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An organization requires a range of IP address to assign one to each of its $1500$ computers. The organization has approached an Internet Service Provider (ISP) for this task. The ISP uses CIDR and serves the requests from the available IP address space $202.61.0.0/17$. The ISP wants to assign an address space to the organization which will minimize the number of routing entries in the ISP’s router using route aggregation. Which of the following address spaces are potential candidates from which the ISP can allot any one of the organization?

  1. $202.61.84.0/21$
  2. $202.61.104.0/21$
  3. $202.61.64.0/21$
  4. $202.61.144.0/21$
  1. $\text{I}$ and $\text{II}$ only
  2. $\text{II}$ and $\text{III}$  only
  3. $\text{III}$ and $\text{IV}$ only
  4. $\text{I}$ and $\text{IV}$ only
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Best answer
56 votes
56 votes

(B) II and III only 


Given IP address space: $202.61.0.0/17$, $17$ bits are in network ID bits(NID) and rest will be host ID bits(HID).

$\underbrace{202.61.0}_{17 \text{ NID } bits}0000000.00000000$

In order to assign $1500$ hosts we need minimum $11$ bits

$\underbrace{202.61.0}_{17 \text{ NID }  bits}\ \underbrace{0000}_{4 \text{ SID } bits}\underbrace{000.00000000}_{11 \ \text{ HID } \ bits}$

We have $4$ subnet bits, eligible networks are those which belongs among possible 16 subnets.

If we expand the given Network bits we can see:

  • $202.61.\textbf{84}.0/21 = 202.61.0\textbf{1010}100.0$
    Not possible as all Host Bits should be zero
  • $202.61.\textbf{104}.0/21 = 202.61.0\textbf{1101}000.0$
    Possible
  • $202.61.\textbf{64}.0/21 = 202.61.0\textbf{1000}000.0$
    Possible
  • $202.61.\textbf{144}.0/21 =202.61.1\textbf{0010}000.0 $
    Not possible as $16^{th}$ bit from right (part of $\text{NID}$ is $0$ and not $1)$
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It can be done in just few seconds if you know the property that to form a supernet the first address of the block must be divisible by the number of demanded ip addresses.

Here the demand is 1500 but we need to give 2048 which is 2^11.

Now test each of the option given.

To be eligible the IP addresses must be divisible by 2^11 that is it must contain 11 zeros from last.

When you check the number of zeros from last only option B and C will be valid.
Answer:

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