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Computer Networks

Chu Chu Informatik Fachabitur Alemán 175 fichas

Computer Networks

Fichas del conjunto (175)

  1. 1Data Rate and Latency 6 telegraph arms eachwith 4 positions 1)How many bits can be transmitted per second when a new adjustment of the tele-graph arms can be performed every 10 seconds? 2)How long it takes to transmit a messagefrom Berlin to Koblenz?1) 4^6= 4096 telegraph arm positions (= states) are possible, with 4096 states, 12 Bits can be encoded. Data rate = 12 Bits / 10 s= 1.2 Bits/s
    2) 62 stations exist⇒just 61 stations need to forward the message⇒61 minutes.
  2. 2What transmission media are used for computer networks?Guided transmission mediaexist and can be Copper cables, where data is transferredas electrical impulses or fiber-optic cables, where data is transferred as light impulses.Wireless transmission can base on radio technology, infrared and laser.
  3. 3Why is not one transmission medium used for all wired and wireless computernetworks?Not all transmission media can be used to cover large distances.
  4. 4Explain the difference between serial data transmission and parallel data trans-mission.A single data line exists, when serial data transmission is used. The bits aretransmitted one after another via the bus.Separate data lines exist, when parallel data transmission is used.
  5. 5Name an advantage of serial data transmission compared with parallel datatransmission.Fewer wires are required
  6. 6Name an advantage of parallel data transmission compared with serial datatransmission.Higher throughput
  7. 7Do computer networks usually implement parallel or serial data transmission?(Explain your answer!)Computer networks usually implement serial data transmission because par-allel data transmission is cost-intensive for long distances.
  8. 8What describes the physical topology of a computer network?It describes the wiring.
  9. 9What describes the logical topology of a computer network?It describes the flow of data between the network devices
  10. 10Cable failure can separate the network in two functioning parts (Bus, Ring, Star, Mesh, Tree and Cellular)Mesh, Tree, Busmay work but itis not ensured
  11. 11Topology contains a single point of failure (A single point of failure can be a device or a cable) (Bus, Ring, Star, Mesh, Tree and Cellular)Bus (the medium!), Ring (the medium!), Star, Cellular
  12. 12Topology used for Thin Ethernet and Thick Ethernet (Bus, Ring, Star, Mesh, Tree and Cellular)Bus
  13. 13Topology contains a performance bottleneck (Bus, Ring, Star, Mesh, Tree and Cellular)Star, Tree (theroot!), Cellular
  14. 14Topology used for WLAN, when no Access Point exists (Bus, Ring, Star, Mesh, Tree and Cellular)Mesh
  15. 15Topology used for Token Ring (logical) (Bus, Ring, Star, Mesh, Tree and Cellular)Ring
  16. 16Topology used for mobile phones (GSM standard) (Bus, Ring, Star, Mesh, Tree and Cellular)Cellular
  17. 17Topology used for Token Ring (physical) (Bus, Ring, Star, Mesh, Tree and Cellular)Star
  18. 18Cable failure leads to complete network failure (Bus, Ring, Star, Mesh, Tree and Cellular)Ring, Bus
  19. 19Topology contains no central component (Bus, Ring, Star, Mesh, Tree and Cellular)Bus, Ring, Mesh
  20. 20Topology used for WLAN, when an Access Point exists (Bus, Ring, Star, Mesh, Tree and Cellular)Cellular
  21. 21Topology used with modern Ethernet standards (Bus, Ring, Star, Mesh, Tree and Cellular)Star
  22. 22With handheld transceivers, often calledwalkie-talkie, twoor more participants can talk to each other. However, at notime, more than one participant can speak. Please describethe reason for this limitation.Only a single channel is used
  23. 23According to what directional dependence principle dowalkie-talkies operate? Simplex, Full-duplex, Half-duplexHalf-duplex
  24. 24Name 2 systems, that operate according to the simplex prin-ciple.Radio, TV, pager, satellite, GPS, radio clock signal
  25. 25Name an advantage and a drawback of communication sy-stems that operate according to the simplex principle?Advantage: When using a wireless network, only a single chan-nel is required. When using a wired network, lesser cabelingeffort is required.Drawback: The information transfer only works in one direc-tion.
  26. 26Name 2 systems, that operate according to the full-duplexprinciple.Ethernet via twisted pair cables, telephone
  27. 27Name an advantage and a drawback of communication sy-stems that operate according to the full-duplex principle?Advantage: The information transfer works in both directionssimultaneously.Drawbacks: When using a wireless network, multiple channelsare required. When using a wired network, the cabeling effortis higher.
