Network Design

A. Symbol rate and noise level

B. Symbol rate and number of symbols

C. Symbol rate and Baud rate

D. Symbol rate and bandwidth

E. Symbol rate and signalling rate

A. Noise power and signal power

B. Signal to noise ratio and signalling rate

C. Bandwidth and signal power

D. Bandwidth and signal to noise ratio

E. Bandwith and number of symbols

A. Hertz (Hz)

B. Bits per second (bps)

C. Baud (Bd)

D. Baud per symbol (bps)

E. Decibels (dB)

A. The power at which the signal is transmitted

B. Attenuation over the distance of transmission

C. Thermal noise

D. Amplification of the received signal

E. Distortion caused by frequency response characteristics of the medium

A. Attenuation is a reduction in signal power over the distance travelled

B. Attenuation is a form of noise and reduces data rates

C. Attenuation the power level at which a signal is transmitted

D. Attenuation the power level at which a signal is received

E. Attenuation is measured in Hertz (Hz)

A. Thermal noise is caused by environmental factors such as switching or mains hum

B. Distortion is cause by other signals in close proximity

C. Interference is caused by vibrations at an atomic level and is always present

D. Crosstalk (inter-symbol interference) is caused by delayed versions of the signal itself

E. Crosstalk (inter-channel interference) is caused by the frequency response characteristics of the medium

A. Bandwidth

B. Signal to noise ratio

C. Data rate

D. Maximum channel capacity

E. Symbol rate

A. The high frequencies used by the signal

B. Thermal noise

C. Resistance and other impedance's

D. The insulating sheath around the cable

E. Interference from other signals in the cable

A. Thermal noise

B. Interference

C. Crosstalk

D. Attenuation

E. Distortion

A. Electrons vibrating and absolute zero

B. Thermal noise and the number of symbols

C. Signalling rates and distortion at high frequencies

D. The Baud criterion and information content

E. Finite bandwidth and the presence of noise

A. Coaxial cable: 2×108 m s-1

B. UTP: 2×108 m s-1

C. Multimode optical fibre: 2× 10 to the power of 8 m s-1

D. Monomode optical fibre: 3× 10 to the power of 8 m s-1

E. Radio: 3× 108 m s-1

A. Reflection

B. Refraction

C. Propagation

D. Diffraction

E. Scattering

A. -10 dB

B. 0 dB

C. +2.78 dB

D. +3 dB

E. +10 dB

A. The wavelength increases

B. The wavelength decreases

C. The wavelength stays the same

D. The wavelength changes phase

E. The wavelength's energy becomes unstable

A. Monomode fibre is made of a single strand of glass, while multimode fibre is made of many strands

B. Monomode fibre can only carry a single signal, while multimode fibre can carry many signals

C. Monomode fibre is made of glass with only one refractive index, while multimode fibre is made of glass with multiple refractive indices

