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Designing IPV4 addressing scheme with subnetting

pjw17279
Level 1
Level 1

I'm currently working through some online cisco labs and have came to the subnetting section and have hit a total wall. Basically I don't know where to begin with subnetting! 

 

I have previously completed a subnetting section where I only had to deal with one or two subnets, but that was a while ago and I've not done subnetting in a long time and I'm not sure where to begin on a problem like this..

 

If anybody could point me in the direction of any handy beginners tutorials or resources that would be greatly appreciated! I've got some documentation open right but I'm having trouble understanding how to apply it to the scale of the lab I'm trying.

 

I've included an image of the problem of the lab I'm attempting, if anybody could walk me through it then that would be greatly appreciated!

 

 

 

2 Replies 2

balaji.bandi
Hall of Fame
Hall of Fame

Keith does for you :

 

https://www.youtube.com/watch?v=UCoVs1Ri1IA

 

BB

***** Rate All Helpful Responses *****

How to Ask The Cisco Community for Help

Start learning cybersecurity with CBT Nuggets. https://courses.cbt.gg/security In this video, Keith Barker covers the basics of IPv4 subnetting. This comprehensive introduction traverses quite a bit of material, including the different IP classes, network masks, and various other aspects of IPv4 ...

Giuseppe Larosa
Hall of Fame
Hall of Fame

Hello @pjw17279 ,

you need to reason in binary knowing that an IPv4 address is made of 32 bits.

You need to imagine the address divided in three portions :  

the network boundary   m bits

the subnet boundary    p bits

the host boundary       n bits

 

in your case m= 16 as the address block actually  a supernet is 195.59.0.0/16

 

p+n+m = 32

p+n= 16

 

now in a subnet with n bits for the host portion there is place for 2^n -2 hosts .

So for example

/ 24 =   255.255.255.0  --->  254 hosts

/25 = 255.255.255.128   ---> 126 hosts

/26 = 255.255.255.192 --->    62 hosts

/27 = 255.255.255.224 --->  30 hosts

 

Looking at your requirements we see that site A LAN segments require two /24 subnets   ( 150 hosts)

Site B LAN segments require two /25 subnets (100 hosts)

Site C LAN segment require two /27

 

So you can use starting from the lowest value in address block ( other approaches are possible)

 

195.59.0.0/ 24   Site A LAN 1

195.59.1.0/24 Site  A LAN 2

now we take the next /24 subnet and we divide it in two parts

195.59.2.0/25   Site B LAN 1

195.59.2.128/25 Site B LAN 2

Now we take next /24 subnet to use for site C and we divide it into 8 sub -subnets and we use the first two

195.59.3.0/27    Site C LAN 1

195.59.3.32/27 Site C LAN 2

 

remember subnetting means moving the subnet boundary to the right. Further subnetting move again and more to the right the subnet boundary reducing the n bits the host portion of the address.

 

Hope to help

Giuseppe

 

 

 

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