Error on Packet Tracer

Alhaji265

Hello,

I am having trouble setting this complex network on Packet tracer. I have came up with this networking scheme -

Create 20 networks
182.47.36.49/21
Formula: 2n >/= 20 (answer 2*2*2*2*2=32)
Range : 0 - 31
32 - 63
64 - 95
96 - 127 and so forth.

On the IP Subnet Calculator it states that the valid host range are 182.47.32.1 to 182.47.63.254.

What did I do wrong? Thanks!

mapletune

I'll bite.


Let's look at everything step by step.


It seems like you want to create 20 subnets from a class B public address 182.47.0.0/16


Just like you calculated, you'd need at least 5 bits reserved as subnet bits in your host mask. with 4 bits, 2^4=16, you'd get 16 subnets. 2^5=32 (32 subnets is enough.)


Now, your base address is a class B ip, thus it's default mask is 182.47.0.0/16 (255.255.0.0) thus we will start subnetting from 11111111.1111111.SSSSSHHH.HHHHHHHH = 11111111.11111111.11111000.00000000 where S are subnet bits and H are host bits. This mask becomes /16 + /5 = /21 Thus, 182.47.0.0/21 ...


This is why you'd get

(EDIT: thnx for pointing out, the following enclosed information is WRONG, dunno what i was smoking...)
/////////////////////////////////////////////////////////////////
!!!!182.47.0.0 (first subnet) (when using zero subnets)
!!!!182.47.32.0 (second subnet)
!!!!182.47.64.0 ....
!!!!...
!!!!182.47.224.0 (last subnet) (32nd subnet when using ip subnet-zero)
//////////////////////////////////////////////////////////////////

=====


I imagine you might know all this already. But just in case, I've reviewed things a bit.


To further diagnose your problem in Packet Tracer, we'd probably need a bit more information.

___

[edit]
thanks to following commenters, i've edited the post to show where i've made a mistake. =p

MrBrian

Guys, with a /21, there is a block size of 8 in the 3rd octet.. that means that every 8 numbers, there's a new network.

I see you wrote out 182.47.36.49/21.. and you said you wanted to create 20 networks. I'm guessing you started with 182.47.0.0/16, and wanted to extend it for 20 subnets, hence why you chose a /21 (2^5 bits for subnets = 32 subnets, where 2^4 bits for subnets = 16, and is too small)

So with a /21 and that network, your subnets would be:
182.47.0.0
182.47.8.0
182.47.16.0
182.47.24.0
....182.47.248.0

That gives you 32 subnets. (The networks can't be .32, .64, .96, etc cause that only gives you 8 subnets! That'd be the case if we had a /19 mask instead though...) hth

Ltat42a

I concur with MrBrian, 182.47.36.49/21 is in the 182.47.32.0 network. Subnets will increment by 8. With 5 bits used for subnetting, you'll get 32 subnets (2046 hosts per subnet).

Network_Engineer

PT has bugs some times... Best to start over on a config, if you cannot resolve after repeated attempts.

mapletune

Well that was embarrassing =p

edited last comment to point out the misinformation. thanks!

MrBrian

Don't worry about it, happens to me all the time!

lantech

It looks like you may have subtracted 21 - 16 to get 5 bits. Using 5 bits to determine your network boundaries gets you an increment of 32 in your interesting octet.

You should subtract your prefix from the next prefix boundary. So it should've been 24 - 21 = 3. 3 bits would give you an increment of 8 for each network in your interesting octet.

Ltat42a

In the original post, he stated he needed 20 networks. 2^5=32. 5 bits are needed (in the 3rd octet) to create the networks (subnets). 2^3 would only produce 8 networks.

Since 5 bits are used to create the subnets, the subnets would increment by 8.

does that help??

lantech

Ltat42a wrote: »

In the original post, he stated he needed 20 networks. 2^5=32. 5 bits are needed (in the 3rd octet) to create the networks (subnets). 2^3 would only produce 8 networks.

Since 5 bits are used to create the subnets, the subnets would increment by 8.

does that help??

I guess I didn't explain it properly. The way I was doing it would find out the subnet boundaries and not the number of subnets.

2^3 would give you the following subnets

182.47.0.0
182.47.8.0
182.47.16.0
and so on until you come to the end.

But it looks like he used 2^5 which would give you the following subnets

182.47.0.0
182.47.32.0
182.47.64.0
and so on until you reach the end.