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Review for Quiz-2 ECE3600 - Fall 2015 Prof. John Copeland 3-29-2015 Computer Networking: A Top Down Approach Featuring the Internet, 5th edition. Jim Kurose, Keith Ross Base material copyright 1996-2006 Addison-Wesley, July 2004. J.F Kurose and K.W. Ross, All Rights Reserved Quiz 2 -1 ECE3600 Quiz-2 Review • Chap. 4 – IP Addresses and Routing • Chap. 5 – LANs, Ethernet Networks • Chap. 6 – Wireless LANs • Chap. 7 – Multimedia Quiz 2 -2 Chapter 4: Network Layer IP Addresses • 4. 1 Introduction • 4.2 Virtual circuit and datagram networks • 4.3 What’s inside a router • 4.4 IP: Internet Protocol – – – – Datagram format IPv4 addressing ICMP IPv6 Quiz 2 -3 Longest prefix matching Prefix Match 11001000 00010111 0001 0 11001000 00010111 0001 1000 11001000 00010111 0001 1 otherwise Link Interface 0 1 2 3 Examples DA: 11001000 00010111 0001 1000 1010 1010 Which interface? DA: 11001000 00010111 0001 0110 1010 0001 Which interface? Matches links 1 & 2 - Link 1 the longest. Only matches Link 0 DA: 11001000 00010111 0001 1100 1010 1010 Only matches Link 2 Q3-4 Quiz 2 -4 IP datagram format IP protocol version number header length (bytes) “type” of data max number remaining hops (decremented at each router) upper layer protocol to deliver payload to how much overhead with TCP? r 20 bytes of TCP r 20 bytes of IP r = 40 bytes + app layer overhead 32 bits type of ver head. len service length fragment 16-bit identifier flgs offset upper time to header layer live checksum total datagram length (bytes) for fragmentation/ reassembly 32 bit source IP address 32 bit destination IP address Options (if any) data (variable length, typically a TCP or UDP segment) E.g. timestamp, record route taken, specify list of routers to visit. Q3-5 Quiz 2 -5 IP Fragmentation and Reassembly Example r 4000 byte datagram r MTU = 1500 bytes 1480 bytes in data field offset = 1480/8 length ID fragflag =4000 =x =0 offset =0 One large datagram becomes several smaller datagrams length =1500 ID fragflag =x =1 offset =0 length =1500 ID fragflag =x =1 offset =185 length =1040 ID fragflag =x =0 offset =370 Quiz 2 -6 Subnets – have a contiguous block of IP addresses which have the first N bits in common (a "/N"). Recipe • To determine the subnets, detach each interface from its host or router, creating islands of isolated networks. Each isolated network is called a subnet. 223.1.1.0/24 223.1.2.0/24 223.1.3.0/24 Subnet mask: /24 (24 1's (32-24) 0's 11111111 11111111 11111111 00000000 Dotted decimal notation: 255.255.255.0 Q3-7 Quiz 2 -7 IP addressing: CIDR CIDR: Classless InterDomain Routing – subnet portion of address of arbitrary length – address format: a.b.c.d/x, where x is # bits in subnet portion of address subnet part Addr. Mask: host part 11001000 00010111 00010000 00000000 11111111 11111111 11111110 00000000 Addr. Mask: 200.23.16.0/23 255.255.254.0 Inverted = 0.0.1.255 Network Address = Host Address (Bitwise AND) Network Mask = A & M Minimum Host Address* = Network Address Maximum Host Address* = Network Address (OR) [Inverted Network Mask] * (reserved, not assigned to host) Quiz 2 -8 IP Address Bitwise Calculations 200.23.16.0/20 1101000 00011001 0001xxxx xxxxxxxx From this, to get Network Mask, 0 or 1 -> 1, x -> 0 11111111 11111111 11110000 00000000 Minimum Host Address: x -> 0 1101000 00011001 00010000 00000000 Maximum Host Address: x -> 1 1101000 00011001 00011111 11111111 Minimum host address is the “Network Address” Maximum host address is the “Broadcast Addr.” Q3-9 Quiz 2 -9 NAT: Network Address Translation 2: NAT router changes datagram source addr from 10.0.0.1, 3345 to 138.76.29.7, 5001, updates table 2 NAT translation table WAN side addr LAN side addr 1: host 10.0.0.1 sends datagram to 128.119.40.186, 80 138.76.29.7, 5001 10.0.0.1, 3345 …… …… S: 10.0.0.1, 3345 D: 128.119.40.186, 80 S: 138.76.29.7, 5001 D: 128.119.40.186, 80 138.76.29.7 S: 128.119.40.186, 80 D: 138.76.29.7, 5001 3 1 10.0.0.4 S: 128.119.40.186, 80 D: 10.0.0.1, 3345 10.0.0.1 10.0.0.2 4 10.0.0.3 4: NAT router 3: