The reconstructed picture labels 1 as the OTU transponder, 2 as the OMU multiplexer, 3 as the ODU demultiplexer, and 4 as the OSC board, matching standard Huawei transmission roles.
Why the others are wrong
OTU to multiplexing: multiplexing is OMU, not transponding.
OMU to demultiplexing: demultiplexing is ODU, the reverse direction.
OSC to traffic conversion: OSC carries only supervision, never client traffic.
H31-341 exam tip
OTU converts, OMU combines, ODU splits, OSC supervises.
2Unlike ROADM, FOADM has some disadvantages. What are the disadvantages of FOAM? (Choose all that apply.)
Site visits are required for node capacity expansion and service adjustment.
Wavelengths are allocated based on planning. Once they have been allocated, it is difficult to rellocate them, making network flexibility poor.
During node capacity expansion, the power budget must be adjusted manually onsite. It cannot be adjusted remotely or automatically.
Wavelength services cannot be provisioned quickly.
Answer: A, B, C, D
The short version
A and B and C and D — fixed OADM is inflexible on every listed point. It needs site visits, fixed planning, manual power budgets, and slow provisioning.
Key concepts in this question
FOADM: fixed optical add/drop multiplexer with hard-wired wavelength add/drop.
ROADM: reconfigurable OADM switchable remotely via WSS.
Flexibility gap: planning, expansion, power balance, and provisioning speed.
Why A and B and C and D are correct
FOADM ports are tied to planned wavelengths, so capacity or service changes need onsite hardware work (A). Wavelengths cannot be easily reallocated once fixed, hurting flexibility (B). Adding channels changes optical power, which must be rebalanced manually onsite rather than by remote automation (C). Together these make fast wavelength provisioning impossible compared with ROADM (D).
Why the others are wrong
No option is excludable: each of A, B, C, and D describes the same fixed-planning limitation from a different angle, so omitting any one would understate the FOADM disadvantage.
H31-341 exam tip
Fixed equals site visit plus manual plus slow; if all four appear, take all four.
3If the mask of an IP subnet is 27 bits, how many hosts does the subnet have?
32
30
14
16
Answer: B
The short version
B — a /27 subnet holds 30 usable hosts. Five host bits give 32 addresses minus network and broadcast.
Key concepts in this question
/27 mask: 27 network bits leave 5 host bits.
Usable hosts: 2^n minus 2 for network and broadcast addresses.
Powers of two: /27 block size is 32.
Why B is correct
Host bits equal 32 minus 27, which is 5. Two to the fifth is 32 total addresses. Subtracting the network and broadcast addresses leaves 30 assignable host addresses, so B is the standard subnet-math result.
Why the others are wrong
A. 32: total addresses before subtracting network and broadcast.
C. 14: usable hosts of a /28, not a /27.
D. 16: total addresses of a /28, not usable hosts of a /27.
H31-341 exam tip
Memorize /27 = 32 total and 30 usable; subtract two every time.
4FILL BLANKIf a trail trace identifier (TTI) in the section monitoring (SM) layer is abnormal, the _____ alarm will be reported. (Enter only letters and special characters.)
SM_TTIM
B2_SD
LOS
LOF
Answer: A
The short version
A — SM_TTIM is reported when the SM-layer TTI is abnormal.
Key concepts in this question
TTI: trail trace identifier carried in overhead to verify connectivity.
SM layer: OTN section-monitoring overhead checking fiber-section continuity.
TTI mismatch: received TTI differs from expected, raising a TIM alarm.
Why A is correct
Option A SM_TTIM fills the blank: Huawei reports SM_TTIM when section-monitoring TTI is abnormal.
Why the others are wrong
B B2_SD is wrong because B2 degradation is a BER defect, not a TTI mismatch.
C LOS is wrong because loss of signal means no light, not a trace-identifier error.
D LOF is wrong because loss of frame is framing failure, not TTI abnormality.
H31-341 exam tip
SM TTI abnormal equals SM_TTIM.
