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CCNA ICND2 Cert Guide: Variable Length Subnet Masks

Dec 19, 2011

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CCNA ICND2 640-816 Official Cert Guide, 3rd Edition

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Exam Preparation Tasks

Review All the Key Topics

Review the most important topics from this chapter, noted with the Key Topics icon in the outer margin of the page. Table 5-8 lists a reference of these key topics and the page numbers on which each is found.

Table 5-8. Key Topics for Chapter 5

Key Topic Element	Description	Page Number
Table 5-2	Classless and classful routing protocols listed and compared	203
List	Steps to analyze an existing design to discover any VLSM overlaps	206
List	Steps to follow when adding a new subnet to an existing VLSM design	209
Paragraph	Statement of the main VLSM subnet assignment strategy or assigning the largest subnets first	214
List	Steps to follow to design a subnet plan using VLSM	218

Complete the Tables and Lists from Memory

Print a copy of Appendix J, “Memory Tables,” (found on the DVD) or at least the section for this chapter, and complete the tables and lists from memory. Appendix K, “Memory Tables Answer Key,” also on the DVD, includes completed tables and lists to check your work.

Definitions of Key Terms

Define the following key terms from this chapter and check your answers in the Glossary:

classful routing protocol, classless routing protocol, overlapping subnets, variable length subnet masks (VLSM)

Read Appendix G Scenarios

Appendix G, “Additional Scenarios,” contains five detailed scenarios that both give you a chance to analyze different designs, problems, and command output and show you how concepts from several different chapters interrelate. Appendix G Scenario 1, Part A, and all of Scenario 5 provide an opportunity to practice and develop skills with VLSM.

Appendix D Practice Problems

Appendix D, “Practice for Chapter 5: Variable Length Subnet Masks,” lists additional practice problems and answers. You can find this appendix on the DVD as a printable PDF.

Answers to Earlier Practice Problems

Answers to Practice Finding VLSM Overlaps

This section lists the answers to the five practice problems in the section, “Practice Finding VLSM Overlaps,” as listed earlier in Table 5-4. Note that the tables that list details of the answer reordered the subnets as part of the process.

In Problem 1, the second and third subnet IDs listed in Table 5-9 happen to overlap. The second subnet’s range completely includes the range of addresses in the third subnet.

Table 5-9. VLSM Overlap Problem 1 Answers (Overlaps Highlighted)

Reference	Original Address and Mask	Subnet ID	Broadcast Address
1	10.1.1.1/20	10.1.0.0	10.1.15.255
2	10.1.17.1/21	`10.1.16.0`	`10.1.23.255`
3	10.1.23.254/22	`10.1.20.0`	`10.1.23.255`
4	10.1.29.101/23	10.1.28.0	10.1.29.255
5	10.1.34.9/22	10.1.32.0	10.1.35.255

In Problem 2, again, the second and third subnet IDs (listed in Table 5-10) happen to overlap, and again, the second subnet’s range completely includes the range of addresses in the third subnet. Also, the second and third subnet IDs are the same value, so the overlap is more obvious.

Table 5-10. VLSM Overlap Problem 2 Answers (Overlaps Highlighted)

Reference	Original Address and Mask	Subnet ID	Broadcast Address
1	172.16.122.1/30	172.16.122.0	172.16.122.3
2	172.16.122.57/27	`172.16.122.32`	`172.16.122.63`
3	172.16.122.33/30	`172.16.122.32`	`172.16.122.35`
4	172.16.126.151/22	172.16.124.0	172.16.127.255
5	172.16.128.151/20	172.16.128.0	172.16.143.255

In Problem 3, three subnets overlap. Subnet 1’s range completely includes the range of addresses in the second and third subnets. Note that the second and third subnets do not overlap with each other, so for the process in this book to find all the overlaps, after you find that the first two subnets overlap, you should compare the next entry in the table (3) with both of the two known-to-overlap entries (1 and 2).