  28. 28—
  29. 29Assign to technical terms „Frames“, „Packets“, „Segments“ and „Signals“ tothe layers of the reference models in the figure (Physical Layer, Data Link Layer, Network Layer, Transport Layer)Signals⇒Physical Layer,
    Frames⇒Data Link Layer,
    Packets⇒Network Layer,
    Segments⇒Transport Layer
  30. 30Why are the Presentation Layer and the Session Layer not intensively used?The functionalities, which are intended for the Session Layer and PresentationLayer, are now part of protocols and services in the Application Layer.
  31. 31Why is the hybrid reference model closer to reality, compared with the TCP/IPreference model?The hybrid reference model illustrates the functioning of computer networksin a realistic way because it distinguishes between the Physical Layer andData Link Layer and it does not subdivide the Application Layer. It combinesthe advantages of the TCP/IP reference model and the OSI reference model,without taking over their drawbacks.
  32. 32Why is the outer conductor (the shield) ofcoaxial cableskept at groundpotential and does completely surround the inner conductor?The shielding of the signal-carrying conductor by the shield, that is kept atground potential, reduces electromagnetic interferences.
  33. 33What is a Transceiver?Via Transceivers, terminal devices are connected with the transmission medi-um.
  34. 34What is the purpose of AUI cables?AUI cables are used to connect terminal devices with Transceivers.
  35. 35Why do modern Ethernet standards use twisted pair cableswith twistedsignal wires and not cables with parallel signal wires?Twisted pairs are better protected against alternating magnetic fields and elec-trostatic interferences from the outside than parallel signal wires.
  36. 36Show by calculationthat regardless of the level of a noise signal,the dif-ference between the payload signal and the complementary signalremains the samewhen using twisted-pair cables. Assume that a signal tobe transmitted has an electrical voltage of 0.5 V. This transmission is affectedby an interfering signal, which has an electrical voltage of 0.25 V.((+Payload Signal) + (Noise))−((−Payload Signal) + (Noise))= 2∗Payload Signal
    (0.5V+ 0.25V)−(−0.5V+ 0.25V) = 0.5V+ 0.25V+ 0.5V−0.25V= 1V
  37. 37Can patch cables, that are wired according to the T568A wiring standard,be used in an computer network infrastructure, which uses the the T568B wiring standard?This is no problem because patch cables are always 1:1 wired. Patch cables donot have an effect on the internal wiring of the jacks.
  38. 38Why is it impossible to connect different buildings with shielded cables?Shields must be electrically grounded on both sides of the cable. If only oneend of a shielded cable is grounded, an antenna effect occurs, which results ina compensation current.
  39. 39Name a benefit and a drawback ofmono-mode (single-mode) fiberscom-pared with multi-mode fibersAdvantage: Can be used for long distances (up to about 70 km).Drawback: Only a single propagation mode.
  40. 40Name a benefit and a drawback ofmulti-mode fiberscompared with mono-mode (single-mode) fibers.Advantage: Several thousand propagation modes.Drawback: Can only be used for short distances (up to about 500 m).
  41. 41The following information come from existing twisted pair network cables. What information is provided about the cable and pair shielding of these cables? E138922 RU AWM 2835 24 AWG 60◦C CSA LL81295 FT2 ETL VERIFIEDEIA/TIA-568A CAT.5 UTP EVERNEW G3C511UTP = Unshielded Twisted Pair
  42. 42What information is provided about the cable and pair shielding of these cables? E188601 (UL) TYPE CM 75◦C LL84201 CSA TYPE CMG FT4 CAT.5E PATCHCABLE TO TIA/EIA 568A STP 26AWG STRANDEDSTP = Shielded Twisted Pair
  43. 43What information is provided about the cable and pair shielding of these cables? E324441 RU AWM 2835 24AWG 60◦C 30V CHANGJIANG TIA/EIA 568B.2 UTPCAT.5eUTP = Unshielded Twisted Pair
  44. 44What information is provided about the cable and pair shielding of these cables? SSTP ENHANCED CAT.5 350MHZ 26AWG X 4P PATCH TYPE CM (UL) C(UL)E200579 CMG CSA LL81924 3P VERIFIEDSSTP = Screened Shielded Twisted Pair
  45. 45What information is provided about the cable and pair shielding of these cables? EC-net 7.5 m 11184406 13/03 PremiumNet 4 PAIR 26AWG S-FTP HF IEC332-1 ENHANCED CATEGORY 5 PATCH CORD EN0173+ISO/IECSFTP = Screened Foiled Twisted Pair
  46. 46What information is provided about the cable and pair shielding of these cables? (UL) E228252 TYPE CM 75◦C 24AWG 4PR UTP C(UL) E228252 CMR 73◦CETL VERIFIED TIA/EIA 568B.2 CAT.5eUTP = Unshielded Twisted Pair
  47. 47What means STRANDED?Stranded cables are typically used to create patch cables because they are veryflexible.
