Transcender question that I dont understand. Halp?
Agent6376
Member Posts: 201
I've just starting exam prepping for this test and I ran across this question while completing my Transcender practice exam. I would like to see if someone could break this down to me because I felt like I had a pretty good hold on subnetting (as far as the basics were concerned)
Question:
You have been hired to design a network structure and an IP addressing scheme for a large university that maintains campuses in several major cities. The network will span 20 locations. Each location will contain between 200 and 250 computers. You must configure each location as a seperate subnet. All computer must be assigned private IP addresses. All of the addresses must be derived from the same classful address range. The network must be fully routed.
A. 169.254.128.0
B. 10.1.2.0
C. 172.20.160.0
D. 192.168.192.0
The answer I chose was D since a /24 mask will include 254 host addresses which covers the amount of hosts in each subnet, but the answer was C.
According to the explanation:
Each classful class C network address range can support only up to 254 hosts because 24 of its leftmost bits contstitute a network ID and only 8 bits can be used for the host ID portion. You could use one class C network address for each subnet, which would require 20 class C network address for the entire network. The scenario requires that all of the IP addresses on the network beling to a single classful address range.
I don't understand exaclty what is being said here. I was under the assumption that if needed, a network could be designed where
192.168.192.0-192.168.192.254
192.168.193.0-192.168.193.254
.
.
192.168.212.0-192.168.212.254
Could someone lend a hand?
Question:
You have been hired to design a network structure and an IP addressing scheme for a large university that maintains campuses in several major cities. The network will span 20 locations. Each location will contain between 200 and 250 computers. You must configure each location as a seperate subnet. All computer must be assigned private IP addresses. All of the addresses must be derived from the same classful address range. The network must be fully routed.
A. 169.254.128.0
B. 10.1.2.0
C. 172.20.160.0
D. 192.168.192.0
The answer I chose was D since a /24 mask will include 254 host addresses which covers the amount of hosts in each subnet, but the answer was C.
According to the explanation:
Each classful class C network address range can support only up to 254 hosts because 24 of its leftmost bits contstitute a network ID and only 8 bits can be used for the host ID portion. You could use one class C network address for each subnet, which would require 20 class C network address for the entire network. The scenario requires that all of the IP addresses on the network beling to a single classful address range.
I don't understand exaclty what is being said here. I was under the assumption that if needed, a network could be designed where
192.168.192.0-192.168.192.254
192.168.193.0-192.168.193.254
.
.
192.168.212.0-192.168.212.254
Could someone lend a hand?
Comments
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ElwoodBlues Member Posts: 117Look for clues in the questions. It's looking for a routable network, the 169 address will not have a default gateway; non routable.
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genXrcist Member Posts: 531Answer C meets all the objectives. It is a Classfull address range that can handle all the numbers thrown at us in the question. 20 sites at 2-250 nodes each can be accomplished with the following:
172.20.160.0/23
D is wrong because when you subnet a Classfull C address, the 4th octet is the interesting octet. Your example is subnetting the 3rd octet, which is not Class C.
C is right because we subnet the 3rd octet, which you can do and still maintain the Class B status.
I think B is wrong because you'll be subnetting the 3rd octet anyway to keep the network managable that you might as well use the Class B.1) CCNP Goal: by August 2012 -
Agent6376 Member Posts: 201I know the 169.254.x.y networks are not routable, I wanted to find out the explanation on why C was the answer over D.
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Agent6376 Member Posts: 201Answer C meets all the objectives. It is a Classfull address range that can handle all the numbers thrown at us in the question. 20 sites at 2-250 nodes each can be accomplished with the following:
172.20.160.0/23
D is wrong because when you subnet a Classfull C address, the 4th octet is the interesting octet. Your example is subnetting the 3rd octet, which is not Class C.
C is right because we subnet the 3rd octet, which you can do and still maintain the Class B status.
I think B is wrong because you'll be subnetting the 3rd octet anyway to keep the network managable that you might as well use the Class B.
This makes sense! Thank you genXrcist, and good luck on your next attempt at this exam. I'd venture to say that if you're teaching others you should be ready to knock it out -
genXrcist Member Posts: 531Thanks Bro, I appreciate that! You've obviously read my woes when it has come to this exam, very humbling indeed. I think 3rd time's gonna be a charm though. In fact, I'm so confident now that I understand & have labbed Sec templates/Certificates so much that I'm actually studying for the 294 already. Shhhh, don't tell anyone.1) CCNP Goal: by August 2012
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Agent6376 Member Posts: 201Today I finished with my actual learning of the info and decided to start exam prepping. I was a little touchy on CAs as well so I setup a lab and it started to make a lot more sense. Transcender has humbled me a bit in the last 2 hours though, and I may wind up reviewing more after I've seen what it's going to throw at me and hit my weaker areas.
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wedge1988 Member Posts: 434 ■■■□□□□□□□Its always easy to forget that you have to include a subnet value with the ip address range. Without it, the ip address range is meaningless.
Just remember that you need one, its easy to get confused when questions are stipidly long! i do from time to time. just read carefully and youll be fine!~ wedge1988 ~ IdioT Certified~
MCSE:2003 ~ MCITP:EA ~ CCNP:R&S ~ CCNA:R&S ~ CCNA:Voice ~ Office 2000 MASTER ~ A+ ~ N+ ~ C&G:IT Diploma ~ Ofqual Entry Japanese