Which two teaming methods should be used to ensure both…

An NSX environment requires physical NIC redundancy for all dvPortGroups when connecting hosts to the
physical network. There are two 10Gb NIC’s per host. Which two teaming methods should be used to ensure
both links are utilized simultaneously? (Choose two.)

An NSX environment requires physical NIC redundancy for all dvPortGroups when connecting hosts to the
physical network. There are two 10Gb NIC’s per host. Which two teaming methods should be used to ensure
both links are utilized simultaneously? (Choose two.)

A.
Virtual Port Channel

B.
LACP Port-Channel

C.
Static Port-Channel

D.
Explicit Failover Order



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Blue Man

Blue Man

Virtual Port Channel is a Cisco Technology and not sure if this is what Vmware is looking for. Besides, Static Port-Channel also provides simultaneous link usage.

RC

RC

B & D are the right answer

Maple

Maple

D’s policy concept is “only Redundancy”. this quiz ask us which policy is for redundancy and high utilization.

if we use D, we can only half of bandwidth (only single port) at same time.
so I think D is incorrect answer.

Ahmed

Ahmed

i think c is wrong as it is called “static EtherChannel”

D!

D!

I would say B and C (even though it should be “Static-Etherchannel” according to VMware documentation): https://kb.vmware.com/s/article/2006129

Quote:
“Only Static mode EtherChannel is supported with Standard vSwitches. LACP is supported only with vSphere Distributed Switches in vSphere 5.1 or with the Cisco Nexus 1000V.”

Virtual Port-channel is when you create a port-channel with different host, and that is a CISCO technology: https://www.cisco.com/c/en/us/products/collateral/switches/nexus-7000-series-switches/white-paper-c11-737022.html

Quote:
“In 2010, Cisco introduced virtual-port-channel (vPC) technology to overcome the limitations of Spanning Tree Protocol. vPC eliminates the spanning-tree blocked ports, provides active-active uplink from the access switches to the aggregation routers, and makes full use of the available bandwidth”

D is wrong as explained before this comment.