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by: Mr. Nathan Borer


Mr. Nathan Borer
GPA 3.6

Daji Qiao

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Daji Qiao
Class Notes
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This 22 page Class Notes was uploaded by Mr. Nathan Borer on Sunday September 27, 2015. The Class Notes belongs to CPR E 489 at Iowa State University taught by Daji Qiao in Fall. Since its upload, it has received 54 views. For similar materials see /class/214526/cpr-e-489-iowa-state-university in Computer Engineering at Iowa State University.

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Date Created: 09/27/15
Cpr E 489 Fall 2011 Midterm 1 Review a Topic 1 Introduction and Layered Networking Architecture 0 Three communication networks Chapters 10 11 I Key differences between three transfer modes a Circuit Switching POTS Plain old telephone service 0 End to end circuit required 0 analog or digital transmission 0 end user simple 0 network complicated a Message Switching Telegraph network 0 Store and forward operation addressing routing and forwarding next best hop digital transmission end user message address 0 network less complicated a Packet Switching Computer network 0 break message into packets 0 packets sent via store and forward 0 end user packet and control info 0 network simple routers 0 Why layering Chapter 20 I Layering simplifies design implementation and testing by partitioning related communication functions into groups that are manageable I Each layer operates according to a protocol I Each layer provides a service to the layer above Benefits of layering a Simplifying the design process a Providing great flexibility in modifying and developing the network 0 7layer OSI reference model Fig 24 Chapter 22 Key functions of each layer example protocol at each layer a Application 0 FTP Telnet and HTTP protocols 0 Presentation FTP Telnet and HTTP protocols o absorbed into Application layer a Session 0 FTP Telnet and HTTP protocols o absorbed into Application layer 0 O O O a Transport 0 Transfers segments from process in one machine to process in another machine endtoend transfer 0 TCP and UDP protocols a Network 0 Transfers packets across multiple links andor multiple networks 0 IP protocol a Data Link 0 Transfers frames across direct connections 0 EthernetPacket RadioPoint to Point a Physical o Transfers bits across a communication link 0 EthernetPacket RadioPoint to Point I Unified views of layers protocols and services Figs 28 29 a Definitions 0 Entity I the process at layern o PDU I Protocol Data Units data to exchange between layern entities I nPDUheadernlPDU o SDU I Service Data Units data to exchange between layern and layern1 entities I nSDUnlPDU o Header I Carries control information such as addresses sequence numbers fag bits length indicators etc 0 Trailer I CRC check bits added at end for error detection 0 SAP I Software port to pass nSDU between layern and layern1 0 Protocol I the set of rules that governs a layern entity 0 Service I provided by layern for use by layern1 a Encapsulation Fig 27 o Layern SDU is encapsulated in the layern PDU o In principle the layern protocol does not interpret or make use of the 0 information contained in layern SDU ie layern1 PDU 0 Layer n1 as a user of the service provided by layer n is only interested o in the correct execution of the layern service required to transfer its 0 PDUs and the details of the implementation of lower layers are irrelevant o Connectionoriented service vs connectionless service example protocol for each Chapter 22 I Connectionoriented 3 phases 0 establish connection btween two layern SAPs I Negotiating connection parameters I Initializing quotstate information Transferring nSDUs using the layern protocol Closing the connection and releasing the various resources allocated to the 0 connection a Example TCP I Connectionless a Connectionless service does not require a connection setup and each SDU is transmitted directly through the SAP 0 Example UDP IP The control information from layern1 to layern must contain the complete address information required to transfer the SDU I Layered services need not be of the same type a TCP operates over IP 0 Other functions Chapter 22 I Segmentation amp Reassembly OO O O O splitting an nSDU to be handled by a smaller sized nPDU Blocking amp Unblocking packing several nSDUs into a larger sized