Huawei Certified ICT Associate - Transmission — Free Practice Questions
10 free sample questions from a bank of 95, with the correct answers and explanations. No signup required — start practising right now.
1NE A and NE B are adjacent NEs.If B3 and lower-order bit errors occur in the A->B direction, and B1, B2, B3, and lower-order bit errors occur in the B->A direction, which of the following performance events are reported by NE A? (Multiple Choice)
HPBBE
MSFEBBE
MSBBE
RSFEBBE
RSBBE
HPFEBBE
Answer: A, C, E, F
The short version
A and C and E and F — near-end BBE for B->A errors plus far-end HP BBE for A->B errors. NE A detects B1/B2/B3 errors arriving from B directly, and reports the A-to-B B3 errors via the far-end return indication.
Key concepts in this question
RSBBE / MSBBE / HPBBE: near-end background block errors at regenerator, multiplex-section, and higher-order path layers.
HPFEBBE / MSFEBBE / RSFEBBE: far-end BBE counts returned by the remote NE for errors it detected.
B-to-A B1 errors are seen by A itself, so A counts RSBBE (E). B-to-A B2 errors are seen by A itself, so A counts MSBBE (C). B-to-A B3 errors are seen by A itself, so A counts HPBBE (A). A-to-B B3 errors are seen by B, which returns a path REI that A counts as HPFEBBE (F). There is no A-to-B B1/B2 error stated, so A reports no RS/MS far-end BBE here.
Why the others are wrong
B. MSFEBBE: would need B2 errors in the A-to-B direction, which are not stated; only B3 plus lower-order errors occur there.
D. RSFEBBE: would need B1 errors in the A-to-B direction, which are not stated.
H31-311 exam tip
Remember B1-near-RS, B2-near-MS, B3-near-HP; errors away from you always come back as FEBBE.
2When the J1 byte mismatch occurs, the () alarm is reported and services may be interrupted.
HP_SLM
RS_TEM
LP_TEM
HP_TIM
Answer: D
The short version
D — J1 mismatch raises HP_TIM. J1 is the higher-order VC-4 path trace byte, so a mismatch means the received trace differs from expected.
Key concepts in this question
J1 byte: higher-order path trace identifier carried in the VC-4 POH.
SLM vs TIM: SLM is signal-label mismatch (C2), TIM is trace-identifier mismatch (J1/J2).
Why D is correct
J1 continuously carries the provisioned path trace string. When transmitter and receiver expectations differ, the receiver declares HP_TIM, and because the path cannot be confirmed, traffic on that VC-4 may be interrupted. RS_TEM and LP_TEM relate to lower-order/section trace handling, not J1, and HP_SLM is driven by the C2 signal-label byte.
Why the others are wrong
A. HP_SLM: reported on C2 signal-label mismatch, not J1 trace mismatch.
B. RS_TEM: regenerator-section trace (J0) related, not the J1 path byte.
C. LP_TEM: low-order trace related (J2), not the higher-order J1 byte.
H31-311 exam tip
Hook J1 to HP_TIM, J2 to LP_TIM, J0 to RS_TIM, and C2 to SLM.
3Which of the following factors determines that the optical signal-to-noise ratio (OSNR) at the receive end decreases to a certain extent After signals are transmitted for a certain distance in a WDM system, the optical signal-to-noise ratio (OSNR) at the receive end decreases to a certain extent.Which of the following is the main cause?
New types of optical fibers are not used.
EDFAs are used.
Fiber attenuation is excessively high.
Multiplexers and demultiplexers are used.
Answer: B
The short version
B — EDFA amplified spontaneous emission is the main OSNR killer. Each amplifier adds ASE noise, so cascaded EDFAs steadily reduce receive-end OSNR.
Key concepts in this question
OSNR: ratio of channel signal power to noise power in the optical reference bandwidth.
EDFA ASE: erbium amplifiers add broadband spontaneous-emission noise along with gain.
Attenuation vs noise: attenuation lowers power but amplifiers restore it at the cost of added noise.
Why B is correct
In a WDM line with cascaded EDFAs, fiber loss is compensated by gain, but every EDFA injects ASE. After several spans the accumulated ASE is what principally degrades OSNR at the receiver. New fiber types, mux/demux insertion loss, or attenuation alone do not inject broadband noise the way EDFA chains do, so B is the main cause.
Why the others are wrong
A. New fiber types not used: fiber type affects loss and nonlinearity, not the dominant accumulated-noise mechanism.
C. Fiber attenuation too high: attenuation is compensated by EDFA gain; the OSNR penalty comes from the accompanying ASE.
