Week 9

Assignment 9

RTP and RTCP

I selected a single captured packet to study the topic of packet sizes. The parsed capture has broken the packet into seperate protocols and given an indication of the header and packet length. Here we can see that the ethernet header is 14 bytes long. We can also determine that the remaining packet, the payload, is 284 bytes long. Thus the entire packet is 298 bytes long deterimined by adding the header to the payload.

[37.00824.00072](0.00045.00977)   CAPTURE-00008 298/298
Number of providers is  4
PROVIDER:ether USER:ip <14+284>
Available bytes:298
src:00-14-4f-80-e5-b8 dst:01-00-5e-01-02-03 type: 2048

The following is a diagram of an ethernet header.

Destination MAC Address
6 bytes

Source MAC Adress
6 bytes

Ethertype/Length
2 bytes

The 284 bytes of payload from the ethernet packet containing 20 bytes of IP header and 264 bytes of data.

PROVIDER:ip USER:udp <20+264>
Available bytes:284
4    src:192.168.1.6 dst:224.1.2.3   hlen:20 len:264 tos:0 id:22371 ttl:1 protocol:17 CheckSum:bdbb

The IP packet header is shown below.

The 264 bytes that comprised the data section of the IP packet actually contain a UDP packet header and data. We can see that the header length for UDP is 8 bytes. The UDP packet header is made of of the source port, the destination port, the length and the checksum.


PROVIDER:udp USER:rtp <8+256>
Available bytes:264
src:42050 dst:42050 length=264 cksum=396c

The remaining 256 bytes of the packet consists of an RTP packet.

PROVIDER:rtp USER:none <0+-1>
  version:2
  pMarked:0
  pType:5 nCsrc:0
  seqNbr:27582
  tStamp:960
  SSRC:2444843577
00000000   ff d8 02 00 0a a9 31 6b 21 d1 89 b3 8d 59 1b 82  ......1k!....Y..
00000010   05 99 19 83 b9 92 19 1b bc 9b 91 05 90 3c 69 03  .............<i.
00000020   91 2d 89 24 93 2b 81 b9 b3 2b 59 1d 03 18 1c 07  .-.$.+...+Y.....
00000030   a1 11 20 a5 1a 2a b8 83 52 40 11 39 35 8b 92 99  .. ..*..R@.95...
00000040   a1 41 70 19 83 36 98 12 39 30 91 9b 90 b1 9d a3  .Ap..6..90......
00000050   14 f9 1b 99 a2 9b 10 0f 9b 80 d1 0a 99 19 d9 da  ................
00000060   8c 03 0b 21 b4 00 c0 10 19 b1 bb ba 19 3a 3d aa  ...!.........:=.
00000070   20 b4 b4 09 40 e9 10 1a c2 b6 89 10 8c 32 a4 09   ...@........2..
00000080   37 11 0c 89 80 a4 b1 10 1b 51 14 3a 83 1d 09 50  7........Q.:...P
00000090   a0 12 c0 1d b2 40 c2 10 84 04 10 5a 3a 91 b9 99  .....@.....Z:...
000000a0   91 30 a3 4b a3 c9 19 37 01 03 03 94 58 15 08 01  .0.K...7....X...
000000b0   39 80 89 a1 b1 33 09 3b 60 81 23 a9 d1 da 9a bc  9....3.;`.#.....
000000c0   c2 ac 19 49 80 9b aa da d9 80 b0 be 01 b9 ae 09  ...I............
000000d0   a8 0b 09 60 12 1a 2e 83 a1 22 b2 bd 10 8b 21 59  ...`....."....!Y
000000e0   11 83 40 41 f1 99 3a ab a9 51 89 3a 15 09 94 13  ..@A..:..Q.:....
000000f0   b6 09 39 93