nPDU Multiplexing amp Demultiplexing Sharing of layern service by multiple layern1 users Multiplexing tag or ID is required in each nPDU to determine which user an n SDU belongs to Splitting amp Recombining In contrast to MultiplexingDemultiplexing Use of several layern services to support a single layern1 user 4layer TCPIP model Chapter 23 Key functions of each layer example protocol at each layer Application 0 ftp telnet http Transport 0 tcpudp Internet 0 i Network Interface 0 ethernet Figs 212 213 214 a Topic 2 Physical Layer Digital transmission vs analog transmission Chapters 30 32 A transmission system makes use of a physical transmission medium or channel that allows the propagation of energy in the form of pulse or variation in voltage current or light intensity Analog Transmission Objective is to transmit a waveform which is a function that varies continuously with time Repeater attempts to restore analog signal and retransmit Distance limited Quality greatly affected by noise errors propagated through and generated by repeaters Digital Transmission Objective is to transmit a symbol which is selected from a finite set of possibilities Repeater recovers original symbols and retransmits Can be designed to keep error probability very small Each regenerated symbol is the same as the original No accumulation of noise Digital transmission is possible over very long distance Benefits 0 Digital transmission systems can operate with lower signal levels or with greater distance between repeaters o This means lower overall system cost The digital networks are suitable for handling many different types of information that can be represented in digital form 0 Transmission channel Chapter 35 Channel bandwidth Nyquist rate Channel bandwidth W o the range of frequencies that is passed by the channel Nyquist rate 0 Rmax 2W pulsessecond a C W log21SNR bps o The fastest rate at which pulses can be transmitted over the channel By going with M 2m amplitude levels we achieve 0 Bit Rate 2W pulsessec m bitspulse 2mW bps o where C is the channel capacity W is frequency SNR is signal to noise ratio I Bit rate Baud rate bit rate 0 ii of bits per second 0 bit rate baud ratex bits per pulse baud rate 0 ii of signal transitions per second 0 depends on channel bandwidth amp coding scheme I Shannon channel capacity C W log2 1 SNR bps Channel Bandwidth W amp Signal to Noise Ratio SNR determine C If transmission rate R gt C reliable communication is not possible If transmission rate R C arbitrarily reliable communication is possible 0 quotArbitrarily reliablequot means the BER can be made arbitrarily small through sufficiently complex coding The relation between Rmax and C is used as a measure of how well a communication system is designed o Line coding Chapter 36 lecture notes I Methods 0 Unipolar NRZ o 1 represented by one physical level 0 0 represented by zero level a Polar NRZ o 1 represented by one physical level 0 0 represented by equal negative level a Bipolar o 1 represented by alternating positive or negative level 0 O is represented by zero level a NRZ Inverted o l is represented by a transition of the physical level 0 O is represented by no transitionsteady signal a Manchester 0 l is represented by a hightolow transition 0 O is represented by a lowto high transition a Differential Manchester 0 l is represented by no transition from previous voltage 0 O is represented by any transition to opposite signal from previous voltage or 0 Manchester coding schemes need bandwidth rich environments They essentially need twice the bandwidth and baud rate to deliver the same bitrate as other coding schemes 0 2BlQ bits Previous Previous voltage voltage 00 A2 AZ O 1 3 A2 3A2 lO A2 A2 l l 3A2 3A2 o Bandwidth requirement Fig 336 unipolar nrz 0 low bandwidth 0 no timing info polar nrz 0 low bw o no timing info bipolar 0 lower bw o no timing info nrzi manchester o symbol rate bit rate differential manchester o symbol rate 2 x bit rate 2blq o symbol rate data rate2 a Topic 3 Error Detection and Recovery 0 Error detection Chapter 39 with corrections on lecture notes I Parity check codes 2outof S code Parity check code 0 Append an overall parity check bit to k information bits Information Bits bl b2 b3 bk Parity Check Bit bkl bl b2 b3 bk mod 2 Codeword bl b2 b3 bk bkl Pattern all codewords have even ii of 1 s Receiver checks whether ii of 1 s is even All errors that change an odd