Table 5-11. VLSM Overlap Problem 3 Answers (Overlaps Highlighted)

Reference	Original Address and Mask	Subnet ID	Broadcast Address
1	192.168.1.113/28	`192.168.1.112`	`192.168.1.127`
2	192.168.1.122/30	`192.168.1.120`	`192.168.1.123`
3	192.168.1.125/30	`192.168.1.124`	`192.168.1.127`
4	192.168.1.245/29	192.168.1.240	192.168.1.247
5	192.168.1.253/30	192.168.1.252	192.168.1.255

Answers to Practice Adding VLSM Subnets

This section lists the answers to the five practice problems in the section, “Practice Adding VLSM Subnets.”

All five problems for this section used the same set of five pre-existing subnets. Table 5-12 lists those subnet IDs and subnet broadcast addresses, which define the lower and higher ends of the range of numbers in each subnet.

Table 5-12. Pre-Existing Subnets for the Add a VLSM Subnet Problems in This Chapter

Subnet	Subnet Number	Broadcast Address
1	10.0.0.0/24	10.0.0.255
2	10.0.1.0/25	10.0.1.127
3	10.0.2.0/26	10.0.2.63
4	10.0.3.0/27	10.0.3.31
5	10.0.6.0/28	10.0.6.15

The rest of the explanations follow the five-step process outlined earlier in the section, “Finding VLSM Subnets,” except that the explanations ignore Step 3 because Step 3’s results in each case are already listed in Table 5-12.

Problem 1

Step 1 The problem statement tells us to use /24.

Step 2 The subnets would be 10.0.0.0, 10.0.1.0, 10.0.2.0, 10.0.3.0, 10.0.4.0, 10.0.5.0, and so on, counting by 1 in the third octet.

Step 3 The first four new possible subnets (10.0.0.0/24, 10.0.1.0/24, 10.0.2.0/24, and 10.0.3.0/24) all overlap with the existing subnets (see Table 5-12). 10.0.6.0/24 also overlaps.

Step 4 10.0.4.0/24 is the numerically lowest new subnet number that does not overlap with the existing subnets.

Problem 2

Step 1 The problem statement tells us to use /23.

Step 2 The subnets would be 10.0.0.0, 10.0.2.0, 10.0.4.0, 10.0.6.0, 10.0.8.0, and so on, counting by 2 in the third octet.

Step 3 Three of the first four new possible subnets (10.0.0.0/23, 10.0.2.0/23, and 10.0.6.0/23) all overlap with existing subnets.

Step 4 10.0.4.0/23 is the numerically lowest new subnet number that does not overlap with the existing subnets.

Problem 3

Step 1 The problem statement tells us to use /22.

Step 2 The subnets would be 10.0.0.0, 10.0.4.0, 10.0.8.0, 10.0.12.0, and so on, counting by 4 in the third octet.

Step 3 The first two new possible subnets (10.0.0.0/22, 10.0.4.0/22) overlap with existing subnets.

Step 4 10.0.8.0/22 is the numerically lowest new subnet number that does not overlap with the existing subnets.

Problem 4

The answer for this problem requires more detail than others, because the /25 mask creates a larger number of subnets that might overlap with the pre-existing subnets. For this problem, at Step 1, you already know to use mask /25. Table 5-13 shows the results of Step 2, listing the first 14 subnets of network 10.0.0.0 when using mask /25. For Step 4, Table 5-13 highlights the overlapped subnets. To complete the task at Step 5, search the table sequentially and find the first non-grayed subnet, 10.0.1.128/25.

Table 5-13. First 14 Subnets of Network 10.0.0.0, Using /25 Mask

Reference	Subnet Number	Broadcast Address
1	`10.0.0.0`	`10.0.0.127`
2	`10.0.0.128`	`10.0.0.255`
3	`10.0.1.0`	`10.0.1.127`
4	10.0.1.128	10.0.1.255
5	`10.0.2.0`	`10.0.2.127`
6	10.0.2.128	10.0.2.255
7	`10.0.3.0`	`10.0.3.127`
8	10.0.3.128	10.0.3.255
9	10.0.4.0	10.0.4.127
10	10.0.4.128	10.0.4.255
11	10.0.5.0	10.0.5.127
12	10.0.5.128	10.0.5.255
13	`10.0.6.0`	`10.0.6.127`
14	10.0.6.128	10.0.6.255

