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"Uneven" Network Configuration with HA/VMotion

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we run a fairly boring and simple scheme where all access layer switches (A,B,C,D) are layer 3 up to the core, local access layer VLANs are not available everywhere.  In order to VM all of the devices we have each of our 4, 5.0 esxi hosts with a trunk cable to each switch:

 

1. 1 NIC - trunk to switch "A"

2. 1 NIC - trunk to switch "B"

3. 1 NIC - trunk to switch "C"

4. 1 NIC - trunk to switch "D"

 

all 4 servers are in a cluster with HA turned on (DRS off).  Everything is running smoothly and has been for about 9 months in this configuraiton.  We've now added switch "E."  The problem is that we cannot add any more network ports to our 4 esxi hosts.  What we were thinking of doing was adding to more ESXI hosts and cabling those to switch "E" and leaving one of the other four switches off.  This would leave us with something like this:

 

1. ESXi trunked to switch "A" "B" "C" and "D"

2. ESXi trunked to switch "A" "B" "C" and "D"

3. ESXi trunked to switch "A" "B" "C" and "D"

4. ESXi trunked to switch "A" "B" "C" and "D"

5. ESXi trunked to switch "A" "B" "C" and "E" (trunked to E instead of D)

6. ESXi trunked to switch "A" "B" "C" and "E" (trunked to E instead of D)

 

i don't have many computers on switch "E" that need to be virtualized, so I don't want to waste a lot of money on vmware licenses and hardware resources.  However, these machines are begging to be virtualized and I cannot move them from switch "E" or change their IP.

 

Is it possible to have all 6 ESXI hosts in the same cluster?  even though failover between ESXi hosts cannot be symmetric becuase 2 of them do not have the same network as the other 4 (or another way, the other 4 do not have the same as the 2).  Is it possible to have failover control based on the network available on a ESXI host?  In other words if I have a VM on ESXI 1 that needs switch "D" it cannot fail to ESXI 5,6 becuase they don't have switch "D."  Yet, a VM on ESXI 5,6 can fail to 1,2,3 or 4 of it is on switch "A" "B" or "C" ?

 

Thanks,

Damon


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