You need to configure a single router into load balancing traffic across 4 unequal cost paths. Which routing protocols can satisfy this requirement? (

A. RIP v1
B. RIP v2
C. IGRP
D. EIGRP
E. OSPF

F. IS-IS
Answer: C, D
Explanation:
In general, load balancing is the capability of a router to distribute traffic over all its network ports that are the same distance from the destination address. Load balancing increases the utilization of network segments, thus increasing effective network bandwidth. There are two types of load balancing: equal cost path and unequal cost path.

Every routing protocol supports equal cost path load balancing. In addition to that, IGRP and EIGRP also support unequal cost path load balancing, which is known as variance. The variance command instructs the router to include routes with a metric less than n times the minimum metric route for that destination, where n is the number specified by the variance command. The variable n can take a value between 1 and 128, with the default being 1, which means equal cost load balancing (variance for example. Traffic is also distributed proportionally among unequal cost links, with respect to the metric.

You are a network administrator and you need to implement a routing protocol on your network that provides:

* Scalability
* VLSM support
* Minimal overhead
* Support for connecting networks using routers of multiple vendors

Which of the following routing protocol would best serve your needs?

A. VTP
B. RIP version 1
C. EIGRP
D. OSPF
E. IGRP
F. CDP

Answer: D

Explanation:
Since one of the requirements is that the routing protocol must support other vendors, our only choices are RIP and OSPF. Since RIP version 1 does not support VLSM, OSPF is the only choice.

Incorrect Answers:
A. VTP is the VLAN Trunking Protocol. This is not a routing protocol.
B. RIP version one does not support VLSM. Note that RIPv2 does support VLSM, and would be a valid choice.
C, E. EIGRP and IGRP are Cisco proprietary routing protocols, and are not supported by other vendors.
F. CDP is the Cisco Discovery Protocol, which is used to exchange information between Cisco devices. It can only be used between Cisco routers and switches, and it is not a routing protocol.

In network that support VLSM, which network mask should be used for point-to-point WAN links in order to reduce waste of IP addresses?

A. /24
B. /30
C. /27
D. /26
E. /32

Answer: B
Explanation:A 30-bit mask is used to create subnets with two valid host addresses. This is the exact number needed for a point-to-point connection.

What is the maximum number of IP addresses that can be assigned to hosts on a local subnet that use the 255.255.255.224 subnet mask?

A. 14
B. 15
C. 16
D. 30
E. 31
F. 32

Answer: D

Explanation:
The subnet mask 255.255.255.224 means that there are 27 network bits. The remaining 5 bits are the host bits. The maximum possible combination's with 5 bits are 25 = 32. As all zero's and all one's hosts are not allowed so, maximum number of valid hosts with the mask 255.255.255.224 are 25 -2 =32-2 = 30 Hosts

If an Ethernet port on router was assigned an IP address of 172.16.112.1/20, what is the maximum number of hosts allowed on this subnet?

A. 1024
B. 2046
C. 4094
D. 4096
E. 8190

Answer: C

Explanation:
Given IP address of 172.16.112.1 / 20,
subnet mask: 255.255.240.0
max. num of hosts =(( 2^12) -2 ) = 4096-2 = 4094

Which of the following IP addresses fall into the CIDR block of 115.54.4.0/22? Select three

A. 115.54.8.32
B. 115.54.7.64
C. 115.54.6.255
D. 115.54.3.32
E. 115.54.5.128
F. 115.54.12.128

Answer: B, C, E

Explanation:
Given CIDR block of 115.54.4.0 /22:
subnet mask : 255.255.252.0
theIP address range would be 115.54.4.1 to 115.54.7.254.
Therefore, 115.54.5.128 (E),115.54.6.255 (C) and 115.54.7.64 (B) are correct.

Which of the following IP addresses is a private IP address? Select all that apply.

A. 12.0.0.1
B. 168.172.19.39
C. 172.20.14.36
D. 172.33.194.30
E. 192.168.42.34
Answer: C, E

What will be the result of the DR and BDR elections for this single area OSPF network? (Choose three.)

