An ISP having an address block 202.16.0.0/15 assigns a block of 6000 IP…

The correct answers are A, B, and C.

Step 1: Determine the Required Block Size

The client requires 6000 IP addresses.

In CIDR, address blocks must contain a power-of-2 number of addresses.

• (2^{12} = 4096) (not sufficient)

• (2^{13} = 8192) (sufficient)

Hence, the client must be allocated a block of 8192 addresses.

Since (8192 = 2^{13}), 13 bits are needed for host addresses.

Therefore, the prefix length is:

32 − 13 = 19

So, the required subnet size is /19.

Step 2: Determine the ISP's Address Range

The ISP owns the block 202.16.0.0/15.

A /15 network covers all addresses from:

202.16.0.0 to 202.17.255.255

All four options lie within this range.

Step 3: Check Valid /19 Network Boundaries

A /19 subnet mask is:

255.255.224.0

Since 224 = 11100000, the third octet increments in blocks of:

256 − 224 = 32

Therefore, valid /19 network addresses must have a third octet equal to:

0, 32, 64, 96, 128, 160, 192, or 224

Step 4: Evaluate Each Option

A. 202.16.0.0/19

• Third octet = 0

• 0 is a valid multiple of 32

Valid

B. 202.17.64.0/19

• Third octet = 64

• 64 is a valid multiple of 32

Valid

C. 202.16.32.0/19

• Third octet = 32

• 32 is a valid multiple of 32

Valid

D. 202.17.24.0/19

• Third octet = 24

• 24 is not a multiple of 32

• Therefore, it is not the starting address of a valid /19 subnet

Invalid

Hence, the correct options are:

A. 202.16.0.0/19

B. 202.17.64.0/19

C. 202.16.32.0/19

D. 202.17.24.0/19