GATE CS 2026 Set 1 — Question 56
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Computer Networks → Network Layer: Addressing → Subnetting & CIDR
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Question
An ISP having an address block assigns a block of 6000 IP addresses to a client, using the classless internet domain routing (CIDR) super-netting approach. Which of the following address blocks can be assigned by the ISP?
Correct answer
(A) 202.16.0.0/19; (B) 202.17.64.0/19; (C) 202.16.32.0/19
Solution
1.Determine the ISP's address range:
The ISP has the block .- The prefix length is 15 bits.
- The second octet (16) in binary is .
- The first 15 bits are fixed:
- The 16th bit can be 0 or 1, meaning the second octet can be () or ().
- The range of addresses is from to .
The client needs 6000 IP addresses.
- The smallest power of 2 that can accommodate 6000 addresses is .
- A block of addresses has a prefix length of .
- Therefore, the client must be assigned a block.
For a block, the starting address must have the last 13 bits as zero.
- Last 13 bits = 8 bits of the 4th octet + 5 bits of the 3rd octet.
- This implies the 3rd octet must be a multiple of (i.e., 0, 32, 64, 96, 128, 160, 192, 224).
- The 2nd octet must be either 16 or 17 to stay within the ISP's range.
- (A) : 2nd octet is 16, 3rd octet is 0 (multiple of 32). Valid.
- (B) : 2nd octet is 17, 3rd octet is 64 (multiple of 32). Valid.
- (C) : 2nd octet is 16, 3rd octet is 32 (multiple of 32). Valid.
- (D) : 3rd octet is 24, which is not a multiple of 32. Invalid.
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