the largest LUN that should be presented to an VMware vSphere 4 host is 2TB – 512B = 4294967295 blocks ( 512B per block ) because 4294967295 is 0xFFFFFFFF in a 32bit field
the largest LUN that should be presented to an VMware vSphere 4 host is 2TB – 512B = 4294967295 blocks ( 512B per block ) because 4294967295 is 0xFFFFFFFF in a 32bit field
Posted in Addon Software, Administration, Technical Info, VMware Info, VMware Solution, VMware vSphere 4 | No Comments »
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