The vSAN Express Storage Architecture (ESA) has specific host count requirements for its data placement schemes. For a RAID-5 erasure coding policy with the default Failures to Tolerate (FTT) of 1, a minimum of four hosts are required. This is because the 2+1 RAID-5 scheme stripes two data components and one parity component across three different hosts, with a fourth host needed for failure tolerance.

Since the administrator's cluster only has three nodes, it does not meet the minimum requirement for RAID-5. The vSAN Auto-Policy Management feature correctly identified this constraint and applied the next best compliant policy, which is RAID-1 (Mirroring), to ensure data availability. RAID-1 only requires three hosts for an FTT of 1.

A. The RAID-5 policy is supported on vSAN ESA, but the cluster in the scenario does not meet the minimum host count requirement for it.

C. Force Provisioning allows VM creation even if the policy is non-compliant, but it does not cause the policy itself to be changed from RAID-5 to RAID-1.

D. Host Rebuild Reserved is a capacity management feature and does not influence the selection of a RAID level based on the number of hosts in the cluster.

1. VMware vSAN Documentation, "vSAN Express Storage Architecture Guide" for vSAN 8.0.

Section: RAID-5/6 Erasure Coding

Content: "RAID-5 erasure coding requires a minimum of four hosts in the cluster. It can tolerate one host failure." This directly confirms that a three-host cluster cannot support a RAID-5 policy.

2. VMware vSAN Documentation, "Administering VMware vSAN" for vSAN 8.0.

Section: vSAN Default Storage Policy

Content: The documentation explains that the default policy is automatically managed based on the cluster's capabilities. "If the cluster does not have enough hosts for RAID-5/6, the default policy uses RAID-1 mirroring." This explains the behavior observed by the administrator.

3. VMware vSAN Design Guide, vSAN 8.0 U1.

Section: 3.2.1, "RAID-5/6 Erasure Coding"

Page: 16

Content: "With vSAN ESA, RAID-5 erasure coding uses a 2+1 scheme, meaning it requires a minimum of 4 hosts for FTT=1." This provides the technical reason for the four-host minimum, reinforcing why the three-node cluster is ineligible.