Introduction to Networking

### Line Coding

Question: A datagram of 3000 bytes (20 bytes of IP header + 2980 bytes IP payload) reached the router and must be forwarded to link with MTU (maximum transmission unit) of 500 bytes. How many fragments will be granted? Also, write MF, offset, and total length values for all.

## Numerical Solution

After the fragmentation of the datagram into fragments (packets), each fragment will carry the same header size, which is the same as the original datagram. So, in this question, 20 bytes will be the compulsory part of each fragment.

As the limit of a link is 500 bytes to transfer, in which 20 bytes is the header size. So, 480 bytes of payload can transfer in each fragment.

### Total Number of Fragments

The total number of fragments can be calculated by considering a ceiling value of the following formula.

So, Total fragment = 2980 / 480 = 7

Descriptive diagram in given below

### Total Length

Total length is the combination of all fragmented payloads. Adding the payload of each fragment will always equal the size of the original payload. As

Total length of payload  = 480+480+480+480+480+480+100 = 2980.

Note: Header length will remain the same 20 bytes at the receiving end. So, 20+2980 bytes = 3000 bytes. It was the original datagram of 3000 bytes.

### More Fragments (MF)

MF value may be 0 or 1.

• When MF bit value is 0 then It tells to the receiver that the current datagram-fragment is the last fragment and no more segment will appear of same datagram.
• When MF bit value is 1 then it tells more fragments are still to come after this fragment.

### Fragment Offset

It uses a scaling factor of 8. Fragment offset can be calculated by using the following formula

Fragment offset for a given fragmented IP datagram = Number of data (payload) bytes ahead of it / 8

• At fragment F1, no byte has already been transferred. So fragment offset = 0/8 B = 0
• At fragment F2, 480 bytes of F1 are already transferred. So fragment offset = 480/8 B = 60
• At fragment F3, 480 bytes of F1 and F2 are already transferred. So fragment offset = 480*2/8 B =120
• At fragment, F4, 480 bytes of F1, F2, and F3 are already transferred. So fragment offset = 480*3/8 B =180
• At fragment F5, 480 bytes of F1,F2,F3 and F4 is already transferred. So fragment offset = 480*4/8 B =240
• At fragment F6, 480 bytes of F1,F2,F3,F4 and F5 is already transferred. So fragment offset = 480*5/8 B =300
• At fragment F7, 480 bytes of F1,F2,F3,F4,F5 and F6 is already transferred. So fragment offset = 480*6/8 B =360