Let's say, that I have a following simple L2 circuit(Martini) configuration:

root@PE4> show configuration protocols l2circuit
neighbor {
    interface ge-0/0/5.777 {
        virtual-circuit-id 777;


This results with the following entry in the l2circuit.0 table:

root@PE4> show route table l2circuit.0

l2circuit.0: 1 destinations, 1 routes (1 active, 0 holddown, 0 hidden)
+ = Active Route, - = Last Active, * = Both
                   *[L2CKT/7] 00:09:46, metric2 2
                    > to via ge-0/0/3.0, label-switched-path PE4->PE2


What table/database does Junos use to associate the FEC with the RSVP-signaled PE4->PE2 transport LSP?


The neighbor statement implies the remote PE (which should be reachable via MPLS), so in short, inet.3.

I'm leaving the config out, but you'd see Remote PE in the output of show l2circuit connections as in your case.

jhead@PE1> show l2circuit connections
Layer-2 Circuit Connections:


Legend for interface status
Up -- operational
Dn -- down
    Interface                 Type  St     Time last up          # Up trans
    ge-3/0/2.0(vc 1)          rmt   Up     Feb  5 18:30:57 2021           1
      Remote PE:, Negotiated control-word: Yes (Null)
      Incoming label: 300432, Outgoing label: 299824
      Negotiated PW status TLV: No
      Local interface: ge-3/0/2.0, Status: Up, Encapsulation: ETHERNET
        Description: to-PE2
      Flow Label Transmit: No, Flow Label Receive: No

Finally, the inet.3 RSVP route toward the remote PE.

jhead@PE1> show route table inet.3

inet.3: 1253 destinations, 1560 routes (192 active, 0 holddown, 1239 hidden)
+ = Active Route, - = Last Active, * = Both      *[RSVP/7/1] 1w1d 17:11:35, metric 5
                 > to via ae1.0, label-switched-path PE1-to-PE2
                 [LDP/9] 14:28:32, metric 1
                 > to via ae1.0, label-switched-path PE1-to-PE2
  • Thanks! I suspected that it's the inet.3 table, but wasn't sure as I had understanding that inet.3 is used strictly for resolving the BGP protocol next-hop addresses. – Martin Feb 7 at 12:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.