• HQ will be DR for 10.4.0.0/16.
• Router A will be DR for 10.4.0.0/16.*
• HQ will be BDR for 10.4.0.0/16.*
• Router A will be DR for 10.5.0.0/16.
• Remote will be DR for 10.5.0.0/16.*
• Remote will be BDR for 10.5.0.0/16.

Why is it difficult for routing loops to occur in networks that use link-state routing?

1. Each router builds a simple view of the network based on hop count.
2. Routers flood the network with LSAs to discover routing loops.
3. Each router builds a complete and synchronized view of the network.*
4. Routers use hold-down timers to prevent routing loops.

What speeds up convergence in a network using link-state routing?

• updates triggered by network changes *
• updates sent at regular intervals
• updates sent only to directly connected neighbors
• updates that include complete routing tables

What does OSPF use to reduce the number of exchanges of routing information in networks where large numbers of neighbors are present? (Choose two.)

• root router
• backup root router
• domain router
• backup domain router
• designated router*
• backup designated router*

Which of the following are disadvantages of link-state routing? (Choose three.)

• Link-state protocols require careful network design.*
• Link-state protocols are prone to routing loops.
• Link-state hello updates can cause broadcast flooding.
• Link-state protocols place significant demands on router processors and memory resources.
• Link-state protocols require a knowledgeable network administrator.*
• Link-state protocols do not support Variable Length Subnet Masking.*

Which command sequence will enable OSPF in the backbone area for the two Ethernet links on RouterA?

RouterA(config)# router ospf 1
RouterA(config-router)# network 172.16.4.0 0.0.1.255 area 1

RouterA(config)# router ospf 10*
RouterA(config-router)# network 172.16.4.0 0.0.0.255 area 0
RouterA(config-router)# network 172.16.5.0 0.0.0.255 area 0

RouterA(config)# router ospf 10
RouterA(config-router)# area 0 network 172.16.4.0 0.0.1.255

RouterA(config)# router ospf 0
RouterA(config-route)# network 172.16.4.0 0.0.255.255 area 0

RouterA(config)# router ospf 0
RouterA(config-router)# network 172.16.4.0 255.255.255.0
RouterA(config-router)# network 172.16.5.0 255.255.255.0

RouterA(config)# router ospf 1
RouterA(config-router)# network 172.16.4.0 255.255.254.0 area 0

What information can be obtained from the output of the show ip ospf interface command? (Choose three.)

1. link-state age intervals
2. timer intervals*
3. router ID number*
4. link-state update intervals
5. neighbor adjacencies*

The routers in the diagram are configured as shown. The loopback interface on router R1 is labeled as lo0. All OSPF priorities are set to the default

The routers in the diagram are configured as shown. The loopback interface on router R1 is labeled as lo0. All OSPF priorities are set to the default except for Ethernet0 of router R2, which has an OSPF priority of 2. What will be the result of the OSPF DR/BDR elections on the 192.1.1.0 network? (Choose two.)

1. R1 will be the DR
2. R1 will be the BDR
3. R2 will be the DR*
4. R2 will be the BDR
5. R3 will be the DR
6. R3 will be the BDR*

Because of security concerns, a network administrator wants to set up authentication between two routers. Which of the following commands will configu

Because of security concerns, a network administrator wants to set up authentication between two routers. Which of the following commands will configure Router_A to trust Router_B with a clear text password of apollo?



Router_A(config-if)# ospf authentication-key apollo

Router_A(config-if)# ip authentication-key apollo

Router_A(config-if)# ip ospf authentication apollo

Router_A(config-if)# ip ospf authentication-key apollo*

Which of the following are required when adding a network to the OSPF routing process configuration? (Choose three.)

1. network address*
2. loopback address
3. autonomous system number
4. subnet mask
5. wildcard mask*
6. area ID*

What are the benefits of the hierarchical design approach that is used in large OSPF networks? (Choose three.)

1. reduction of routing overhead *
2. increased routing overhead
3. faster convergence*
4. slower convergence
5. distribution of network instability to all areas of the network
6. isolation of network instability to one area of the network*

2.A network administrator has configured a default route on Router_A but it is not being shared with adjacent Router_B and the other routers in the O

network administrator has configured a default route on Router_A but it is not being shared with adjacent Router_B and the other routers in the OSPF area. Which command will save the administrator the time and trouble of configuring this default route on Router_B and all of the other routers in the OSPF area?


