VMware 3V0-12.26 Exam Dumps - Actual Questions Answers

Term

Correct Definition

Must Have

Non-negotiable.

Should Have

Important but are not necessary to success.

Could Have

Desirable but not necessary for success.

Won ' t Have

Will not be delivered as part of this scope of work.

The MoSCoW prioritization method classifies requirements according to how critical they are to the success and scope of a solution.

A Must Have requirement is non-negotiable . Failure to satisfy it means the solution does not meet an essential business or technical requirement.

A Should Have requirement is important and normally expected, but the project can still succeed without it if necessary. Such requirements may be deferred when schedule, cost, or resource constraints demand prioritization.

A Could Have requirement is desirable but has lower business impact. These capabilities are implemented when time and resources permit and can be removed from the current delivery with comparatively little effect on overall success.

A Won ' t Have requirement is explicitly excluded from the current scope. It is not necessarily permanently rejected; it may be reconsidered in a later phase or release.

For VCF architecture work, MoSCoW helps convert workshop findings into prioritized requirements and prevents optional functionality from competing with mandatory architectural objectives.

Study Guide References/Topics: Requirements Gathering; MoSCoW Prioritization; Conceptual Design; Business and Technical Requirements; Scope Management.

B and E are correct because the customer has supplied VM counts and service-level objectives, but has not supplied the actual workload resource profile required to translate those VM counts into physical compute and storage capacity.

E establishes the average VM footprint—principally vCPU, RAM, and storage . Without these values, an architect cannot calculate aggregate capacity for either the initial 2,000-VM state or the projected 8,000-VM state. VMware capacity-management metrics explicitly account for allocated vCPU, memory, and VM disk consumption when evaluating cluster capacity.

B is equally important because provisioned resources are not the same as actual demand. Average utilization determines realistic CPU and memory consumption, consolidation ratios, oversubscription assumptions, and required growth headroom. Broadcom ' s guidance states that VM CPU and memory should be sized against real workload demand measured over representative periods , while VCF Operations uses utilization/demand information for capacity planning.

A may affect individual application compatibility but is not the primary aggregate sizing assumption. C is a configuration/design decision rather than a workload-sizing input. D is relevant to achieving the two-hour RPO, but the specific backup product does not establish the core compute, memory, and storage capacity required for 2,000–8,000 VMs.

The official VCF Architect exam guide explicitly includes capacity planning, scalability, and documenting assumptions as core architect competencies. ( Broadcom Docs )

Study Guide References/Topics: Capacity Planning; Workload Profiling; Assumptions and Constraints; VM Resource Sizing; Scalability; VCF Physical Design.

RAID-6 with FTT=2 is the appropriate vSAN storage policy for Site A because it satisfies both requirements: tolerance of two simultaneous host failures and maximum storage efficiency.

In vSAN, Failures to Tolerate (FTT)=2 means that a storage object remains accessible after two concurrent failures within the applicable fault-domain model. Both RAID-1 FTT=2 and RAID-6 FTT=2 can provide this level of protection, but their capacity overhead differs significantly.

RAID-1 FTT=2 uses three mirrored copies of the data, resulting in approximately 3× raw-capacity consumption for the protected object. By comparison, RAID-6 FTT=2 uses erasure coding with dual parity, providing two-failure tolerance with substantially better usable-capacity efficiency.

Site A contains 16 ESX hosts , providing more than enough hosts/fault domains to support RAID-6 FTT=2 placement requirements. This makes erasure coding the preferred choice where the design explicitly prioritizes storage efficiency while maintaining dual-failure protection.

A and D provide only FTT=1 and therefore cannot tolerate two simultaneous host failures. B provides the required resilience but consumes considerably more storage capacity than RAID-6.

Study Guide References/Topics: vSAN Storage Policy-Based Management; Failures to Tolerate; RAID-1 Mirroring; RAID-5/RAID-6 Erasure Coding; Capacity Efficiency; vSAN Cluster Sizing and Fault Domains.