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ccna_glossary

CCNA Glossary

Questions for CCNA:

1 of 28 Chapter 1: Internetworking Identify the possible causes of LAN traffic congestion. Too many hosts in a broadcast domain, broadcast storms, multicasting, and low bandwidth are all possible causes of LAN traffic congestion.

Chapter 1: Internetworking

  • Define the OSI layers, understand the function of each, and describe how devices and networking protocols can be mapped to each layer.
    • You must remember the seven layers of the OSI model and what function each layer provides. The Application, Presentation, and Session layers are upper layers and are responsible for communicating from a user interface to an application. The Transport layer provides segmentation, sequencing, and virtual circuits. The Network layer provides logical network addressing and routing through an internetwork. The Data Link layer provides framing and placing of data on the network medium. The Physical layer is responsible for taking 1s and 0s and encoding them into a digital signal for transmission on the network segment.
  • Describe the difference between a collision domain and a broadcast domain.
    • Collision domain is an Ethernet term used to describe a network collection of devices in which one particular device sends a packet on a network segment, forcing every other device on that same segment to pay attention to it. With a broadcast domain, a set of all devices on a network hear all broadcasts sent on all segments.
  • Differentiate a MAC address and an IP address and describe how and when each address type is used in a network.
    • A MAC address is a hexadecimal number identifying the physical connection of a host. MAC addresses are said to operate on layer 2 of the OSI model. IP addresses, which can be expressed in binary or decimal format, are logical identifiers that are said to be on layer 3 of the OSI model. Hosts on the same physical segment locate one another with MAC addresses, while IP addresses are used when they reside on different LAN segments or subnets.
  • Understand the difference between a hub, a bridge, a switch, and a router.
    • A hub creates one collision domain and one broadcast domain. A bridge breaks up collision domains but creates one large broadcast domain. They use hardware addresses to filter the network. Switches are really just multiple‐port bridges with more intelligence; they break up collision domains but creates one large broadcast domain by default. Bridges and switches use hardware addresses to filter the network. Routers break up broadcast domains (and collision domains) and use logical addressing to filter the network.
  • Identify the functions and advantages of routers.
    • Routers perform packet switching, filtering, and path selection, and they facilitate internetwork communication. One advantage of routers is that they reduce broadcast traffic.
  • Differentiate connection‐oriented and connectionless network services and describe how each is handled during network communications.
    • Connection‐oriented services use acknowledgments and flow control to create a reliable session. More overhead is used than in a connectionless network service. Connectionless services are used to send data with no acknowledgments or flow control. This is considered unreliable.
  • Define the OSI layers, understand the function of each, and describe how devices and networking protocols can be mapped to each layer.
    • You must remember the seven layers of the OSI model and what function each layer provides. The Application, Presentation, and Session layers are upper layers and are responsible for communicating from a user interface to an application. The Transport layer provides segmentation, sequencing, and virtual circuits. The Network layer provides logical network addressing and routing through an internetwork. The Data Link layer provides framing and placing of data on the network medium. The Physical layer is responsible for taking 1s and 0s and encoding them into a digital signal for transmission on the network segment.