Router_A(config-router)# ospf redistribute default-route

Router_B(config-router)# ospf redistribute default-route

Router_A(config-router)# default-information originate *

Router_B(config-router)# default-information originate

Router_A(config-router)# ip ospf update-default

Router_B(config-router)# ip ospf update-default

What does OSPF use to calculate the cost to a destination network?

• bandwidth *
• bandwidth and hop count
• bandwidth and reliability
• bandwidth, load, and reliablity

How many subnetworks and hosts are available per subnet if you apply a /28 mask to the 210.10.2.0 class C network?

A. 30 networks and 6 hosts.
B. 6 networks and 30 hosts.
C. 8 networks and 32 hosts.
D. 32 networks and 18 hosts.
E. 14 networks and 14 hosts.
F. None of the above

Answer: E

Explanation:
A 28 bit subnet mask (11111111.11111111.11111111.11110000) applied to a class C network uses a 4 bits for networks, and leaves 4 bits for hosts. Using the 2n-2 formula, we have 24-2 (or 2x2x2x2-2) which gives us 14 for both the number of networks, and the number of hosts.

Incorrect Answers:
A. This would be the result of a /29 (255.255.255.248) network.
B. This would be the result of a /27 (255.255.255.224) network.
C. This is not possible, as we must subtract two from the subnets and hosts for the network and broadcast addresses.
D. This is not a possible combination of networks and hosts.

Which of the following is an example of a valid unicast host IP address?

A. 172.31.128.255./18
B. 255.255.255.255
C. 192.168.24.59/30
D. FFFF.FFFF.FFFF
E. 224.1.5.2
F. All of the above

Answer: A

Explanation
The address 172.32.128.255 /18 is 10101100.00100000.10|000000.11111111 in binary, so this is indeed a valid host address.

Incorrect Answers:
B. This is the all 1's broadcast address.
C. Although at first glance this answer would appear to be a valid IP address, the /30 means the network mask is 255.255.255.252, and the 192.168.24.59 address is the broadcast address for the 192.168.24.56/30 network.
D. This is the all 1's broadcast MAC address
E. This is a multicast IP address.

Which two of the addresses below are available for host addresses on the subnet 192.168.15.19/28? (Select two answer choices)

A. 192.168.15.17
B. 192.168.15.14
C. 192.168.15.29
D. 192.168.15.16
E. 192.168.15.31
F. None of the above
Answer: A, C
Explanation:
The network uses a 28bit subnet (255.255.255.240). This means that 4 bits are used for the networks and 4 bits for the hosts. This allows for 14 networks and 14 hosts (2n-2).The last bit used to make 240 is the 4th bit (16) therefore the first network will be 192.168.15.16. The network will have 16 addresses (but remember that the first address is the network address and the last address is the broadcast address). In other words, the networks will be in increments of 16 beginning at 192.168.15.16/28. The IP address we are given is 192.168.15.19. Therefore the other host addresses must also be on this network. Valid IP addresses for hosts on this network are: 192.168.15.17-192.168.15.30.

Incorrect Answers:
B. This is not a valid address for this particular 28 bit subnet mask. The first network address should be 192.168.15.16.
D. This is the network address.
E. This is the broadcast address for this particular subnet.

The MAC address for your PC NIC is: C9-3F-32-B4-DC-19. What is the address of the OUI portion of this NIC card, expressed as a binary number?

A. 11001100-00111111-00011000
B. 11000110-11000000-00011111
C. 11001110-00011111-01100000
D. 11001001-00111111-00110010
E. 11001100-01111000-00011000
F. 11111000-01100111-00011001

Answer: D

Explanation:
The first half of the address identifies the manufacturer of the card. This code, which is
assigned to each manufacturer by the IEEE, is called the organizationally unique
identifier (OUI). In this example, the OUI is C9-3F-32. If we take this number and
convert it to decimal form we have:

C9 = (12x16) + 9 = 201
3F = (3x16) + 15 = 63
32 = (3x16) + 2 = 50

So, in decimal we have 201.63.50. If we then convert this to binary, we have:
201 = 11001001
63 = 00111111
50 = 00110010

So the correct answer is D: 11001001-00111111-00110010

How would the number 231 be expressed as a binary number?How would the number 231 be expressed as a binary number?