Reply arrives changes datagram dest. address: dest addr from 138.76.29.7, 5001 138.76.29.7, 5001 to 10.0.0.1, 3345 NAT Table for Internal Servers must be configured manually (Port Forwarding). Quiz 2 -10 ICMP: Internet Control Message Protocol • • • used by hosts & routers to communicate network-level information – error reporting: unreachable host, network, port, protocol – echo request/reply (used by ping) network-layer “above” IP: – ICMP msgs carried in IP datagrams ICMP message: type, code plus first 8 bytes of IP datagram causing error Type Code description 0 0 echo reply (ping) 3 3 3 3 3 3 0 1 2 3 6 7 dest. network unreachable dest host unreachable dest protocol unreachable dest port unreachable dest network unknown dest host unknown 4 0 8 9 10 11 12 0 0 0 0 0 source quench (congestion control - not used) echo request (ping) route advertisement router discovery TTL expired bad IP header Quiz 2 -11 Traceroute and ICMP • Source sends series of UDP segments to dest – First has TTL =1 – Second has TTL=2, etc. – Unlikely port number • When nth datagram arrives to nth router: – Router discards datagram – And sends to source an ICMP message (type 11, code 0) – Message includes name of router& IP address • When ICMP message arrives, source calculates RTT • Traceroute does this 3 times Stopping criterion • UDP segment eventually arrives at destination host • Destination returns ICMP “host unreachable” packet (type 3, code 3) • When source gets this ICMP, it stops. Quiz 2 -12 IPv6 Header (Cont) New: “flow label”, “longer addresses” Missing: fragmentation (flags, ID, offset) Priority: identify priority among datagrams in flow Flow Label: identify datagrams in same “flow.” (concept of“flow” not well defined). Next header: identify upper layer protocol for data “6to4 Translation” 4-byte IPv4 -> 16-byte IPv6 A.B.C.D -> :2002:aabb:ccdd/80 “:aa:”=“A in 2-char hex” “:bb:”=“B in 2-char hex” etc. IPv4 address can become an IPv6 sub-net with 80 bits for “host” addresses (1e24 hosts) http://en.wikipedia.org/wiki/6to4 Quiz 2 -13 Example – 6to4 convert 130.207.17.25 to IPv6 address. Convert the decimal byte-representations to hex: 130 = 0x82 207 = 0xCF 17 = 0x11 25 = 0x19 IPv6 addresses are written with colons separating every 16 bits (4 hex characters). :0000: can be written :: The first 16 bits 0x2002 are a reserved /16 block of addresses reserved for IPv4 translations. :2002::::::::/16 “6to4 Translation” Next add the IPv4 32 bits: 4-byte IPv4 -> 16-byte IPv6 :2002:82CF:1119::::::/48 A.B.C.D -> :2002:aabb:ccdd/32 This is not just a single IPv6 “:aa:”=“A in 2-char hex” “:bb:”=“B in 2-char hex” address, but a block of 2^80 etc. possible host addresses that IPv4 address can become an IPv6 can replace private subnet sub-net with 80 bits for “host” addresses, like 192.168.0.0/16's. addresses (1e24 hosts) http://en.wikipedia.org/wiki/6to4 Quiz 2 -14 Chapter 4 - IP Routing • 4.5-6 Routing – Distance Vector (RIP) – Link state (OSPF) – Hierarchical routing (BGP) • 4.7 Broadcast and multicast routing Some material copyright 1996-2006 J.F Kurose and K.W. Ross, All Rights Reserved Computer Networking: A Top Down Approach Featuring the Internet, 5th edition. Jim Kurose, Keith Ross Addison-Wesley 4/3/15 Quiz 2 -15 Distance Vector: link cost changes Link cost changes: good news travels fast bad news travels slow “count to infinity” problem! 44 iterations before algorithm stabilizes: see text Poisoned reverse: If Z routes through Y to get to X : Z tells Y its (Z’s) distance to X is infinite (so Y won’t route to X via Z) will this completely solve count to infinity problem? 60 4 x y 50 1 z Y advertises X in 4 hops Z sends datagrams for X to Y Z advertises "X in 5 hops". Y-X link cost goes to 60 Y thinks Z can route in 5 hops, so Y advertises "X in 6", sends datagrams back to Z. Z sends datagrams back to Y, advertises "X in 7". Y sends datagrams back to Z, advertises "X in 8". Quiz 2 -16 RIP (Distance-Vector Algorithm) B M 128.230.0.0/16 Y Router A Table Prefix Distance Port 128.230. 2 X 130.207. 6 N 209.196. 7 X 24.56. 