D. Monomode fibre has a single coloured coating, while multimode fibre has a striped coating

E. Monomode fibre has a much narrower core than multimode fibre, so power losses are lower

A. The twisting of the pairs reduces the impact of external noise through cancellation

B. A protective jacket surrounds the twisted pairs of wire

C. Each pair is twisted by a different amount to reduce crosstalk

D. It has a braid or foil shield to reduce interference

E. Different grades are available with different frequency and attenuation characteristics

A. The plastic insulating sheath prevents interference reaching the copper conductors inside

B. The braiding prevents interference reaching the internal conductor

C. The dielectric between the braiding and the conductor is the main means of reducing interference

D. The twisting of the cable uses cancellation to reduce interference

E. The copper meterial of the internal conductor is resistant to interference

A. Straight-through cable must be used

B. Crossover cables must be used

C. At least one device must have an internal crossover in the interface

D. An odd number of crossovers is required, which may be present in cable and/or the device interfaces

E. An even number of crossovers is required, which may be present in cable and/or the device interfaces

A. Fibre has higher bandwidth

B. Fibre has lower attenuation

C. Fibre infrastructure is easier to install

D. Fibre is more resistant to interference

E. Fibre does not require grounding on inter-building runs

A. Infrastructure for wireless infrastructure is generally cheaper, since cables are not required

B. Attenuation increases with the square of the distance, rather than distance

C. Radio is less prone to interference and noise

D. Since radio bandwidth is infinite, data throughput is generally higher with wireless media

E. There is no limit to transmit power on unlicensed radio bands

A. The signals are travelling at a faster speed from transmitter to receiver

B. The data is being transmitted at a faster rate (in bits per second)

C. A carrier wave is being modulated, rather than a signal being sent directly on the medium

D. A `local loop' phone line is being used to transmit the data

E. One end of the transmission path is in somebody's home

A. x and y

B. y and z

C. x and z

D. x and t

E. π and y

A. By the direction (positive or negative) of the transition

B. By the presence or absence of an additional transition

C. By the transition being in the same or the opposite direction as the previous one

D. By the amplitude of the transition

E. By varying the exact time of the regular transition

A. 1

B. 2

C. 4

D. 8

E. 16

A. The number of symbols

B. The number of bits per symbol

C. The number of symbols per bit

D. The number of bits

E. The number of modulation systems used

A. 3 symbols, differing in amplitude

B. 3 symbols, differing in attenuation

C. 8 symbols, differing in amplitude

D. 8 symbols, differing in attenustion

E. 8 symbols, differing in phase

A. Compresses 4 bytes into 5 bits to increase data transfer speeds

B. Transmits 4 bits, then pauses for 5 bits to allow other stations to transmit

C. Encodes 4 bits as 5 symbols to increase data throughput

D. Encodes 4 bits as 5 bits to include a parity bit for error detection

E. Encodes 4 bits as 5 bits in order to increase the number of transitions for better synchronisationE. Encodes 4 bits as 5 bits in order to increase the number of transitions for better synchronisation

