail-drop deltas establish both classification and queue loss during this isolated test. The full queue supplies the context for the tail-drop outcome.

Incorrect Answers

Answer A is incorrect because the stem excludes a class policer and identifies the queue tail-drop counter. Meter drops and queue drops locate different enforcement stages.

Answer B is incorrect because a larger buffer may absorb a finite burst but cannot by itself resolve indefinitely excessive arrivals. These counters identify queue loss without proving that a buffer-only remedy is sufficient.

Answer D is incorrect because these are class queue tail drops, not evidence from physical error counters. A physical fault could exist elsewhere, but these observations do not establish one as the cause of these drops.

Answer E is incorrect because a failed match would not explain tail drops within the selected BUSINESS queue. The counters describe loss after selection rather than an unmatched-packet count.

 

Question 22

A dedicated test flow is the only traffic matching an ingress policer class. During one interval its exceed-action drop counter increases by 4,000 packets. The subsequent egress queue reports zero drops and no backlog during that same interval. Which next action follows most directly from the evidence without assuming why the ingress traffic exceeds the meter?

1. Compare the ingress rate and burst settings with the offered profile and the approved traffic contract.
2. Investigate only downstream congestion because the local output has no drops.
3. Increase egress queue depth to retain the 4,000 packets discarded by the ingress policer.
4. Disable ingress excess drops because the counter proves that the configured contract is too restrictive.
5. Raise the egress bandwidth guarantee to correct the ingress token replenishment rate.

Correct Answer: A

 

Correct Answer

Answer A is correct because the isolated counter attributes these recorded drops to the ingress meter. Comparing configuration, demand and entitlement can distinguish a misconfiguration from legitimate enforcement without assuming either.

Incorrect Answers

Answer B is incorrect because downstream checks may still matter for other loss, but the local ingress counter already identifies these recorded discards. Zero egress drops does not cancel evidence from an earlier stage.

Answer C is incorrect because those packets were dropped before reaching the output queue. Extra output storage cannot recover them, and no output backlog is observed.

Answer D is incorrect because the counter proves enforcement, not that the contract or settings are wrong. Disabling enforcement before comparing entitlement and traffic profile would assume the missing conclusion.

Answer E is incorrect because egress scheduling and ingress policing are separate mechanisms. The stated bandwidth change would not adjust the earlier meter.

 

Question 23

An IOS XE output policy uses priority 2000, with units of kb/s and documented congestion-related LLQ excess dropping. During a long congested test, its matched voice class offers 3 Mb/s; priority excess-drop counters rise after the permitted burst credit is exhausted. The class has no separate explicit policer. Which interpretation is best supported?

1. The configured 2000 value is a packet queue limit, so only queue-depth tuning is relevant.
2. The rising counter means the voice classifier failed and the flow used class-default.
3. The matched class exceeds its 2-Mb/s priority allocation under the stated congestion behavior.
4. Raising only the burst allowance will sustain the full 3 Mb/s indefinitely at the unchanged rate.
5. The absence of a separate police command means these drops cannot be related to the priority allocation.

Correct Answer: C

 

Correct Answer

Answer C is correct because 2000 kb/s is 2 Mb/s, below the 3-Mb/s sustained offer. Once burst credit is exhausted, the documented LLQ excess treatment explains these priority-class drops.

Incorrect Answers

Answer A is incorrect because the supplied units identify a rate allocation. Interpreting it as a packet count directs the remedy toward a different configuration parameter.

Answer B is incorrect because the stem identifies matched voice traffic and the priority class’s own excess counter. A classifier miss would not explain that class-specific evidence.

Answer D is incorrect because more initial credit can accommodate a finite burst but does not cover a continuing 1-Mb/s excess. The test deliberately extends beyond burst effects.

Answer E is incorrect because the documented LLQ behavior supplies congestion-related excess treatment for priority traffic. A separate explicit policer is not required for that stated behavior.

 

Question 24

A team raises a nonpriority class’s bandwidth guarantee from 10% to 30%. Policy verification confirms the change on the actual path. Before and after, the output is lightly loaded with zero queue depth; synchronized transaction traces place nearly all response delay in server processing after request delivery. Response time barely changes. Which interpretation is justified?

1. Increasing the queue limit is the next evidenced remedy for the delayed response.
2. The change does not address the measured dominant delay because there is no output contention to relieve.
3. Converting the class to strict priority is the next evidenced remedy for the server processing delay.
4. The unchanged response time demonstrates that the new policy is inactive.
5. Changing a downstream DSCP match is justified because unchanged latency establishes a marking mismatch.

Correct Answer: B

 

Correct Answer

Answer B is correct because the scheduling guarantee helps when classes compete for output service. The observed requests are already delivered promptly, and the trace locates most delay in subsequent server processing.

Incorrect Answers

Answer A is incorrect because additional queue capacity addresses waiting traffic or loss under pressure. No queue buildup is present, and the measured delay occurs beyond delivery.

Answer C is incorrect because strict priority addresses competition at the output scheduler. It cannot shorten the measured server work after delivery, and the observed queue is already empty.

Answer D is incorrect because the stem directly verifies the active policy. Lack of benefit at an uncongested queue is not evidence that the setting failed to take effect.

Answer E is incorrect because latency alone does not establish that mismatch. The given trace and zero queues identify a different measured constraint; a marking claim would need separate configuration or capture evidence.

 

Question 25

A synchronized voice test verifies the intended output policy on the actual egress interface. An isolated test flow increments its intended class counter; all relevant class queue and policer drop counters remain zero during the interval. End-to-end loss still occurs. These observations do not cover interface hardware errors or other hops. Which THREE conclusions are justified? Choose THREE.

1. The intended voice class selects the isolated test packets.
2. The zero class-drop counters locate the loss on a downstream hop rather than this router.
3. The local priority allocation should be increased because end-to-end loss establishes that it is too low.
4. Further loss localization must include evidence beyond these class queue and policer counters.
5. The intended output attachment is verified for this test path.
6. The local class match establishes that downstream hops use the intended voice queue.

Correct Answers: A, D, E

 

Correct Answers

Answer A is correct because the correlated matches are from the isolated test flow, so unrelated historical or concurrent class traffic is not needed to explain them.

Answer D is correct because the end-to-end observation remains unexplained by these zero counters. Other local drop points, interface errors and other hops remain within the investigation scope.

Answer E is correct because the interface and direction were checked against actual forwarding. This establishes the local attachment during the test, rather than relying only on a policy definition.

Incorrect Answers

Answer B is incorrect because the stem excludes other local drop points and hardware errors from the observations. These specific zero counters cannot clear the entire router or identify a downstream location.

Answer C is incorrect because the correlated local excess-drop evidence does not support that inference. The end-to-end symptom alone does not identify this allocation as the cause.

Answer F is incorrect because local selection provides no evidence about another hop’s classifier or scheduler. Downstream policy agreement must be checked separately.