webrtc - Which ICE candidate am I using and why? -

[questions in bold below]

i have setup kurento media server 5.1.3 in datacenter behind firewall running os ubuntu 14.04. has 2 network cards:

• 222.222.222.222 (eth0 - private ip)
• 111.111.111.111 (eth1 - public ip)

attached below sdp (setremotedescription) when browser connected kurento media server

type: answer, sdp: v=0 o=- 5487318114793304426 0 in ip4 0.0.0.0 s=kurento media server c=in ip4 0.0.0.0 t=0 0 a=group:bundle audio video m=audio 59068 rtp/savpf 111 0 c=in ip4 111.111.111.111 a=rtpmap:111 opus/48000/2 a=rtpmap:0 pcmu/8000 a=sendrecv a=rtcp:59068 in ip4 111.111.111.111 a=rtcp-mux a=ssrc:669011897 cname:user39019747@host-6e83e4c2 a=extmap:3 http://www.webrtc.org/experiments/rtp-hdrext/abs-send-time a=mid:audio b=as:20 a=ice-ufrag:ymdk a=ice-pwd:lylifk5ueqzpwm91ddj37e a=fingerprint:sha-256 ff:0f:81:8c:41:4e:b4:b6:c6:d8:36:f3:d6:5f:09:fd:5f:af:13:b3:9d:fc:12:66:ac:f3:56:d6:5b:0a:73:5d a=candidate:1 1 udp 2013266431 111.111.111.111 55239 typ host a=candidate:2 1 udp 2013266431 222.222.222.222 59068 typ host a=candidate:4 1 udp 1677721855 111.111.111.111 59068 typ srflx raddr 222.222.222.222 rport 59068 m=video 59068 rtp/savpf 100 c=in ip4 111.111.111.111 b=as:100 a=rtpmap:100 vp8/90000 a=sendrecv a=rtcp-fb:100 ccm fir a=rtcp-fb:100 nack a=rtcp-fb:100 nack pli a=rtcp-fb:100 goog-remb a=rtcp:59068 in ip4 111.111.111.111 a=rtcp-mux a=ssrc:138242433 cname:user39019747@host-6e83e4c2 a=extmap:3 http://www.webrtc.org/experiments/rtp-hdrext/abs-send-time a=mid:video a=ice-ufrag:ymdk a=ice-pwd:lylifk5ueqzpwm91ddj37e a=fingerprint:sha-256 ff:0f:81:8c:41:4e:b4:b6:c6:d8:36:f3:d6:5f:09:fd:5f:af:13:b3:9d:fc:12:66:ac:f3:56:d6:5b:0a:73:5d a=candidate:1 1 udp 2013266431 111.111.111.111 55239 typ host a=candidate:2 1 udp 2013266431 222.222.222.222 59068 typ host a=candidate:4 1 udp 1677721855 111.111.111.111 59068 typ srflx raddr 222.222.222.222 rport 59068 

i not sure, seems using following candidate:

a=candidate:4 1 udp 1677721855 111.111.111.111 59068 typ srflx raddr 222.222.222.222 rport 59068 

am right?

but given fact ip 222.222.222.222 internal ip why appears in ice candidate?

why doesn't pick "a=candidate:1 1 udp 2013266431 111.111.111.111 55239 typ host"? since ip publicly accessible.

when using tools "nload" check traffic, eth0 doesn't have traffic, , can noticed eth1 got lot of traffic (video , audio stream)

what "a=candidate:4 1 udp 1677721855 111.111.111.111 59068 typ srflx raddr 222.222.222.222 rport 59068" means?

the webrtc client tries ice candidates until finds 1 works. there priority queue ice candidates of potential addresses added to. webrtc tries these 1 @ time, , once finds 1 works, uses candidate media. webrtc not know address public , private, tries candidates until 1 succeeds or of them fail.

ice designed create connections in presence of nat issues.