Question 62

1. Total data to fragment:
   UDP user data = 8880 bytes.
   UDP header = 8 bytes.
   Total IP payload = $8880 + 8 = 8888\text{ bytes}$.

2. Maximum payload per fragment:
   MTU = 1500 bytes.
   IP header = 20 bytes.
   Maximum payload per fragment = $1500 - 20 = 1480\text{ bytes}$.
   Notice that 1480 is divisible by 8 ($1480 / 8 = 185$).

3. Fragment generation:

• Fragment 1: 1480 bytes, Offset = $0 / 8 = 0$, MF = 1
• Fragment 2: 1480 bytes, Offset = $1480 / 8 = 185$, MF = 1
• Fragment 3: 1480 bytes, Offset = $2960 / 8 = 370$, MF = 1
• Fragment 4: 1480 bytes, Offset = $4440 / 8 = 555$, MF = 1
• Fragment 5: 1480 bytes, Offset = $5920 / 8 = 740$, MF = 1
• Fragment 6: 1480 bytes, Offset = $7400 / 8 = 925$, MF = 1
  Total data in first 6 fragments = $6 \times 1480 = 8880\text{ bytes}$.

• Fragment 7 (Last fragment):
  • Payload size = $8888 - 8880 = 8\text{ bytes}$.
  • Offset = $\dfrac{8880}{8} = 1110$.
  • MF = 0.

Total fragments = 7, and offset in the last fragment = 1110 (option C).