All other Cisco networking related discussions. balaji31d
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### Subnetting doubt

Hi, Below is the question with the explanation. I still dont understand how and why they subtracted 7 from 30 (30 – 7 = 23.). Can somebody help to explain please.

A. 10.10.0.0/18 subnetted with mask 255.255.255.252
B. 10.10.0.0/25 subnetted with mask 255.255.255.252
C. 10.10.0.0/24 subnetted with mask 255.255.255.252
D. 10.10.0.0/23 subnetted with mask 255.255.255.252
E. 10.10.0.0/16 subnetted with mask 255.255.255.252

Explanation

We need 113 point-to-point links which equal to 113 sub-networks < 128 so we need to borrow 7 bits (because 2^7 = 128).

The network used for point-to-point connection should be /30.
So our initial network should be 30 – 7 = 23.

So 10.10.0.0/23 is the correct answer.

You can understand it more clearly when writing it in binary form:

/23 = 1111 1111.1111 1110.0000 0000
/30 = 1111 1111.1111 1111.1111 1100 (borrow 7 bits)
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Balaji M
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### Re: Subnetting doubt

32 - 2 ( for the host bits) = 30
30 - 7 ( for the network bits) = 23

it's a round about way to get there.
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### Re: Subnetting doubt

Sorry for being a noob. I still do not understand. Is it a formula?

* We need 113 point-to-point links which equal to 113 sub-networks < 128 so we need to borrow 7 bits (because 2^7 = 128). - i understand
* The network used for point-to-point connection should be /30 - Also understand.
* So our initial network should be 30 – 7 = 23. (This is the part which i dont understand.)
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Balaji M
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### Re: Subnetting doubt

I don't know why people make subnetting more difficult than it needs to be. These point to point links are going to be /30s. How many addresses are there per /30? 4. 2 host addresses, 1 subnet and 1 broadcast. So 4 * 113 = 452. Hence you'll need a /23 as a /23 gives you 512 addresses

If they really wanted to save address space, they could use /31s and only need a single /24 for all of them.

Fred
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### Re: Subnetting doubt

mellowd wrote:I don't know why people make subnetting more difficult than it needs to be. These point to point links are going to be /30s. How many addresses are there per /30? 4. 2 host addresses, 1 subnet and 1 broadcast. So 4 * 113 = 452. Hence you'll need a /23 as a /23 gives you 512 addresses

Indeed. This is the math I would have used.

Once you understand that you need /30's, you can restate the problem: "How many /30's are needed to handle 113 links?" The answer should be obvious: You need 113 of them.

So how many IP's does that take? Well, one /30 requires 4 IP's, so that's 113*4=452 IP addresses.

So the hard part is: "What power of 2 will accommodate 452 IP addresses?" At that point, it's important to know your powers of 2. 2, 4, 8, 16, 32, 64, 128, 256, 512... If you can't do that off the top of your head, work on it. So 512 is the first power of 2 that will accommodate your subnets. Which power is it? Count them. It's the 9th power of 2, so 2^9=512. So you need 9 total bits to accommodate your subnets. IPv4 addresses are 32-bits, and you need 9 bits to accommodate your networks: 32-9=23. So you can fit them all in a /23. balaji31d
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### Re: Subnetting doubt

mellowd wrote:I don't know why people make subnetting more difficult than it needs to be. These point to point links are going to be /30s. How many addresses are there per /30? 4. 2 host addresses, 1 subnet and 1 broadcast. So 4 * 113 = 452. Hence you'll need a /23 as a /23 gives you 512 addresses.

Fred wrote:So the hard part is: "What power of 2 will accommodate 452 IP addresses?" At that point, it's important to know your powers of 2. 2, 4, 8, 16, 32, 64, 128, 256, 512... If you can't do that off the top of your head, work on it. So 512 is the first power of 2 that will accommodate your subnets. Which power is it? Count them. It's the 9th power of 2, so 2^9=512. So you need 9 total bits to accommodate your subnets. IPv4 addresses are 32-bits, and you need 9 bits to accommodate your networks: 32-9=23. So you can fit them all in a /23.

Perfect!!!!!!! This is the exact way of me understanding the problem. Thank you and god bless you.

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Balaji M
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Balaji M
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Nawras
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### Re: Subnetting doubt

hi why make a complex calculation i think and actually i understand it like this and in this simple:

from host perceptive we just need 2 addresses so 2^2=4
two of it for broadcast and native
AND
that's it ...
so
"xxxxxxxx.xxxxxxxx.xxxxxxxx.xxxxxx|00"
the last two bit for point to point "host ID"

Now for 113 subnet you have to understand just that if you want 113 so you can take it from 2^7

2^1=2
2^2=4
2^3=8
2^4=16
2^5=32
2^6=64
2^7=128 which is cover 113

so we need 7 bit to reach this number "113"

so two bit for host id and 7 bit for net id in total 9 bit

wich is /23

32 all bits - 23 t= 9
so

in very and simple detail

SebastianWilliams
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### Re: Subnetting doubt

It is really confusing.

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