An ISP having Address Block 202.16.0.0/15. Assign a Block of 6000 IP Ad dress to a client using the Classless Addressing. Which of the following Address Block can be assigned by the ISP?

The correct option is 202.16.0.0/19.

Let's understand why this address block is the correct choice by breaking down the requirements and evaluating the properties of the address blocks step-by-step.

Step 1: Determine the required size of the address block
The client requires 6000 IP addresses.
In Classless Inter-Domain Routing (CIDR) / Classless Addressing, the number of addresses in a block must be a power of 2. Specifically, a block with a prefix length of N contains 2^{32 - N} addresses.
We need to find the smallest power of 2 that can accommodate 6000 IP addresses:
2¹¹ = 2048 (not enough)
2¹² = 4096 (not enough)
2¹³ = 8192 (sufficient)
Therefore, the allocated block must contain 8192 addresses, which corresponds to:
32 - N = 13 ⇒ N = 19
So, the subnet mask prefix must be /19. All four given options have a /19 prefix, which matches this requirement.

Step 2: Understand the ISP's address block range
The ISP has the address block 202.16.0.0/15.
A /15 prefix means that the first 15 bits are fixed, representing the network prefix of the ISP.
Let's represent the second octet of the ISP IP address (16) in binary:
16 in decimal = 00010000 in binary.
Since the prefix is /15, the first 15 bits of any allocated address must match the first 15 bits of the ISP's block:
- First octet: 202 (8 bits fixed: 11001010)
- Second octet: The first 7 bits are fixed (15 - 8 = 7 bits).
The binary representation of 16 is 00010000. The first 7 bits are 0001000. The 8th bit of this octet can be either 0 or 1.
Thus, the second octet can range from 00010000 (16) to 00010001 (17) in decimal.
So, any valid IP address assigned by this ISP must start with 202.16.x.y or 202.17.x.y.

Step 3: Analyze the boundary/alignment rules in CIDR
In CIDR, a block of size 2^{32 - N} must start at an address that is divisible by the block size.
For a /19 block, the block size is 8192 addresses.
Since 8192 addresses span exactly 32 blocks of size 256 (8192 / 256 = 32), the starting address must have its third octet (the 'x' in 202.16.x.0) be a multiple of 32.
Let's check the starting third-octet values for the given options:
- For 202.16.32.0/19: The third octet is 32, which is a multiple of 32 (32 × 1). This is a valid boundary.
- For 202.16.0.0/19: The third octet is 0, which is a multiple of 32 (32 × 0). This is a valid boundary.
- For 202.17.24.0/19: The third octet is 24, which is not a multiple of 32. This is an invalid boundary.
- For 202.17.64.0/19: The third octet is 64, which is a multiple of 32 (32 × 2). This is a valid boundary.

Step 4: Identify the correct option
Among the options, the block 202.16.0.0/19 starts at a valid boundary (0 is a multiple of 32), lies fully within the ISP's address space (202.16.0.0 to 202.17.255.255), and provides the correct number of addresses (/19). Therefore, 202.16.0.0/19 is a valid block that can be assigned by the ISP.