Question

Consider a network that uses Ethernet and IPv4. Assume that IPv4 headers do not use any options field. Each Ethernet frame can carry a maximum of 1500 bytes in its data field. A UDP segment is transmitted. The payload (data) in the UDP segment is 7488 bytes. Which ONE of the following choices has the CORRECT total number of fragments transmitted and the size of the last fragment including IPv4 header?

• A. 5 fragments, 1488 bytes
• B. 6 fragments, 88 bytes
• C. 6 fragments, 108 bytes
• D. 6 fragments, 116 bytes

Hint:

Remember: Max IP data per Ethernet frame = 1500 – 20 = 1480 bytes, and fragment size must be a multiple of 8 bytes.

Answer 1

The Correct Option is D

Step 1: Compute total size of the IP datagram before fragmentation.
Given:
- UDP payload = 7488 bytes
- UDP header = 8 bytes
Total data carried by IP = \( 7488 + 8 = 7496 \) bytes.
Since IPv4 has no options, IP header = 20 bytes. So total original IP packet size = \( 7496 + 20 = 7516 \) bytes.
Step 2: Determine maximum data per fragment.
Ethernet MTU = 1500 bytes (includes IP header + data).
Maximum data per IP fragment = \[ 1500 - 20 = 1480 \text{ bytes} \] (Note: 1480 is divisible by 8, which is required for IP fragmentation.)
Step 3: Compute number of fragments.
Total data to fragment = 7496 bytes. Number of full fragments: \[ \left\lfloor \frac{7496}{1480} \right\rfloor = 5 \] Remaining bytes for last fragment: \[ 7496 - (5 \times 1480) = 96 \text{ bytes} \] So total fragments = \( 5 + 1 = \boxed{6} \).
Step 4: Compute size of last fragment.
Last fragment payload = 96 bytes. Add IP header (20 bytes): \[ 96 + 20 = \boxed{116 \text{ bytes}} \] Step 5: Analyze the options.
(A) 5 fragments, 1488 bytes: Incorrect — needs 6 fragments.
(B) 6 fragments, 88 bytes: Incorrect last size.
(C) 6 fragments, 108 bytes: Incorrect last size.
(D) 6 fragments, 116 bytes: Correct.
Step 6: Conclusion.
The correct answer is 6 fragments, 116 bytes.