Subnetting Confusion
I was ok with subnetting untill I read this post on internet.
#-Link-Snipped-#
Here they have tried to subnet a IP 192.168.1.0/24
I have pasted that part below if you dont want to search for it.
"We will try and subnet 192.168.1.0 /24 to sooth this network which allows a total number of 254 hosts I recommend you get familiar with this table below. I never leave home without it!
Lets start with HQ with 50 hosts, using the table above:
We are borrowing 2 bits with value of 64. This is the closest we can get for 50 hosts.
HQ - 192.168.1.0 /26 Network address
HQ = 192.168.1.1 Gateway address
192.168.1.2, First usable address
192.168.1.62- Last usable address. Total address space -192.168.1.2 to 192.168.1.62
192.168.1.63 will be the broadcast address (remember to reserve the first and last address for the Network and Broadcast)
HQ Network Mask 255.255.255.192 - we got the 192 by adding the bit value from the left to the value we borrowed = 128+64=192
HQ address will look like this 192.168.1.0 /26"
Now its clear that he asked us to memorize a table. No problem with that.
Now he said "
Lets start with HQ with 50 hosts, using the table above:
We are borrowing 2 bits with value of 64. This is the closest we can get for 50 hosts.
"
My question :-
1) We are borrowing 2 bits ? WHY ???
I know according to his table its correct. But Dont we have to use the formula 2^n -2 = No: of Hosts ? By using this we get totally different number.
I know he is trying to calculate for 50 hosts and not subnets. Is the above table hold good in case if we are calculating the no: of subnets ?
I mean to ask can we use the same table to calculate the number subnets required ?
Lets consider we want to divide 199.131.20.0/24 in to 5 subnets.
So the formula 2^3 holds good. so we need to borrow 3 bits.
How do we infer the above statement using the given table above ?
#-Link-Snipped-#
Here they have tried to subnet a IP 192.168.1.0/24
I have pasted that part below if you dont want to search for it.
"We will try and subnet 192.168.1.0 /24 to sooth this network which allows a total number of 254 hosts I recommend you get familiar with this table below. I never leave home without it!
Bit Value
128
64
32
16
8
4
2
1
Bits Borrowed
1
2
3
4
5
6
7
8
Subnet mask
128
192
224
240
248
252
254
255
Subnet Prefix /CIDR
/25
/26
/27
/28
/29
/30
Lets start with HQ with 50 hosts, using the table above:
We are borrowing 2 bits with value of 64. This is the closest we can get for 50 hosts.
HQ - 192.168.1.0 /26 Network address
HQ = 192.168.1.1 Gateway address
192.168.1.2, First usable address
192.168.1.62- Last usable address. Total address space -192.168.1.2 to 192.168.1.62
192.168.1.63 will be the broadcast address (remember to reserve the first and last address for the Network and Broadcast)
HQ Network Mask 255.255.255.192 - we got the 192 by adding the bit value from the left to the value we borrowed = 128+64=192
HQ address will look like this 192.168.1.0 /26"
Now its clear that he asked us to memorize a table. No problem with that.
Now he said "
Lets start with HQ with 50 hosts, using the table above:
We are borrowing 2 bits with value of 64. This is the closest we can get for 50 hosts.
"
My question :-
1) We are borrowing 2 bits ? WHY ???
I know according to his table its correct. But Dont we have to use the formula 2^n -2 = No: of Hosts ? By using this we get totally different number.
I know he is trying to calculate for 50 hosts and not subnets. Is the above table hold good in case if we are calculating the no: of subnets ?
I mean to ask can we use the same table to calculate the number subnets required ?
Lets consider we want to divide 199.131.20.0/24 in to 5 subnets.
So the formula 2^3 holds good. so we need to borrow 3 bits.
How do we infer the above statement using the given table above ?
Replies
-
durga chok ! 😁
if you would like to go through this blog post of mine, you might understand why he has assumed the bits to be borriwed as 2 - > #-Link-Snipped-#
please remember that
subnetting - we keep a tab on the number of networks
vlsm- we keep a tab on number of hosts -
jeethuGreat!!!
Actually I thought he is trying to do subnetting,
I went through your post. It was great!
But at the end its given "next VLSM"
But I cant find VLSM there. Please give me the link to your VLSM post.
Thanks!! -
durga ch:-|, thats becasue I have not yet come up with VLSM post. I am currently working on NAT and PAT and supernetting as they are confusing. Once I comeup with VLSM will let you know.
but the logic for VLSM woul be simple
See you generally count your hosts from right most bit right? so your increment count also would be from right . where as in Subnetting , you would consider the leftmost bits for subnets , that is the reason why you would consider the increment from left. rst all is same.
try this and let me know
You are reading an archived discussion.
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