  • Which layer chooses and determines the availability of communicating partners along with the resources necessary to make the connection; coordinates partnering applications; and forms a consensus on procedures for controlling data integrity and error recovery?
    • The Application layer is responsible for finding the network resources broadcast from a server and adding flow control and error control (if the application developer chooses).
  • Which layer is responsible for converting data packets from the Data Link layer into electrical signals?
    • The Physical layer takes frames from the Data Link layer and encodes the 1s and 0s into a digital signal for transmission on the network medium.
  • At which layer is routing implemented, enabling connections and path selection between two end systems?
    • The Network layer provides routing through an internetwork and logical addressing.
  • Which layer defines how data is formatted, presented, encoded, and converted for use on the network?
    • The Presentation layer makes sure that data is in a readable format for the Application layer.
  • Which layer is responsible for creating, managing, and terminating sessions between applications?
    • The Session layer sets up, maintains, and terminates sessions between applications.
  • Which layer ensures the trustworthy transmission of data across a physical link and is primarily concerned with physical addressing, line discipline, network topology, error notification, ordered delivery of frames, and flow control?
    • PDUs at the Data Link layer are called frames and provide physical addressing plus other options to place packets on the network medium.
  • Which layer is used for reliable communication between end nodes over the network and provides mechanisms for establishing, maintaining, and terminating virtual circuits; transport‐fault detection and recovery; and controlling the flow of information?
    • The Transport layer uses virtual circuits to create a reliable connection between two hosts.
  • Which layer provides logical addressing that routers will use for path determination?
    • The Network layer provides logical addressing, typically IP addressing and routing.
  • Which layer specifies voltage, wire speed, and cable pinouts and moves bits between devices?
    • The Physical layer is responsible for the electrical and mechanical connections between devices.
  • Which layer combines bits into bytes and bytes into frames, uses MAC addressing, and provides error detection?
    • The Data Link layer is responsible for the framing of data packets.
  • Which layer is responsible for keeping the data from different applications separate on the network?
    • The Session layer creates sessions between different hosts' applications.
  • Which layer is represented by frames?
    • The Data Link layer frames packets received from the Network layer.
  • Which layer is represented by segments?
    • The Transport layer segments user data.
  • Which layer is represented by packets?
    • The Network layer creates packets out of segments handed down from the Transport layer.
  • Which layer is represented by bits?
    • The Physical layer is responsible for transporting 1s and 0s (bits) in a digital signal.
  • Put the following in order of encapsulation: Packets, Frames, Bits, Segments
    • Segments, packets, frames, bits
  • Which layer segments and reassembles data into a data stream?
    • Transport
  • Which layer provides the physical transmission of the data and handles error notification, network topology, and flow control?
    • Data Link
  • Which layer manages logical device addressing, tracks the location of devices on the internetwork, and determines the best way to move data?
    • Network
  • What is the bit length and expression form of a MAC address?
    • 48 bits (6 bytes) expressed as a hexadecimal number
  • In the work area below draw a line from the OSI model layer to its PDU.
    • Physical to bits; Datalink to frame; Network to packet; Transport to segment.

