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Question: Given: Class B IP address 169.202.0.0.

Network when divided into 4 sub networks, each one requires 16,000 IP addresses. They need to be assigned sequentially without gaps.

Find the sub networks.

Here is my solution:

Total IP addresses in 1 subnet are 16000 = 2^14 – 384 Bits required for subnet: 4 = 2^2 = 2 bits

Bits Required for Host id: 16000 = 214 =14 bits

Subnet mask: 255.255.192.0

Subnet 1: 169.202.0.0 To 169.202.62.127 = (62x256) +128=16000

Subnet 2: 169.202.62.128 To 169.202.125.255

Subnet 3: 169.202.126.0 To 169.202.188.127

However in the solution booklet something else was mentioned( solution given below) which confused me. Please help/guide me.

What do them mean by wasting few IP addresses and how are those particular ones selected. Moreover how did we get the solution given in the image below.

enter image description hereSubnet 4: 169.202.10 111100.1000000 = 169.202.188.128 To 169.202.11 111011.1111111 = 169.202.251.255

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    $\begingroup$ Can you clarify what about it confused you? What specifically is your question? What parts of it do you understand, and what is the first part you don't understand? Please edit your post to make your question more specific. $\endgroup$ – D.W. Dec 20 '17 at 17:13
  • $\begingroup$ The title you have chosen is not well suited to representing your question. Please take some time to improve it; we have collected some advice here. Thank you! $\endgroup$ – Raphael Dec 21 '17 at 10:28
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After subnetting the the network Id for any subnet will consists of two parts:

  1. First part consisting of 1st 16 bits(from left side)
  2. Subnet identifier which will be of 2 bits.

All addresses within a particular subnet must have the same network Id part.

Now consider subnet 2(created by u)and consider the 3rd octet from left.

1st address - 00111110

last address - 01111101

You can see in the 1st address subnet Id is 00 but in the last address subnet id is 01. This cannot be accepted.

So to remove this problem we will give each subnet all the addresses that can be made with the remaining 14 bits. This is the right way.

The advantage of this scheme is that it allows u to expand ur network since there are always some unused addresses.

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  • $\begingroup$ What do them mean by wasting few IP addresses and how are those particular ones selected. Moreover how did we get the solution given in the image below. $\endgroup$ – Kamya Joshi Dec 22 '17 at 6:50
  • $\begingroup$ A subnet in this case needs only 16000 addresses. But If I use the last 14 bits for host id the subnet will get $2^{14}=16384 $ addresses. Those 384 addresses will get wasted. $\endgroup$ – Kishan Kumar Dec 22 '17 at 7:28
  • $\begingroup$ Talking about the solution : For the 1st subnet keep the subnet id bits 00. For the 1st address of this network i,e the network id address keep the host id bits to 0. The complete address will be 169.202.0.0. Now for the last address of this subnet set all bits in the host id part to 1. The address will be 169.202.00111111.11111111 (The last two bytes are in binary). Similarly for the 2nd subnet change the subnet Id from 00 to 01. For addresses use the same procedure as for subnet 1. $\endgroup$ – Kishan Kumar Dec 22 '17 at 7:28
  • $\begingroup$ @KamyaJoshi U should accept the answer whenever u are satisfied with it. This shows that the answer was helpful and thus helps future users.This is one of the main guidelines for using this community. $\endgroup$ – Kishan Kumar Jan 3 '18 at 14:20

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