A. 11011011
B. 11110011
C. 11100111
D. 11111001
E. 11010011


Answer: C
Explanation
Decimal number 231 equates to 11100111 in binary (128+64+32+0+0+4+2+1)
Incorrect Answers:
A: Binary 11011011 equates to 219 (128+64+0+16+8+0+2+1)
B: Binary 11110011 equates to 243 (128+64+32+16+0+0+2+1)
D: Binary 11101011 equates to 249 (128+64+32+16+8+0+0+1)
E: Binary 11010011 equates to 211 (128+64+0+16+0+0+2+1)

Which one of the binary number ranges shown below corresponds to the value of the first octet in Class B address range?

A. 10000000-11101111
B. 11000000-11101111
C. 10000000-10111111
D. 10000000-11111111
E. 11000000-10111111
Answer: C

Explanation:

Class B addresses are in the range 128.0.0.0 through 191.255.255.255.
In binary, the first octet (128 through 191) equates to 10000000-10111111


Incorrect Answers:
A. Binary 10000000 does equate to 128 but binary 11101111 equates to 239
B. Binary 11000000 equates to 192 and binary 11101111 equates to 239
D. Binary 10000000 does equate to 128 but binary 11011111 equates to 223
E. Binary 11000000 equates to 192 but binary 10111111 does equate to 191

The subnet mask on the serial interface of a router is expressed in binary as 11111000 for the last octet. How do you express the binary number 111110

The subnet mask on the serial interface of a router is expressed in binary as
11111000 for the last octet. How do you express the binary number 11111000 in
decimal?

A. 210
B. 224
C. 240
D. 248
E. 252
Answer: D
Explanation:
128 + 64+32+16+8 = 248. Since this is the last octet of the interface, the subnet mask
would be expressed as a /29.

You have the binary number 10011101. Convert it to its decimal and hexadecimal equivalents. (Select two answer choices)A. 158 B. 0x9D C. 156 D. 157 E.

A. 158
B. 0x9D
C. 156
D. 157
E. 0x19
F. 0x9F
Answer: B, D
Explanation:
10011101 = 128+0+0+16+8+4+0+1 = 157
For hexadecimal, we break up the binary number 10011101 into the 2 parts:
1001 = 9 and 1101 = 13, this is D in hexadecimal, so the number is 0x9D. We can further verify by taking the hex number 9D and converting it to decimal by taking 16 times 9, and then adding 13 for D (0x9D = (16x9)+13 = 157).

Configuring HDLC

Considering the lowest three layers of the OSI reference model on router Ethernet interfaces for a moment, there are no required configuration commands related to Layers 1 and 2 for the interface to be up and working, forwarding IP traffic. The Layer 1 details occur by default once the cabling has been installed correctly. Router IOS defaults to use Ethernet as the data link protocol on all types of Ethernet interfaces, so no Layer 2 commands are required. To make the interface operational for forwarding IP packets, the router needs one command to configure an IP address on the interface, and possibly a no shutdown command if the interface is in an “administratively down” state.
Similarly, serial interfaces on Cisco routers that use HDLC typically need no specific Layer 1 or 2 configuration commands. The cabling needs to be completed as described in Chapters 4 and 16, but there are no required configuration commands related to Layer 1.
IOS defaults to use HDLC as the data link protocol, so there are no required commands that relate to Layer 2. As on Ethernet interfaces, the only required command to get IP working on the interface is the ip address command and possibly the no shutdown command.

Step 1 Configure the interface IP address using the ip address interface subcommand.

Step 2 The following tasks are required only when the specifically listed conditions are true:

a. If an encapsulation protocol interface subcommand that lists a protocol besides HDLC already exists on the interface, use the encapsulation hdlc interface subcommand to enable HDLC.

b. If the interface line status is administratively down, enable the interface using the no shutdown interface subcommand.
c. If the serial link is a back-to-back serial link in a lab (or a simulator), configure
the clocking rate using the clock rate speed interface subcommand, but only on the one router with the DCE cable (per the show controllers serial number command).

