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  1. Junior Member Registered Member
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    #1

    Default microsoft press book, need help on some question

    any1 who have the microsoft press 70-291 book, please turn to page 2-52 and page 2-53,
    Case Scenario exercise question 2: What is the configuration error and question 3:What is the configuration error. What is the step should be used to find out the answer, although the answer is given at the end of the chapter.

    I though i am confident on the subnetting, but i still can't don't know how to find the answer.thanks.
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  3. Member CompuTron99's Avatar
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    #2
    I went back and looked. I Had number 1 and number 3 correct.

    #2 I've gone over this a few times and I am stumped as to why the correct answer is what MS has in the book. I checked the site that has the corrections, but that wasn't listed as a mistake.

    #3 The router IP address for Clients C and D is 131.107.120.1/21 (255.255.248.0). This means that for the computers to be on the same network, they must range between 131.107.120.1 - 131.107.127.254. Client D had an IP address of 131.107.128.1/21, which is not in the same network.
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  4. Junior Member Registered Member
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    #3
    thanks for answers,CompuTron99. that means q2 has no answer?
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  5. Junior Member Registered Member
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    #4

    Default Sometimes books are wrong.

    Im not saying the book is wrong this time but I will say this! error on 2-53 I did figure out and the book is right. the sub-nets are separated by a value of 8 as indicated by the mask .248 client C range is 127.1 to 127.6 and Client B is range is 128.1 to 128.6

    I spent an hour and a half looking at 2-52 config error and using 224 seems ta work as well as 240 but Im sure im missing something. After all After nearly passing the 70-291 the first time out, I have now failed it more times than I care to say but I will not stop. Using MS Press, Measureup, Exam cram and also have Server 2003
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  6. Senior Member Devilsbane's Avatar
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  7. Junior Member Registered Member
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    #6
    I know it's been a long time, but the way the thread is now, Q2 is left unanswered. You are understanding Q1 & Q3, so I'll just look at Q2. [By the way, DB, 2nd ed has Q2 on 2-54 & Q3 on 2-55, so I assume near... was using 1st ed, but what is spoken of fits 2nd ed except for the page numbers.]
    Actually, to explain Q2, I'll point out that there are multiple ways to fix Q3: Fix Client D's IP, or change the router interface's IP and Client C's IP, and also the default gateway/router on both C & D. For simplicity they tell you to fix Client D's IP - and the real test will always want what MS considers the simplest (in the sense of least changes, etc), at least when it matches the MS Best Practices papers.
    So, on to Q2. But first, more explanation: skip to the non-italics if you don't need this extra detail.
    Clients A-D are all using a /27, which is the same as a SNM of 255.255.255.224. /24 would have the last byte as 0, /25's last byte would be 128, /26 would be 192, /27 - 224, /28 - 240. [If you don't understand why, ask, because this is clearer if you understand the basics. Somebody on here is sure to explain.] If the SNM is 128 you have a max of two subnets the 0 (in this example 192.168.2.0) and the 128 [192.168.2.128]. If the SNM is 192 you have a max of four subnets the 0 (in this example 192.168.2.0), the 64 (192.168.2.64), the 128 [192.168.2.128], the 192 (192.168.2.192), and the 128 [192.168.2.128]. Etc. In other words, if SNM=128, starting with 0 and going to 128, count by 128s. If SNM=192, count by 64s going from 0 to 192 (0, 64, 128, 192). If SNM=224, count by 32s from 0 to 224 (0, 32, 64, 96, 128, 160, 192, 224), If SNM=240, count by 16s from 0 to 240. Etc.
    OK, notice that /27 (SNM=224) means each subnet starts at a multiple of 32, so possible subnets include 0-31 (with available IPs of 1-30 because 0 is the network ID and 31 is the broadcast for that subnet), etc. Another possible subnet would be the 160 network, with 161-190 being available IPs and 191 being the broadcast. Clients A-D, and both internal interfaces on the router are in that 161-190 range. So they must ALL be on the same router interface - but the diagram shows they aren't. Here are two ways to fix the problem.
    1) Change the IPs for Clients A & B & the router interface, plus the DGW for Clients A & B, to numbers that fit on another subnet (say the 128 subnet, in other words, between 129 and 158. [Yes, you could do it for Clients C & D & their router interface. So change my answer 1 to answer 1a, and call changing C & D option 1b]
    2) Or change the SNM to 240 (=/28 on everything except the public router interface. Now Clients A & B are on the 176 network (IPs 177-190) and clients C & D are on the 160 network (clients 161-174).
    MS considers answer 2 to be more elegant, meaning easier to change 1 digit on each of 6 spots rather than 5 entire last bytes. You could argue that answer 1 is more elegant & I would neither agree nor disagree, but if you want to get credit for the 70-291, you'd be better off choosing answer 2. MS calls it the "best" answer, and though both work, and sometimes in the real world you need answer 1 (because it's not 2 clients on each but 20 or 30), in other cases in the real world answer 2 would be better (smaller subnets, less traffic, etc). But for the test, the answer the MS Press book gives is "best."
    I hope you read through all this and see that the MS Press book DOES give a valid answer, but only time can get you to the place that you can accept that (at least for the MS test) the MS answer is the "best" of the options given (though even they can make mistakes on the real test).
    Last edited by JastMe; 02-16-2013 at 03:45 AM. Reason: To fix a couple typos from 8) turning into emoticons
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