GATE 2026 CS (CS1) – Question 56
An ISP having an address block 202.16.0.0/15 assigns a block of 6000 IP addresses to a client, using the classless inter-domain routing (CIDR) super-netting approach. Which of the following address blocks can be assigned by the ISP?
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Correct answer: (A) 202.16.0.0/19
Explanation
1. Number of addresses needed = 6000.
In CIDR, block sizes must be powers of 2. The smallest power of 2 greater than or equal to 6000 is $2^{13} = 8192$. (Since $2^{12} = 4096 < 6000$).
Thus, $h = 13$ host bits are required, which gives a prefix length of $32 - 13 = 19$.
2. In a $/19$ subnet, the third octet mask has $19 - 16 = 3$ network bits. The block size in the third octet is $2^{8 - 3} = 2^5 = 32$.
Hence, the third octet of any valid $/19$ subnet must be an integer multiple of 32: $\{0, 32, 64, 96, 128, 160, 192, 224\}$.
3. The ISP's allocated range `202.16.0.0/15` spans `202.16.0.0` to `202.17.255.255`.
Among the options:
- `202.16.0.0/19`: Third octet is 0 (multiple of 32), valid network address within range.
- Note: `202.16.0.0/19` is the standard initial contiguous block assigned from the base of the ISP's range.
Therefore, option (A) is correct.