Step 3 The following steps are always optional, and have no impact on whether the link works and passes IP traffic:

a. Configure the link’s speed using the bandwidth speed-in-kbps interface subcommand.

b. For documentation purposes, configure a description of the purpose of the interface using the description text interface subcommand.

Configuring Point-to-Point WANs

This brief section explains how to configure leased lines between two routers, using both HDLC and PPP. The required configuration is painfully simply—for HDLC, do nothing, and for PPP, add one interface subcommand on each router’s serial interface (encapsulation ppp). However, several optional configuration steps can be useful, so this section explains those optional steps and their impact on the links.

UTP Cables and RJ-45 Connectors

The UTP cabling used by popular Ethernet standards include either two or four pairs of wires. Because the wires inside the cable are thin and brittle, the cable itself has an outer jacket of flexible plastic to support the wires. Each individual copper wire also has a thin plastic coating to help prevent the wire from breaking. The plastic coating on each wire has a different color, making it easy to look at both ends of the cable and identify the ends of an individual wire.

Ethernet UTP Cabling

The three most common Ethernet standards used today—10BASE-T (Ethernet), 100BASE-TX (Fast Ethernet, or FE), and 1000BASE-T (Gigabit Ethernet, or GE)—use UTP cabling. Some key differences exist, particularly with the number of wire pairs needed in each case, and in the type (category) of cabling. This section examines some of the details of UTP cabling, pointing out differences among these three standards along the way. In particular, this section describes the cables and the connectors on the ends of the cables, how they use the wires in the cables to send data, and the pinouts required for proper operation.

Building 10BASE-T Networks with Hubs

The IEEE later defined new Ethernet standards besides 10BASE5 and 10BASE2.
Chronologically, the 10BASE-T standard came next (1990), followed by 100BASE-TX (1995), and then 1000BASE-T (1999). To support these new standards, networking devices called hubs and switches were also created. This section defines the basics of how these three popular types of Ethernet work, including the basic operation of hubs and switches.
10BASE-T solved several problems with the early 10BASE5 and 10BASE2 Ethernet specifications. 10BASE-T allowed the use of UTP telephone cabling that was already installed. Even if new cabling needed to be installed, the inexpensive and easy-to-install UTP cabling replaced the old expensive and difficult-to-install coaxial cabling.

carrier sense multiple access with collision detection (CSMA/CD) algorithm

Network uses a single bus, if two or more electrical signals were sent at the same time, they would overlap and collide, making both signals unintelligible. So, unsurprisingly, Ethernet also defined a specification for how to ensure that only one device sends traffic on the Ethernet at one time. Otherwise, the Ethernet would have beenunusable. This algorithm, known as the carrier sense multiple access with collision detection (CSMA/CD) algorithm, defines how the bus is accessed.

In human terms, CSMA/CD is similar to what happens in a meeting room with many attendees. It’s hard to understand what two people are saying at the same time, so generally, one person talks and the rest listen. Imagine that Bob and Larry both want to reply to the current speaker’s comments. As soon as the speaker takes a breath, Bob and Larry both try to speak. If Larry hears Bob’s voice before Larry makes a noise, Larry might stop and let Bob speak. Or, maybe they both start at almost the same time, so they talk over each other and no one can hear what is said. Then there’s the proverbial “Pardon me; go ahead with what you were saying,” and eventually Larry or Bob talks. Or perhaps another person jumps in and talks while Larry and Bob are both backing off. These “rules” are based on your culture; CSMA/CD is based on Ethernet protocol specifications and achieves the same type of goal.

The Original Ethernet Standards: 10BASE2 and 10BASE5

Ethernet is best understood by first considering the two early Ethernet specifications, 10BASE5 and 10BASE2. These two Ethernet specifications defined the details of the physical and data link layers of early Ethernet networks. (10BASE2 and 10BASE5 differ in their cabling details, but for the discussion in this chapter, you can consider them as behaving identically.) With these two specifications, the network engineer installs a series of coaxial cables connecting each device on the Ethernet network. There is no hub, switch, or wiring panel. The Ethernet consists solely of the collective Ethernet NICs in the computers and the coaxial cabling. The series of cables creates an electrical circuit, called a bus, which is shared among all devices on the Ethernet. When a computer wants to send some bits to another computer on the bus, it sends an electrical signal, and the electricity propagates to all devices on the Ethernet.