9 X A X C Router B Table Prefix Distance Port 128.230. 2 X 130.207. 6 X 209.196. 5 M 24.56. 11 X Z Router C Table Prefix Distance Port 128.230. 2 X 130.207. 4 X 209.196. 7 X 24.56. 11 P Construct the Routing Table for Router X. Use "L" for the port to the local LAN. Router X Table Prefix Distance Port 128.230. 1 L 130.207. 5 C 209.196. 6 B 24.56. 10 A Using Poison Reverse, construct the Updates sent from Router X to A, B, and C. (infinity -> 15). Update X to A Table Prefix Distance 128.230. 1 130.207. 5 209.196. 6 24.56. 15 Update X to B Table Prefix Distance 128.230. 1 130.207. 5 209.196. 15 24.56. 10 Update X to C Table Prefix Distance 128.230. 1 130.207. 15 209.196. 6 24.56. 10 “Poison Reverse” prevents “ping-pong” routes. Quiz 2 -17 Graph abstraction 5 2 u v 2 1 x Graph: G = (N,E) 3 w 3 1 5 z 1 y 2 N = set of routers = { u, v, w, x, y, z } E = set of links ={ (u,v), (u,x), (v,x), (v,w), (x,w), (x,y), (w,y), (w,z), (y,z) } Remark: Graph abstraction is useful in other network contexts Example: P2P, where N is set of peers and E is set of TCP connections Quiz 2 -18 Graphical Method (animated - keep clicking) 5 5 3 v 2 2 w 5 4 3 u 3 2 1 2 x 8 1 3 1 5 z 4 2 y 1 Next Slide Quiz 2 -19 Hierarchical OSPF Only this router seen outside Area 1, where it represents all the subnets in Area 1 Quiz 2 -20 Hierarchical Routing (BGP) Our routing study thus far - idealization all routers identical network “flat” … not true in practice scale: with 200 million destinations: • can’t store all dest’s in routing tables! • routing table exchange would swamp links! administrative autonomy • internet = network of networks (Autonomous Systems) • each network admin may want to control routing in its own network Quiz 2 -21 Interconnected ASes 3c 3a 3b AS3 2a 1c 1a 1d 1b AS1 Intra-AS Routing algorithm Inter-AS Routing algorithm Forwarding table 2c AS2 2b • Forwarding table is configured by both intra- and inter-AS routing algorithm – Intra-AS sets entries for internal dests – Inter-AS & Intra-As sets entries for external dests Quiz 2 -22 Broadcast Routing • Deliver packets from source to all other nodes • Source duplication is inefficient: duplicate duplicate creation/transmission R1 R1 duplicate R2 R2 R3 R4 source duplication R3 R4 in-network duplication Source duplication: how does source determine recipient addresses? Quiz 2 -23 In-network duplication • Flooding: when node receives brdcst pckt, sends copy to all neighbors – Problems: cycles & broadcast storm • Controlled flooding: node only brdcsts pkt if it hasn’t brdcst same packet before – Node keeps track of pckt ids already brdcsted – Or reverse path forwarding (RPF): only forward pckt if it arrived on shortest path between node and source • Spanning tree – No redundant packets received by any node Quiz 2 -24 Chap. 6 - Wireless LANs What is: Wireless Access Point (AP) IEEE 802.11a, b, g - WiFi IEEE 802.15 WiMax CDMA/CA (Collision Avoidance) "Hidden Terminal" problem How helped by "CA": RTS, CTS packets Modulation Techniques TDMA - Time slots CDMA - Chip codes MAC Packet Header 3rd Address needed on radio side - why Same as Ethernet on wired side (no AP MAC address) Chap. 6,7,8 Review Quiz 2 -25 802.11 frame: addressing R1 router H1 Internet AP R1 MAC addr H1 MAC addr dest. address AP MAC addr H1 MAC addr R1 MAC addr address 1 address 2 source address 802.3 frame (Ethernet) address 3 802.11 frame Chap. 6,7,8 Review Quiz 2 -26 Cellular Networks Modulation Techniques Combined FDMA & TDMA - Freq. Div &Time slots CDMA - Chip codes Hexagonal Cells Why Hex, where is antenna located (serves 3 cells) Parade of Technolgies AMPS (Advanced Mobile Phone Sys.): FM, 1 call/freq. 0.8 GHz 2G (2nd Generation) IS-136 - Combined FDMA & TDMA GSM - Combined FDMA & TDMA IS-95 - CDMA 2.5G - Enhanced data rates: GPRS, EDGE - (144 to 380 Kbps) 3G - GSM with CDMA Chap. 6,7,8 Review Quiz 2 -27 Mobility • Handoff – used first in Cellular Telephone Systems – Connection stretches along path through previous cells Chap. 6,7,8 Review Quiz 2 -28 Chap. 7 - Multimedia What is: QoS - Quality of