新技术作业(all)

Chapter 11. Give the full names of the following acronyms:LAN ,WAN , MANA: LAN- Local Area Network;MAN-Metropolitan Area NetworkWAN-Wide Area Network;2. What types are networks classified into on behalf of applications/ coverage area/ transmission media/switching mode A: In term of applications, networks can be classified into three types: for voice, video and data;In term of area coverage, networks can be classified into LAN, MAN and WAN;By transmission medium, networks can be classified into two types: wired and wireless networks. By switching mode, networks can be classified into circuit switching networks and packet switching networks. 3. Think about the advantages and disadvangtages of wired transmission by comparing with wireless. A: For wired networks, the advantages are : Broadband, easy to configure, fast, long-distance; Disadvantages: less mobility;But wirless networks feature in their mobility but with disadvantages of less bandwidth, limit transmission distance, and also with less security. Chapter 21. What are the layers of the OSI model?The 7-layer OSI reference Module is as followsApplication/presentation/Session/transportation/network/data link/physical 2. Which layer determines path selection in an Internet ?A: Network layer 3. What types of things are defined at the physical layer?A: The physical layer define procedures for physical connection and execution, it includes types of functions :v Electrical: voltage level, timing and coding of voltage level v mechanical : physical connecting and cabling v Functional: bit-rate, max.transmission distance 4. Which includes more overhead, connection-oriented or connectionless services?A: connection-oriented5. Give a brief description about the two types of VC.A: There are two kinds of virtual connections : perminant virtual connection(PVC) and switched virtual connetion(SVC) (1) PVC: a permanent virtual circuit established for a user by network. PVC can be used at any time with guaranteed QoS, but it suffers the drawbacks of: fixed end-end connection and inefficiency bandwidth. . (2) SVC: a temporarily established virtual circuit with 3-implementation phrases; it features in flexible end-end and higher efficiency in bandwidth, but with negotiated QoS guarantee. 6. What are CS/SAR, VC, TDM/STDM, CBR/VBRCS/SAR : convergence sublayer/segmentation and reassembly sublayer TDM/STDM: Time division multiplexing/Statistical division multiplexing;CBR/VBR: Constant Bit rate/variable bit rate.7. Present 2 policies of flow control.Implicit flow control and explicit flow control.8. Tell the types of switching and multiplexingA: there are two types of switching modes including circuit switching and packet switching, and there are three types of multiplexing: frequency division multiplexing, time division multiplexing and code division multiplexing. 9. what is the basic voice rate?A: 64kb/sChapter 3QWhat are the two bit-rates of ISDN PRI services? AIn North America and Japan, 23 B channels (1.472 Mbps) plus 1 D channel (64 kbps); in Europe and Australia, 31 B channels (1.984 Mbps) plus 1 D channel (64 kbps). QWhat are the main services supplied by ISDN?A- ISDN was developed for integrated-services, so it can be used for all three kinds servicesvoice, video and data with 128kb/s bandwidth, like 64kb/s voice, email, telemetry, fax, teleconference, and so on. Q- What are the bit rates of D-channel, B-Channel in BRI of ISDN? A-In BRI, D-channel is 16kb/s and B-Channel is 64kb/s. Q- What Multiplexing is used by ISDN in the physical layer?A- TDMQ- Tell the main difference between Modem and ISDN?A- The signal transmitted over local loop is digital -In ISDN whereas analog in conventional modem technology, so that voice and internet s can not be accesses at the same time.Q-What switching is employed in ISDN?A- Either circuit switch or packet switch.Q-Is D-Channel a out-band/ in-band signaling system?A- out-bandQ-What is the DLL layers protocol used in ISDNs D channel.A- To deliver the Q931 massage error-freely across error-prone physical links.Q-what is the network layers protocol used in ISDNA-To establish, maintain and terminate the logical network connections between two ends.Chapter 41. Sketch the PSTNs hierarchical topology structure. A-The PSTNs hierarchical topology structure can be described as three hierarchies, including the regional switch, tandem switch and local loop. It can be sketched as: 21543Regional officeFully meshedLocalloopTandem Offices 2. Sketch the SS7s hierarchical topology structure, and point out the relative point components. A- The he SS7s hierarchical topology structure is presented as SCPSTPSTPSTPSTPSSPSSPSCP-local loop- SS7 linkPBXpacket switchesThere are three components involved in SS7 networks, they are STP(signal transfer point), SSP(signal switch point) and SCP (signal control point). 3. Sketch the SS7s OSI reference layerA- OSIphysicalData linkNetworkapplicationMTP layer 1MTP layer 2SCCP ISUPMTP layer 3SS7Chapter 51. What kind of technology is Frame Relay? A- It is communication Technology for interconnecting LANs in wide area. 2. Sketch the layer2s PDU formation and point out the main functions of fields named DLCI, DE,EA,FECN and BECN.A:Flag Add. Data FCS Flag The 2-byte address filed is as followsEA1DEBECNFECNDLCIEA0C/RDLCIThe core functions done in a FR note is :1) Frame delimiting, alignment, and flag transparenc by using flag;2) Virtual circuit multiplexing and demultiplexing by updating DCLI;3) Check of maximum and minimum frame size ; 4) Detection of transmission, format, and operational errors with FCS.FECN-Forward-explicit congestion notification ( 1bit)BECN-Backward-explicit congestion notification(1bit) FECN /BECN=1 indicates congestion occurs in the transmission direction or in the opposite direction of transmission. DAExtended address. EA=1 means the current byte is determined to be the last DLCI octet. DE -Discard eligibility. ED=0 means the frame will be guaranteed to be transmitted in any cases.ED=1 means the frame will be discarded before other frames in a congested network. 