Solution

The IP packet has:

• Header size = 20 bytes
• Payload = 4000 bytes
• MTU = 1500 bytes

The maximum payload that can fit in one fragment is:

$1500 - 20 = 1480$ bytes.

Since IPv4 fragment offsets are measured in units of 8 bytes, 1480 bytes is valid because $1480/8=185$.

Therefore the 4000-byte payload is divided as:

• Fragment 1: 1480 bytes payload
• Fragment 2: 1480 bytes payload
• Fragment 3: 1040 bytes payload

The fragment offsets are:

• Fragment 1: $0/8=0$
• Fragment 2: $1480/8=185$
• Fragment 3: $(1480+1480)/8=370$

Fragment	Payload	Total Length	Fragment Offset	MF bit
1	1480 bytes	1500 bytes	0	1
2	1480 bytes	1500 bytes	185	1
3	1040 bytes	1060 bytes	370	0

The MF (More Fragments) bit is 1 for the first two fragments and 0 for the final fragment.

Thus, the original packet is transmitted as three IPv4 fragments with offsets 0, 185, and 370.