Data Communication and Networking - Chapter 4: Communication Networks - Vo Que Son

Nội dung text: Data Communication and Networking - Chapter 4: Communication Networks - Vo Que Son

1. Data Communication and Networking Dr. –Ing. Vo Que Son Email: sonvq@hcmut.edu.vn Telecomm. Dept. DCN 1 Faculty of EEE HCMUT
2. Local Area Network  A local area network (LAN) is a computer network that interconnects computers in a limited area such as a home, school, computer laboratory, or office building using network media  Medium:  Coaxial cable  Twisted-pair line Telecomm. Dept. DCN 3 Faculty of EEE HCMUT
3. Physical Topology Bus Extended Star Ring Hierarchical Star Mesh Telecomm. Dept. DCN 5 Faculty of EEE HCMUT
4. Project 802  Modules in Project 802  LLC: based on HDLC protocol  Multiplexing protocols transmitted over the MAC layer (when transmitting) and decoding them (when receiving).  Providing node-to-node flow and error control  MAC: provides addressing and channel access control mechanisms that make it possible for several terminals or network nodes to communicate within a multiple access network that incorporates a shared medium Telecomm. Dept. DCN 7 Faculty of EEE HCMUT
5. Ethernet Topology: bus, star, ring Media Access Control: Deterministic, Non-deterministic Addressing: Every computer has a unique way of identifying itself : MAC address or physical address. The physical address is located on the Network Interface Card (NIC). MAC addresses have no structure, and are considered flat address spaces. MAC addresses are sometimes referred to as burned-in addresses (BIAs) because they are burned into read-only memory (ROM) and are copied into random-access memory (RAM) when the NIC initializes. • 0000.0c12.3456 or 00-00-0c-12-34-56 • If MAC is all bits 1: broadcast address Telecomm. Dept. DCN 9 Faculty of EEE HCMUT
6. Ethernet: 802.3  10Base-2: 50Ω Thin cable, 185m.  10Base-5: 50Ω Thick cable, 500m.  10Base-T: 100Ω UTP cable, 100m.  10Base-F: Fiber optic cable, 1000m.  100Base-TX: 100Ω UTP/STP cable, 100m.  100Base-T4: 100Ω UTP (4p) cable, 100m.  100Base-FX: Fiber optic cable, 400m.  1000Base-T: 100Ω UTP/STP cable, 100m. Telecomm. Dept. DCN 11 Faculty of EEE HCMUT
7. Ethernet: 10BASE-2  Thin Ethernet  50Ω Thin cable, 185m. Telecomm. Dept. DCN 13 Faculty of EEE HCMUT
8. Ethernet Operation Specified by the technology being used. Determine who can transmit and when. Two types: Deterministic: “Let’s take turns”. • Token-Ring, FDDI. Non-deterministic: “First come, first serve”. • Ethernet : CSMA/CD. Carrier Sense Multiple Access with Collision Detection (CSMA/CD). Telecomm. Dept. DCN 15 Faculty of EEE HCMUT
9. Ethernet MAC Ethernet is a shared-media broadcast technology. The access method CSMA/CD used in Ethernet performs three functions: Transmitting and receiving data packets Decoding data packets and checking them for valid addresses before passing them to the upper layers of the OSI model Detecting errors within data packets or on the network Telecomm. Dept. DCN 17 Faculty of EEE HCMUT
10. Fast Ethernet  Speed: 100 Mbps  Distance: <250m  Backward compatible with Ethernet 10 Mbps  Uses Twisted-pair or Fiber Telecomm. Dept. DCN 19 Faculty of EEE HCMUT
11. Token Ring  Token ring LAN technology was a protocol which resided at the data link layer (DLL) of the OSI model. It used a special three-byte frame called a token that travels around the ring. Token- possession grants the possessor permission to transmit on the medium. Token ring frames travel completely around the loop.  Initially used only in IBM computers, it was eventually standardized with protocol IEEE 802.5.  Physical ring topology Telecomm. Dept. DCN 21 Faculty of EEE HCMUT
12. Token Bus  Also use token as Token Ring  Physical Bus topology  Token bus was standardized by IEEE standard 802.4. It is mainly used for industrial applications  Due to difficulties handling device failures and adding new stations to a network, token bus gained a reputation for being unreliable and difficult to upgrade Telecomm. Dept. DCN 23 Faculty of EEE HCMUT
13. IP Header IP protocol version 32 bits total datagram number header length head. type of length (bytes) ver length (bytes) len service for “type” of data fragment 16-bit identifier Flags fragmentation/ offset reassembly max number time to upper header remaining hops live layer checksum (decremented at 32 bit source IP address each router) 32 bit destination IP address upper layer protocol to deliver payload to Options (if any) E.g. timestamp, record route how much overhead with data taken, specify TCP? (variable length, list of routers 20 bytes of TCP typically a TCP to visit. 20 bytes of IP or UDP segment) = 40 bytes + app layer overhead Telecomm. Dept. DCN 25 Faculty of EEE HCMUT