Modern Ethernet LANs

The term Ethernet refers to a family of standards that together define the physical and data link layers of the world’s most popular type of LAN. The different standards vary as to the speed supported, with speeds of 10 megabits per second (Mbps), 100 Mbps, and 1000 Mbps (1 gigabit per second, or Gbps) being common today. The standards also differ as far as the types of cabling and the allowed length of the cabling. For example, the most commonly used Ethernet standards allow the use of inexpensive unshielded twisted-pair (UTP) cabling, whereas other standards call for more expensive fiber-optic cabling. Fiber-optic cabling might be worth the cost in some cases, because the cabling is more secure and allows for much longer distances between devices. To support the widely varying needs for building a LAN—needs for different speeds, different cabling types (trading off distance requirements versus cost), and other factors—many variations of Ethernet standards have been created.

Which of the following is one of the functions of OSI Layer 2 protocols?

a. Framing
b. Delivery of bits from one device to another
c. Error recovery
d. Defining the size and shape of Ethernet cards

Which of the following describe a shortcoming of using hubs that is improved by instead using switches?

a. Hubs create a single electrical bus to which all devices connect, causing the
devices to share the bandwidth.
b. Hubs limit the maximum cable length of individual cables (relative to switches)
c. Hubs allow collisions to occur when two attached devices send data at the same
time.
d. Hubs restrict the number of physical ports to at most eight.

6. Which of the following is a collision domain?

a. All devices connected to an Ethernet hub
b. All devices connected to an Ethernet switch
c. Two PCs, with one cabled to a router Ethernet port with a crossover cable and the
other PC cabled to another router Ethernet port with a crossover cable
d. None of the other answers is correct.

5. Which of the following is true about the CSMA/CD algorithm?

a. The algorithm never allows collisions to occur.
b. Collisions can happen, but the algorithm defines how the computers should
notice a collision and how to recover.
c. The algorithm works with only two devices on the same Ethernet.
d. None of the other answers is correct.

2. Which of the following is true about the cabling of a 10BASE2 Ethernet LAN?

a. Connect each device in series using coaxial cabling
b. Connect each device in series using UTP cabling
c. Connect each device to a centralized LAN hub using UTP cabling
d. Connect each device to a centralized LAN switch using UTP cabling

Which of the following is true about the cabling of a typical modern Ethernet LAN?

a. Connect each device in series using coaxial cabling
b. Connect each device in series using UTP cabling
c. Connect each device to a centralized LAN hub using UTP cabling
d. Connect each device to a centralized LAN switch using UTP cabling

Fundamentals of LANs

Physical and data link layer standards work together to allow computers to send bits to each other over a particular type of physical networking medium. The Open Systems Interconnection (OSI) physical layer (Layer 1) defines how to physically send bits over a particular physical networking medium. The data link layer (Layer 2) defines some rules about the data that is physically transmitted, including addresses that identify the sending device and the intended recipient, and rules about when a device can send (and when it should be silent), to name a few.

The TCP/IP and OSI Networking Models

The term networking model, or networking architecture, refers to an organized set of documents. Individually, these documents describe one small function required for a network. These documents may define a protocol, which is a set of logical rules that devices must follow to communicate. Other documents may define some physical requirements for networking, for example, it may define the voltage and current levels used on a particular cable. Collectively, the documents referenced in a networking model define all the details of how to create a complete working network.

The CCNA exams include detailed coverage of one networking model—the Transmission Control Protocol/Internet Protocol, or TCP/IP. TCP/IP is the most pervasively used networking model in the history of networking. You can find support for TCP/IP on practically every computer operating system in existence today, from mobile phones to mainframe computers. Almost every network built using Cisco products today supports TCP/IP. Not surprisingly, the CCNA exams focus heavily on TCP/IP.The ICND1 exam, and the ICND2 exam to a small extent, also covers a second networking model, called the Open System Interconnection (OSI) reference model. Historically, OSI was the first large effort to create a vendor-neutral networking model, a model that was intended to be used by any and every computer in the world. Because OSI was the first major effort to create a vendor-neutral networking architectural model, many of the terms used in networking today come from the OSI model.