GATE CSE 2020 | Question: 38

Question

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. $202.61.144.0/21$

• A. $\text{I}$ and $\text{II}$ only
• B. $\text{II}$ and $\text{III}$  only
• C. $\text{III}$ and $\text{IV}$ only
• D. $\text{I}$ and $\text{IV}$ only

Comments

How did u get that it requires 4 subnet bits??

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1. Formula NetworkID +SubnetID=Subnet Mask (Continuous run of 1's)

  - 1500 IP addresses = Express in powers of 2 then take log(value), we get 11 bits

  - No of prefix bits for subnet is 32-11=21 bits as in subnetting bits are taken from network portion,

  -Using 17 + SubnetID=21(Subnet mask)

  -We get 21-17=4 bits taken to form subnets in addition to network bits

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 What they want to say by this line :- ?

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. 

Answer 1

(B) II and III only 

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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 $1)$

Comments

I did not understand why the $(iv)$ options were wrong here. please explain in a simple way.

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@Hira Thakur option(iv) is not even part of the network

given the ISP has 202.61.0.0/17=202.61.00000000.0 

option (iv) is 202.61.144.0/21 = 202.61.10010000.0 

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@Sachin Mittal 1 sir i thnk
(I) is not a valid address space

Reason:-
Let’s first see, from routing table,how to decide which interface an incoming packet will be routed to?
- we basically do AND ops between the subnet mask of the interfaces and the incoming packet's IP,
the result of this AND is the network id of the interface when the packet should be routed.
Below is a an example of this

now, let's go back to our qstn
(I)'s address space =$202.61.84.0/21$
                                 =$202.61.01010 |$ $100.0$ (with /21 subnet mask)

Now, as per the qstn any IP inside the address space $202.61.0.0/17$ can be a valid IP(cuz the organisation request for IP to the ISP,which  serves the requests from the available IP address space 202.61.0.0/17 ),so any incoming packet having IP addresses inside this address space should be routed to the organisation .

But see in the below photo, a packet came with ip $202.61.01010$ $111.0/21$ [this is inside the 202.61.0.0/17 address space], but it cannot be routed to (I)'s address space!!, cuz after the AND ops, the network id won't match(i think this happens due to this 1 [ 202.61.01010 |100.0 ]  inside (I)'s address space)

Thereby we can say no valid packet can ever be routed to (I) address space

NB:- I also set that 6th 1 in 3rd octet to 0, i.e 202.61.01010 | 000.0 (with /21 subnet mask),which i denoted as interface B, now and any valid packet can route to (I) address space

Answer 2

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.

Comments

dutta18 According to your logic in option (D) last $11$ bits are $0$. it is also matched.

$202.61.144.0\implies 202.61.10010000.00000000$

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at 17^th bit should be 0 it is given /17 that’s why see after 17^th bit