Part 2 - Ethernet Networking and Data

  • Describe the operation of Carrier Sense Multiple Access with Collision Detection (CSMA/CD).
    • CSMA/CD is a protocol that helps devices share the bandwidth evenly without having two devices transmit at the same time on the network medium. Although it does not eliminate collisions, it helps to greatly reduce them, which reduces retransmissions, resulting in a more efficient transmission of data for all devices.
  • Differentiate half‐duplex and full‐duplex communication and define the requirements to utilize each method.
    • Full‐duplex Ethernet uses two pairs of wires at the same time instead of one wire pair like half‐duplex. Full‐duplex allows for sending and receiving at the same time, using different wires to eliminate collisions, while half‐duplex can send or receive but not at the same time and still can suffer collisions. To use full‐duplex, the devices at both ends of the cable must be capable of and configured to perform full‐duplex.
  • Describe the sections of a MAC address and the information contained in each section.
    • The MAC, or hardware, address is a 48‐bit (6‐byte) address written in a hexadecimal format. The first 24 bits, or 3 bytes, are called the organizationally unique identifier (OUI), which is assigned by the IEEE to the manufacturer of the NIC. The balance of the number uniquely identifies the NIC.
  • Identify the binary and hexadecimal equivalent of a decimal number.
    • Any number expressed in one format can also be expressed in the other two. The ability to perform this conversion is critical to understanding IP addressing and subnetting.
  • Identify the fields in the Data Link portion of an Ethernet frame.
    • The fields in the Data Link portion of a frame include the preamble, Start Frame Delimiter, destination MAC address, source MAC address, Length or Type, Data, and Frame Check Sequence.
  • Identify the IEEE physical standards for Ethernet cabling.
    • These standards describe the capabilities and physical characteristics of various cable types and include but are not limited to 10Base‐2, 10Base‐5, 10Base‐T, 100Base-T, and 1000Base-T.
  • Differentiate types of Ethernet cabling and identify their proper application.
    • The three types of cables that can be created from an Ethernet cable are straight‐through (to connect a PC's or router's Ethernet interface to a hub or switch), crossover (to connect hub to hub, hub to switch, switch to switch, or PC to PC), and rolled (for a console connection from a PC to a router or switch).
  • Describe the data encapsulation process and the role it plays in packet creation.
    • Data encapsulation is a process whereby information is added to the frame from each layer of the OSI model. This is also called packet creation. Each layer communicates only with its peer layer on the receiving device.
  • Understand how to connect a console cable from a PC to a router and switch.
    • Take a rolled cable and connect it from the COM port of the host to the console port of a router. Start your emulations program such as putty or SecureCRT and set the bits per second to 9600 and flow control to None.
  • Identify the layers in the Cisco three‐layer model and describe the ideal function of each layer.
    • The three layers in the Cisco hierarchical model are the core (responsible for transporting large amounts of traffic both reliably and quickly), distribution (provides routing, filtering, and WAN access), and access (workgroup connectivity into the distribution layer).
  • For each of the following situations, determine whether a straight-through, crossover, or rolled cable would be used: Host to host, Host to switch or hub, Router direct to host, Switch to switch, Router to switch or hub, Hub to hub, Hub to switch, Host to a router console serial.
    • Host to host - Crossover, Host to switch or hub - Straight-through, Router direct to host - Crossover, Switch to switch - Crossover, Router to switch or hub - Straight-through, Hub to hub - Crossover, Hub to switch - Crossover, Host to a router console serial - Rolled

Part 3 - Introduction to TCP/IP

  • Differentiate the DoD and the OSI network models.
    • The DoD model is a condensed version of the OSI model, composed of four layers instead of seven, but is nonetheless like the OSI model in that it can be used to describe packet creation and devices and protocols can be mapped to its layers.
  • Identify Process/Application layer protocols.
    • 'Telnet is a terminal emulation program that allows you to log into a remote host and run programs. File Transfer Protocol (FTP) is a connection‐oriented service that allows you to transfer files. Trivial FTP (TFTP) is a connectionless file transfer program. Simple Mail Transfer Protocol (SMTP) is a sendmail program.
  • Identify Host‐to‐Host layer protocols.
    • Transmission Control Protocol (TCP) is a connection‐oriented protocol that provides reliable network service by using acknowledgments and flow control. User Datagram Protocol (UDP) is a connectionless protocol that provides low overhead and is considered unreliable.
  • Identify Internet layer protocols.
    • 'Internet Protocol (IP) is a connectionless protocol that provides network address and routing through an internetwork.
    • Address Resolution Protocol (ARP) finds a hardware address from a known IP address.
    • Reverse ARP (RARP) finds an IP address from a known hardware address.
    • Internet Control Message Protocol (ICMP) provides diagnostics and destination unreachable messages.
  • Describe the functions of DNS and DHCP in the network.
    • Dynamic Host Configuration Protocol (DHCP) provides network configuration information (including IP addresses) to hosts, eliminating the need to perform the configurations manually.
    • Domain Name Service (DNS) resolves hostnames—both Internet names such as www.microsoft.com and device names such as Workstation 2—to IP addresses, eliminating the need to know the IP address of a device for connection purposes.
  • Identify what is contained in the TCP header of a connection‐oriented transmission.
    • The fields in the TCP header include the source port, destination port, sequence number, acknowledgment number, header length, a field reserved for future use, code bits, window size, checksum, urgent pointer, options field, and finally, the data field.
  • Identify what is contained in the UDP header of a connectionless transmission.

