[10 marks] Refer to Frame 3. By fragmenting the carried IP datagram to two fragments with about equal length, form two corresponding Ethernet-II frames.

Please make sense of this and explain the answer.

Frame 3 0000 78 8d 17 b 0 6d 72 00 1d 0 9 41 cc 59 08 00 45 00 x. .mr. .A.Y. .E. 0010 00 30 2f f2 40 00 80 0 6 91 3a co a8 00 11 3d 81 .0 0020 3b 61 4 2a 13 88 33 55 aa 51 00 00 00 00 70 02 ;a. ..3U.0. ..p. 0030 ff ff 54 f 3 00 00 02 04 04 ec 01 0 1 04 02 First frame (IP total length of 36 0000 78 8d 17 bo 6d 72 00 1d 09 41 cc 59 08 00 45 00 x. mar A. Y. E. 0010 00 24 2f f2 50 00 80 06 81 46 c 0 a 00 11 3d 81 .0/.0. 0020 3b 61 04 2 a 13 88 33 55 aa 51 00 00 00 00 70 02 ;a. 30.0 p. 0030 In the frame above 24 in red is the total length of the 1 fragment The 1" three bits of 50 in red, are the three flags for fragmentation In the D.datauhaol2015wwceg3185Msolutions assi gnment4_ceg3185 2015w.doc Page 2 of original IP datagram. don't fragment flag is set However, in order to finish the question we ignore this flag. more is 1. 8146 in red is the new checksum Second frame (IP total length of 32) 0000 78 8d 17 b 0 6d 72 00 1d 09 41 cc 59 08 00 45 00 x. mar A. Y. E. 0010 00 20 2f f 2 40 00 80 06 91 4a co a8 00 11 3d 81 0 0020 3b 61. 0030 54 f 3 00 00 02 04 04 ec 01 01 04 02 In the frame above 20 in red is the total length of the 2 fragment The 1 t three bits of 40 in red, are the three flags for fragmentation In the original IP datagram, don't fragment flag is set However, in order to finish the question we ignore this flag. more is set to 0. 914A in red is the new checksum Frame 3 0000 78 8d 17 b 0 6d 72 00 1d 0 9 41 cc 59 08 00 45 00 x. .mr. .A.Y. .E. 0010 00 30 2f f2 40 00 80 0 6 91 3a co a8 00 11 3d 81 .0 0020 3b 61 4 2a 13 88 33 55 aa 51 00 00 00 00 70 02 ;a. ..3U.0. ..p. 0030 ff ff 54 f 3 00 00 02 04 04 ec 01 0 1 04 02 First frame (IP total length of 36 0000 78 8d 17 bo 6d 72 00 1d 09 41 cc 59 08 00 45 00 x. mar A. Y. E. 0010 00 24 2f f2 50 00 80 06 81 46 c 0 a 00 11 3d 81 .0/.0. 0020 3b 61 04 2 a 13 88 33 55 aa 51 00 00 00 00 70 02 ;a. 30.0 p. 0030 In the frame above 24 in red is the total length of the 1 fragment The 1" three bits of 50 in red, are the three flags for fragmentation In the D.datauhaol2015wwceg3185Msolutions assi gnment4_ceg3185 2015w.doc Page 2 of original IP datagram. don't fragment flag is set However, in order to finish the question we ignore this flag. more is 1. 8146 in red is the new checksum Second frame (IP total length of 32) 0000 78 8d 17 b 0 6d 72 00 1d 09 41 cc 59 08 00 45 00 x. mar A. Y. E. 0010 00 20 2f f 2 40 00 80 06 91 4a co a8 00 11 3d 81 0 0020 3b 61. 0030 54 f 3 00 00 02 04 04 ec 01 01 04 02 In the frame above 20 in red is the total length of the 2 fragment The 1 t three bits of 40 in red, are the three flags for fragmentation In the original IP datagram, don't fragment flag is set However, in order to finish the question we ignore this flag. more is set to 0. 914A in red is the new checksum