A. They generally use two signal levels, one positive and one negative

B. They can use any number of signal levels, all of which must be positive

C. They use three signal levels, including zero, which cannot be repeated more than once

D. They only use a single signal level, which is not zero

E. They can use any number of signal levels, but when a zero is repeated, a line code is required

A. 1

B. 2

C. 3

D. 4

E. 8

A. 100 Mbps Ethernet over copper

B. IEEE 802.11 over a radio physical layer

C. Dialup modem access to the Internet

D. Radio and television broadcast

E. ADSL home Internet access

A. The number of bits expected to be in error in a typical PDU

B. The number of bits expected to be in error per second

C. The proportion of bits expected to be in error

D. The probability of an PDU containing an error

E. The base-2 logarithm of the packet length

A. Error detection codes and error correction codes

B. Idle-RQ and continuous RQ

C. Selective repeat and go-back-N

D. Feedback error correct and forward error correction

E. Checksums and CRCs

A. 18%

B. 37%

C. 80%

D. 100% media utilisation

E. It depends on the normalised delay-bandwidth product

A. Data transmission rate

B. Typical PDU length

C. Physical length of the transmission link

D. Propagation speed of medium

E. Bandwidth of the physical channel

A. The speed at which errors can be detected or corrected

B. The number of PDUs which can be checked each second

C. The number of bits used for the code in each PDU sent

D. The proportion of errors which can be detected or corrected using the code chosen

E. The proportion of the number of bits of original data to the number of bits of data plus code sent

A. Idle RQ

B. Cyclic Redundancy Check (CRC)

C. 2D parity

D. Forward error correction

E. Checksums

A. They are based on polynomial division

B. They can only detect errors with an odd number of incorrect bits

C. They are used in LAN protocols

D. They are used in WAN protocols

E. They can be implemented in hardware

A. They can detect single bit errors

B. They can correct single bit errors

C. They can detect 2-bit errors

D. They are robust in the face of multiple errors

E. They cannot be implemented in hardware

A. The transmission time

B. The propagation time

C. The normalised delay-bandwidth product

D. The sliding window size

E. The length of an average PDU

A. Standard and idle-RQ

B. Go-back-N and selective repeat

C. Sliding window and synchronised window

D. Error detection only and error correcting mode

E. Synchronisation and acknowledgement

A. Reservation Aloha

B. CSMA

C. Token Passing

D. CSMA/CA

E. CSMA/CD

A. CSMA always has a higher media utilisation

B. CSMA checks that no other stations are transmitting before transmitting data

C. CSMA has a collision detection function

D. CSMA cannot operate in a radio environment

E. CSMA requires that the value of a is known

A. The sum tTX + tPROP

B. The difference tTX - tPROP

C. The product tPROP × tTX

D. The quotient tPROP / tTX

E. The quotient tTX / tPROP

A. 1-persistent CSMA

B. p-persistent CSMA

C. Nonpersistent CSMA

D. CSMA/CD

E. CSMA/CA

A. CSMA, because collisions will prevent correct communication

B. CSMA/CA, because collisions are unavoidable with radio

C. CSMA/CA, because of random backoff errors

D. CSMA/CD, because of hidden nodes and the duplex transmission requirement

E. CSMA/CD, because JAM signals cannot be transmitted on a radio link

A. Unslotted versions generally perform better than slotted versions

B. Persistent CSMA performs best at high loads

C. Nonpersistent CSMA outperforms slotted Aloha at high loads

D. CMSA performs best when a is large

E. CSMA always performs better than token passing

A. The reserved phase of Reservation Aloha

B. TDMA

C. FDMA

D. CDMA

E. CSMA

A. It waits for a random backoff period after a collision is detected

B. It waits for a random backoff period after the medium becomes free

C. It can anticipate when other stations are about to transmit, and holds back data until it has seen these transmissions take place