The fields in the UDP header include only the source port, destination port, length, checksum, and data. The smaller number of fields as compared to the TCP header comes at the expense of providing none of the more advanced functions of the TCP frame.

  • Identify what is contained in the IP header.

The fields of an IP header include version, header length, priority or type of service, total length, identification, flags, fragment offset, time to live, protocol, header checksum, source IP address, destination IP address, options, and finally, data.

  • Compare and contrast UDP and TCP characteristics and features.

TCP is connection‐oriented, acknowledged, and sequenced and has flow and error control, while UDP is connectionless, unacknowledged, and not sequenced and provides no error or flow control.

  • Understand the role of port numbers.

Port numbers are used to identify the protocol or service that is to be used in the transmission.

  • Identify the role of ICMP.

Internet Control Message Protocol (ICMP) works at the Network layer and is used by IP for many different services. ICMP is a management protocol and messaging service provider for IP.

  • Define the Class A IP address range.

The IP range for a Class A network is 1–126. This provides 8 bits of network addressing and 24 bits of host addressing by default.

  • Define the Class B IP address range.

The IP range for a Class B network is 128–191. Class B addressing provides 16 bits of network addressing and 16 bits of host addressing by default.

  • Define the Class C IP address range.

The IP range for a Class C network is 192 through 223. Class C addressing provides 24 bits of network addressing and 8 bits of host addressing by default.

  • Identify the private IP ranges.

The Class A private address range is 10.0.0.0 through 10.255.255.255. The Class B private address range is 172.16.0.0 through 172.31.255.255. The Class C private address range is 192.168.0.0 through 192.168.255.255.

  • Understand the difference between a broadcast, unicast, and multicast address.

A broadcast is to all devices in a subnet, a unicast is to one device, and a multicast is to some but not all devices.

  • What is the Class C address range in decimal and in binary?

192 through 223, 110 xxxxx

  • What layer of the DoD model is equivalent to the Transport layer of the OSI model?

Host‐to‐host

  • What is the valid range of a Class A network address?

1 through 126

  • What is the 127.0.0.1 address used for?

Loopback or diagnostics

  • How do you find the network address from a listed IP address?

Turn all host bits off.

  • How do you find the broadcast address from a listed IP address?

Turn all host bits on.

  • What is the Class A private IP address space?

10.0.0.0 through 10.255.255.255

  • What is the Class B private IP address space?

172.16.0.0 through 172.31.255.255

  • What is the Class C private IP address space?

192.168.0.0 through 192.168.255.255

  • What are all the available characters that you can use in hexadecimal addressing?

0 through 9 and A , B , C , D , E , and F

  • What are the four layers of the DoD model

Process/Application Host‐to‐Host Internet Network Access

  • Layer of the DoD model on which few Protocols operate
    • IP to DoD Internet
    • Telnet to DoD Process/Application
    • FTP to DoD Process/Application
    • SNMP to DoD Process/Application
    • DNS to DoD Process/Application
    • DHCP/BOOTP to DoD Process/Application
    • X Window to DoD Process/Application
    • NFS to DoD Process/Application
    • TFTP to DoD Process/Application
    • SMTP to DoD Process/Application
    • LPD to DoD Process/Application
    • IP to DoD Interner
    • ARP to DoD Internet
    • ICMP to DoD Internet
    • RARP to DoD Internet
    • Proxy ARP to DoD Internet
    • TCP to DoD Host-to-Host
    • UDP to DoD Host-to-Host

Part 4 - Subnetting

  • Identify the advantages of subnetting.

Benefits of subnetting a physical network include reduced network traffic, optimized network performance, simplified management, and facilitated spanning of large geographical distances.

  • Describe the effect of the ip subnet‐zero command.

This command allows you to use the first and last subnet in your network design.

  • Identify the steps to subnet a classful network.

