Question:medium
An ISP having an address block 202.16.0.0/15 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?
An ISP having an address block 202.16.0.0/15 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?
Show Hint
6000 addresses need a /19 block since \(2^{13}=8192 \geq 6000\); check that each candidate's third octet is a multiple of 32 (so the block is properly aligned) and that it falls inside 202.16.0.0 to 202.17.255.255.
Updated On: Jul 22, 2026
- 202.16.0.0/19
- 202.17.64.0/19
- 202.16.32.0/19
- 202.17.24.0/19
Show Solution
The Correct Option is A, B, C
Solution and Explanation
Step 1: Work directly in binary. The parent block 202.16.0.0/15 in binary is 11001010.0001000x.xxxxxxxx.xxxxxxxx, so the second octet's last bit is free (16 or 17) and the third and fourth octets are completely free.
Step 2: 6000 addresses rounds up to the next power of two, 8192, which needs 13 host bits, so the assigned block must be a /19, meaning bits 1-19 are network and bits 20-32 are host (13 host bits, matching 8192).
Step 3: For a /19 network address, writing the third octet in binary as 3 network bits followed by 5 host bits, those 5 host bits must all be 0, which is the same as saying the third octet's value is a multiple of 32.
(A) 0 = 00000 000 in binary form (multiple of 32), and 0/19 boundary aligned. Valid.
(B) 64 = 010 00000, bottom 5 bits are 0, aligned. Valid.
(C) 32 = 001 00000, bottom 5 bits are 0, aligned. Valid.
(D) 24 = 000 11000, bottom 5 bits are 11000 (not zero), not aligned to a /19 boundary. Invalid.
(A) 0 = 00000 000 in binary form (multiple of 32), and 0/19 boundary aligned. Valid.
(B) 64 = 010 00000, bottom 5 bits are 0, aligned. Valid.
(C) 32 = 001 00000, bottom 5 bits are 0, aligned. Valid.
(D) 24 = 000 11000, bottom 5 bits are 11000 (not zero), not aligned to a /19 boundary. Invalid.
Step 4: All of A, B, C also have second octet 16 or 17 and are therefore within the ISP's 202.16.0.0/15 pool, so they are legally assignable; D fails the alignment check.
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