IPv4 Fragment: Position and Payload Byte Numbers

Given: M bit = 0, HLEN = 10, Total Length = 400, Fragment Offset = 300.

Step 1: Header size

HLEN is counted in 32-bit (4-byte) words:

$$\text{Header size} = \text{HLEN} \times 4 = 10 \times 4 = 40\ \text{bytes}$$

Step 2: Payload carried by this fragment

$$\text{Payload} = \text{Total Length} - \text{Header} = 400 - 40 = 360\ \text{bytes}$$

Step 3: Position of the fragment in the original datagram

The Fragment Offset field is expressed in units of 8 bytes:

$$\text{start offset} = 300 \times 8 = 2400\ \text{bytes}$$

So the payload of this datagram begins at byte 2400 of the original datagram's data field.

Step 4: Interpretation of M = 0

M = 0 (no more fragments) means this is the last fragment of the original datagram.

Step 5: Sequence numbers of the first and last payload bytes

Counting bytes from 0 at the start of the original data field:

$$\text{first byte} = 2400$$

$$\text{last byte} = 2400 + 360 - 1 = 2759$$

Hence the original datagram's payload was $2759 + 1 = 2760$ bytes long (bytes 0 to 2759), and this fragment covers bytes 2400 to 2759. (If numbered from 1, the answers are 2401 and 2760.)

Final answer: position = last fragment starting at byte offset $300 \times 8 = 2400$; payload sequence numbers: first byte = 2400, last byte = 2759 (0-based).