  48. 48Do cables exist which are not STRANDEDSolid cables are less flexible than stranded cables and they are typically usedfor permanent installation (structured wiring in a building).
  49. 49What is the meaning of PATCH?Patch cables are used to connect network devices.
  50. 50Do cables exist which are not PATCH?Crossover cables exist.
  51. 51What is the difference between PATCH cables and other cables?Crossover cables are used to connect two network devices directly, without aHub or Switch. A crossover cable contains crossing wires, which connect theoutput pins with the input pins of both devices.
  52. 52What is the meaning of the information 24AWG or 26AWG?American wire gauge (AWG) is a standardized wire gauge system, which isused mainly in the United States and Canada for the diameters of electricallyconducting wires.
    AWG-24 = 0.51054 mm diameter.
    AWG-26 = 0.405 mm diameter.
  53. 53What is the meaning of the information UL CM FT1/FT4 together with a degreevalue (e.g.60◦Cor75◦C)?FT1 and FT4 are vertical flame tests.
  54. 54What is the purpose of Repeatersin computer networks?A Repeater retransmits all received signals at a higher level or higher power,so that the signal can cover longer distances.
  55. 55What is the major difference between Repeaters and Hubs?Hubs are Repeaters with >2 interfaces.
  56. 56Why do Repeaters and Hubs not requirephysical or logical addresses?Repeaters and Hubs have no physical or logical network addresses becausethey just forward the received signals.
  57. 57Whatnetwork topology(s) do Hubs implement?Physical topology: Star network because of the cabling.
    Logical topology: Bus network, because equal to a long cable, were all net-work devices are connected with, a Hub forwards incoming signals to all otherinterfaces.
  58. 58Name twoadvantages of using a Hubcompared to a physical Bus network.Better reliability, because the failure of individual cable segments does notresult in a complete network failure.
    Adding or removing network devices does not cause network interruptions.
  59. 59What is a collision domain?The collision domain is a network or a section of a network where multiplenetwork devices use a shared transmission medium. It includes all networkdevices which compete for accessing a shared transmission medium.
  60. 60What says the 5-4-3 rule?In a collision domain, not more than 5 segments can be connected. For this, amaximum of 4 Repeaters are used. Only at 3 segments, active senders (terminaldevices) can be connected.
  61. 61Why does the 5-4-3 rule exist?Hubs cannot be cascaded infinitely. The round-trip time (RTT) must not beexceeded. If the network is is too large, the RTT will become too high. Thencollisions occur more frequent and undetected collisions are possible.
  62. 62Why are line codes necessary in computer networks?Computers are digital machines. Transmision mediums work analogous. Theline codes specify the conversion of binary data (⇒binary numbers) intosignals (encoding).
  63. 63Several different line codes exist. Why is it impossible to use one single line code for each network technology?Different transmission mediums are used for computer networks. Different numbers of signal levels are used.
  64. 64The most simple line code is Non-Return-To-Zero(NRZ). What mode ofoperation does it implement?It represents logical 0 and 1 is by using different voltage levels.
  65. 65What two problem scan occur, when NRZ is used to encode data?Baseline Wander and Clock Recovery.
  66. 66Explain the problem Baseline WanderBaseline Wander = shift of the average when using NRZ. The receiver distin-guishes the physical signal levels by using the average of a certain numberof received signals. Signals far below the average, interprets the receiver aslogical 0 bit. Signals significantly above the average, interprets the receiver aslogical 1 bit. When transmitting a long series of logical 0 bits or 1 bits, theaverage can shift so much, making it difficult to detect a significant change inthe physical signal.
  67. 67Explain the problem Clock RecoveryClock Recovery when using NRZ. Even if the processes for encoding and deco-ding run on different computers, they need to be controlled by the same clock.In each clock cycle, the sender transmits a bit and the receiver receives a bit.If the clocks of sender and receiver drift apart, the receiver may lose countduring a long sequence of logic 0 bits or 1 bits.
