MS Self-Pace 70-291 Book - IP Addressing Troubleshooting

JLL

I am currently working the Case Scenario Exercise on page 2-51 of the Microsoft Self-Pace 70-291 Book and would like to know if there is some sort of methodical approach in troubleshooting IP Addressing issues. I have a pretty good handle on IP Addressing, but when a diagram is presented that has all the computers, hubs, router(s) along with their current IP configuration, then asked to identify the configuration error, it's a handful.

Any help is greatly appreciated.

Thanks in advance.

JLuna

blagger

Have a closer look at client C on the lower side of the router. It's only a subtle error, but check the ip setting for the default gateway. Let me know what you see!

You have to remember that a router pretty much segregates a network into separate broadcast zones, so the clients on one section of the router will acces it via one default gateway address, and the clients on the other side will access it via a different address.

JLL

Hey Blagger,

That scenario is pretty easy to figure out. The gateway address is not pointing to the correct interface on the router. I think number 2 and 3 are more interesting.

Thanks!

JLuna

blagger

Yep, fair play to MS, this chapter is very vague.

The second case scenario is just b****ks in my opinion, but the third one is soluble.

Just use the equation to determine if the two addresses are on the same subnet. You've got 131.107.127.1 and 131.107.128.1 on a /21 network. the /21 becomes 255.255.248.0.

256 - 248 = 8. so 127 / 8 = 15 (disregard remainder), and 128 / 8 = 16. Because the numbers are different, then they are on different subnets. So the clients are on the same segment, but different subnets.

grey fox

The second case scenario is just b****ks in my opinion

Thanks for that. I thought I was the only person that was having trouble with this question.

JLL

Yes, thank you for the detailed explanation blagger. You've been very helpful.

Thanks to grey fox for posting that as well. I was beginning to think that I was the only one a bit confused about it.

Here is an excellent white paper I came across the Internet while looking for some more information on Variable Length Subnet Mask (VLSM). It is lengthy, but definitely worth a read.

http://searchnetworking.techtarget.com/searchNetworking/downloads/IPAddressing.pdf

grey fox

Thanks for the link.

chicken13

grey fox wrote:

The second case scenario is just b****ks in my opinion

Thanks for that. I thought I was the only person that was having trouble with this question.

No I thought exactly the same thing this morning when I looked at it!

Danman32

Another thing to get familiar with is the common subnet boundaries in the octets. 128 is a 1 in the MSb, where that bit will be 0 for 127 (the rest of the bits in the octet are all 1's). Since the most significant bit is different for both numbers in the third octet, any subnet mask where the third octet is not 0 means that the MSB is significant in specifying the network, therefore they are on different nets.

Binary placeholders in an 8 bit number are 128,64,32,16,8,4,2,1
Subnet mask #s would be 128,192,224,240,248,252,254,255.

If you can't remember the possible subnet mask #s, you can reconstruct them before you actually start the exam or once you start the exam, you can use the calculator to help you. To generate the subnet #'s start with 128, and start adding the placeholder value on the right. 128+64=192, 192+32=224, etc.

I didn't have the luxury of any calculator in the CCNA exam so I had to memorize the common numbers.

Keep playing with the numbers, you'll start seeing the common patterns.

mikey_b

Yeah, Client C has the wrong Default Gateway, the router interface for that segment has an IP of 192.168.1.129. Client C (and apparently Client D) are pointing to 192.168.1.126 and 192.168.1.128 respectively. That is not their Default Gateway.

JLL

Danman32,

"If you can't remember the possible subnet mask #s, you can reconstruct them before you actually start the exam or once you start the exam, you can use the calculator to help you. To generate the subnet #'s start with 128, and start adding the placeholder value on the right. 128+64=192, 192+32=224, etc."

This is an excellent suggestion and plan on doing it when I take the exam!

Thank!

JLuna