D. It uses separate frequencies for transmitting and acknowledgments

E. It retransmits immediately with probability p

A. Collisions may not be noticed at higher transmission rates

B. The interframe gap becomes too short for the processor speed to keep up with

C. Hidden nodes are more likely at higher data transmission rates

D. Random numbers required for the backoff time calculation cannot be produced quickly enough

E. The increase in a means that utilisation of the medium will fall unless propagation times are also reduced

A. Token passing has a higher media utilisation than Aloha

B. Token Ring is a well-known implementation of token passing

C. Token passing is a deterministic media access protocol

D. Stations may only transmit when they are in possession of an electronic `token', which is passed in a special PDU

E. Token passing is less complex than the CSMA or Aloha protocols

A. IP unicast address

B. IPv6 unicast address

C. X.121 address used by X.25

D. Data link circuit identifier (DLCI) used by Frame Relay

E. IEEE MAC address used by Ethernet and WiFi

A. DNS address

B. IEEE MAC address

C. IPX address

D. IPv4 address

E. Appletalk address

A. ARP is not required in IPX networks

B. ARP maps level 3 addresses such as IP into level 2 addresses such as Ethernet

C. ARP is used only when addressing is dynamic

D. ARP requests are normally broadcast

E. Information from ARP replies can be cached in an ARP table

A. The A record maps IP addresses to textual names

B. The SOA (start of authority) record specifies the authoritative DNS server and timing information for a DNS zone

C. The AAAA record is used to look up IPv6 addresses

D. The MX record specifies the mail exchanger (mail server) for DNS domain name along with a priority value

E. The NS record provides a name server address for a DNS zone

A. 12.3.4.5

B. 123.45.67.8

C. 23.45.67.89

D. 234.56.7.89

E. 56.78.9.0

A. 1.2.3.4

B. 1.234.56.7

C. 12.34.56.67

D. 12.345.67.89

E. 123.45.67.89

A. 123.45.67.0/22

B. 123.45.67.16/28

C. 123.45.67.40/23

D. 123.45.67.100/26

E. 123.45.67.255/24

A. 5.5.5.1

B. 10.10.1.1

C. 127.127.1.1

D. 172.27.27.27

E. 225.225.1.1

A. It should be sent anywhere

B. It should be sent to all members of a group

C. It should be sent to any one member of a group

D. It should be sent in the direction away from the anycast address

E. It should be encapsulated in the anycast protocol before sending on

A. An anycast address

B. A directed broadcast address

C. A unicast address

D. A local scope address

E. A multicast address

A. Bridges perform the same function as layer-2 switches

B. Routers perform the same function as layer-3 switches

C. Switches are generally faster than bridges and routers since more functionality is hardware-based

D. Bridges and routers both perform path determination for PDUs

E. Bridges and routers use addresses from the same OSI layer to make path-determination decisions

A. A physical loop free topology

B. Spanning Tree Protocol running on all bridges

C. A loop-free topology, either physical or logical after some links are disabled by Spanning Tree Protocol

D. A routing table to be maintained on every bridge

E. No repeaters (hubs) in the network

A. The layer 2 source address is added (or updated) in the bridging table

B. The layer 2 destination address is added (or updated) in the bridging table

C. The layer 3 source address is added (or updated) in the bridging table

D. The layer 3 destination address is added (or updated) in the routing table

E. The layer 3 source address is added (or updated) in the routing table

A. Those with broadcast or unknown layer 2 source addresses

B. Those with broadcast or unknown layer 2 destination addresses

C. Only those with unknown layer 2 source addresses

D. Only those with broadcast 2 destination addresses

E. Those with unknown layer 2 source or destination addresses

A. Open vs. proprietary

B. Classful vs. classless

C. Multiprotocol vs. single protocol

D. Interior vs. exterior

E. Link-state vs. distance-vector

A. Alternate

B. Discarding

C. Disabled

D. Forwarding

E. Learning

A. Backup

B. Designated

C. Discarding

D. Edge

E. Root

A. It depends on every router having complete tolopogy information about all links in the network

B. It is based on receiving periodic updates from neighbours with copies of their routing table

C. It is usually faster to converge than distance-vector routing

D. It uses Dijkstra's algorithm

E. Examples of link state protocols include OSPF and IS-IS

A. It uses the Bellman-Ford algorithm

B. It uses information from neighbour's routing tables to build up its own routing information

C. It normally uses periodic updates

D. RIPv2 is a distance vector routing protocols for IPv4

E. It can only be used in loop-free networks

A. It cannot be used with IPv4, only with IPv6

B. It increases precision of routing information

C. It can be used with classful routing protocols

D. It reduces load and bandwidth of routing protocols to provide scalability

E. It is the opposite of route aggregation

A. Link Aggregation Control Protocol (LACP)

B. Open Shortest Path First (OSPF)

C. Rapid Spanning Tree Protocol (RSTP)

D. Routing Information Protocol (RIP)

E. Virtual Router Redundancy Protocol (VRRP)

A. Link Aggregation Control Protocol (LACP)

Protocol (LACP)

C. Rapid Spanning Tree Protocol (RSTP)

D. Routing Information Protocol (RIP)

E. Virtual Router Redundancy Protocol (VRRP)

A. Running a routing protocol, such as RIP, on the station

B. Using a gateway redundancy protocol, such as VRRP, between the routers

C. Using DHCP with very short lease times and changing the router information provided

D. Using ICMP router discovery to allow stations to find out routing information

E. Running Spanning Tree Protocol on the stations themselves

A. VLANs can reduce the impact of broadcast traffic in a network

B. VLANs can improve the security of a network

C. Traffic between VLANs must pass through a router

D. Traffic within VLANs must pass through a router

E. Traffic for multiple VLANs can be carried over a trunked link using IEEE 802.1Q tagging

A. Multiple links between switches bundled together as a single logical link

B. Load balancing traffic from multiple VLANs over multiple links

C. Sending traffic to multiple destinations over a single VLAN

D. Sending traffic on multiple VLANs over a single link

E. Using multiple links between switches to carry traffic for the same number of VLANs

A. MST allows VLANs and STP to operate together by creating a separate STP instance for every VLAN

B. MST allows VLANs and STP to operate together by creating multiple STP instances, each of which can carry traffic for a number of VLANs

C. MST allows trunking and STP to operate together by combining STP instances over trunked links

D. MST allows multiple VLANs to be carried over a single link using tagging

E. MST is a non-standard method of running STP with a faster convergence time for higher performance

A. Availability

B. Bandwidth

C. Jitter

D. Latency

E. Reliability

A. VoIP has specific reliability requirements, but no delay requirements

B. Video streaming has specific reliability requirements, but no jitter requirements

C. Video conferencing has specific bandwidth requirements, but no delay requirements

D. VoIP has specific delay and jitter requirements

E. Data transfer has specific jitter and delay requirements

A. Queues can decrease the bandwidth available to applications

B. Queues can cause increased delay and jitter in data transferred

C. Queues can decrease the success of applications such as file transfer

D. Queues decrease the latency of the network which causes reliability problems

E. Queues require extra memory on the end nodes in a network

A. Integrated services classifies traffic on entry to the network and treats it differently at each node

B. Differentiated services requires a QoS request to be made to the network which may be accomodated if resources are available

