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
- 202.16.0.0/19
- 202.17.64.0/19
- 202.16.32.0/19
- 202.17.24.0/19
The Correct Option is A, B, C
Solution and Explanation
The ISP owns the address block \(202.16.0.0/15\). We need to find which candidate blocks of size 6000 addresses (rounded up to a valid power-of-2 block) can legally be carved out of this block using CIDR super-netting.
Step 1: Find the full range owned by the ISP.
A \(/15\) prefix has \(32-15=17\) host bits, so it spans \(2^{17}=131072\) addresses. This merges two consecutive \(/16\) blocks, so the ISP's entire range is:
\[202.16.0.0 \text{ to } 202.17.255.255\]Step 2: Find the prefix length needed for 6000 addresses.
Since \(2^{12}=4096 < 6000\) and \(2^{13}=8192 \geq 6000\), the client needs 13 host bits, i.e. a \(/19\) block (since \(32-13=19\)), which gives 8192 addresses (more than the requested 6000, so it satisfies the requirement).
Step 3: Check each candidate block for two conditions.
(a) It must lie completely inside \(202.16.0.0 - 202.17.255.255\).
(b) Its network address must fall on a valid \(/19\) boundary. Since each \(/19\) block covers 8192 addresses \(= 32\) units of the third octet, valid starting values of the third octet are \(0, 32, 64, 96, 128, 160, 192, 224\).
Option A: \(202.16.0.0/19\) - third octet is \(0\), a valid boundary, and lies inside the ISP range. Valid.
Option B: \(202.17.64.0/19\) - third octet is \(64\), a valid boundary, and \(202.17.x\) lies inside the ISP's \(/15\) range. Valid.
Option C: \(202.16.32.0/19\) - third octet is \(32\), a valid boundary, inside the ISP range. Valid.
Option D: \(202.17.24.0/19\) - third octet is \(24\), which is NOT a multiple of \(32\), so this is not a legitimate \(/19\) network address (it actually falls inside the \(202.17.0.0/19\) block, not at its own boundary). Invalid.
Final Answer:
\[\boxed{\text{A, B, C: } 202.16.0.0/19,\ 202.17.64.0/19,\ 202.16.32.0/19}\]Top GATE CS Computer Science and IT Engineering Questions
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