14. Classful IP addressing  Network Bits:  Provided by Internet Network Information Center  Define the network the remote device wants to connect to  Defined by subnet- mask  Host Bits:  Managed by Network Administrator Class C Network address: 192.168.10.100/255.255.255.0 (or /24)  Define the device IP Addr 11000000. 101010000. 00001010. 01100100 address in the network that the N . N . N . H remote device wants AND operation to connect to Default 11111111. 11111111. 11111111. 00000000 subnet  All bits are 0: no use Mask  All bits are 1: Network 11000000. 101010000. 00001010. 00000000 broadcast address addr Telecomm. Dept. DCN 27 Faculty of EEE HCMUT
15. Classful IP addressing: Example Find the class of each address. a. 00000001 00001011 00001011 11101111 b. 11000001 10000011 00011011 11111111 c. 14.23.120.8 d. 252.5.15.111 Solution a. The first bit is 0. This is a class A address. b. The first 2 bits are 1; the third bit is 0. This is a class C address. c. The first byte is 14; the class is A. d. The first byte is 252; the class is E. Telecomm. Dept. DCN 29 Faculty of EEE HCMUT
16. Subnetting  Non-subnetted network  Subnetted network into separate smaller networks Telecomm. Dept. DCN 31 Faculty of EEE HCMUT
17. Subnetting Example  Given network 172.16.0.0: We need 6 usable subnets and up to 8100 hosts on each subnet.  Define class: class B, 255.255.0.0  6 usable subnets: borrow 3 bits  Determine the subnets from 3 borrowed bits from the host portion (last 2 bytes):  0 subnet: .00000000.00000000  1st subnet: .00100000.00000000(32=25)  2nd subnet: .01000000.00000000  3rd subnet: .0110000.00000000  4th subnet: .10000000.00000000  5th subnet: .10100000.00000000  6th subnet: .11000000.00000000(6x25)  subnet: .11100000.00000000 Telecomm. Dept. DCN 33 Faculty of EEE HCMUT
18. Subnetting Addresses are loose by subnetting. Network administrator must strike a balance between the number of subnets required, the hosts per subnet that is acceptable, and the resulting waste of addresses. Telecomm. Dept. DCN 35 Faculty of EEE HCMUT
19. Supernetting Summarize a contiguous group of class C addresses using variable-length mask Example: 150.158.16.0/20 IP Address (150.158.16.0) & mask length (20) IP add = 10010110 10011110 00010000 00000000 Mask = 11111111 11111111 11110000 00000000 Contains 16 Class C blocks: From 10010110 10011110 00010000 00000000 • i.e. 150.158.16.0 Up to 10010110 10011110 00011111 00000000 • i.e. 150.158.31.0 Telecomm. Dept. DCN 37 Faculty of EEE HCMUT
20. Subnetting: Example 1 For example, consider a relatively small company with a Class C network, 201.45.222.0/24. They have six subnetworks in their network. The first four subnets (S1, S2, S3 and S4) are relatively small, containing only 10 hosts each. However, one of them (S5) is for their production floor and has 50 hosts, and the last (S6) is their development and engineering group, which has 100 hosts. Classfull IP Subneting? Telecomm. Dept. DCN 39 Faculty of EEE HCMUT
21. VLSM Example Given the Class C network of 204.15.5.0/24, subnet the network in order to create the network in the figure below, with the host requirements shown:  NetA: 14 hosts  NetB: 28 hosts  NetC: 2 hosts  NetD: 7 hosts  NetE: 28 hosts Solution? Classful subnetting? VLSM? What do we benefit when using VLSM? Telecomm. Dept. DCN 41 Faculty of EEE HCMUT
22. VLSM Example Solution: VSLM assign the largest first •netB: 204.15.5.0/27 host address range 1 to 30 •netE: 204.15.5.32/27 host address range 33 to 62 •netA: 204.15.5.64/28 host address range 65 to 78 •netD: 204.15.5.80/28 host address range 81 to 94 •netC: 204.15.5.96/30 host address range 97 to 98 NetB: 28 hosts (using a block of 32) NetE: 28 hosts (using a block of 32) NetA: 14 hosts (using a block of (16) NetD: 7 hosts (using a block of (16) NetC: 2 hosts (using a block of 4) Telecomm. Dept. DCN 43 Faculty of EEE HCMUT
23. A Working VLSM Example VLSM solution Telecomm. Dept. DCN 45 Faculty of EEE HCMUT
24. A Working VLSM Example Telecomm. Dept. DCN 47 Faculty of EEE HCMUT
25. Network Devices  Hub/Repeater  Switch/Bridge  Router/3-layer switch  Gateway Telecomm. Dept. DCN 49 Faculty of EEE HCMUT
