Question

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?

• A. 202.16.0.0/19
• B. 202.17.64.0/19
• C. 202.16.32.0/19
• D. 202.17.24.0/19

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.

Answer 1

The Correct Option is A, B, C

Step 1: Determine the prefix length needed for 6000 addresses.
A CIDR block with prefix length /p has \(2^{32-p}\) addresses. We need the smallest power of two at least 6000. Since \(2^{12}=4096 < 6000\) and \(2^{13}=8192 \geq 6000\), the client needs a block of 8192 addresses, so the number of host bits is 13 and the prefix length is \(p = 32-13 = 19\). So a valid assignment must be a /19 block.

Step 2: Determine the span of the parent /15 block.
202.16.0.0/15 fixes the entire first octet as 202 and the top 7 bits of the second octet. Since 16 in binary is 00010000, the free last bit of the second octet lets it be 00010000 (16) or 00010001 (17). So the parent block spans 202.16.0.0 through 202.17.255.255.

Step 3: Determine the alignment rule for a /19 sub-block.
A /19 mask leaves the bottom 5 bits of the third octet plus the whole fourth octet as host bits (5+8=13 bits, matching \(2^{13}=8192\)). So a legal /19 network address must have its third octet be an exact multiple of \(2^5=32\), and its fourth octet must be 0.

Step 4: Test each option.
(A) 202.16.0.0/19: third octet 0, and \(0/32=0\) is an integer, second octet 16 is inside 16-17. Valid.
(B) 202.17.64.0/19: third octet 64, and \(64/32=2\) is an integer, second octet 17 is inside 16-17. Valid.
(C) 202.16.32.0/19: third octet 32, and \(32/32=1\) is an integer, second octet 16 is inside 16-17. Valid.
(D) 202.17.24.0/19: third octet 24, and \(24/32=0.75\) is not an integer, so 202.17.24.0 is not a legal /19 network start, it actually falls inside the 202.17.0.0/19 block. Invalid.

Step 5: Conclusion, the ISP can assign (A), (B) and (C).

\[ \boxed{\text{A, B and C}} \]