what is true about traffic from a virtual machine connected to a port group on a vNetwork Standard Switch with

Assuming that VLANs are not configured, what is true about traffic from a virtual machine connected to a port group on a vNetwork Standard Switch with NO uplinks?

A.
The virtual switch will drop the packets of no uplink is present.

B.
Virtual machines on any vNetwork Standard Switch on the same ESXi host can receive the traffic.

C.
Virtual machines in any port group on the virtual switch can receive the traffic.

D.
Only virtual machines in the same port group on the virtual switch can receive the traffic.

Explanation:

Page 13 from vsphere-esxi-vcenter-server-50-networking-guide.pdf

6 Comments on “what is true about traffic from a virtual machine connected to a port group on a vNetwork Standard Switch with

  1. Magnus says:

    Hi,
    the correct answer should be “Virtual machines in any port group not using VLAN ID on the virtual switch can receive the traffic”

    thanks.
    /Magnus

    1. Daniel says:

      No its not. A vSwitch with multiple uplinks and multiple portgroups (no VLANS) any tracffic is able to be received by anyone else on that switch (on the same host)

  2. mr_tienvu says:

    Ref: Page 15 from vsphere-esxi-vcenter-server-50-networking-guide.pdf

    You can create a new standard switch with or without Ethernet adapters.
    If you create a standard switch without physical network adapters, all traffic on that switch is confined to that switch. No other hosts on the physical network or virtual machines on other standard switches can send or receive traffic over this standard switch. You might create a standard switch without physical network adapters if you want a group of virtual machines to be able to communicate with each other, but not with other hosts or with virtual machines outside the group.

  3. costin says:

    Tested this in lab: 2 machines, same switch with no uplinks, different portgroups each(no Vlans): they can communicate with eachother. So correct is C indeed.


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