C. Quality services requires both a protocol such as RSVP and specific queueing disciplines at each node

D. Differentiated services classifies traffic on entry to the network and treats it differently at each node

E. Integrated services requires the underlying network to support differentiated services in order to operate correctly

A. Authentication

B. Authorisation

C. Confidentiality

D. Integrity

E. Privacy

A. Asymmetric algorithm

B. Digital signature algorithm

C. One-way hash algorithm

D. Public key algorithm

E. Secret key algorithm

A. It stands for Cryptographic Handshake Authorisation Protocol

B. It is an authentication protocol usually used with Wireless LANs

C. It sends the password in the clear

D. It involves a cryptographic challenge being sent, and the response is a one-way hash of the challenge with a secret

E. It is a confidentiality protocol used to protect data from eavesdropping

A. Access point

B. Authentication server

C. Authenticator

D. Supplicant

E. Switch authentication centre (SAC)

A. They are defined in an ITU standard

B. They use a hierarchical trust model

C. Certificate Authorities (CAs) sign them using their private keys

D. They contain public keys which are signed to confirm their authenticity

E. They allows secure distribution of private keys

A. WEP (Wired Equivalent Privacy) is the most recent and secure standard

B. WEP (Wired Equivalent Privacy) is a standard used on both wireless and wired networks

C. WPA (WiFi Protected Access) is defined by the IEEE

D. IEEE 802.11i implements the WPA2 security model

E. WPA2 is not recommended for secure installations owing to weaknesses in the Integrity Check Value (ICV)

A. Internet Key Exchange (IKE) provides secure key exchange using the Diffie- Hellman algorithm

B. Encapsulating Security Payload (ESP) packets provide authentication, integrity and confidentiality

C. Authentication Header (AH) packets provide authentication, integrity and confidentiality

D. Integrity is provided by the Secure Hash Algorithm 1 (SHA1)

E. Only one encryption algorithm for providing confidentiality is available with IPSec

A. It can only be used for securing web pages, where the URL changes from http: to https:

B. It operates netween the network and transport layer, hence its alternative name, Transport Layer Security (TLS)

C. It provides only an authentication service for a range of applications

D. It provides authentication and privacy above the transport layer for a number of different applications

E. It operates on top of TCP, using the same well-known port number as the unsecured service, but with a different URL

A. Packets may be filtered based only on source and destination IP addresses and TCP or UDP port numbers

B. Packet filters are purely hardware-based devices

C. Packet filters are also known as proxy servers

D. Packet filters may form a component of a firewall

E. Packet filtering alone is a sufficient defence against all types of network attack

A. Controlling access to certain locations based on unsuitable content, for example in schools

B. Encrypting content of pages as they cross the public Internet

C. Increasing performance by caching frequently accessed remote information

D. Screening content for malware such as viruses

E. Spying on which network resources users are accessing

A. Fault management

B. Configuration management

C. Addressing management

D. Performance management

E. Security management

A. All SNMP messages are sent over UDP port 161

B. SNMP messages are sent over TCP (for commands and responses) and UDP (for notifications)

C. SNMP uses UDP port 161 for commands and responses, and UDP port 162 for notifications

D. SNMP runs over TCP, ports 161 (for notifications) and 162 (for commands and responses)

E. SNMP runs directly over IP (as protocol 161), without a transport layer being present

A. All SNMP messages are sent over UDP port 161

B. SNMP messages are sent over TCP (for commands and responses) and UDP (for notifications)

C. SNMP uses UDP port 161 for commands and responses, and UDP port 162 for notifications

D. SNMP runs over TCP, ports 161 (for notifications) and 162 (for commands and responses)

E. SNMP runs directly over IP (as protocol 161), without a transport layer being present

A. Abstract Syntax Notation 1 (ASN.1)

B. Management Information Base (MIB)

C. User-based Security Model (USM)

D. View-based Access Control Model (VACM)

E. XML-Structured Management Information (XSMI)

A. All devices use the same MIB structure

B. All devices of the same type have the same MIB structure

C. All devices from the same manufacturer have the same MIB structure

D. Devices will have varying MIB structure, depending on which version of SNMP is used to access them

E. Devices will have varying MIB structure, depending on function, hardware, software, configuration and manufacturer

A. A priority from 0-7

B. Facility, describing the source of the message

C. Hostname from which the message originates

D. Timestamp of the event causing the message

E. The user who was running the software which caused the message

A. Both protocols performs authentication

B. Both protocols are specified in RFCs

C. Only DIAMETER performs accounting

D. Only DIAMETER supports roaming in mobile networks

E. Only DIAMETER supports failover from one server to another

A. Eyes

B. Cable testers

C. The show controller in Cisco IOS

D. The ifconfig command in Linux

E. The route command in Windows

A. Checking the L3 address on the interfaces with the ifconfig (or similar) command

B. Checking the routing table with route command

C. Ensuring DNS name resolution is working using the nslookup (or similar) command

D. Hop-by-hop checking using the traceroute command

E. Using the ping command so send ICMP packets

A. ping

B. netstat

C. traceroute

D. route

E. arp

A. 404 Not found

B. 501 Internal server error

C. Address not found

D. Blank page being returned

E. Timeout