Consider an IP packet with a length of 4500 bytes that includes a 20-byte IPv4…
2021
Consider an IP packet with a length of 4500 bytes that includes a 20-byte IPv4 header and a 40-byte TCP header. The packet is forwarded to an IPv4 router that supports a Maximum Transmission Unit (MTU) of 600 bytes. Assume that the length of the IP header in all the outgoing fragments of this packet is 20 bytes. Assume that the fragmentation offset value stored in the first fragment is 0. The fragmentation offset value stored in the third fragment is
- A.
144
- B.
255
- C.
102
- D.
124
- E.
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Correct answer: A
✅ Given
Total IP packet size = 4500 bytes
IP header = 20 bytes
So, data (payload) = 4500 − 20 = 4480 bytes
(MTU applies to total packet size including IP header)
🔹 Step 1: Maximum data per fragment
MTU = 600 bytes
Each fragment has a 20-byte IP header
So max data per fragment:
600−20=580 bytes
But fragmentation requires data size to be a multiple of 8 bytes (because of Fragment Offset units).
Nearest multiple of 8 ≤ 580:
576 bytes
🔹 Step 2: Fragment sizes
Each fragment carries 576 bytes of data (except possibly last).
🔹 Step 3: Fragment offsets
Offset is in units of 8 bytes, not bytes.
🔸 First fragment
Offset = 0
🔸 Second fragment
Data before it = 576 bytes
576/8 = 72
🔸 Third fragment
Data before it = 576 + 576 = 1152 bytes
1152/8 = 144
✅ Final Answer
A. 144