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Questions (August 30, 2002)
Answers (August 30, 2002)

This week's questions cover topics for Exam 70-216: Implementing and Administering a Microsoft Windows 2000 Network Infrastructure.

Questions (August 30, 2002)

Question 1
You work for an ISP, and you're in charge of allocating public IP addresses to clients. The ISP, NerdsAGoGo, has the following clients that require a specific number of IP addresses:

  • NZSA: 12 IP addresses.
  • EnriousTech: 27 IP addresses
  • Flower Mound Organic Farm Collective: 45 IP addresses
  • TruthAddict: 81 IP addresses

NerdsAGoGo has only the following subnets available:

  • Subnet Alpha: 10.250.213.0 /26
  • Subnet Beta: 10.250.214.0 /27
  • Subnet Gamma: 10.250.215.0 /28
  • Subnet Delta: 10.250.216.0 /29
  • Subnet Iota: 10.250.217.0 /24
  • Subnet Omega: 10.250.218.0 /25

Which of the following allocations should you make so as to use the fewest number of IP addresses? (Select all that apply; you can only allocate subnets once.)

  1. NZSA: subnet Gamma
  2. NZSA: subnet Beta
  3. EnriousTech: subnet Beta
  4. EnriousTech: subnet Alpha
  5. Flower Mound Organic Farm Collective: subnet Omega
  6. Flower Mound Organic Farm Collective: subnet Alpha
  7. TruthAddict: subnet Omega
  8. TruthAddict: subnet Alpha

Question 2
You work for the ISP NerdsAGoGo. You have a list of IP addresses of hosts on the ISP's network that have been hosting FTP servers that distribute content of questionable artistic merit. The Association for More Aesthetically Pleasing Art has convinced NerdsAGoGo to do something about these FTP servers\] about these FTP servers. Your job is to disregard any qualms you have about censorship and track down these servers.

You look through the list of IP addresses and try to determine which hosts are from subnet Unicron, which has the range 10.250.213.128 /27. The hosts are as follows.

  • 10.250.213.132 rodimus.nerdsagogo.com
  • 10.250.213.126 prime.nerdsagogo.com
  • 10.250.213.142 grimlock.nerdsagogo.com
  • 10.250.213.148 snarl.nerdsagogo.com
  • 10.250.213.156 swoop.nerdsagogo.com
  • 10.250.213.158 sludge.nerdsagogo.com
  • 10.250.213.164 slag.nerdsagogo.com
  • 10.250.213.180 soundwave.nerdsagogo.com

Which of these hosts are on subnet Unicron? (Select all that apply.)

  1. rodimus.nerdsagogo.com
  2. prime.nerdsagogo.com
  3. soundwave.nerdsagogo.com
  4. slag.nerdsagogo.com
  5. swoop.nerdsagogo.com
  6. sludge.nerdsagogo.com
  7. snarl.nerdsagogo.com
  8. grimlock.nerdsagogo.com

Question 3
You're looking into different subnetting schemes for your ISP, NerdsAGoGo. You're trying to break up the C range, which has the address 210.56.100.0 /24. One client requires 54 hosts on one subnet, a second client requires 114 hosts on one subnet, a third requires 25 hosts on one subnet, and a fourth requires two subnets with 12 hosts each. Which of the following subnetting schemes should you implement to satisfy these client requirements?

  1. 210.56.100.0 /25
    210.56.100.128 /25
    210.56.100.144 /26
    210.56.100.182 /27
    210.56.100.194 /27
  2. 210.56.100.228 /25  > 114
    210.56.100.64 / 26  > 54
    210.56.100.0 / 28 > 12
    210.56.100.116 / 28 > 12
    210.56.100.32 / 27 > 25
  3. 210.56.100.0 / 26
    210.56.100.64 / 26
    210.56.100.128 / 26
    210.56.100.192 / 26
    210.56.100.224 / 26
  4. 210.56.100.128 /25  > 114
    210.56.100.64 / 26  > 54
    210.56.100.0 / 28 > 12
    210.56.100.16 / 28 > 12
    210.56.100.32 / 27 > 25

Answers (August 30, 2002)

Answer to Question 1
The correct answers are A—NZSA: subnet Gamma; C—EnriousTech: subnet Beta; F—Flower Mound Organic Farm Collective: subnet Alpha; and G—TruthAddict: subnet Omega.

First, calculate the number of hosts that each subnet can support using the formula (2^(32-n))-2, where n is the number after the IP address, following the forward slash (/) character. The subnets can support the following hosts

  • Subnet Alpha: 10.250.213.0 /26 = (2^(32-n))-2 = 62
  • Subnet Beta: 10.250.214.0 /27 = (2^(32-n))-2 = 30
  • Subnet Gamma: 10.250.215.0 /28 = (2^(32-n))-2 = 14
  • Subnet Delta: 10.250.216.0 /29 = (2^(32-n))-2 = 6
  • Subnet Iota: 10.250.217.0 /24 = (2^(32-n))-2 = 254
  • Subnet Omega: 10.250.218.0 /25 = (2^(32-n))-2 = 126

After you've performed this calculation, you can see that

  • NZSA, 12 IP addresses (Gamma wastes two)
  • EnriousTech, 27 IP addresses (Beta wastes three)
  • Flower Mound Organic Farm Collective, 45 IP addresses (Alpha wastes 17)
  • TruthAddict, 81 IP addresses (Omega Wastes 45)

Answer to Question 2
The correct answers are A—rodimus.nerdsagogo.com, E—swoop.nerdsagogo.com; F—sludge.nerdsagogo.com; G—snarl.nerdsagogo.com, and H—grimlock.nerdsagogo.com. The simplest way to answer this question is to work out the range of subnet Unicron. To do so, calculate the number of hosts:

/27 = (2^(32-n))-2 = 30 hosts

Because 10.250.213.128 is the network address, it's not addressable. Therefore, the first of the 30 hosts will be 10.250.213.129, and the next subnet will start at 10.250.213.160. The broadcast address for subnet Unicron is 10.250.213.159. The addressable range will be 129 to 158, inclusive. Any hosts outside of this range aren't on this subnet.

Answer to Question 3

The correct answer is D—210.56.100.128 /25 > 114, 210.56.100.64 / 26 > 54,210.56.100.0 / 28 > 12, 210.56.100.16 / 28 > 12, 210.56.100.32 / 27 > 25. Which of the possible answers has the correct subnet masks that will allow the right number of hosts on each subnet? Remember the formula

Hosts = (2^(32-n))-2 \[where /n = n\]
/25 = 126 hosts
/26 = 62 hosts
/27 = 30 hosts
/28 = 14 hosts

One subnet is required at /25, one at /26, one at /27, and two at /28. If you get this far, you'll have reduced the options to two possibilities. The next step is to determine whether the listed subnets fit together or whether they overlap—there's no point in designing a subnetting scheme where the same host address overlaps on different subnets!

210.56.100.128 /25 = 210.56.100.129 -> 210.56.100.254 (addressable)
210.56.100.64 / 26 = 210.56.100.65 -> 210.56.100.126 (addressable)
210.56.100.0 / 28 = 210.56.100.1 -> 210.56.100.14 (addressable)
210.56.100.16 / 28 = 210.56.100.17 -> 210.56.100.30 (addressable)
210.56.100.32 / 27 = 210.56.100.33 -> 210.56.100.62 (addressable)