D. Mux/demux used: they add insertion loss and filtering effects but are passive and do not generate ASE.
H31-311 exam tip
Any question asking what mainly degrades OSNR points to EDFA ASE accumulation.
4A MAC address consists of 46 bits and is globally unique.Which of the following is the meaning of the MAC address whose eighth bit is 1?
Bicast address
Multicast address
Broadcast address
Unicast address
Answer: B
The short version
B — I/G bit 1 means a group (multicast) address. The eighth transmitted bit of the first octet distinguishes individual from group addresses.
Key concepts in this question
MAC I/G bit: least significant bit of the first octet; 0 is unicast, 1 is group/multicast.
48-bit MAC: six octets, globally unique per OUI assignment (stem's 46 is a typo).
Broadcast: the special all-ones group address, a subset of multicast handling.
Why B is correct
Ethernet transmits the first octet LSB first, so bit 8 in wire order is the I/G bit. A value of 1 marks the address as a group address, which exam wording calls a multicast address. Broadcast FF-FF-FF-FF-FF-FF is one special group address, but the general meaning of I/G = 1 is multicast/group, hence B.
Why the others are wrong
A. Bicast address: not a valid Ethernet address type; distractor wording.
C. Broadcast address: only the all-ones address is broadcast; I/G = 1 covers all group addresses.
D. Unicast address: requires I/G = 0, the opposite of the stated bit value.
H31-311 exam tip
I/G = 0 individual, I/G = 1 group; second bit (U/L) is a different question.
5When severely errored seconds (SESs) are present in the PM section at the ODUk layer, which of the following alarms is reported on the NMS?
ODUk_PM_SES
ODUk_PM_UAS
ODUk_PM_SESR
ODUk_PM_ES
Answer: A
The short version
A — PM-layer SES is reported as ODUk_PM_SES. Severely errored seconds counted in path monitoring raise the matching PM SES event.
Key concepts in this question
ODUk PM: path monitoring overhead tracking end-to-end ODUk signal quality.
SES: severely errored second declared when the error ratio within a second exceeds threshold.
UAS / SESR / ES: unavailable seconds, SES ratio, and errored seconds are separate counters.
Why A is correct
The NMS names performance events after the layer and type that produced them. SES detected by ODUk path monitoring is therefore reported as ODUk_PM_SES. UAS, SESR, and ES are related but distinct PM statistics with their own thresholds and event names, so only A matches the stated SES condition.
Why the others are wrong
B. ODUk_PM_UAS: unavailable seconds after consecutive SES, not the SES event itself.
C. ODUk_PM_SESR: the SES ratio statistic, not the SES occurrence alarm.
D. ODUk_PM_ES: errored second (lighter defect), not the severe SES threshold.
H31-311 exam tip
Match the suffix exactly: SES, ES, UAS, and SESR are four different PM counters.
6Which of the following methods can be used to multiplex lower-order ODUk to higher-order ODUk?
Wavelength division multiplexing
Time division multiplexing
Frequency division multiplexing
Space division multiplexing
Answer: B
The short version
B — ODUk multiplexing is time-division multiplexing. Lower-order ODU tributaries are interleaved into fixed time slots of the higher-order ODU.
Key concepts in this question
ODU multiplexing: mapping several lower-rate ODUk signals into one higher-rate ODUk.
Time-division multiplexing: tributaries occupy assigned tributary slots in the OPU/ODU frame.
WDM vs TDM: WDM stacks wavelengths; ODU multiplexing stacks time slots within one wavelength.
Why B is correct
G.709 OTN multiplexes lower-order ODUj into higher-order ODUk through the OPUk tributary-slot structure. Each tributary gets periodic time-slice capacity, which is the definition of time-division multiplexing. Wavelength, frequency, and space division describe parallel optical paths, not the ODU digital hierarchy.
Why the others are wrong
A. Wavelength division multiplexing: combines whole wavelengths on fiber, not ODUk signals inside one wavelength.
C. Frequency division multiplexing: radio/frequency-channel concept, not the OTN tributary-slot mechanism.
D. Space division multiplexing: uses separate fibers or cores, unrelated to ODUk hierarchy.
H31-311 exam tip
See ODU multiplexing and answer TDM; see wavelengths on fiber and answer WDM.
7A PAUSE frame uses a reserved multicast address and is not forwarded by a bridge or switch. In this way, the PAUSE frame does not generate any additional information.
