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Home > Archive > Server 2003 > September 2004 > 70-293 Subnetting Question - Problem with QOD answer
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70-293 Subnetting Question - Problem with QOD answer
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| theprofessor 2004-09-02, 6:58 pm |
| Here's the QOD question and their answer - my comments follow:
You are a Server specialist hired by Techno Research Inc to implement and manage a network. Windows 2003 is newly introduced. You need to make sure that your network infrastructure can accommodate the new OS and that it can work with other OS. You need to modify the existing TCP/IP addressing infrastructure. You are given the following network information about your office.
Network Number 192.168.1.128
Subnet mask 255.255.255.192
You need to assign IP address to the client computers in your office. Which of the following are NOT the valid addresses that can be used for the client computers?
A) 192.168.1.129
B) 192.168.1.138
C) 192.168.1.169
D) 192.168.1.253
E) 192.168.1.254
F) None of the choices.
Correct answer is (NOT): F
Explanation: According to the technical information found in MS KB, each TCP/IP host is identified by a logical IP address, and a unique IP address is required for each host and network component that communicates using TCP/IP. Each IP address defines the network ID and host ID. An IP address is 32 bits long and is composed of four 8-bit fields, called octets. There are five address classes. Microsoft supports Class A, B, and C addresses assigned to hosts. Each address class can accommodate networks of different sizes. There are several guidelines you should follow to make sure you assign valid IP addresses. All hosts on a given network must have the same network ID to communicate with each other. All TCP/IP hosts, including interfaces to routers, require unique host IDs.
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Theprofessor's notes
Unfortunately, the QOD supplied answer is wrong and the explanation, while a brief but succinct overview of TCP/IP addressing, does not address this question at all. This is a straight forward subnetting question. The correct answers are 192.168.1.253 (D) and 192.168.1.254 (E) Let's take a look at why.
We are given the 102.168.1.128 network with a subnet mask of 255.255.255.192. The 192 in the subnet mask tells us that this is a subnet (If the subnet mask octets are anything but 255 or 0 the we are dealing with a subnet - you can take that to the bank  ).
Lets count the number of ones in the subnet mask. A 255 equals 11111111 or 8 ones so we have 8 + 8 + 8 + 2 (192 equals 11000000) - that makes 26 ones. A standard class C subnet mask (255.255.255.0) has 24 ones. That means we have two ones (or two bits) that we can use for subnetting. With two bits we can divide a network into 4 subnets. Let's divide the 192.168.1.0 network into 4 subnets - visualize it like this:
|---192.168.1.255 (end of the fourth subnet)
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| Foruth Subnet
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|---192.168.1.192 (Start of the fourth subnet)
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| Third Subnet
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|---192.168.1.128 (Start of the third subnet)
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| Second Subnet
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|---192.168.1.64 (Start of the second subnet)
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| First Subnet
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|---192.168.1.0 (Start of the first subnet)
Now, the question states that we have the 192.168.1.128 network. That puts us squarely on the third subnet. The 192.168.1.128 address (the lowest address on this subnet) is the network identifier so we can't assign that as a host addres. The highest address on the third subnet is 192.168.1.191 (one less than the start of the fourth subnet) and that's the broadcast address - we can't assign that as a host address. That leaves the addresses 192.168.1.129 thru 192.168.1.190 that we can assign to hosts on the third subnet. Answers A, B & C fall into this range and so are valid addresses that can be used for the client computers on this subnet. Addresses 192.168.1.253 (D) 192.168.1.254 (E) are valid on the fourth subnet. But we are on the third subnet, so they are NOT the valid addresses that can be used for the client computers. This is what the question is asking so the correct answers are D & E. | |
| sandy7000 2004-09-07, 11:29 pm |
| Wow! Thanks, Professor. I appreciate the time you've taken to post this. | |
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| sandy7000 2004-09-08, 6:34 pm |
| I do appreciate the extra info. Just a note that I have adult ADD & struggle w/ more than a paragraph at a time...(not a reflection on what you've said.)
If you could give more visual pictures or break the paragraphs into smaller ones, it would help the comprehension level.
Thanks! Still a good explanation. |
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) so time is at a bit of a premium. Also, because I am in a university, I have to come up with my own exams in these courses. You guys think taking tests is hard, try making them up 
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. I like to keep things light and the problem with this type of communication is that you cannot see the smirk on my face (avatar notwithstanding) when I say stuff.