06-27-2023
09:10 AM
- last edited on
07-04-2023
11:37 PM
by
Translator
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?
Solved! Go to Solution.
06-27-2023
10:01 AM
- last edited on
07-04-2023
11:50 PM
by
Translator
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.
06-27-2023
09:28 AM
- last edited on
07-04-2023
11:44 PM
by
Translator
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
.
.
.
06-27-2023
09:36 AM
- last edited on
07-04-2023
11:46 PM
by
Translator
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?
06-27-2023 09:43 AM
Correct. Either you go with 64 or 32 which is 30 hosts.
Either /26 or /27
06-27-2023 09:44 AM
Ahaa, is there a reason for this?
06-27-2023 09:58 AM
Himmm....I would say due the 8 bits.
06-27-2023
10:01 AM
- last edited on
07-04-2023
11:50 PM
by
Translator
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.
06-27-2023 10:20 AM
Thank you! This made sense!
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