  68. 68How to avoid Baseline Wander and Clock RecoveryIn order to prevent Baseline Wander, when using a line code with 2 physicalsignal levels, the usage of both signal levels must be equally distributed.
    One way to avoid the clock recovery problem is by using a separate line,which transmits just the clock. In computer networks, a separate signal linejust for the clock is not practical because of the cabling effort. Instead, it isrecommended to increase the number of guaranteed signal level changes toenable the clock recovery from the data stream.
  69. 69Name at least 5 line codes that use 2 signals levels.NRZ, NRZI, Unipolar RZ, Manchester, Manchester II, Differential Manchester.
  70. 70Name at least 3 line codes that use 3 signal levels.MLT-3, RZ, AMI, B8ZS.
  71. 71Which line codes ensure asignal level changefor each logical 1 bit?NRZI, MLT-3, Unipolar RZ, AMI, B8ZS.
  72. 72Which line codes ensure asignal level changefor each transmitted bit?RZ, Manchester, Manchester II, Differential Manchester
  73. 73Why do not all line codes ensure asignal level change for each transmitted bit?Lack of efficiency
  74. 74Which line codes ensure that the signal levels are equally distributed?AMI, B8ZS, Manchester, Manchester II.
  75. 75Why is it important for the receiver of signals, which are encoded accordingto theDifferential Manchester Encoding, to know the initial signal level?Depending on the initial signal level, two signal sequences, inverse to eachother, are possible.
  76. 76What is a scrambler?A scrambler is a device, which modifies a data stream according to a simplealgorithm in a way that it is easy to reverse.
  77. 77Why are scramblers used?When the AMI line code is used, clock recovery is impossible for the receiver,when series of logical 0 bits are transmitted. In AMI case, scramblers are used,to interrupt long series of logic 0 bits. This makes the clock recovery for thereceiver possible.
  78. 78All line codes have drawbacks. What can be done toavoid the problems,that can result from these drawbacks?Modern network technologies encode the bit stream first with a line code thatworks efficient on the one hand, but also ensures clock recovery and avoidsbaseline wander. These encodings improve the bit stream in a way, that afurther encoding with the line codes NRZ, NRZI and MLT-3 does not resultin any problems. Examples of line codes, which improve the bit stream first,are 4B5B, 5B6B and 8B10B. These line codes encode fixed-size input blocksinto fixed-size output blocks.
  79. 79Which line code maps groups of 4 payload bits onto groups of 5 code bits?4B5B
  80. 80Which line code maps groups of 5 payload bits onto groups of 6 code bits?5B6B
  81. 81Why do some line codes, that map groups of payload bits onto groups of codebits, implement variants with neutral inequality, positive inequality and negative inequality?Variants with positive or negative inequality alternate to prevent BaselineWander.
  82. 82How is the efficiency of a line code calculated?Efficiency = ratio of bit rate (payload in bits per time) and baud rate (signalchanges per second).
  83. 83What is the purpose of Bridges in computer networks?For connecting different physical networks, Bridges are required because theyforward frames from one physical network to another one.
    Bridges and Switches check the correctness of the frames via checksums.
  84. 84How manyinterfaces(„Ports“) provides a Bridge?2 ports.
  85. 85What is the major difference between Bridges and Layer-2-Switches?Bridges with >2 ports are called Multiport Bridge or Layer-2-Switch.
  86. 86Why do Bridges and Layer-2-Switches not require physical or logical addresses?Bridges do not need addresses for filtering and forwarding the frames, becausethey do not actively participate in the communication. They work transparent,just like the devices of the Physical Layer.
  87. 87Name at least two examples of Bridge implementations.WLAN Bridges and Laser Bridges.
  88. 88What is the advantage ofLearning Bridgesin contrast to „dumb“ Bridges?Learning Bridges learn which network devices are accessible via which port.
  89. 89What information is stored in the forwarding tablesof Bridges?The information, which network devices are accessible via which port in localforwarding tables.
  90. 90What happens, if for a network device, no entry exists in the forwarding table of a Bridge?This is not a problem because the table is only used for optimization. If fora network device no entry in the forwarding table exists, the Bridge forwardsthe frame to every port, which is connected to a physical network.
  91. 91Why do Bridges try to avoid loops?Loops can cause malfunctions and reduce the performance of the network oreven lead to a network failure.