Service. Guarantees: Packet delay less than a Maximum Bandwidth greater than a Minimum Defined by SLA, Service Level Agreement, w ISP Policing Token Bucket algorithm Streaming Media Out-of-Band Control Channel (RTSP) Client Buffering - smoothes out delay variations Interactive Media requires short round-trip time limits use of large buffers less delay variation tolerance Chap. 6,7,8 Review Quiz 2 -29 Bandwidth Needed for Media • Streaming Media - Not compressed • Bandwidth = (Sampling Rate)*log2(Levels) • Telephone - (PCM) – 8000 samples/s, 256 levels (8 bits)=64 kb/s – Compressed voice = 5kb/s to 12kb/s • CD Music, Stereo (2 channels) – 44,100 samples/s, 65k levels (16 bits)= 1.5 Mb/s – Compressed (MP3) 96 to 160 kb/s • Video – Compressed -(MPEG4) < 1 Mbps) Chap. 6,7,8 Review Quiz 2 -30 Streaming Multimedia: UDP or TCP? UDP • server sends at rate appropriate for client (oblivious to network congestion !) – often send rate = encoding rate = constant rate – then, fill rate = constant rate - packet loss • short playout delay (2-5 seconds) to compensate for network delay jitter • error recover: time permitting TCP • • • • send at maximum possible rate under TCP fill rate fluctuates due to TCP congestion control larger playout delay: smooth TCP delivery rate HTTP/TCP passes more easily through firewalls Chap. 6,7,8 Review Quiz 2 -31 Video - Recovery from Packet Loss • TCP - too slow • UDP - Requires (one or all) – FEC (Forward Error Correction) – Interleaving • lone dropouts -> many short dropouts – Application-Level Recovery • Freeze picture • Extrapolate across gap Chap. 6,7,8 Review Quiz 2 -32 VoIP Setting up a Connection • SIP - Session Initiation Protocol • Finds "callee" by IP, email address, or telephone no. • Call Management – add new media streams – conference calls – transfer and hold Chap. 6,7,8 Review Quiz 2 -33 Compare SIP with H.323 • H.323 is another signaling protocol for real-time, interactive • H.323 is a complete, vertically integrated suite of protocols for multimedia conferencing: signaling, registration, admission control, transport and codecs. • SIP is a single component. Works with RTP, but does not mandate it. Can be combined with other protocols and services. • H.323 comes from the ITU (telephony). • SIP comes from IETF: Borrows much of its concepts from HTTP. SIP has a Web flavor, whereas H.323 has a telephony flavor. • SIP uses the KISS principle: Keep it simple stupid. Chap. 6,7,8 Review Quiz 2 -34 Content distribution networks (CDNs) Content replication • Challenging to stream large files (e.g., video) from single origin server in real time origin server in North America • Solution: replicate content at hundreds of servers CDN distribution node throughout Internet – content downloaded to CDN servers ahead of time – placing content “close” to user avoids impairments (loss, delay) of sending content over long paths CDN server – CDN server typically in CDN server in S. America CDN server in Asia edge/access network in Europe Chap. 6,7,8 Review Quiz 2 -35 Scheduling And Policing Mechanisms • scheduling: choose next packet to send on link • FIFO (first in first out) scheduling: send in order of arrival to queue – real-world example? – discard policy: if packet arrives to full queue: who to discard? • Tail drop: drop arriving packet • priority: drop/remove on priority basis • random: drop/remove randomly Chap. 6,7,8 Review Quiz 2 -36 Policing Mechanisms Token Bucket: limit input to specified Burst Size and Average Rate. • bucket can hold b tokens • tokens generated at rate r token/sec unless bucket full • over interval of length t: number of packets admitted less than or equal to (r t + b). Chap. 6,7,8 Review Quiz 2 -37 IETF Differentiated Services • Edge Router – Marks packets as "in-profile" or "out-profile" – Marks packets as to "Class of Service" • Core Routers – "Per Class" traffic management – Assured Forwarding of "in-profile" packets RSVP - Resource Reservation Protocol • Reserves resources (bandwidth, delay limits, …) along an end-to-end path. Chap. 6,7,8 Review Quiz 2 -38