Understand how IP addressing and subnetting work. First, determine your block size by using the 256‐subnet mask math. Then count your subnets and determine the broadcast address of each subnet—it is always the number right before the next subnet. Your valid hosts are the numbers between the subnet address and the broadcast address.

  • Determine possible block sizes.

This is an important part of understanding IP addressing and subnetting. The valid block sizes are always 2, 4, 8, 16, 32, 64, 128, etc. You can determine your block size by using the 256‐subnet mask math.

  • Describe the role of a subnet mask in IP addressing.

A subnet mask is a 32‐bit value that allows the recipient of IP packets to distinguish the network ID portion of the IP address from the host ID portion of the IP address.

  • Understand and apply the 2 x – 2 formula.

Use this formula to determine the proper subnet mask for a particular size network given the application of that subnet mask to a particular classful network.

  • Explain the impact of Classless Inter‐Domain Routing (CIDR).

CIDR allows the creation of networks of a size other than those allowed with the classful subnetting by allowing more than the three classful subnet masks.

  • What is the subnet, broadcast address, and a valid host range for 192.168.100.25/30

192.168.100.25/30. A /30 is 255.255.255.252. The valid subnet is 192.168.100.24, broadcast is 192.168.100.27, and valid hosts are 192.168.100.25 and 26.

  • What is the subnet, broadcast address, and a valid host range for 192.168.100.37/28

192.168.100.37/28. A /28 is 255.255.255.240. The fourth octet is a block size of 16. Just count by 16s until you pass 37. 0, 16, 32, 48. The host is in the 32 subnet, with a broadcast address of 47. Valid hosts 33–46.

  • What is the subnet, broadcast address, and a valid host range for 192.168.100.66/27

A /27 is 255.255.255.224. The fourth octet is a block size of 32. Count by 32s until you pass the host address of 66. 0, 32, 64, 96. The host is in the 64 subnet, and the broadcast address is 95. Valid host range is 65–94.

  • What is the subnet, broadcast address, and a valid host range for 192.168.100.17/29

192.168.100.17/29. A /29 is 255.255.255.248. The fourth octet is a block size of 8. 0, 8, 16, 24. The host is in the 16 subnet, broadcast of 23. Valid hosts 17–22.

  • What is the subnet, broadcast address, and a valid host range for 192.168.100.99/26

192.168.100.99/26. A /26 is 255.255.255.192. The fourth octet has a block size of 64. 0, 64, 128. The host is in the 64 subnet, broadcast of 127. Valid hosts 65–126.

  • What is the subnet, broadcast address, and a valid host range for 192.168.100.99/25

192.168.100.99/25. A /25 is 255.255.255.128. The fourth octet is a block size of 128. 0, 128. The host is in the 0 subnet, broadcast of 127. Valid hosts 1–126.

  • You have a Class B network and need 29 subnets. What is your mask?

A default Class B is 255.255.0.0. A Class B 255.255.255.0 mask is 256 subnets, each with 254 hosts. We need fewer subnets. If we used 255.255.240.0, this provides 16 subnets. Let's add one more subnet bit. 255.255.248.0. This is 5 bits of subnetting, which provides 32 subnets. This is our best answer, a /21.

  • What is the broadcast address of 192.168.192.10/29?

A /29 is 255.255.255.248. This is a block size of 8 in the fourth octet. 0, 8, 16. The host is in the 8 subnet, broadcast is 15.

  • How many hosts are available with a Class C /29 mask?

A /29 is 255.255.255.248, which is 5 subnet bits and 3 host bits. This is only 6 hosts per subnet.

  • What is the subnet for host ID 10.16.3.65/23?

A /23 is 255.255.254.0. The third octet is a block size of 2. 0, 2, 4. The subnet is in the 16.2.0 subnet; the broadcast address is 16.3.255.

Part 5 -

ccna_glossary.txt · Last modified: 2016/08/11 16:41 by Mike J. Kreuzer PhD MCSE MCT Microsoft Cloud Ecosystem