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B420M3 w/1240+PE+1280+ IOM 2208+ESXi 5.5

Alberta0101
Level 1
Level 1

Hi All,

I have B420M3 4CPU's w/1240+Port Expander+1280+ IOM 2208+ESXi 5.5

based on my understanding I should get 40GB+40Gb+40GB+40GB=160GB total bandwidth per blade, see attached diagram.

How many vNICs should be created in the service profile in order to achieve 160GB of bandwidth?

if 4 vNICs, how should vNIC/vHBA be mapped to vCONs?

 

Thanks in advance.

1 Accepted Solution

Accepted Solutions

Walter Dey
VIP Alumni
VIP Alumni

Correct: its 80G full duplex per fabric

my 2c

- between each I/O adaptor and IOM there is a pc of 4 10G links; therefore per stream you have a 10G limitiation

- consider the oversupscription Northbound, IOM - FI, and FI Northbound

- with 2 I/O adaptors, you have 2 vCON's

- if you create vnic's, use manual placement to assign them to fabric A or B, and vCON 1 and 2.

- min number of vnic's is 4 (vCON 1 & 2 Fabric A, (vCON 1 & 2 Fabric B)

Walter.

View solution in original post

1 Reply 1

Walter Dey
VIP Alumni
VIP Alumni

Correct: its 80G full duplex per fabric

my 2c

- between each I/O adaptor and IOM there is a pc of 4 10G links; therefore per stream you have a 10G limitiation

- consider the oversupscription Northbound, IOM - FI, and FI Northbound

- with 2 I/O adaptors, you have 2 vCON's

- if you create vnic's, use manual placement to assign them to fabric A or B, and vCON 1 and 2.

- min number of vnic's is 4 (vCON 1 & 2 Fabric A, (vCON 1 & 2 Fabric B)

Walter.

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