  92. 92What protocol use Bridges to handle loops?Spanning Tree Protocol (STP).
  93. 93What is a spanning tree?It is a subgraph of the graph, which covers all nodes, but it is cycle-free,because edges have been removed.
  94. 94What information contains the Bridge ID according to the IEEE?The Bridge ID consists of the Bridge priority (2 bytes) and MAC address (6bytes) of the Bridge port with the lowest port ID.
  95. 95What is the difference between theBridge IDaccording to the IEEE and the Cisco extended versionof the Bridge ID?Cisco subdives the original 2 bytes long part for the Bridge priority. 4 bits nowrepresent the Bridge priority. The remaining 12 bits are used to encode theVLAN ID.
  96. 96How many priority values can be encoded with the Bridge ID according to the IEEE?65,536 priority values can be represented.
  97. 97How many priority values can be encoded with the Cisco extended versionof the Bridge ID?4 bits represent the Bridge priority ⇒only 16 values can be represented.
  98. 98What is a Bridge Protocol Data Unit(BPDU) and for what is it used?Bridges exchange information about Bridge IDs and path costs via specialdata frames, called Bridge Protocol Data Unit (BPDU).
  99. 99What is the selection criteria for determining, whether a Bridge becomes the Root Bridge?First, the Bridges have to determine the Bridge with the lowest Bridge Priorityin the Bridge ID. This Bridge is the Root Bridge of the spanning tree to begenerated.
  100. 100What is a Designated Bridge and what is its task?For each physical network, a single one of the directly connected Bridges needsto be selected as responsible for forwarding the frames towards in the directionof the Root Bridge. This Bridge is called Designated Bridge for this network.
  101. 101How many Designated Bridges does a computer network contain?For each physical network, a single Designated Bridge exists.
  102. 102What is the selection criteria for determining, whether a Bridge becomes a Designated Bridge?The Bridge with the lowest path costs to the Root Bridge is selected as Desi-gnated Bridge.
  103. 103What is the impact of Bridges and Layer-2-Switches on the collision domain?If a physical network is subdivided via a Bridge or Switch, also the collisiondomain is divided and the number of collisions decreases.For Bridges and Switches, each port forms its own collision domain.
  104. 104What is a switched network?In a switched network, each port of the switches is connected with just a singlenetwork device.
  105. 105Name an advantage of a switched network.Such a network is free from collisions and state of the art.
  106. 106The format of wha taddresses is defined by Data Link Layer protocols? physical network addresses, logical network addressesphysical network addresses
  107. 107How are physical network addresses called?MAC addresses (Media Access Control).
  108. 108What protocol uses Ethernet for the address resolution?Address Resolution Protocol (ARP).
  109. 109Who receives a frame with the destination address FF-FF-FF-FF-FF-FF?This address is the MAC broadcast address. Every participant in the physicalnetwork receives this frame
  110. 110What is MAC spoofing?MAC addresses can be modified via software. The method is called MAC spoofing.
  111. 111One way to mark the frames’ borders is via character count in the frameheader. Name a potential issue that can arise from this method.If the field, which contains the number of bytes payload inside the frame ismodified during transmission, the receiver is unable to correctly detect the endof the frame.
  112. 112One way to mark the frames’ borders is via Byte Stuffing. Name a drawbackof this method.The strong relationship with the ASCII character encoding.
  113. 113Why work up-to-date Data Link Layer protocols, such as Ethernet and WLAN,bit-oriented and not byte-oriented?Because this allows using any character encoding.
  114. 114What information contains anEthernet frame?Sender MAC address, Preamble to synchronize the receiver, CRC checksum, VLAN tag, Receiver MAC address, Information about the Network Layer protocol used
  115. 115Why do computer networks use protocols for media access control?With Ethernet and WLAN, the network devices or stations use a shared trans-mission medium. To coordinate media access and to avoid collisions, mediaaccess control methods are required
  116. 116Why use Ethernet and WLAN different media access control methods?With wireless networks, it is not guaranteed that all stations can detect allcollisions.In wired networks with a shared transmission medium, each participant recei-ves the transmissions of all other participants.
  117. 117How do Ethernet devices react, when they detect acollision?If a collision is detected, the sender stops the frame transmission and sends thejam signal to announce the collision. If the maximum number of transmissionattempts is not yet reached, the sender tries to transmit the frame again aftera random time.
