 | 1 | initial version |
Long time ago , as u thk , tcp will retrans all packets even if only one lost, but later designer thk this way is tool slow and waste lots of time, so they design some options to heal this prob, like sack h1-->h2 send 1460 bytes [num 1 ] h1-->h2 send 1460 bytes [num 2 ] h1-->h2 send 1460 bytes [num 3 ] h1-->h2 send 1460 bytes [num 4 ] (lost !!) h2-->h1 send ack=3(that means i receive num 3 packet and packets which number lower than 3) h1-->h2 send 1460 bytes [num 5 ] h1-->h2 send 1460 bytes [num 6 ] h2-->h1 send ack=6(i receive num 6 and num 5), sack=5-6,dup_ack=4(i didnt receive num 4, pls resent) h1-->h2 send 1460 bytes [num 7 ] h2-->h1 send ack=7(i receive num 7) sack=5-7,dup_ack=4(i didnt receive num 4, pls resent) h1-->h2 send 1460 bytes [num 8 ] h2-->h1 send ack=8(i receive num 8), sack=5-8,dup_ack=4(i didnt receive num 4, pls resent) h1-->h2 send 1460 bytes [num 4 ] h1-->h2 send 1460 bytes [num 9] h1-->h2 send 1460 bytes [num 10 ] continue ......
| | 2 | No.2 Revision |
Long time ago , as u thk , tcp will retrans all packets even if only one lost, but later designer thk this way is tool slow and waste lots of time, so they design some options to heal this prob, like sack
sack
h1-->h2 send 1460 bytes [num 1 ]
h1-->h2 send 1460 bytes [num 2 ]
h1-->h2 send 1460 bytes [num 3 ]
h1-->h2 send 1460 bytes [num 4 ] (lost !!)
h2-->h1 send ack=3(that means i receive num 3 packet and packets which number lower than 3)
h1-->h2 send 1460 bytes [num 5 ]
h1-->h2 send 1460 bytes [num 6 ]
h2-->h1 send ack=6(i receive num 6 and num 5), sack=5-6,dup_ack=4(i didnt receive num 4, pls resent)
h1-->h2 send 1460 bytes [num 7 ]
h2-->h1 send ack=7(i receive num 7) sack=5-7,dup_ack=4(i didnt receive num 4, pls resent)
h1-->h2 send 1460 bytes [num 8 ]
h2-->h1 send ack=8(i receive num 8), sack=5-8,dup_ack=4(i didnt receive num 4, pls resent)
h1-->h2 send 1460 bytes [num 4 ]
h1-->h2 send 1460 bytes [num 9]
h1-->h2 send 1460 bytes [num 10 ]
continue ......