number of bits are detectable All evennumbered errors are undetectable o No error correction 2out ofS code 0 an encoding scheme which uses five digits consisting of exactly three Os and two ls o ten possible combinations enough to represent the digits 0 9 o This scheme can detect all single biterrors and all odd numbered bit errors However it still cannot correct for these errors 0000000 I Pattern redundancy undetectable errors Parity bit cannot detect even number of errors Pattern all code words have even number of ones I Models of transmission errors probability oferror detection failure I Internet checksum add all the words as unsigned numbers in the packet if the words are size n then we are using modulo 2Nl arithmetic sum mod 2nl x additive inverse of x in mod 2nl arithmetic is the checksum mam 5m mummy my zua zan mrnemmrvuwmmi ilxlNvmsemmekedbizkxivs mmm manigm nmbnhm mmmmnmumwm mmmmmmwmw miubmdmmmu gemmrmmnquq mmshmmmsmrumu lehmisimwmn dummy miumnumnl Pox 06 mmpowmmmmmumvm xuxxummq xumm mwmvenurmbr dwsnnm mdrenmmermnzmn 1 mmmmeumaxxum bx xlorlxlsx wx ox oroxm 9mm 5 1mm1mam1 murmmmumw Mmmmn a wammapmw enurburs mm mmmmummum Mixramurnsnzannmdwneelxmmmeorw mmmaumum s Forenws mwnmluiuom indlmherep exquuwukxmmay MIXavnmmvevuwmmmwnhdqreeMWJGNm m mm xlzindemui doubbermrsii W is mm a mammmm mmauwmmrm swimquot em manmmms mu soxxuxamaqummmu I If gx has x l as a factor it cannot divide ex that has odd number of terms so it can detect all odd number of errors a How to tell whether gx has x l as a factor 0 Evaluate gx atx 1 o fgxxl gl 0 YES else NO 0 error bursts I For Error Bursts of Length L a For error burst starting at bit location i L and ending at bit location i l o ex xiLl xi xi dx where dx xLl l I gx has degree n k a L lt n k 1 o gx cannot divide dx because degdx lt degglxn 0 Can detect all such error bursts I L n k l a dx is divisible by gx ifand only if dx gx a Fraction of such error bursts that are undetectable is 2nk1L gt n k l a Fraction of such error bursts that are undetectable is mink 0 Special Topic Sockets Programming Lecture notes I What is a socket What is the network byte order a Network byte order is big endian MSB first a A socket is a communication endpoint that allows programmers to write application programs easily without worrying about the underlying network architecture I In TCP sockets programming how to establish a TCP connection int cl isock cient socket descriptor Int n struct sockaddrin remoteaddr address of remote server char buffer256 buffer for reading data from the socket ifargc lt 2 printfquotnvalid number of argumentsnquot exitl create client socket ifclisock socketPFINET SOCKSTREAM 0 lt O there was a problem opening the socket perrorquotsocket error nquot exitl exit with an error char addrl argvl char addr2 argv2 set up address of remote server remoteaddrsinfamily PFNET remoteaddrsinport htons5003 remoteaddrsinaddrsaddr inetaddraddrl connect to socket using the remote address connectclisock struct sockaddr ampremoteaddr sizeofremoteaddr n readclisock buffer 255 printfquots snquot addrl buffer cosecisock struct sockaddrin remoteaddr2 address of remote server remoteaddr2sinfamily PFNET remoteaddr2sinport htons5003 remoteaddr2sinaddrsaddr inetaddrader int clisock2 create client socket ifcisock2 socketPFINET SOCKSTREAM 0 lt 0 there was a problem opening the socket perrorquotsocket error nquot exitl exit with an error connectclisock2 struct sockaddr ampremoteaddr2 sizeofremoteaddr2 n readclisock buffer 255 printfquots snquot addr2 buffer cosecisock2 What are the functions of socket bind listen accept connect system calls a socket int creates an endpoint for communication and returns a socket descriptor a bind attach socket to a particular network interface a listen marks the socket as a passive socket used to accept incoming connections 0 for success 1 for error accept accept an incoming connection on a socket This blocks the caller until a connection arrives if the socket is not marked as quotnonblocking If it is marked such as What is clientse a Paradig O as quotnonblocking a call to acceptfails with errors connect initiate a connection on a socket for connectionbased protocols TCP rver paradigm m Computers that provide services are servers Computers that request and enjoy their services are clients In most cases server applications wait passively for client application to initiate communication Sockets explicitly