Switch – Can default vlan 1 traffic goes to native vlan 2

switchtrunkvlan

I have 2 switches running on 802.1Q trunking mode. On
both switches the default vlan is 1.

  1. If I were to configure the native vlan to 2, can vlan 1 traffic still
    travel from switch to switch?
  2. After I configure the native vlan, do i need to allow it in the trunk?
    like switchport trunk allowed vlan 2?
  3. Does that means all untagged traffic like vlan 1 connected pc's will
    go to vlan 2 under native vlan in trunk? Do i need to configure vlan
    2 access mode?
  4. Does it happen like this: In first switch, all untagged frames like
    vlan 1 will goes under native vlan 2 in the trunk. Then they
    will travel across the trunk. After reach switch 2, these untagged
    frame belongs to vlan 2. These frame can talk to those untagged vlan
    2 (vlan 2 access port). Am i right?

Best Answer

If I were to configure the native vlan to 2, can vlan 1 traffic still travel from switch to switch?

Yes, now VLAN 1 is tagged. VLAN 2 is untagged.

After I configure the native vlan, do i need to allow it in the trunk? like switchport trunk allowed vlan 2?

By default, all VLANs are allowed on the trunk. If it was allowed before, it still is.

Does that means all untagged traffic like vlan 1 connected pc's will go to vlan 2 under native vlan in trunk? Do i need to configure vlan 2 access mode?

VLAN 1 is now tagged on the trunk port. The access ports work the same.

Does it happen like this: In first switch, all untagged frames like vlan 1 will goes under native vlan 2 in the trunk. Then they will travel across the trunk. After reach switch 2, these untagged frame belongs to vlan 2. These frame can talk to those untagged vlan 2 (vlan 2 access port). Am i right?

No, you're a little confused here. Frames are only tagged on trunk ports. Since you have set the native VLAN to 2, all other VLANs will be tagged. VLAN 1 is still VLAN 1, whether it's tagged or not.

I think you're confused by the term 'default.' All that means is that VLAN 1 exists on the switch(es) without you having to explicitly create it.