GATE 2024 CS1 Q.65: solution and answer

Official answer: 6Source: Official answer key, GATE 2024 CS1 (IISc Bengaluru)

1. The datagram carries $1420-20=1400$ bytes of data. Every fragment has its own 20-byte header, and the data in every fragment except the last must be a multiple of 8 bytes (the fragment offset counts in units of 8 bytes).

2. First link, MTU $542$: each fragment has room for $542-20=522$ bytes of data. The largest multiple of 8 not above 522 is $\lfloor 522/8\rfloor \times 8=65\times 8=520$ bytes.

3. On the first link, $1400=520+520+360$. The router therefore receives 3 fragments, with data sizes $520$, $520$ and $360$ bytes.

4. Second link, MTU $360$: each fragment has room for $360-20=340$ bytes of data. The largest multiple of 8 not above 340 is $\lfloor 340/8\rfloor \times 8=42\times 8=336$ bytes.

5. A router only fragments; it never reassembles. Each incoming fragment is split on its own. $520=336+184$ gives 2 fragments, and $184=23\times 8$ is a multiple of 8 even though it is not the last fragment overall. The second $520$ also gives 2 fragments. $360=336+24$ gives 2 fragments, the last of which carries the final 24 bytes of the datagram.

6. The receiver gets $2+2+2=6$ fragments, with data sizes $336,184,336,184,336,24$. These sum to $1400$ bytes, which matches the original data.