created used and released by both client and server applications Client initiates a conversation by opening a connection to the server Once the connection is set up the client sends request to server Server sends response back to client a Iterative server vs concurrent server 0 He said we didn t need to study this for the test I Error handling EXAM 2 NOTES BEGIN HERE waMuV quotmum n mam s m sequence m Wm veva pm am my sendAEan m gem 5m vecenevwmdaw uv1 n mam 5 am msmuence m n my discard packet Sender send current packet in sequence wait for ACK if ACK received send next packet in sequence slide send window up else timeout resend current packet GoBackN ARQ mbit sequence numbering 2m 2 N1 Receiver wait for packet arrival if packet is NOT error freeCRC discard packet else crc passed shift register circuit ifpacket arrived next in sequence report data to the upper layer next in sequence ACKnext in sequence else discard packet ack next in sequence endif endif Sender Wait for ACK ifACK received before timeout of oldest outstanding frame if ACK is error free ifRnext element of Slast 1Srecent 1 Slast Rnext else Discard ACK en else Discard ACK en i else timeout retransmit entire send window of size n Slast Slast1Slastn1 endif Selective Repeat ARQ mbit sequence numbering 2m 2 WsWr Receiver Wait for packet ifpacket is error free ifSrecent element of Rnext Rnext 1 ifSrecent Rnext Report Rnext and all inorder frames to upper layer Rnext sequence of highest report packet 1 ACKRnext else buffer Srecent NAKRnext end if else discard packet ACKRnext end if discard frame wait for ACKNAK ifACKNAK within Slast timeout ifpacket is error free ifpacket is ACK ifRnext element of Slast 1 Srecent 1 Slast Rnext else discard ACK endif else Slast Rnext retransmit Slast ONLY endif el discard ACKNAK en else retransmit Slast ON LY end if 0 Why SampW gt GBN gt SR a Topic 5 Data Link Layer 0 Framing ch 54 I Byte stuffing in characteroriented framing Data 0 integer of characters a Flags 0 Start of Frame DLE STX 0X10 0X02 0 End of Frame DLE ETX 0X10 0X03 a Special Treatment 0 insert DLE in front of each DLE in data I Byte stuffing in PPP a Data 0 integer of bytes a ag 0 Start and End 0X7E a Special treatment 7E gt 7D SE 0 7D gt 7D 5D I Bit stuffing in HDLC o arbitrary number of bits 39 0 01111110 0 Special Treatment insert an extra 0 after five consecutive 1 s a unstuffing example 0 011111 I next bit 0 stuff bit I nextbit 10 flag 11 error I CRObased Framing GFP 0 Medium Access Control MAC Schemes ch 60 61 62 I ulnerable Period a Data frame arrives at A at time to a A schedules transmission of this frame according to MAC rules a VP for As Frame transmission is time period during which 0 data frame arrives at B and B schedules according to MAC rules 0 Es frame transmission collides with As I ALOHA Slotted ALOHA a VP for ALOHA o t0x t0x 0 Length 2x a ALOHA Load vs Throughput o S throughput 02 when G load 05 a VP for Slotted ALOHA x 0 Length X a Slotted ALOHA Load vs Throughput o S throughput 04 when G load 1 I CSMA CSMACD a Smaller a value tpropX better throughput a Vulnerable Period 0 t0tprop t0tprop Length 2tprop a CSMA options 0 1persistent 0 non persistent o p persistent a How does CD work minimum frame size 0 abort transmission as soon as a collision is detected 0 Min frame size I Takes up to 2tprop time to find out whether it has captured channe I increase min frame size or reduce max distance bw stations I Psuccessful transmission np1pn1 I Where n is the number of stations contending for the channel a p the probability that a station will attempt to transmit during a given minislot a This is maximum when p 1n o psuccess gt 1e a Load vs Throughput atpropx o 1pers I better than ALOHA amp Slotted ALOHA for small a I worse than ALOHA when a gt 1 worse than Slotted ALOHA when a gt 05 0 non persistent I higherthroughputthan1pers I worse than ALOHA agt1 I worse than Slotted ALOHA agt05 I Why ALOHA gt Slotted ALOHA gt CSMA gt CSMACD a To increase the efficiency of these protocols the goal is to reduce the vulnerable period and increase the throughput 0 LAN Structure ch 661 NIC parallel to serial conversion a ram 0 buffer a rom o MACphysical address unique a can run in promiscuous mode 0 Ethernet LAN and IEEE 8023 ch 67 I IEEE 8023 MAC protocol a CSMACD I IEEE 8023 frame format 0 Repeater bridge router gateway Ch 6110 6111 I ater a physical layer a hub a rebroadcast packet to all ports I