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Home > Archive > 70-218 > August 2002 > CIDR Q for a 3 subnet LAN.
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CIDR Q for a 3 subnet LAN.
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| CyberDude 2002-08-13, 1:00 pm |
| After my CCNA I thought I was cool with TCP/IP, until I came across this Q from Transcender. Can someone shed some light, as I have always believed that if all other computers have one subnet mask on your segment, and you have a different one, then you would not be able to connect with anyone, let alone anyone on a different segment across a router. Has the big T messed this Q up, or has may grey matter gone squishy?  | |
| KScheler 2002-08-13, 2:19 pm |
| I just tried it in my lab. I have one computer with 192.168.135.100 - 255.255.255.0 and one with 192.168.135.101 - 255.255.0.0 The one with 192.168.135.101 - 255.255.0.0 can successfully ping the other but 192.168.135.100 can not ping 192.168.135.101. It must have something to do with the net ID. | |
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| Unless I am missing something, you are right. With computer1 having a different subnetmask as computer2 and the router it shouldn.t be able to communicate with any of them. | |
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| scratch that computer1 can ping #2 by broadcast and the transcender reasoning for subnet 3seems right although to communicate back to subnet one I do not believe that the /18 mask will find it's way to router1
if that makes any sense | |
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| edit. too lazy to make sure I am 100 percent technically right so post removed. | |
| CyberDude 2002-08-14, 3:33 am |
| So what you are saying is that icmp acts like netbios (ie. as they both broadcast). I thought that a subnet mask is what made routing at the network layer possible, along with the default gateway?
I just do not see how a computer with one subnet mask can communicate with other computers that have another subnet mask, especially when they are on the same subnet.
The two main TCP/IP config properties are IP address and subnet mask (if not routing). The subnet mask determines which network and the IP address determines which node.
I still believe that this Q is wrong. | |
| twister166 2002-08-15, 10:27 pm |
| I think trancender is correct, let me see if I can explain it a bit better, with /18 it will see the seg1 and seg2 as local trafiic with the /19. Seg3 will appears as remote traffic to PC1. Thus, PC1 think it is local traffic to PC2, it will just resolve the IP to MAC and transfer data, because it is local, no router are involved. PC1 still thinks that the PC3 and PC4, but ARP will no longer resolve it as local because it has a router in the middle, thus local to local resolution fails and cannot talk to any PC on seg2. Seg3 is consider as remote network (PC1's network is from 128.0 to 191.255) so the data will be forward to the router and assuming that the router address is correct, the data gets foward. Bearing in mind that when it you are local to local you are just resolving IP to MAC and transfer. Think about when you ping 147.45.1.5, you just ping the address you don't supply the mask with it. I hope I explained it clearly. | |
| CyberDude 2002-08-16, 1:46 am |
|  I am sorry but I am still baffled about this one.
If all you needed for IP connectivity was an address, then why is it that whatever TCP/IP info you read, you are always told that you need at least an IP address AND a subnet mask?
PC1 is the only /18 computer on the whole network, and in my eyes it should not be able to communicate with any other nodes.
If as everyone is saying is true, that PC1 can communicate with its own segment and segment two, even though the other nodes are /19 as well as crossing a router, then why can it not contact segment three which is also /19 and accross a router?
Each router has the routes for the other segments, so if PC1 CAN travel to segment 2, then it should be able to travel to segment 3 as well, because all routers have the routeplan. | |
| twister166 2002-08-16, 8:46 am |
| Combine the IP and OSI for a sec. Think network layer, IP is a connectionless just want to find a route to destination, it also decied if the packet are destinate to local network or remote network, right? If if local, it uses ARP to resolve the IP address and MAC. Once it is got the MAC of the distination, it can now transmitt data, IP is no longer in play. Since PC1 has /18, anything in the 3rd octect in range of 128 to 191 will be consider local. It will use ARP to get the MAC, the destination PC will responde with a matching MAC. It then (assume) ethernet, convert packet to frame to bit and hits the wire. The is why PC1 can talk to PC2, because it is local to local traffic. Yet PC3 and PC4 also appear to be local to PC1, but it is segmented by router, PC1 will not get the MAC respond because PC3 and PC4 is in a different broadcast / collision domain. So, PC3 and PC4 will not talk to PC1. Since PC5 and PC6 are consider as remote traffic because the 3rd octect is out side of PC1's network, PC1 determines that it is remote and foward the packet/frame/bits to router. Once IP determine that it is local to remote traffic, it forward data to the default gate's ip address. And now PC5 and PC6 talks to PC1.