  118. 118Why is it important that the transmission of a frame is not completed whena collision occurs in an Ethernet network?Otherwise, the network device might already be finished with the transmissionand believes the transmission was successful.
  119. 119What is done to ensure that the transmission of a frame is not completed whena collision occurs in anEthernetnetwork?Each frame must have a certain minimum length. It must be dimensioned ina way, that the transmission duration for a frame with minimum length doesnot fall below the maximum RTT (round trip time).This ensures that a collision reaches the sender before its transmission is fi-nished. If a sender detects a collision, it knows that its frame has not arrivedcorrectly at the receiver, and can try the transmission again later.
  120. 120Which twospecial characteristicsof the transmission medium inwirelessnetworkscauseundetected collisionsat the receiver?Hidden terminal problem and Fading
  121. 121What is Hidden terminal problem and Fading.Hidden terminal problem (problem caused by invisible or hidden terminal devi-ce). Because of obstacles, not all stations can detect all transmissions, althoughthey interfere each other at the Access Point.
    Fading (decreasing signal strength). The electromagnetic waves of the wirelessnetwork are weakened by obstacles and in free space. Caused by the positionsof stations to each other, their signals are so weak, that the stations cannotdetect each others transmissions.
  122. 122What is theNetwork Allocation Vector(NAV) for what purpose is it used?The NAV is a counter variable which is maintained by each node itself. Itcontains the expected time when the transmission medium will be occupied.It reduces the number of collisions when CSMA/CA is used.
  123. 123What is the Contention Window(CW) and for what purpose is it used?If the NAV and another DIFS with an idle transmission medium has expired, abackoff time is created from the CW. The backoff time is calculated by using arandom value between the minimum CW and maximum CW and multiplyingthis random value with the slot time. After the backoff time has expired, theframe is transmitted. The CW prevents that all stations which wait for a freetransmission medium, start their transmissions at the same time.
  124. 124Name a benefit and a drawback of using the control framesRequest To Send(RTS) andClear To Send(CTS)?Advantage: It reduces collisions because it solves the problem of hidden ter-minals.
    Drawbacks: Delays occur, which are caused by the reservation of the trans-mission medium. The RTS and CTS frames, which are used to reserve thetransmission medium, are overhead.
  125. 125What is the function of the Address Resolution Protocol?The Address Resolution Protocol (ARP) is used to convert IP address of theNetwork Layer to MAC address of the Data Link Layer.
  126. 126What is the ARP cache?The ARP cache is a table, which contains IP addresses and MAC addresses,that belong together. It is used to speed up the address resolution.
  127. 127What is the purpose ofRoutersin computer networks? (Also explain the difference to Layer-3-Switches.)They forward packets between networks with different logical address rangesand provide a WAN interface.
  128. 128What is the purpose of Layer-3-Switches in computer networks? (Also explain the difference to Routers.)They are Routers too which means they forward packets between networkswith different logical address ranges but they do not provide a WAN interface.
  129. 129What is the purpose of Gateways in computer networks?They enable communication between networks, which base on different proto-cols.
  130. 130Why are Gateways in the network layer of computer networks seldom required nowadays?Modern computer networks operate almost exclusively with the Internet Pro-tocol (IP). For this reason, a protocol conversion at the Network Layer ismostly not required.
  131. 131Which devices divide the collision domain?Bridge, Layer-2-Switch, Layer-3-Switch, Router
  132. 132Which devicesdivide the broadcast domain?Router, Layer-3-Switch
  133. 133What is the meaning of Unicast in the network layer of computer networks?An IP address is assigned to a single receiver.
  134. 134What is the meaning of Broadcast in the network layer of computer networks?An IP address is assigned to all receivers in the subnet.
  135. 135What is the meaning of Anycast in the network layer of computer networks?An IP address is used to reach a single device of a group of devices.
  136. 136What is the meaning of Multicast in the network layer of computer networks?An IP address is assigned to a group of receivers
  137. 137Why contains the IPv4 address space only 4,294,967,296 addresses?IPv4 addresses have a length of 32 bits (4 bytes). Thus, the address spacecontains232= 4,294,967,296possible addresses.
  138. 138Why was Classless Interdomain Routing (CIDR)introduced?Because with address classes, many addresses are wasted and it is impossibleto dynamically adjust address classes.
