Blackboard

Hierarchical Addressing

Byte by Byte, bit by bit, sets and subsets

Subnetting

Supernetting

128.171.10.0/24
128.171.11.0/24

the above two networks can be combined into a single network, with network number

128.171.10.0/23

The netmask for this is

255.255.254.0

Fragmentation

First fragmentation calculation:

MTU = 1,000
n = 10,000
h = 20
n / (MTU - h) = 10,000 / 980 = 10.x
therefore, 11 fragments

Another way to see that 11 fragments are needed is that the first 10 fragments can carry at most 9,800 bytes of payload, so an 11th fragment is needed.

Router Fragmentation example

Note that a fragment offset of 400 would be encoded as 50 (since 400 / 8 = 50)

Router Fragmentation example 2

MTU is 504

The received fragment has 20 bytes of header, 980 of data

The first fragment has 20 bytes of header, 480 of data (cannot have 484 since that is not a multiple of 8)(

The second fragment has fragment offset 480, which is encoded as 60 (since 480/8 = 60)

IP Header Bytes

Used for the in-class exercise.

In-Class Exercise

Here is the original IP packet.

45 00 00 1F
90 AB 01 00
11 11 -- --
80 AB 0A 01
80 AB 0A 02
01 02 03 04
05 06 07 08
09 0A 0B

The first fragment is:

45 00 00 1C
90 AB 21 00
10 11 -- --
80 AB 0A 01
80 AB 0A 02
01 02 03 04
05 06 07 08

The three flag bits are: 0, DF, and MF. Since MF is 1, the first 4 bits are 0010, which is 2 in hexadecimal.

The second fragment is:

45 00 00 17
90 AB 01 01
10 11 -- --
80 AB 0A 01
80 AB 0A 02
09 0A 0B

Management of holes during reassembly

IP minimum MTU

The minimum MTU for carrying IP is defined to be 576 bytes. Like all MTUs, this includes the IP header, so the maximum "safe" IP payload size is 556 bytes. Safe means we can be pretty sure it won't need fragmentation.

IPv6 optional headers

IPv4 has options and a protocol field. IPv6 has the next header field, which may identify a next header or the next higher-level protocol.