The bottom line, PC1 to PC2, PC3, PC4 are consider as local to local traffic, they should be in the same broadcast domain, since PC3 and PC4 are not, PC3 and PC4 fails. And PC1 to PC5 and PC6 are local to remote traffic and should involve the default gateway, and it does. So, it works.
If you are still confused, send an e-mail to me at twister166@hotmail.com, and leave your number. I will call you. | |
| twister166 2002-08-16, 8:53 am |
| Oops, I just realized that you are in Germany, forgot that you can be any where, the phone call would be hard to make now. | |
| CyberDude 2002-08-16, 11:41 am |
| Thanks twister for allowing the penny to drop. | |
| twister166 2002-08-16, 2:02 pm |
| Welcome Dude, not too sure if I explained it clear enough. I just passed 70-218, I am going to take the N+ and Server+ next week, after that CTT+, here I go. I am trying to get the explainations in and get some pratice for teaching! :-) | |
| Zaraspook 2002-08-16, 3:16 pm |
| Here's my crack at it:
Once computer 1 obtains and IP address via DNS, WINS, Broadcast, etc. it checks it’s Network ID to determine if the destination computer it wants to communicate with is local or remote. If it deems that it is remote, it sends out a special broadcast on the local wire that can be answered only by the router serving the default gateway role. The router sends computer 1 back its MAC address and communication can proceed.
On the other hand, if computer 1 deems that the computer it wants to communicate with is local, it sends out a local broadcast on the wire to obtain the MAC address for a computer located on the same local physical segment. Only a destination computer located on the same local physical segment can return its MAC address in response to this broadcast. A router cannot and will not respond to this particular broadcast.
When computer 1 is sending out this local broadcast, no computers located on the same local physical segment are able to respond, since computers 3 & 4 are actually located on a different physical segment. What is happening in this scenario, is computer 1 is seeing computers 3 & 4 as being local, since if we break down their IP addresses into binary, the first 18 bits corresponds to its own subnet mask, which is the left most 18 bits.
Subnet 1: 10010011 00111100 101
Subnet 2: 10010011 00111100 100
Subnet 3: 10010011 00111100 110
10000000 to 10111111 = 128 to 191 in third octet using /19 subnet mask is seen as being local by computer 1.
Therefore, any IP addresses that fall in the range of 128 to 191 in the third octet and using a /19 subnet mask will be seen as local by computer 1, regardless if it’s located on the same local physical segment or not, and any local broadcast it sends out on the local wire to obtain the MAC addresses for computers 3 & 4 is destined to fail, because, again, in this scenario computers 3 & 4 are actually located on a different physical segment. You must also remember to keep in mind that the router will be unable to respond to this broadcast.
Further, computer 1 will be able to communicate successfully with computers 5 & 6, since their IP addresses in the third octet fall outside the 128 to 191 range, and it will see computers 5 & 6 as being remote.
147.60.218.57 /19 = 10010011 00111100 11011010 00111001
147.60.197.66 /19 = 10010011 00111100 11000101 01000010
Makes sense?  | |
| CyberDude 2002-08-17, 6:17 am |
| Thank you as well zaraspook. It is a taxing Q which had my grey matter boiling. I reread through my CCNA notes on TCP/IP and have had a good refresh. Thank you again guys for helping out a dude in need. |
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