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Help with ip addressing


rv31
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If you have a LAN with more than 255 PCs with the first PC having an IP of 10.0.0.2 and the routers ip is 10.0.0.1, after you reach the ip 10.0.0.254 what would be the next ip on the LAN, "10.0.1.0" ??

How come (in packet tracer simulation package) i can have 2 pc connected to a switch with ips PC1 = 10.0.0.1 and PC2 = 10.0.2.2 for example and the PCs they can ping each other whereas if i use the ips PC1 = 192.168.1.1 and PC2 = 192.168.2.1 then they can't ping each other...?

Edited by rv31
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welcome to subnetting 101.

10.x.x.x. default subnet

255.0.0.0

192.168.x.x default subnet

255.255.255.0

short but sweet, everything above the 255s has to be the same. So in your 192 scenario whatever you place in third octet has to be the same on all boxes in your network. or drop to a class b subnet (255.255.0.0) on your boxes and use 192.168.xxx.xxx

Edited by iamtheky
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welcome to subnetting 101.

10.x.x.x. default subnet

255.0.0.0

192.168.x.x default subnet

255.255.255.0

short but sweet, everything above the 255s has to be the same. So in your 192 scenario whatever you place in third octet has to be the same on all boxes in your network. or drop to a class b subnet (255.255.0.0) on your boxes and use 192.168.xxx.xxx

Well explained, got it. Ok so you can choose your own subnet to suit your LAN ips needs. What i need to really understand then is how subnetting works, for examaple i think i have seen subnets like 255.255.214.0 so what is the deal there....

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think of 255.255.255.0 = 11111111.11111111.11111111.00000000

all the bits that are 'on' have to be identical - (which is why in the 255 examples the numbers must be exactly the same as all bits are on)

Think of it like so:

My home network has a subnet of 255.255.255.248 to allow for only 5 network devices (not counting the router at .1)

11111000=248

so regardless of what my IP is everything above the 1s must be the same, so lets take your example 192.168.2.1

192.168.2 must all be the same because 255 is all 1s. so we will isolate the subnet .248

1=00000001

248=11111000

so the first 5 bits are locked in just like the 192.168.2 was. Leaving us the last 3 to play with.

00000000=all 0s network address - unusable

00000001=router

00000010=2

00000011=3

00000100=4

00000101=5

00000110=6

00000111=broadcast address - unusable

same subnet of .248 with a more complex final IP octet like .166 and determining the available IPs in the network

10100000=network address

10100001=161

10100010=162

10100011=163

10100100=164

10100101=165

10100110=166

10100111=167=broadcast

Edited by iamtheky
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If you have a LAN with more than 255 PCs with the first PC having an IP of 10.0.0.2 and the routers ip is 10.0.0.1, after you reach the ip 10.0.0.254 what would be the next ip on the LAN, "10.0.1.0" ??

How come (in packet tracer simulation package) i can have 2 pc connected to a switch with ips PC1 = 10.0.0.1 and PC2 = 10.0.2.2 for example and the PCs they can ping each other whereas if i use the ips PC1 = 192.168.1.1 and PC2 = 192.168.2.1 then they can't ping each other...?

You probably have a subnet mask of 255.255.252.0. To answer your question your next IP Address would be 10.0.0.255.

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To answer your question your next IP Address would be 10.0.0.255

10.0.0.255 is a broadcast address B)

that's wqhat i was meaning. So the ip could be say 10.0.1.0 if the subnet allowed for it, the key thing i have got is that it depends on what the subnet is. I am ok with it if it is a default subnet like 255.255.255.0 or 255.255.0.0 but haven't quite clicked onto what ip range you can have how you have a subnet like 255.255.214.0 yet. (VLSM)

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You get to play with the bits that are 'off', regardless of configuration. So in 255.255.214.0 (11010110=214)

pick any IP - we will go with 192.168.134.240 - I want to know what other IPs I could talk to

Knowns:

255=must have 192 in the 1st octet

255=must have 168 in the 2nd octet

214=????

0=can have any number 1-254 (0 if its not the network) in the 4th octet.

3rd octet of IPAddress= 134 = 10000110

3rd octet of subnet = 214 = 11010110

everything above a one is locked (bits 2,3 and 8 are locked in the on, bits 5 and 7 in the off) so modifying the remaining bits you get the following options for the third octet

134=10000110

135=10000111

142=10001110

143=10001111

166=10100110

167=10100111

174=10101110

175=10101111

so [192].[168].[any from the list].[1-254] would work for the subnet 255.255.214.0

imho :)

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You get to play with the bits that are 'off', regardless of configuration. So in 255.255.214.0 (11010110=214)

pick any IP - we will go with 192.168.134.240 - I want to know what other IPs I could talk to

Knowns:

255=must have 192 in the 1st octet

255=must have 168 in the 2nd octet

214=????

0=can have any number 1-254 (0 if its not the network) in the 4th octet.

3rd octet of IPAddress= 134 = 10000110

3rd octet of subnet = 214 = 11010110

everything above a one is locked (bits 2,3 and 8 are locked in the on, bits 5 and 7 in the off) so modifying the remaining bits you get the following options for the third octet

134=10000110

135=10000111

142=10001110

143=10001111

166=10100110

167=10100111

174=10101110

175=10101111

so [192].[168].[any from the list].[1-254] would work for the subnet 255.255.214.0

imho :)

OK i think i've got it, thanks. My next question will tell if i have maybe.

So why do we need 3 private address ranges, 10.x.x.x, 172.16.x.x -172.31.x.x and 192.168.x.x – 192.168.x.x etc...

And from what i've read you shouldn't really have more than 500 hosts on any network and all of these private address ranges can host huge amounts of hosts according to how you tweak the subnet.

And then if you needed more than 500 hosts all you need to do is add another router and tweak the subnet for the 10.x.x.x so that the second network is on another subnet etc....

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class a, b, and c. if you need over 254 objects you would use a class b private range, if you needed more than a million you would use a class a?

http://tools.ietf.org/html/rfc1918

And from what i've read you shouldn't really have more than 500 hosts on any network

I would think that you read more than 500 objects on a segment if you are talking dividing with a router, and i wouldnt go over 50 without dropping at least a switch. I have never seen a recommendation for another 'network', or am even sure i completely understand what you mean by that.

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