Problem 5

Like Problem 4, the answer for Problem 5 requires more detail, because the /26 mask creates a larger number of subnets that might overlap with the pre-existing subnets. For this problem, at Step 1, you already know to use mask /26. Table 5-14 shows the results of Step 2, listing the first 12 subnets of network 10.0.0.0 when using mask /26. For Step 4, Table 5-14 highlights the overlapped subnets. To complete the task at Step 5, search the table sequentially and find the first non-grayed subnet, 10.0.1.128/26.

Table 5-14. First 12 Subnets of Network 10.0.0.0, Using /26 Mask

Reference	Subnet Number	Broadcast Address
1	`10.0.0.0`	`10.0.0.63`
2	`10.0.0.64`	`10.0.0.127`
3	`10.0.0.128`	`10.0.0.191`
4	`10.0.0.192`	`10.0.0.255`
5	`10.0.1.0`	`10.0.1.63`
6	`10.0.1.64`	`10.0.1.127`
7	10.0.1.128	10.0.1.191
8	10.0.1.192	10.0.1.255
9	`10.0.2.0`	`10.0.2.63`
10	10.0.2.64	10.0.2.127
11	10.0.2.128	10.0.2.191
12	10.0.2.192	10.0.2.255

Answers to Practice Designing VLSM Subnets

This section lists the answers to the two practice problems in the section, “Practice Designing VLSM Subnets.”

Answers for VLSM Subnet Design, Problem 1

For Problem 1, subnetting network 172.20.0.0 with mask /22 means that the subnets will all be multiples of 4 in the third octet: 172.20.0.0, 172.20.4.0, 172.20.8.0, and so on, through 172.20.252.0. Following the rule to choose the numerically lowest subnet IDs, you would allocate or use 172.20.0.0/22, 172.20.4.0/22, and 172.20.8.0/22. You would also then mark the next subnet, 172.20.12.0/22, to be sub-subnetted.

For the next mask, /25, all the subnet IDs will be either 0 or 128 in the last octet, and increments of 1 in the third octet. Starting at 172.20.12.0 per the previous paragraph, the first four such subnets are 172.20.12.0/25, 172.20.12.128/25, 172.20.13.0/25, and 172.20.13.128/25. Of these, you need to use three, so mark the first three as used. The fourth will be sub-subnetted at the next step.

For the third and final mask, /30, all the subnet IDs will increment by 4 in the fourth octet. Starting with the subnet ID that will be sub-subnetted (172.20.13.128), the next /30 subnet IDs are 172.20.13.128, 172.20.13.132, 172.20.13.136, 172.20.13.140, and so on. The first three in this list will be the three used per the requirements and rules for Problem 1.

Answers for VLSM Subnet Design, Problem 2

For Problem 1, subnetting network 192.168.1.0 with mask /27 means that the subnets will all be multiples of 32 in the fourth octet: 192.168.1.0, 192.168.1.32, 192.168.1.64, 192.168.1.96, and so on, through 192.168.1.224. Following the rule to choose the numerically lowest subnet IDs, you would allocate or use 192.168.1.0/27, 192.168.1.32/27, and 192.168.1.64/27. You would also then mark the next subnet, 192.168.1.96/27, to be sub-subnetted.

For the next mask, /28, all the subnet IDs will be multiples of 16 in the last octet. Starting at 192.168.1.96 per the previous paragraph, the first four such subnets are 192.168.1.96, 192.168.1.112, 192.168.1.128, and 192.168.1.144. Of these, you need to use three, so mark the first three as used. The fourth will be sub-subnetted at the next step.

For the third and final mask, /30, all the subnet IDs will increment by 4 in the fourth octet. Starting with the subnet ID that will be sub-subnetted (192.168.1.144), the next /30 subnet IDs are 192.168.1.144, 192.168.1.148, 192.168.1.152, 192.168.1.156, and so on. The first three in this list will be the three used per the requirements and rules for Problem 1.

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