  139. 139Describe in simple words thefunctioning of CIDR. Focus on the way, IP addresses are treated and subnets are created.Since the introduction of CIDR, the address class of an IPv4 address is nolonger important. All hosts in a network have a subnet mask assigned, whichhas a length of 32 bits (4 bytes) and is used to specify the number of subnetsand hosts. The network mask splits the host ID of an IP address into subnetID and host ID. 1-bits in the subnet mask indicate, which part of the addressspace is used for subnet IDs and 0-bits indicate, which part of the addressspace is used for host IDs.
  140. 140Name the three private IPv4 address spaces10.0.0.0/8
    172.16.0.0/12
    192.168.0.0/16
  141. 141Which two major classes of routing protocols exist?Distance Vector Routing Protocols and Link State Routing Protocols
  142. 142Which algorithms for best path calculation implement the routing pro-tocol classes from subtask 1?Distance Vector Routing Protocols implement the Bellman-Ford algorithm. Link State Routing Protocols implement the Dijkstra algorithm
  143. 143What is an autonomous system?Each AS consists of a group of logical networks, which use the Internet Pro-tocol, are operated and managed by the same organization (e.g. an InternetService Provider, a corporation or university) and use the same routing pro-tocol.
  144. 144The Border Gateway Protocol(BGP) is a protocol for. .Inter-AS routing
  145. 145Which routing protocol class from subtask 1 implements the BGP?Link state routing.
  146. 146Open Shortest Path First(OSPF) is a protocol for. .Intra-AS routing
  147. 147Whichrouting protocol classfrom subtask 1 implements OSPF?Link state routing
  148. 148The Routing Information Protocol(RIP) is a protocol for. . .Intra-AS routing
  149. 149Which routing protocol classfrom subtask 1 implements the RIP?Distance vector routing.
  150. 150When RIP is used, each Router communicates only with its direct neighbors. What are the advantages and drawbacks of method?Advantage: The network is not flooded⇒protocol causes little overhead.
    Drawback: Long convergence time because updates propagate slowly.
  151. 151When RIP is used, the path cost (metric) depend only on the number of Rou-ters (hops), which need to be passed on the way to the destination network. What is the drawback of this method?The metric hop count often results in routes, which are not optimal, becauseall network segments have an equal weight
  152. 152When OSPF is used, all Routers communicate with each other. What are the advantages and drawbacks of method?Advantage: Short convergence time.
    Drawback: The network is flooded⇒protocol causes strong overhead.
  153. 153What is the function of the Internet Control Message Protocol (ICMP)?It is used for the exchange of diagnostic and control messages, as well as error messages.
  154. 154Give two examples for command line tools, which use ICMP.ping, tranceroute
  155. 155Explain thedifferencesbetween TCP and UDP.UDP:
    -Connectionless Transport Layer protocol. Transmissions take placewithout previous connection establishment.–More simple protocol in contrast to the connection-oriented TCP. Only responsible for addressing of the segments. Does not secure thedata transmission.–The receiver does not acknowledge transmissions at the sender. Seg-ments can get lost during transmission.
    TCP:
    –Connection-oriented Transport Layer protocol.–Makes connections via IP reliable in a way that is desired or simplynecessary for many applications.–Guarantees that segments reach their destination completely and thecorrect order. Lost or unacknowledged TCP segments are requestedby the receiver at the sender.
  156. 156Describetwo examples, where using the Transport Layer protocol TCP ma-kes sense.TCP is used for Email transmission, file transmission and web page transmis-sion because no part of the information is allowed to get lost
  157. 157Describetwo examples, where using the Transport Layer protocol UDP ma-kes sense.If UDP is used for video transmission or video telephony, the only consequenceof losing a segment is losing an image
  158. 158What is a socket?Sockets are the platform-independent, standardized interface between the im-plementation of the network protocols in the operating system and the appli-cations.A socket consists of a port number and an IP address.
  159. 159What specifies theSeq numberin an TCP segment?The sequence number of a segment is the position of the segments first bytein the data stream
  160. 160What specifies theAck numberin an TCP segment?The sequence number of the next expected segment
  161. 161Describe thesilly window syndrome and its effectThe Silly window syndrome is a problem where a large number of packets issent, which increases the protocol overhead.Scenario: A receiver is overloaded and his receive buffer is completely filled.Once the application has read a few bytes (e.g. 1 byte) from the receive buffer,the receiver sends a segment with the free storage capacity of the receivebuffer. For this reason, the sender transmits a segment which contains just1 byte payload.Overhead: At least 40 bytes for the TCP/IP headers, 40 bytes for the acknow-ledgement and 40 bytes for the segment with notifies about the current freestorage capacity in the receive window.