5OTUCn and ODUCn implement the functions of the digital section and channel layer.
TRUE
FALSE
Answer: A
The short version
A — TRUE; OTUCn is the section layer and ODUCn the channel layer. Beyond-100G OTN keeps the same digital-layer split at higher rates.
Key concepts in this question
OTUCn: optical transport unit Cn providing section-layer framing and FEC.
ODUCn: optical data unit Cn providing channel/path-layer mapping and monitoring.
Digital vs optical layers: OTU/ODU are digital layers above OCh/OMS/OTS.
Why A is correct
G.709 beyond-100G retains OTUCn for transport-section functions such as framing, scrambling, and FEC, while ODUCn carries client mapping, tandem monitoring, and path PM. The statement correctly assigns section duties to OTUCn and channel duties to ODUCn, so TRUE is banked.
Why the others are wrong
B. FALSE: would deny the standard OTU-section and ODU-channel division carried into OTUCn/ODUCn.
H31-341 exam tip
OTU always pairs with section, ODU always pairs with channel, at any rate.
6Which of the following statements about the TMB1AST2 board are true?
The TM1 optical port must be connected to the TMI1 optical port.
The TMB1AST2 board supports the IEEE 1588v2 clock processing and OTDR functions.
When the board is working with a DFIU03/DFIU04 board, the OSC optical port must be configured with an optical module that supports the 1511 nm wavelength.
When the board is working with a DSFIU01/DSFIU02 board, the TM1/RM1 optical port must be configured with an optical module that supports the 1511 nm wavelength, and the TM2/RM2 optical port must be configured with an optical module that supports the 1491 nm wavelength.
Answer: A, B, C, D
The short version
A and B and C and D — all four TMB1AST2 statements are banked true. Port pairing, 1588v2/OTDR support, and the two FIU wavelength rules all hold.
IEEE 1588v2 and OTDR: clock processing and fiber-measurement functions.
FIU pairing: OSC and tributary optics must match the supervisory wavelengths of the peer FIU.
Why A and B and C and D are correct
The banked key treats TM1-to-TMI1 pairing as required (A), credits the board with 1588v2 clock plus OTDR capability (B), and requires 1511 nm OSC optics with DFIU03/04 (C) alongside the split 1511/1491 nm tributary optics with DSFIU01/02 (D). Taken together they describe correct installation and feature practice for this board family.
Why the others are wrong
No option is excludable under the banked key: dropping any of A, B, C, or D would contradict the specified pairing, feature, or wavelength requirement.
H31-341 exam tip
For board-feature all-of-the-above items, verify each wavelength clause separately.
7OOS overheads are optical-layer overheads. Which of the following statements about OOS overheads is false?
They include OCh overheads.
They include OTS overheads.
They include OTN overheads.
They include OMS overheads.
Answer: C
The short version
C — OTN overheads are not part of OOS. OOS covers the optical layers OCh, OMS, and OTS.
Key concepts in this question
OOS: optical-layer overheads outside the digital OTN frame.
OCh / OMS / OTS: channel, multiplex-section, and transmission-section optical overheads.
OTN overheads: digital SM/PM/TCM bytes inside OTUk/ODUk.
Why C is correct
The question asks which statement is false. OOS comprises the optical supervisory overheads for the channel, multiplex section, and transmission section. Claiming OOS includes OTN digital overheads confuses the optical supervisory channel with the OTUk/ODUk frame overhead, so C is the false statement and the correct choice.
Why the others are wrong
A. OCh overheads: genuinely part of the optical channel layer under OOS.
B. OTS overheads: genuinely part of the optical transmission section under OOS.
D. OMS overheads: genuinely part of the optical multiplex section under OOS.
H31-341 exam tip
OOS equals three optical layers only: OCh plus OMS plus OTS.
8Which of the following statements about the OptiXtrans E6600/OptiX OSN 1800 system architecture Is true?