Bridge a MAC or Data Link layer a switch a keeps table of MAC addresses per port to avoid broadcasting I Router a network layer a keeps a table of known networks and their next hop a default route for unknown networks I Gateway a higher layer a application based protocol translation a interconnects networks of different protocols 0 Wireless LAN and IEEE 80211 Lecture notes I CSMACA protocol n nodes sense for a jammed or busy signal before attempting transmission a if the channel is busy the node backs off a random time and senses a if the channel is clear nodes use a jamming signal on the carrier to show its intent to transmit before actually transmitting I Truncated stops with a max exponent of 10 Exponential Backoff If a collision n occurs for a particular packet transmission then the host will chose a random minislot time period of 2Tprop in the future timeinterval 0 2m 1 where m minn10 to wait for For WiFi and Ethernet LAN the hosts will give up after 16 frame collisions I CSMACA vs CSMACD a CSMACD does NOT request exclusive access to the channel before a Topic 6 O O attempting to transmit Once it senses that the channel is idle it simply starts transmitting If a collision is detected the signal on the channel is not the same as the signal we are trying to send then a short jamming signal is sent usually all 1 s to notify the other hosts that a collision has occurred CSMACA with exponential backoff WIFI establishes a handshake between the sender and receiverto obtain exclusive access to the channel Only once sender gets confirmation from the receiver that it is ready to receive packets will transmission begin lfthere are contentions between multiple hosts trying to gain sole control over the channel then the hosts perform the exponential backoff algorithm to reschedule a time to sense the channel WiFi uses this because the hosts in the network are not in constant communication to with the channel as compared to a wired ethernet LAN where the individual hosts can constantly sense the channel for idlebusy periods as well as collisions Network layer IP Addressing Name vs Address Name DNS Name what meaningful and easy to remember variable and hard to process sl a n ookup I IP address IPv4 where fixed length easy to process 4 binary octets Addressing schemes Classfull Addressing Subnet Addressing part of orig hostid becomes subnetid Supernet Addressing CIDR a supernet is a block of contiguous subnetworks addressed as a single subnet in the larger network addresses converted to binary and aligned group all same bits group of same bits is new network address and mask is the length of that group IP Routing how does a router look up its routing table to forward an IP packet 2 3 4 Whether the destination IP address appears in one of the table entries Whether the destination network address appears in one of the table entries with help of network mask The default router entry If none of above searches is successful declare packet undeliverable send ICMP Host Unreachable Error packet back to the sender FINAL EXAM NOTES BEGIN HERE Yawc a Nmam um mmmms Wyamucmvmsau am a mmmwu an sang IE u 215 x I I Vusm IHL TVPl S39rwcc mame um mm 72 Fragmgm Wm WM MW Hummm Swune PAddvess mm w My u OPEn5 Padllng nmmwemu Wms nmexceededsend EMVvawd mwwzucunzu mmmssmv cmmcwsum cw m mums vevsmmldam mm cmnmmmmwmm mmmm quupmmm mam AC Vvavmenummexsembw mum MW mquot mam 5 mm mmmmwmm s rm Mummmaumm mcaumuvvevhudevs ms mm H mmm Ewan 2mm ma pmmvmm a u amers aveaHawedm ivmeM 3mm MFbM I Exampl O o 1 More Fragments afterthis 0 O e MTU 576 bytes IP header 20 bytes data part 1484 bytes Max Payload of layer 2 frame is 576 bytes as per MTU Max data each IP fragment can carry is 576 20for the header 556 byte 0 But this needs to be a multiple of 8 bytes for the offset 0 This means max data each IP fragment can deliver is 552 bytes First Fragment 001 Offset 0 20 bytes for header 552 bytes for data 0 Data bytes left to send after this fragment 1484 552 932 Second fragment Flag 001 Offset 5528 69 o 20 bytes for header 552 bytes for data 0 Data bytes left to send after this fragment 932 552 380 Third fragment Flag 000 no fragments left after this Offset 552 5528 138 20 bytes for the header 380 bytes for the data 000 O O O a Shortestpath routing Chapters 751 752 754 0 Distance Vector routing I BellmanFord algorithm a I Initialization destination node d is distance 0 from itself Di for all i I a 2 