  162. 162Describe the functioning ofsilly window syndrome avoidanceThe receiver notifies the sender about free storage capacity in the receivewindow not before 25% of the reception buffer is free or a segment size of sizeMSS can be received.
  163. 163Which two possiblereasonsfor the occurrence of congestion in computernetworks exist?Receiver capacity. The receiver can not process the received data fast enoughand therefore its receive buffer becomes full.Network capacity. Congestion of the network occurs
  164. 164Why does the sender maintaintwo windowswhen using TCP and not justa single one?The Advertised Receive Window avoids congestion of the receiver.The Congestion Window avoids congestion of the network
  165. 165What is theslow-startphase?The exponential growth phase
  166. 166What is thecongestion avoidance phase?The linear growth phase
  167. 167Describe what fast retransmit is?After three duplicate ACKs arrived, the lost segment is sent again
  168. 168Describe whatfast recovery is?The slow-start phase after three duplicate ACKs arrived is avoided. If threeduplicate ACKs arrive, the congestion window is set directly on the thresholdvalue.
  169. 169The concept of TCP congestion control is calledAIMD(= Additive Increase/ Multiplicative Decrease).Describe the reasonfor the aggressive reductionand conservative increase of the congestion windowhe consequences of a congestion window which is too large in size are worsethan for a window which is too small.If the window is too small in size, available bandwidth remains unused. If thewindow is too large in size, segments will get lost and must be transmittedagain. This increases the congestion of the network even more!The congestion state must be left as fast as possible. Therefore, the size of thecongestion window is reduced significantly.
  170. 170Describe the functioning of a Denial-of-Service attack viaSYN floodA client sends many connection requests (SYN), but does not respond to theacknowledgments (SYN ACK) of the server viaACK. The server waits sometime for the acknowledgment of the clients because the delay of the confirma-tion could be caused by a network issue. During this period, the address ofthe client and the status of incomplete connection are stored in the memoryof the network stack.By flooding the server with connection requests, the table which stores theTCP connections in the network stack is completely filled. This causes theserver to become unable to establish new connections. The memory consump-tion at the server may become this large that the main memory gets completelyfilled and the server crashes.
  171. 171What network devices are used in computer networks?Modem, Repeater, Hub (Multiport Repeater), Bridge, Layer-2-Switch, Router,Layer-3-Switch, Gateway
  172. 172Assign the devices to the layers of the hybrid reference modelPhysical Layer: Modem, Repeater, Hub (Multiport Repeater)Data Link Layer: Bridge, Layer-2-SwitchNetwork Layer: Router, Layer-3-Switch, Gateway
  173. 173What network device(s) is (are) used to. . . 1. connect networks with different logical address ranges? 2. transmit signals over long distances by modulating them to a carrier frequencyin the ultra low frequency band? 3. connect physical networks? 4. extend the range of LANs? 5. connect wireless network devices in the infrastructure mode? 6. enable communication between networks, which use different protocols?1.Router or Layer-3-Switch
    2.Modem
    3.Bridge or Switch
    4.Repeater or Hub (Multiport Repeater)
    5.Access Point
    6.Gateway
  174. 174Which protocol is used to. . . 1. provide congestion control and flow control? 2. resolves logical addresses into physical addresses? 3. avoid collisions inside physical networks? 4. provide routing within autonomous systems via the Bellman-Ford algorithm? 5. remote control computers in an encrypted way? 6. provide routing within autonomous systems via the Dijkstra algorithm? 7. assign the network configuration to network devices?1.TCP
    2.ARP
    3.CSMA/CA
    4.RIP
    5.SSH
    6.OSPF
    7.DHCP
  175. 175What network device(s) is (are) used to. . . 8. remote control computers in a unencrypted way 9. realize connectionless inter-process communication? 10. resolves domain names into logical addresses? 11. detect collisions inside physical networks? 12. download and upload files in an unencrypted way? 13. exchange (deliver) emails? 14. exchange diagnostic and control messages? 15. reduce a computer network to a loop-free tree?8.Telnet
    9.UDP
    10.DNS
    11.CSMA/CD
    12.FTP
    13.SMTP or POP3
    14.ICMP
    15.Spanning Tree Protocol (STP)

Información

Descripción
Fachabitur Informatik Fichas sobre Computer Networks, creado por Chu Chu el 27/02/2021.
Creado por
Chu Chu · feb 2021
Actualizado
2 mar 2021
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