The OptiXtrans E6608T/E6608/E6616 and OptiX OSN 1800 II TP/OSN 1800 II Pro/OSN 1800 V Pro use the L0+L1+L2 system architecture.
The OptiXtrans E6608T/E6616 and OptiX OSN 1800 II TP/OSN 1800 V Pro use an L0+L1+L2 system architecture.
The OptiXtrans E6616 and OptiX OSN 1800 V Pro use an L0+L1+L2 system architecture.
The OptiXtrans E6608T/E6608 and OptiX OSN 1800 II TP/OSN 1800 II Pro use an L0+L1+L2 system architecture.
Answer: B
The short version
B — E6608T/E6616 plus OSN 1800 II TP/V Pro use L0+L1+L2. That exact model grouping is the banked architecture statement.
Key concepts in this question
L0+L1+L2: joint optical, OTN/TDM, and packet switching architecture.
E6600 series: E6608T, E6608, and E6616 chassis variants.
OSN 1800 series: II TP, II Pro, and V Pro packet-optical variants.
Why B is correct
Huawei groups the E6608T and E6616 with the OSN 1800 II TP and V Pro as the L0+L1+L2 convergent platforms. The other groupings either add the E6608 or II Pro where the banked documentation does not credit the full three-layer architecture, so only B matches the tested combination.
Why the others are wrong
A. E6608T/E6608/E6616 plus II TP/II Pro/V Pro: over-includes E6608 and II Pro beyond the banked set.
C. E6616 plus V Pro only: under-includes E6608T and II TP that also qualify.
D. E6608T/E6608 plus II TP/II Pro: swaps in E6608 and II Pro while omitting E6616 and V Pro.
H31-341 exam tip
Memorize the exact four-model string: E6608T, E6616, II TP, V Pro.
9Which of the following statements about trail deletion on the NMS is false?
If a path is deleted from the NE layer, the WDM paths at both the network layer and NE side are deleted.
Deleting a path at either the network or NE layer interrupts services.
If a path is deleted from the network layer, only the WDM path at the network layer is deleted on the NMS.
Deleting a path at the network layer does not interrupt services.
Answer: B
The short version
B — network-or-NE deletion both interrupting service is the false claim. Network-layer-only deletion is logical and non-disruptive.
Service impact: only NE-layer changes touch live traffic.
Why B is correct
The question asks for the false statement. Deleting from the NE layer removes device cross-connections at both views and interrupts service, while deleting from the network layer removes only the NMS logical record without touching hardware. Because network-layer deletion does not interrupt traffic, the blanket claim in B that either layer interrupts service is false.
Why the others are wrong
A. NE-layer delete removes both views: true consequence of touching hardware config.
C. Network-layer delete removes only network record: true description of logical-only deletion.
D. Network-layer delete does not interrupt: true because hardware cross-connections remain.
H31-341 exam tip
NE layer equals hardware impact; network layer equals paper deletion.
10A multiframe alignment signal is used for multiframe alignment. When some OTUk and ODUk overhead signals need to be transmitted across multiple frames, the MFAS must be used to implement multi-frame alignment.
TRUE
FALSE
Answer: A
The short version
A — TRUE; MFAS provides multiframe alignment for multi-frame overhead. OTUk/ODUk overheads spanning frames need the sequence number.
Key concepts in this question
MFAS: multiframe alignment signal byte counting OTUk/ODUk frames.
Multi-frame overhead: TTI, TCM, and other signals spread across consecutive frames.
Alignment need: receiver must know position within the multiframe cycle.
Why A is correct
Several OTUk and ODUk overhead fields are too large for one frame and are transmitted over 16 or 256 frames. MFAS carries the incrementing frame count so the receiver can reassemble them in order. The statement correctly describes that purpose, so TRUE is banked.
Why the others are wrong
B. FALSE: would deny the documented MFAS role in aligning multi-frame OTN overhead.
H31-341 exam tip
See multiframe or MFAS and think frame counter for spread-out overhead.