FALSE
TRUE
Answer: B
The short version
B — TRUE; PAUSE uses a non-forwarded reserved multicast address. Bridges consume it for hop-by-hop flow control instead of flooding it.
Key concepts in this question
PAUSE frame: IEEE 802.3x flow-control frame with EtherType 0x8808.
Reserved multicast 01-80-C2-00-00-01: link-constrained address bridges do not forward.
Hop-by-hop backpressure: PAUSE only pauses the directly connected transmitter for a quanta time.
Why B is correct
PAUSE is sent to the reserved MAC 01-80-C2-00-00-01 precisely so bridges and switches act on it locally rather than forwarding it. Because it is terminated on the link, it pauses the neighbor without generating forwarded traffic elsewhere, making the statement true.
Why the others are wrong
A. FALSE: would mean PAUSE is flooded like ordinary multicast, contradicting 802.3x link-constrained behavior.
H31-311 exam tip
PAUSE equals 01-80-C2-00-00-01 plus do-not-forward; that pairing always means TRUE.
8Which of the following alarms does not trigger the AU-AIS alarm?
R-LOS
MS-AIS
B2-EXC
AU-LOP
Answer: D
The short version
D — AU-LOP itself does not source AU-AIS. AU-AIS is inserted upstream on LOS/LOF/MS-AIS/MS excess-BER defects.
Key concepts in this question
AU-AIS: administrative-unit alarm indication signal sent downstream to suppress cascade alarms.
AU-LOP: loss of AU pointer, itself a detected defect rather than a trigger for AU-AIS.
Why D is correct
When the regenerator or multiplex section fails (R-LOS, MS-AIS, excessive B2 errors), the downstream AU is filled with AIS so lower layers raise only the consequential alarm. AU-LOP means the pointer processor already lost the AU; it is the reported defect at that AU and generates HP-side consequences such as RDI, not a fresh AU-AIS in the same AU.
Why the others are wrong
A. R-LOS: loss of signal causes downstream MS-AIS and AU-AIS insertion.
B. MS-AIS: multiplex-section AIS directly forces AU-AIS downstream.
C. B2-EXC: excessive section BIP errors trigger MS-AIS/AU-AIS protection actions.
H31-311 exam tip
Upstream section failures point down as AU-AIS; AU-LOP is the odd one out.
9Which of the following is the maximum number of OSN 3500 subracks that can be installed in a 2.2 m cabinet?
3
2
1
4
Answer: B
The short version
B — two OSN 3500 subracks fit in a 2.2 m cabinet. The 3500 subrack height plus power, heat, and cabling space limits it to two.
Key concepts in this question
OSN 3500 subrack: large-capacity SDH/SONET platform with tall chassis.
2.2 m ETS cabinet: standard Huawei cabinet with limited usable height.
Installation planning: fiber, power, and ventilation clearances reduce usable slots.
Why B is correct
Huawei installation practice allows two OSN 3500 subracks in one 2.2 m cabinet. One subrack wastes cabinet capacity while three or four exceed the available height once power distribution, cable routing, and heat dissipation are included, so the banked maximum of 2 is the practical and tested value.
Why the others are wrong
A. 3: exceeds usable 2.2 m space for two tall 3500 chassis with accessories.
C. 1: understates the cabinet capacity; two do fit.
D. 4: only possible with smaller platforms such as OSN 2500, not the 3500.
H31-311 exam tip
Associate OSN 3500 plus 2.2 m cabinet with the number 2.
10When synchronization is performed on the U2000, the NE data that is the same as that on the U2000 is not uploaded to the U2000, and the data that exists on the U2000 but not on the NE is deleted.
FALSE
TRUE
Answer: A
The short version
A — FALSE; the description of U2000 synchronization is inaccurate. Synchronization reconciles NMS and NE data rather than behaving exactly as stated.
Key concepts in this question
U2000 synchronization: uploads and compares NE configuration against the NMS database.
Upload vs delete: sync updates the NMS view from the NE; identical entries are still compared, not silently skipped.
Exact-wording traps: TRUE/FALSE items fail on any inaccurate clause.
Why A is correct
U2000 synchronization polls the NE, compares each entry with the NMS database, and updates mismatches so the NMS matches NE reality. The stem's absolute claim that identical data is never uploaded and that every NMS-only entry is simply deleted misstates that compare-and-update process, so the statement as worded is judged FALSE per the banked key.
Why the others are wrong
B. TRUE: would require the stem to describe the compare, upload, and delete behavior exactly, which it does not.
H31-311 exam tip
On U2000 sync wording, treat absolute never/always claims as FALSE unless exact.