3. What is the data-link connection identifier (DLCI)? A - DLCI denotes data link connection identifier 4. Frame relay is used as a LAN/MAN/WAM/Internet?A- WAN 5. EA= (0,1,2,3) means this is the last byte of the address field. A- 16. The frame Relay Switches are located in : a users home/at the central office of a PSTN/at a government office/at a corporationA- at the central office of a PSTN7. A LAN can be connected to a Frame relay network by a peater/ FRAD/switch/gateway.A- FRAD.Chapter 6 (SDH) 1. Present the layer Stacks of SDHA-2. Present the STS-1 frame of SDH? Calculate the interface bit-rates of STS-1, STM-1,STM-4,STM-16.9 rowsSOHLOHSPE(Synchronous payloadEnvelope)POH87-column3-column125usSTS-1: 90x9x8x8000=51.840Mb/s;STM-1: 90x9x8x8000*3=51.840*3=155.52Mb/s;STM-4: 90*9*8*8000*12=622.08Mb/s;STM_16: 90*9*8*8000*48=2.4883Gb/s; 3. What are the functions of the pointers in LOH of SDH frames.There are 3 bytes used as pointers to perform two functions: The first 2-byte called H1 and H2 used for indicating the payload offset which allows the users payload floating in the SPE. The third byte H3 is used to overcome the problem caused by the timing difference among different networks. In that case, SDH can perform asynchronous function in the TDM networks. 4. What are the bit-rate interfaces of STS-1/basic voice bitrate/T1/E1 A- STS-1: 51.84mb/s; Voice bit-rate: 64kb/s;T1: 1.544Mb/s;E1: 2.048Mb/s.5. By what network topology can SDH be featured with self-healing ability.A-Ring /With the ring architecture of topology. 6. What do C/VC/VT/TUG mean in SHD ?AC: Container; VC: Virtual Container; VT- Virtual Tributary; TUG: Tributary unit grope.7. Which container is used for accommodating T1 bit-rate traffic? AC11 Chapter 7 (ATM) 1. Explain why the fixed and short cell of ATM can support CBR traffic.A- We know that the package time of a long packet may exceed the limit of the over all transmission delay for a voice traffic, so that to save the package time, a short packet is required for delivery a CTB traffic. Moreover, to save the switching time by using hardware switches, fixed length packet is also required. So that, ATM proposed a 53-byte short cell to serve CTB traffic. 2. Depict the ATM cell format, and answer how many bits in ATMs header are used for VCI and VPI, and What functions else are executed in ATM layer.A-48 bytePayload5 byteHeaderThere are 5 bytes in the header of a ATM cell, in which 28 bits are used as VCI and VPI. The main functions of ATM layer is as follows:At the edge switch node: Cell header insertion/remove At intermediate switch nodes: Cell header read/rewrite Implicit flow control cell discarderror detection /1-bit correctionAt all switches: Cell relay/switching according to VPI/VCI3. Depict the frame format of FR. What are the core functions executed in FRs DLL layer.A-A:Flag Add. Data FCS Flag The 2-byte address filed is as followsEA1DEBECNFECNDLCIEA0C/RDLCIThe core functions done in a FR note is :1) Frame delimiting, alignment, and flag transparenc by using flag;2) Virtual circuit multiplexing and demultiplexing by updating DCLI;3) Check of maximum and minimum frame size ; 4) Detection of transmission, format, and operational errors with FCS.4. Tell the most significant differences between IP and ATM. A-The main difference between IP and ATM is that IP employs connectionless packet switch and ATM employs connection oriented packet switch. 5. Tell the main differences between FR and ATM. A-ATM employs fixed length packet, and hardware in switch nodes, so it can be used for RT traffic, like voice and video; whereas FR employs flexible length packet, and execute store- delivery scheme in switch nodes, so can not be used for RT traffic. Both are considered as fast switch technologies, and execute the minimum L2 functions. Chapter8 Assignments1. Sketch the network architecture of GSM, and indicate the components of BSS, and NSS respectively. 1Answer: BSS: MS，BS/BTS，BSC；NSS：MSC，HLR，VLR，AC，EIR.2. Sketch the frame of Um interface of GSM. BS MS76543210357126157T B Data f Training f Data TB GB 38.25A slot: 0.577ms; there are 156.25bits in the slot; (x8=4.615ms);The length of a frame of Um: 0.577ms x8=4.615ms; 156.25x8=1250bits;3Tell the working spectrum of GSM 900, and tell how much channels are there in the upstream band, and how much is the channel space, and how much is the frequency gap between the up and down channels. Upstream: 890-915Mhz; Downstream: 935-960Mhz;There are 125 Channels in total in up-/down-stream band. The channel space between two adjacent channels is 200Khz; and the frequency gap between a pair of up/downstream channels is 45Mhz. 4. If the code rate is 3480kb/s, and the information bit rate is 15kb/s, how much is the spreading factor?spreading factor=3480/15=232.5. What does GSM refer to as? GSM: global system for mobile communication. 6. Tell the three international protocols of 3G.WCDMA; CDMA2000; TD-SCDMAChapter 9 Broadband Access1. There are two ways to realize broadband access, what are they?A:The two ways to realize broadband access are : (1) by employing new modulation/coding scheme over old transmission media; (2) using optical fiber. 2. Sketch the network architecture of ADSL, and tell the frequency allocations of voice, uplink dada, and downlink dada. A: The network topology isResidenceThe frequency allocation is3. Sketch the network architecture of HFC, and tell the frequency allocations of broadcast video,