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Spid's Fri (10/18) Win2K Pro. QoD
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| Happy Friday everyone! This one will give you something to "noodle" over for the weekend, and there's no better time than the present to learn a little about basic subnetting. 
You have been brought on board to troubleshoot a companies network. Users an Client-A are unable to connect to resources on Server-B. The IP configuration is listed below: (subnet mask is /22 or 255.255.252.0 if you are not familiar with /n notation). Router-A is a multihomed Windows 2000 Server with 3 NICs attached to the 3 network segments A,B and C). Client-A resides on Segment-A, Client-B resides on Segement-B, etc...
Client-A: IP Address - 172.16.31.101/22
Default Gateway - 172.16.28.1
Server-B: IP Address - 172.16.32.53/22
Default Gateway - 172.16.36.1
Client-B: IP Address - 172.16.37.115/22
Default Gateway - 172.16.36.1
Router-A to Segment-A interface: 172.16.28.1/22
Router-A to Segment-B interface: 172.16.36.1/22
Router-A to Segment-C interface: 172.16.40.1/22
Why is Client-A unable to access resources on Server-B?
A. The IP address of Client-A is incorrect.
B. The IP address of Server-B is incorrect.
C. The IP address of Router-A to Segment-B interface is incorrect.
D. The IP address of Router-A to Segement-A interface is incorrect.
Bonus points if you can tell me whether or not Client-B can access resources on Server-B as well.
Tip - you'll probably need to do some "ANDing" to determine what actual networks the Devices live on. Grab Freak's subnetting guide at www.mcsefreak.com/subnetting.htm if you need some help | |
| Slinky 2002-10-18, 11:34 am |
| B. My bonus answer is that client B will be able to access resources on server B. | |
| denis_baribeau 2002-10-18, 12:38 pm |
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| tweetgirl 2002-10-18, 1:19 pm |
| It is Friday and we should be celebrating but no Spid is making us think. He wants us to "noodle". Hmmmm. I am going to have to do this one when I get home from work. | |
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Originally posted by tweetgirl
It is Friday and we should be celebrating but no Spid is making us think. He wants us to "noodle". Hmmmm. I am going to have to do this one when I get home from work.
Well I've got to continue studying for the 70-216 exam this weekend, so you lucky people get to "suffer" as well. http://www.stopstart.fsnet.co.uk/smilie/whinging.gif
Someday I'll get off of my procrastinating butt and take that blasted exam. | |
| Linux_Hawk 2002-10-18, 5:04 pm |
| You can tell this by the default Gateway.
Thanks for the mind bender. | |
| hazz_bin 2002-10-19, 4:35 am |
| Whew. After noodling a bit I agree with Linux_hawk that it is B. (Okay Spid, I will now proceed to not think for the next two days.) | |
| heck2000 2002-10-19, 6:28 pm |
| Hey Spid, from what I read from the link you've posted, there should be no problem between any of the clients or servers in this network because their Network ID is typical for all, judeging by the subnet mask.
Open to comments!!!! | |
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Originally posted by heck2000
Hey Spid, from what I read from the link you've posted, there should be no problem between any of the clients or servers in this network because their Network ID is typical for all, judeging by the subnet mask.
Open to comments!!!!
No tricks heck2000.
As things stand currently, there is a problem with the network configuration on one of the devices. One of the answers given to choose from does identify what the issue is.  | |
| tweetgirl 2002-10-20, 11:43 am |
| Well, Spid you did it. My head is a noodle but I am going with B and for the bonus, client B should be able to access resources. | |
| Tronn 2002-10-20, 12:10 pm |
| I say 'B' is the answer and no the client on B will not be able to contact the server on B.
This is because the client on B will check the IP address of server B against it's own subnet mask. Doing this will tell client B that server B is on a different subnet so it will send its frames to it's default gateway instead.
TRONN | |
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| quote:
Originally posted by Spid
Happy Friday everyone! This one will give you something to "noodle" over for the weekend, and there's no better time than the present to learn a little about basic subnetting.
You have been brought on board to troubleshoot a companies network. Users an Client-A are unable to connect to resources on Server-B. The IP configuration is listed below: (subnet mask is /22 or 255.255.252.0 if you are not familiar with /n notation). Router-A is a multihomed Windows 2000 Server with 3 NICs attached to the 3 network segments A,B and C). Client-A resides on Segment-A, Client-B resides on Segement-B, etc...
Client-A: IP Address - 172.16.31.101/22
Default Gateway - 172.16.28.1
Server-B: IP Address - 172.16.32.53/22
Default Gateway - 172.16.36.1
Client-B: IP Address - 172.16.37.115/22
Default Gateway - 172.16.36.1
Router-A to Segment-A interface: 172.16.28.1/22
Router-A to Segment-B interface: 172.16.36.1/22
Router-A to Segment-C interface: 172.16.40.1/22
Why is Client-A unable to access resources on Server-B?
A. The IP address of Client-A is incorrect.
B. The IP address of Server-B is incorrect.
C. The IP address of Router-A to Segment-B interface is incorrect.
D. The IP address of Router-A to Segement-A interface is incorrect.
Bonus points if you can tell me whether or not Client-B can access resources on Server-B as well.
Tip - you'll probably need to do some "ANDing" to determine what actual networks the Devices live on. Grab Freak's subnetting guide at www.mcsefreak.com/subnetting.htm if you need some help
And the answer is.....B 
The IP address of Server-B is not within the same IP network as the Router-A to Segment-B interface.
If we "AND" the IP address of Server-B with the subnet mask 255.255.252.0, we find that Server-B is on the 172.16.32.0 network.
If we "AND" the IP address of the Router-A to Segment-B interface with the subnet mask of 255.255.252.0, we find that it is on the 172.16.36.0 network.
Because Server-B and the Router-A to Segement-B interface have different IP network IDs, they are on seperate IP subnetworks. Each subnet in a routed IP network must have at least one router interface that resides on the same subnet.
Doing the same calculations for Client-B and the router interface connected to Segment-B results in Client-B and the router interface residing on the same (172.16.36.0) IP subnetwork. Because Server-B doesn't match the IP network ID for Client-B or the corresponding router interface, the IP address for Server-B is incorrect.
Change the IP address of Server-B to a unique address in the 172.16.36.1 - 172.16.39.254 range and everything will be happy. (Of course you can't use 172.16.36.1 because it is already used )
I know this question was a little beyond the scope of the 210 exam, but it is something that you'll need to know for the 70-216 exam. Even though CIDR is becoming more widely used, Classful subnetting is still used quite a bit.
Nice job to the people who worked through this! |
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