GATE CS 2020 Set 1 — Question 48

To determine the potential candidates, we must satisfy the host requirement and ensure the block is a valid sub-allocation from the ISP's range that allows for route aggregation (i.e., a single CIDR block). The organization needs 1500 IP addresses. The smallest power of 2 that can accommodate 1500 hosts is $2^{11} = 2048$. A block of 2048 addresses corresponds to a $/21$ subnet mask ($32 - 11 = 21$). The ISP's range is $202.61.0.0/17$. In a $/17$ network, the first 17 bits are fixed. The 17th bit is the first bit of the 3rd octet. For $202.61.0.0$, this bit is $0$. Thus, all sub-allocated blocks must have the first bit of their 3rd octet as $0$. This limits the 3rd octet to the decimal range $[0, 127]$. For an address to be a valid start of a $/21$ block, the last $32 - 21 = 11$ bits must be zero. This means the last 3 bits of the 3rd octet must be zero. In decimal, the 3rd octet must be a multiple of $2^3 = 8$.

• I. $202.61.84.0/21$: The 3rd octet is $84$. $84 = 10 \times 8 + 4$. Since it is not a multiple of 8, it is not a valid $/21$ network address.
• II. $202.61.104.0/21$: The 3rd octet is $104$. $104 = 13 \times 8$. This is a valid $/21$ network address and falls within the ISP's range ($104 < 128$).
• III. $202.61.64.0/21$: The 3rd octet is $64$. $64 = 8 \times 8$. This is a valid $/21$ network address and falls within the ISP's range ($64 < 128$).
• IV. $202.61.144.0/21$: The 3rd octet is $144$. While $144$ is a multiple of 8, it is outside the ISP's range ($144 > 127$).

4. Conclusion:
Only candidates II and III are valid $/21$ blocks that can be carved out of the ISP's $/17$ space. Assigning either of these as a single block minimizes the routing entries to one.Therefore, the correct option is (B).