Updating find minimum distance to destination through neighbors for each i I d o DiminjCJDj for all ji a repeat step 2 until convergence I Routing loop Split Horizon Split Horizon with Poisoned Reverse Path Vector a Routing loop 0 counting to infinity problem 0 ex 2 nodes both think the shortest path is through the other keep iterating and updating cost infinitely a Split horizon o the minimum cost to a given destination is not sent to a neighbor if the neighbor is the next node along the shortest path a Split horizon with poisoned reverse 0 allows node to send the minimum costs to all its neighbors however the min cost to a given destination is set to infinity if the neighbor is the next node along the shortest path a Path Vector o LoopFree o routing protocol which maintains the path information that gets updated dynamically Updates which have looped through the network and returned to the same node are easily detected and discarded Each entry in the routing table contains the destination network the next router and the path to reach the destination 0 0 Link State routing Dijkstra s algorithm a shortest path from a source node to all other nodes on a network a 1 Initialization o s o DECS for all js o DS0 a 2Finding the next closest node Find node i not in N such that o Dimin 0 Add i to N o If N contains all nodes stop a 3 Updating minimum costs after node i added to N for each nodej not in N o DjminDJDCij 0 Go to step 2 0 Link State vs Distance Vector I Link state will respond faster to linkfailures because it floods its information to all nodes in the network I Distance Vector will use less bandwidth because it does not need to flood the entire network only neighbors I RIP OSPF BGPv4 Chapter 860 0 RIP I Routing Information Protocol max hops 15 16 hops is considered infinity Sends update of distance to each destination to neighbors every 30 seconds a after 180 seconds with no update from neighbor sets cost to 16 infinity a Split horizon with poisoned reverse 0 OSPF I Open Shortest Path First a enables each router to learn the complete topology a each router monitors the cost ofthe link to each of its neighbors and then floods the link state information to other routers in the network linkstate protocol 0 BGPv4 Topic 7 Transport Layer Transmission Control Protocol Seqnobyte index TCP ACK is cumulative ACK No NAK only ACK Duplicate ACKs may imply packetloss TCP Error Control a SRARQ i No NAK ii Retransmit upon 1 RTO 91590N7 2 4th ACK with same Seqno 3 dup ACKs TCP segment format 0 see book page 605 TCP connection establishmenttermination 3way handshake 0 Host A sends a connection request to host B by setting the SYN bit Host A also registers its initial sequence number to use Seqnox Host B acknowledges the request by setting the ACK bit and indicating the next data byte to receive Ackno x1 The plus one is needed because the SYN bit consumes one sequence number At the Same time host B also sends a request by setting the SYN bit and registering its initial sequence number to use Seqnoy Host A acknowledges the request from B by setting the ACK bit and confirming the next data byte to receive Acknoy1 Note that the sequence number is set x1 On receipt at B the connection is established a TCP ACK and error control RTO Karn s algorithm a TCP flow control difference bw flow amp congestion control 0 SWND lt RWND a TCP congestion control Slow Start Congestion Avoidane TCP Tahoe TCP Reno 0 step 1 probe the network capacity pipe size I Slow Start SS a start with CWND1 o in units of Max Segment Size 536 bytes a each time a non dup ack is received 0 CWND CWND 1 a exponential growth of CWND I Congestion Avoidance CA a SSThresh a enter CA phase when CWND gt SSThresh O O O O O O a each time a non dup ack is received 0 CWND CWND 1CWND o wait for whole window before increasing CWND by 1 step 2 congestion detection I RTO I 3 Duplicate ACKs 4th with same Seqno step 3 on detection I slowdown transmission TCP Tahoe I SSThresh floor CWNDZ I Retransmit ost segment I Start with 88 TOP Reno I SSThresh floor CWNDZ I Retransmit ost packet a number of outstanding frames CWND a each dup ACK gt 1 outstanding frame exits pipe 0 count number of dup ACKs I After cwnd2 dup ACKs a resume transmission 1 per dup ack I non dup ACK received a CWNDSSthresh a start from CA I On RTO a ssthresh foorcwnd2 a retransmit a cwnd1 a 88 then CA


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