For use on Problem 4

4" (20%) Consider these 3 IP frames (we know they are IP v4 frames from the 08 00 in byte 12 and byte 13) (a) (7%) How many ICMP frames are there? (it can be 0, 1, 2, or 3). Explain your reason. (b) (796) How many ICMP echo request frames are there? Explain your reason. (c) (6%) How many private IP addresses in the block 192.168.0.0/16 are there as the source IP address of these 3 frames? Explain! Analysis of Wireshark captured IP suite frames. (20%) In the following, there are 3 data frames captured by Wireshark: Frame A 0020 b1 94 08 00 4d 3e 00 01 00 1d 61 62 63 64 65 66 _. . .M>.. ..abcdef 030 67 68 69 6a 6b 6c 6d 6e 6f 70 71 72 73 74 75 76 ghijklmn opqrstuv 0040 77 61 62 63 64 65 66 67 68 69 wabcdefg hi Frame B 020 01 0a 0f 0a 0b 35 72 79 52 8a 00 00 00 00 a0 02 .. . . .Sry R....... 0040 ec 32 00 00 00 00 01 03 03 02 Frame C boe ffffffffffff ea, al-51 a3 22 af 98 00 4500 Q E. 0010 00 c9 a7 a6 40 00 40 11 d0 d4 ce a8 01 01 ff ff .. . .4.6 0020 ff ff 04 13 1c ff 00 b5 77 57 4b 41 4e 4e 4f 55 0030 25 4e 00 00 00 0 0 04 al 51 a3 22 af 57 4e 44 N...... .Q.".WND 0040 52 33 38 30 30 00 00 00 00 00 00 00 00 57 4e 44 R3800... .. . . WND 050 52 33 38 30 30 00 00 00 00 00 00 00 00 00 00 00 R3800... .. . . . . 0060 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00.. . . . . . . .. . . . . . . 0070 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00........ 0080 00 31 2e 30 32 2e 30 32 00 00 00 00 00 00 00 00 1.02.02 . . . :0090 00 00 00 00 00 00 00 01 00 00 00 01 00 00 00 02_.. .. .. . .. . . . 00a0 00 00 00 02 00 00 00 00 00 00 00 00 00 00 00 00 00b0 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00.. . . .. . . . . . . . . .. 00c0 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 _. . . . .. 00de 00 00 00 00 00 00 00