Subnetting (noob)

[mblackshore]

Hello folks in Cisco community,
I'm going threw a CCNA 200-301 course for future certification.

I'm confused regarding this course subnetting process.

Example
1. Corporation has a network for 5 stores.
2. All these stores are to have their own subnet.

Corporates network is

192.168.5.0/24

Their example of solution:

/24 = 24/8 = 3 = 255.255.255.0
5 stores = 5 bits (11100000) = 128+64+32= 224
New netmask for stores: 255.255.255.224
Use lowest bit as icrementation (11100000)= 32
1 host
1 broadcast
30 hosts.

I'm questioning if this is the right way of doing this?
My hypotetical solution:

Since there is 8 bits creating a binary value of 256.
I thought I could just take 256 / 5 = 51,2 ~ 50
256-50=206 (starts with 0 so it will be 205)
Binary conversion= 11001110
New netmask: 255.255.255.205
1 host
1 broadcast
48 hosts

Am I missing something?

[Flavio Miranda]

Hi

 It is not ok

192.168.5.0/26

192.168.5.64/26

192.168.5.128/26

192.168.5.192/26

For 50 host and /24, this what you can get

You push the mask /27,.then

192.168.5.0/27

192.168.5.32/27

192.167.5.64/27

.

.

.

[mblackshore]

Kindly expand on that.

Correct me if I'm wrong
The /26 translates to

255.255.255.192

You have

0-63 = 64 adresses (1 host, 1 broadcast, 62 hosts)

no?

[Flavio Miranda]

Correct. Either you go with 64 or 32 which is 30 hosts.

 Either /26 or /27

[mblackshore]

Ahaa, is there a reason for this?

[Flavio Miranda]

Himmm....I would say due the 8 bits.

[Kanan Huseynli · Accepted solution]

Hi,

you need to choose number of subnets which is higher or equal to 5 and is divisible by 2, which is 8.

So, number of subnets 8, you need

borrow S

bit from host part.

2^S=8 => S =3

New mask is network

bits+ S

bits (borrowed bits) which results

24+3=27 so /27

(or

255.255.255.224

in Decimal Dotted Notation). Number of hosts will be

2^H

(where H is host bits that is zero's in mask),

H=32-27=5

so all these 8 subnets have

2^H=2^5=32

hosts in it..

8 subnets each with 32 hosts (one ID, one broadcast, 30 usable). Take 5 of them use for required sites. 3 x /27 will be free, non-used.

This is symmetrical division. You can also use VLSM.

[mblackshore]

Thank you! This made sense!