cancel
Showing results for
Search instead for
Did you mean:
cancel
352
Views
5
Helpful
2
Replies   Beginner

## Sub netting Question

Question:

Use the IP address 172.16.35.5/20 to subnet across the following subnets, each subnet to have the following number of hosts, A – 600, B – 200, C – 100, D – 31.

For each subnet, derive, showing all working

1. The network address (NID)
2. The lowest host IP addresses
3. The highest host IP addresses
4. The subnet mask.

Please help if anyone knows how to work this out.

1 ACCEPTED SOLUTION

Accepted Solutions    Hall of Fame Master

Hello @Tkay ,

you need to find the number of hosts bits N for each subnet noting that

S+N = 32

And 2^N -2 > n. hosts in the subnet

for example to be able to support 600 hosts you would need

N=10  so that 2^10 -2 = 1024 -2 and this means the first subnet should be a /22

To accomodate 200 hosts N=8 is fine 2^8= 256 -2 = 254 > 200.  ---> /24

N=7 is fine for the 100 hosts subnets 2^8 -2 = 126 > 100       ---> 25

N=6 is fine for 31 hosts subnet as 2^6 -2 = 62                       ---> 26

First of all we find the base network of the prefix

172.16.35.5/20

We write 35 in binary the third that is the interesting byte

00100011      the boundary line is at the 20th bit

0010  || 0011

the base address of the prefix is 172.16.32.0/20     as we need to put all zeros in the host portion of the address.

From the base prefix we can find four /22 subnets

172.16.32.0/22

172.16.36.0/22

172.16.40.0/22

172.16.44.0/22

The first one 172.16.32.0/22 accomodates the needs of the first subnet

base address 172.16.32.0

first IP 172.16.32.1

lat usable IP IP 172.16.35.254

subnet mask /22 = 255.255.252.0

Then we take the 172.16.36.0/22 and we further divide it in 4 /24 to accomodate the second subnet

172.16.36.0/24

first IP 172.16.36.1

last IP 172.16.36.254

subnet mask /24 = 255.255.255.0

At this point we can go on and we take 172.16.37.0/24 and we first divide it in two two subnets to accomodate the third required subnet

172.16.37.0/25

first IP 172.16.37.1

last IP 172.16.37.126

/25 = 255.255.255.128

Finally we can take the other /25 172.16.37128/25 subnet and split it in two

172.16.37.128/26

first IP 172.16.37.129

last IP 172.16.37.191

/26 = 255.255.255.192

note other solutions are possible but the suggested method is to proceed from the main address block and then proceed to split in subnets starting to accomodate the larger subnets and then going to accomodate the subnets with less hosts.

Some binary math on the interesting byte is required. Other methods to perform calculations are possible.

Hope to help

Giuseppe

2 REPLIES 2    Hall of Fame Master

Hello @Tkay ,

you need to find the number of hosts bits N for each subnet noting that

S+N = 32

And 2^N -2 > n. hosts in the subnet

for example to be able to support 600 hosts you would need

N=10  so that 2^10 -2 = 1024 -2 and this means the first subnet should be a /22

To accomodate 200 hosts N=8 is fine 2^8= 256 -2 = 254 > 200.  ---> /24

N=7 is fine for the 100 hosts subnets 2^8 -2 = 126 > 100       ---> 25

N=6 is fine for 31 hosts subnet as 2^6 -2 = 62                       ---> 26

First of all we find the base network of the prefix

172.16.35.5/20

We write 35 in binary the third that is the interesting byte

00100011      the boundary line is at the 20th bit

0010  || 0011

the base address of the prefix is 172.16.32.0/20     as we need to put all zeros in the host portion of the address.

From the base prefix we can find four /22 subnets

172.16.32.0/22

172.16.36.0/22

172.16.40.0/22

172.16.44.0/22

The first one 172.16.32.0/22 accomodates the needs of the first subnet

base address 172.16.32.0

first IP 172.16.32.1

lat usable IP IP 172.16.35.254

subnet mask /22 = 255.255.252.0

Then we take the 172.16.36.0/22 and we further divide it in 4 /24 to accomodate the second subnet

172.16.36.0/24

first IP 172.16.36.1

last IP 172.16.36.254

subnet mask /24 = 255.255.255.0

At this point we can go on and we take 172.16.37.0/24 and we first divide it in two two subnets to accomodate the third required subnet

172.16.37.0/25

first IP 172.16.37.1

last IP 172.16.37.126

/25 = 255.255.255.128

Finally we can take the other /25 172.16.37128/25 subnet and split it in two

172.16.37.128/26

first IP 172.16.37.129

last IP 172.16.37.191

/26 = 255.255.255.192

note other solutions are possible but the suggested method is to proceed from the main address block and then proceed to split in subnets starting to accomodate the larger subnets and then going to accomodate the subnets with less hosts.

Some binary math on the interesting byte is required. Other methods to perform calculations are possible.

Hope to help

Giuseppe   Beginner

Thank you so so much. You have helped a lot. I had been asking some of my colleagues and they were not serious. Create Content to create content
This widget could not be displayed.