26. Network Devices Switch A multi port bridge Support full duplex Hardware switching Dynamically builds and maintains a Content- Addressable Memory (CAM) table  Segmenting Collision Domain  Can not prevent layer 2 broadcast domain Telecomm. Dept. DCN 51 Faculty of EEE HCMUT
27. Network Devices Gateway: Operate at Application layer in OSI model Convert data between networks with different protocols Telecomm. Dept. DCN 53 Faculty of EEE HCMUT
28. Switching  allows multiple simultaneous A transmissions  hosts have dedicated, direct C’ B connection to switch  switches buffer packets 1 2 3  Ethernet protocol used on each 6 incoming link, but no collisions; 5 4 full duplex C  each link is its own collision domain B’  switching: A-to-A’ and B-to-B’ A’ simultaneously, without switch with six interfaces collisions (1,2,3,4,5,6)  not possible with dumb hub Telecomm. Dept. DCN 55 Faculty of EEE HCMUT
29. Switch: self-learning Source: A Dest: A’  switch learns which hosts A A A’ can be reached through which interfaces C’ B  when frame received, switch “learns” location of 1 2 sender: incoming LAN 6 3 segment 5 4  records sender/location pair in switch table C B’ A’ MAC addr interface TTL A 1 60 Switch table (initially empty) Telecomm. Dept. DCN 57 Faculty of EEE HCMUT
30. Loop problem in learning MAC addresses Telecomm. Dept. DCN 59 Faculty of EEE HCMUT
31. Routing Each router needs running a routing protocol to update the its routing table Delay of packets Queuing delay Processing delay Propagation delay Telecomm. Dept. DCN 61 Faculty of EEE HCMUT
32. Routing Algorithms: Link State Dijkstra’s algorithm Notation:  net topology, link costs  c(x,y): link cost from node x known to all nodes to y; = ∞ if not direct  accomplished via “link state neighbors broadcast”  D(v): current value of cost of  all nodes have same info path from source to dest. v  computes least cost paths from one node (‘source”) to  p(v): predecessor node along all other nodes path from source to v  gives forwarding table for that  N': set of nodes whose least node cost path definitively known  iterative: after k iterations, know least cost path to k dest.’s Telecomm. Dept. DCN 63 Faculty of EEE HCMUT
33. Dijkstra’s algorithm: example Resulting shortest-path tree from u: v w u z x y Resulting forwarding table in u: destination link v (u,v) x (u,x) y (u,x) w (u,x) z (u,x) Telecomm. Dept. DCN 65 Faculty of EEE HCMUT
34. Dijkstra’s algorithm: Example Given a network topology with the link cost of distance (km). Assuming v=2.108 m/s: Find the Shortest Path from node A to other nodes so that the end-to-end delay is smallest? If the packet processing time at each node is 1us, how can you modify the algorithm? Telecomm. Dept. DCN 67 Faculty of EEE HCMUT
35. Bellman-Ford example 5 Clearly, dv(z) = 5, dx(z) = 3, dw(z) = 3 v 3 w B-F equation says: 2 5 u 2 1 d (z) = min { c(u,v) + d (z), 3 z u v 1 c(u,x) + dx(z), x y 2 1 c(u,w) + dw(z) } = min {2 + 5, 1 + 3, 5 + 3} = 4 Node that achieves minimum is next hop in shortest path ➜ forwarding table Telecomm. Dept. DCN 69 Faculty of EEE HCMUT
36. Distance Vector Algorithm Basic idea: From time-to-time, each node sends its own distance vector estimate to neighbors Asynchronous When a node x receives new DV estimate from neighbor, it updates its own DV using B-F equation: Dx(y) ← minv{c(x,v) + Dv(y)} for each node y ∊ N Under minor, natural conditions, the estimate Dx(y) converge to the actual least cost dx(y) Telecomm. Dept. DCN 71 Faculty of EEE HCMUT
37. D (z) = min{c(x,y) + Dx(y) = min{c(x,y) + Dy(y), c(x,z) + Dz(y)} x = min{2+0 , 7+1} = 2 Dy(z), c(x,z) + Dz(z)} node x table = min{2+1 , 7+0} = 3 cost to cost to x y z x y z x 0 2 7 x 0 2 3 y ∞ ∞ ∞ y 2 0 1 from from z ∞ ∞ ∞ z 7 1 0 node y table cost to x y z y 2 1 x ∞ ∞ ∞ x z y 2 0 1 7 from z ∞ ∞ ∞ node z table cost to x y z x ∞ ∞ ∞ y ∞ ∞ ∞ from Network Layer time z 7 1 0 Telecomm. Dept. DCN 73 Faculty of EEE HCMUT
38. Distance Vector: link cost changes Link cost changes:  node detects local link cost change 1 y  updates routing info, recalculates 4 1 distance vector x z 50  if DV changes, notify neighbors At time t0, y detects the link-cost change, updates its DV, “good and informs its neighbors. news At time t1, z receives the update from y and updates its table. It computes a new least cost to x and sends its neighbors its DV. travels At time t2, y receives z’s update and updates its distance table. fast” y’s least costs do not change and hence y does not send any message to z. Telecomm. Dept. DCN 75 Faculty of EEE HCMUT