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Exam Prep Questions

Question 1

One of the advantages of variable-length subnet masking is the capability to structure your network in hierarchical fashion. Which of the following represents the benefits of hierarchical addressing? (Choose two.)

  1. Reducing the amount of information stored in routing tables

  2. Enabling the use of addressing with discontiguous networks

  3. Allocating IP addresses more efficiently

  4. Allowing for classful routing algorithms

Answers A and C are correct. Hierarchical addressing schemes make routing tables smaller and allocate addresses more efficiently. Answer B is incorrect because discontiguous networks are problematic and not an advantage of a hierarchical structure. Answer D is incorrect because, although related, classful algorithms are not a direct benefit of hierarchical addressing schemes.

Question 2

Which mechanism would you use to reduce the number of routing table updates generated from implementing a large number of Class C addresses?

  1. VLSM

  2. The ip helper command

  3. CIDR

  4. OSPF

Answer C is correct. CIDR reduces the number of updates by allocating a contiguous block of addresses. Answer A is incorrect because VLSM enables you to further subnet a subnetwork. Answer B is incorrect because the ip helper command allows you to break the router rule that prevents the forwarding of broadcast packets. Answer D is incorrect because OSPF is a link-state routing protocol and not necessarily a mechanism for classless IP addressing.

Question 3

CIDR is predominantly used in which of the following scenarios?

  1. OSPF only

  2. EIGRP only

  3. Route summarization

  4. Classless routing

Answer D is correct. CIDR stands for Classless Interdomain Routing and is predominately used in classless routing scenarios. Answers A and B are incorrect because CIDR can be used with OSPF and EIGRP, among others. Answer C is incorrect because classful protocols also summarize at major network boundaries by default.

Question 4

You are presented with an IP address with a prefix of /22. How many more subnets can you add to your design if you further subnet with a VLSM mask of /27?

  1. 256

  2. 64

  3. 32

  4. 16

Answer C is correct. 27 bits – 22 bits = 5 bits. 25 = 32 subnetworks gained. Answer A is incorrect because there would need to be 8 bits separating the original subnet bit and the VLSM bit to generate 256 subnets. Answer B is incorrect because there would need to be 6 bits separating the original subnet bit and the VLSM bit to generate 64 subnets. Answer D is incorrect because there would need to be 4 bits separating the original subnet bit and the VLSM bit to generate 16 subnets as opposed to the 5 bits that generate 32 subnets.

Question 5

Which one of the following is not an address reserved for private networking according to RFC 1918?





Answer D is correct. is not within a reserved private address range. Answer A is incorrect because it is within the private Class A range of– Answer B is incorrect because it is within the private Class B range of– Answer C is incorrect because it is within the private Class C range of–

Question 6

Which formula should be used to determine the number of subnets needed on a Cisco router running Cisco IOS release 12.0 or higher?

  1. n–2

  2. 2n–2

  3. 2n

  4. 2n/2

Answer C is correct. The standard formula for subnets now is 2n subnets available. The all-ones subnet has always been legal according to the RFC, and subnet zero is enabled by default on all Cisco routers with a Cisco IOS release of 12.0 and higher. Answer A is incorrect because the math is simply wrong. Answer B is incorrect because that is the formula for calculating the available hosts. Answer D is incorrect because the math is wrong.

Question 7

Which of the following are valid terms relating to the general concept of grouping contiguous classless addresses? (Choose all that apply.)

  1. Supernetting

  2. Subnetting

  3. Aggregation

  4. Summarization

Answers A, C, and D are correct. These are all terms related to CIDR and the grouping of classless addresses. Answer B is incorrect because subnetting is the specific process of arbitrarily dividing up networks to provide a more hierarchical routing design.

Question 8

Which one of the following prefixes is most often implemented on WAN point-to-point serial links with VLSM?

  1. /16

  2. /24

  3. /28

  4. /30

Answer D is correct. This particular VLSM mask provides for only two hosts, which is perfect for WAN connections. Answer A is incorrect because it is a standard Class B mask of 16 bits of CIDR subnetting. Answer B is incorrect because it is a standard Class C mask of 24 bits of CIDR subnetting. Answer C is incorrect because, although it provides a good subnet for breaking up larger subnets, /30 is the most efficient choice for WAN links.

Question 9

Which of the following protocols supports VLSM in a Cisco routing environment? (Choose all that apply.)

  1. Static routing

  2. OSPF

  3. IS-IS

  4. EIGRP

  5. RIPv2

Answers A, B, C, D, and E are correct. Static routing, OSPF, IS-IS, EIGRP, and RIP version 2 all support VLSM. BGP4 also supports VLSM. RIPv1 and IGRP do not support VLSM.

Question 10

Which of the following statements relating to VLSM and route summarization are true? (Choose all that apply.)

  1. OSPF must have the no auto-summary entry configured because OSPF automatically summarizes IP networks.

  2. You can only further subnet a subnetwork that is currently unused.

  3. The routing protocol being used must transport the prefix length as well as use a 32-bit address.

  4. To properly implement summarization in a network, the multiple addresses must share similar low-order bits.

Answers B and C are correct. You cannot further subnet a network using VLSM if the address space is already in use. Also, VLSM and route summarization demand a 32-bit addressing routing protocol that carries the prefix length (subnet mask information). Answer A is incorrect because EIGRP, not OSPF, must have the no auto-summary entry configured. Answer D is incorrect because the multiple addresses must share similar high-order bits.


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