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hw 2 numerical questions

actual answers

recording my actual answers to they’re easy to re-enter…​

Note
 2017-09-13 23:42:18-04:00 got a 60/65 points; yet to see what was wrong with these answers…​
Table 1. numerical questions
no. question answer

1

Find the source hardware address.

B9 AA B8 AO OF C3

2

Find the destination hardware address.

C4 BC 6E 3A 5C E7

3

What type of frame is this?

08 00

4

What fragment Flag(s) are set?

011

5

What is the fragment offset?

6145 = 1 1000 0000 0001

6

What is the TTL count?

0x83 = 131

7

What is the Header Checksum?

EB 40

8

If the header includes no options or padding, what are the first five bytes of the datagram data?

41 8E B7 C2 CB

9

Find the destination IP address.

B3 DC 88 4D

10

What class is the destination IP address?

Class B; B = 1011 → net: B3 DC ^ host: 88 4D

11

What is the network ID in the destination address?

B3 DC (see #10)

12

What is the host ID in the destination address?

88 4D (see #10)

13

Write the destination IP address in dotted decimal notation.

179.220.136.77

14

Find the source IP address.

CC 51 2D BB

15

What class is the source IP address?

Class C; C = 1100 → netid: CC 51 2D ^ host: BB

16

What is the network ID in the source address?

CC 51 2D (see #15)

17

What is the host ID in the source address?

BB (see #15)

18

Write the source IP address in dotted decimal notation.

204.81.45.187

19

Can this message be delivered directly by the source to the destination, or will it require routers to handle the message.

Will require routers, as they’re on different netids (CC 51 2D != B3 DC).

hw #2 scratch work

Warning
 below was with a ton of misunderstanding. above section are my answers, but the work is on paper.

scratch work done for hw #2 while waiting for lecture to start

Note
 ocr detected free online, for hw 2’s ip datagram
original
C4 BC 6E 3A 5C E7 B9 AA B8 AO OF C3 08 00 45 F9
00 72 E4 D2 78 01 83 CD EB 40 CC 51 2D BB B3 DC
88 4D 41 8E B7 C2 CB 95 C9 68 BD 7B D2 3F D6 DD
OF 28 20 DD 21 FO ED 32 27 F1 50 3F 89 04 C7 2A
F2 34 CC D9 34 71 35 F4 98 74 06 BA 93 32 11 E0
54 06 90 FA 85 FE DF 78 C1 B9 E6 5A 93 75 1C AD
FB 88 10 C2 AA 5F C5 7D 93 97 C1 62 94 3B 89 7F
FC 34 BB 4B 27 6D 92 6C 46 75 OA 86 40 37 4B CA
wrapped at 4 bytes
C4 BC 6E 3A
5C E7       # (1)
      B9 AA
B8 AO OF C3 # (2)
08 00       # (3)

            # (4)
      45    # (5)
         F9 # (6)
00 72       # (7)
      E4 D2 # (8)
7           # (9)
 8 01 83 CD # (10)
EB          # (11)
   40       # (12)
      CC 51 # (13)
2D BB B3 DC # (14)
88 4D 41 8E # (15)
            # (16) (17) (18)
            # (19)
B7 C2 CB 95
C9 68 BD 7B
D2 3F D6 DD
OF 28 20 DD
21 FO ED 32
27 F1 50 3F
89 04 C7 2A
F2 34 CC D9
34 71 35 F4
98 74 06 BA
93 32 11 E0
54 06 90 FA
85 FE DF 78
C1 B9 E6 5A
93 75 1C AD
FB 88 10 C2
AA 5F C5 7D
93 97 C1 62
94 3B 89 7F
FC 34 BB 4B
27 6D 92 6C
46 75 OA 86
40 37 4B CA

  1. frame: end of dst hw address

  2. frame: end of src hw address

  3. frame: end of frame type

  4. "ip datagram" ie: start of "frame data" 46-1500 bytes

  5. version & hlen

  6. type of service

  7. total length including payload

  8. identification

  9. fragment flags

  10. fragment offset

  11. ttl

  12. type

  13. header checksum

  14. src ip address

  15. dest ip address

  16. …​

  17. …​

  18. …​

  19. end of ip datagram header