Question 65

The original IP datagram has a $20\text{-byte}$ header and $1420 - 20 = 1400\text{ bytes}$ of data payload.

1. Fragmentation on the first link ($\text{MTU}_1 = 542\text{ bytes}$):

   • Maximum data per fragment must be a multiple of $8\text{ bytes}$:
     $$\left\lfloor \frac{542 - 20}{8} \right\rfloor \times 8 = \left\lfloor \frac{522}{8} \right\rfloor \times 8 = 65 \times 8 = 520\text{ bytes}$$
   • The $1400\text{-byte}$ payload is split into 3 fragments with payload sizes:
     • Fragment 1: $520\text{ bytes}$ data ($540\text{ bytes}$ total)
     • Fragment 2: $520\text{ bytes}$ data ($540\text{ bytes}$ total)
     • Fragment 3: $1400 - 1040 = 360\text{ bytes}$ data ($380\text{ bytes}$ total)

2. Fragmentation on the second link ($\text{MTU}_2 = 360\text{ bytes}$):

   • Maximum data per fragment:
     $$\left\lfloor \frac{360 - 20}{8} \right\rfloor \times 8 = \left\lfloor \frac{340}{8} \right\rfloor \times 8 = 42 \times 8 = 336\text{ bytes}$$
   • Each of the 3 fragments arriving at the router exceeds $\text{MTU}_2 = 360\text{ bytes}$ (even Fragment 3 has total size $360 + 20 = 380\text{ bytes} > 360\text{ bytes}$!) and is further fragmented:
     • Fragment 1 ($520\text{ B}$ data) $\rightarrow 2$ fragments ($336\text{ B} + 184\text{ B}$ data)
     • Fragment 2 ($520\text{ B}$ data) $\rightarrow 2$ fragments ($336\text{ B} + 184\text{ B}$ data)
     • Fragment 3 ($360\text{ B}$ data) $\rightarrow 2$ fragments ($336\text{ B} + 24\text{ B}$ data)

Total fragments delivered at the receiver $= 2 + 2 + 2 = 6$.