GATE 2020 CS Q.38: solution and answer

Official answer: BSource: Official answer key, GATE 2020 CS (IIT Delhi)

1. Hosts needed: $1500\le 2^{11}=2048$ and $2^{10}=1024<1500$, so 11 host bits are needed and the prefix length is $32-11=21$. Every candidate is a /21, which matches (a /21 holds 2048 addresses, 2046 of them usable, enough for 1500 computers).

2. The ISP's pool 202.61.0.0/17 has 17 network bits: the first two octets (16 bits) plus the top bit of the third octet, which is fixed at 0. The other 7 bits of the third octet and all of the fourth octet are free, so the pool spans 202.61.0.0 to 202.61.127.255, i.e. third octet $0$ to $127$.

3. To keep the ISP's routing table small by route aggregation, the allotted block must lie inside the ISP's /17, so that the ISP still advertises one /17 entry for it. The block must also be a valid /21 network address: a /21 leaves 3 bits of the third octet for hosts, so the third octet must be a multiple of $2^3=8$.

4. I. 202.61.84.0/21: $84=10\times 8+4$, not a multiple of 8, so 202.61.84.0 is not a /21 network address (the /21 containing it starts at 202.61.80.0). Not valid.

5. II. 202.61.104.0/21: $104=13\times 8$ and $104<128$. Valid and inside the pool. It covers 202.61.104.0 to 202.61.111.255.

6. III. 202.61.64.0/21: $64=8\times 8$ and $64<128$. Valid and inside the pool. It covers 202.61.64.0 to 202.61.71.255.

7. IV. 202.61.144.0/21: $144=18\times 8$ is a valid /21, but $144\ge 128$, so it lies outside 202.61.0.0/17 and is not part of the ISP's pool. Not available.

8. Only II and III qualify: option B.