GATE CS 2025 Set 2 Q13 — Fragmentation and IP Addressing, IPv4, CIDR

To find the number of fragments and the size of the last fragment, we follow these steps:

1. Determine the total IP payload:

   The UDP segment consists of the UDP payload and the UDP header (8 bytes):

   $$\text{Total IP Payload} = 7488 \text{ bytes (UDP payload)} + 8 \text{ bytes (UDP header)} = 7496 \text{ bytes}$$
2. Determine the maximum IP payload per fragment:

   The maximum transmission unit (MTU) of Ethernet is 1500 bytes. Excluding the 20-byte IPv4 header (without options):

   $$\text{Max IP Payload} = 1500 - 20 = 1480 \text{ bytes}$$

   Since 1480 is a multiple of 8, it is a valid fragment size.

3. Calculate the number of fragments: $$\text{Number of fragments} = \left\lceil \frac{7496}{1480} \right\rceil = \lceil 5.065 \rceil = 6 \text{ fragments}$$
4. Calculate the size of the last fragment:

   The first 5 fragments each carry 1480 bytes of IP payload. The remaining payload for the $6^{\text{th}}$ fragment is:

   $$\text{Last fragment payload} = 7496 - (5 \times 1480) = 96 \text{ bytes}$$

   Including the 20-byte IPv4 header, the total size of the last fragment is:

   $$\text{Last fragment size} = 96 + 20 = 116 \text{ bytes}$$

Thus, there are 6 fragments and the last fragment size is 116 bytes.