rv31 Posted February 10, 2009 Share Posted February 10, 2009 (edited) 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 February 10, 2009 by rv31 Link to comment Share on other sites More sharing options...
iamtheky Posted February 10, 2009 Share Posted February 10, 2009 (edited) welcome to subnetting 101.10.x.x.x. default subnet 255.0.0.0192.168.x.x default subnet 255.255.255.0short 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 February 10, 2009 by iamtheky Link to comment Share on other sites More sharing options...
rv31 Posted February 10, 2009 Author Share Posted February 10, 2009 welcome to subnetting 101.10.x.x.x. default subnet 255.0.0.0192.168.x.x default subnet 255.255.255.0short 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.xxxWell 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.... Link to comment Share on other sites More sharing options...
iamtheky Posted February 10, 2009 Share Posted February 10, 2009 (edited) think of 255.255.255.0 = 11111111.11111111.11111111.00000000all 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=248so regardless of what my IP is everything above the 1s must be the same, so lets take your example 192.168.2.1192.168.2 must all be the same because 255 is all 1s. so we will isolate the subnet .248 1=00000001248=11111000so 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 - unusable00000001=router00000010=200000011=300000100=400000101=500000110=600000111=broadcast address - unusablesame subnet of .248 with a more complex final IP octet like .166 and determining the available IPs in the network10100000=network address10100001=16110100010=16210100011=16310100100=16410100101=16510100110=16610100111=167=broadcast Edited February 10, 2009 by iamtheky Link to comment Share on other sites More sharing options...
Noise Posted February 10, 2009 Share Posted February 10, 2009 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. Link to comment Share on other sites More sharing options...
FAT64 Posted February 11, 2009 Share Posted February 11, 2009 To answer your question your next IP Address would be 10.0.0.25510.0.0.255 is a broadcast address B) Link to comment Share on other sites More sharing options...
rv31 Posted February 11, 2009 Author Share Posted February 11, 2009 To answer your question your next IP Address would be 10.0.0.25510.0.0.255 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) Link to comment Share on other sites More sharing options...
iamtheky Posted February 11, 2009 Share Posted February 11, 2009 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 toKnowns:255=must have 192 in the 1st octet255=must have 168 in the 2nd octet214=????0=can have any number 1-254 (0 if its not the network) in the 4th octet.3rd octet of IPAddress= 134 = 100001103rd octet of subnet = 214 = 11010110everything 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 octet134=10000110135=10000111142=10001110143=10001111166=10100110167=10100111174=10101110175=10101111so [192].[168].[any from the list].[1-254] would work for the subnet 255.255.214.0imho Link to comment Share on other sites More sharing options...
rv31 Posted February 12, 2009 Author Share Posted February 12, 2009 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 toKnowns:255=must have 192 in the 1st octet255=must have 168 in the 2nd octet214=????0=can have any number 1-254 (0 if its not the network) in the 4th octet.3rd octet of IPAddress= 134 = 100001103rd octet of subnet = 214 = 11010110everything 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 octet134=10000110135=10000111142=10001110143=10001111166=10100110167=10100111174=10101110175=10101111so [192].[168].[any from the list].[1-254] would work for the subnet 255.255.214.0imho 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.... Link to comment Share on other sites More sharing options...
iamtheky Posted February 12, 2009 Share Posted February 12, 2009 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/rfc1918And from what i've read you shouldn't really have more than 500 hosts on any networkI 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. Link to comment Share on other sites More sharing options...
GeneralMandible Posted February 14, 2009 Share Posted February 14, 2009 Here is a great whitepaper on IP addressing.http://www.3com.com/other/pdfs/infra/corpi...n_US/501302.pdf Link to comment Share on other sites More sharing options...
Recommended Posts
Create an account or sign in to comment
You need to be a member in order to leave a comment
Create an account
Sign up for a new account in our community. It's easy!
Register a new accountSign in
Already have an account? Sign in here.
Sign In Now