Question 48

Host and prefix requirements:

• The organization needs at least 1500 addresses.
• The smallest power of 2 greater than or equal to 1500 is $2^{11} = 2048$.
• A block of 2048 addresses requires 11 host bits, leaving $32 - 11 = 21$ network prefix bits (a /21 CIDR block).

ISP address space:

• Available space: 202.61.0.0/17.
• A /17 mask means the first 17 bits are fixed: 202.61.00000000.0 to 202.61.01111111.255 (third octet ranges from 0 to 127).

CIDR boundary alignment for /21:

• In a /21 block, the third octet has $21 - 16 = 5$ network bits, so the block size in the third octet is $2^{8 - 5} = 2^3 = 8$.
• Hence, the third octet of any valid /21 network address must be an integer multiple of 8, and must lie in the range $[0, 127]$ to fall inside the ISP's /17 space.

Evaluate candidates:

• I. 202.61.84.0/21: 84 is not divisible by 8 ($84 = 10 \times 8 + 4$). Invalid network address.
• II. 202.61.104.0/21: $104 = 13 \times 8$ (divisible by 8) and $104 \le 127$. Valid candidate.
• III. 202.61.64.0/21: $64 = 8 \times 8$ (divisible by 8) and $64 \le 127$. Valid candidate.
• IV. 202.61.144.0/21: $144 > 127$, which is outside the ISP's allocated /17 address space.
• Thus, candidates II and III only are potential candidates.