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Lecture1, TCP/IP
Lecture1, TCP/IP
Lecture1, TCP/IP
Lecture1, TCP/IP
Lecture1, TCP/IP
Lecture1, TCP/IP
Lecture1, TCP/IP
Lecture1, TCP/IP
Lecture1, TCP/IP
Lecture1, TCP/IP
Lecture1, TCP/IP
Lecture1, TCP/IP
Lecture1, TCP/IP
Lecture1, TCP/IP
Lecture1, TCP/IP
Lecture1, TCP/IP
Lecture1, TCP/IP
Lecture1, TCP/IP
Lecture1, TCP/IP
Lecture1, TCP/IP
Lecture1, TCP/IP
Lecture1, TCP/IP
Lecture1, TCP/IP
Lecture1, TCP/IP
Lecture1, TCP/IP
Lecture1, TCP/IP
Lecture1, TCP/IP
Lecture1, TCP/IP
Lecture1, TCP/IP
Lecture1, TCP/IP
Lecture1, TCP/IP
Lecture1, TCP/IP
Lecture1, TCP/IP
Lecture1, TCP/IP
Lecture1, TCP/IP
Lecture1, TCP/IP
Lecture1, TCP/IP
Lecture1, TCP/IP
Lecture1, TCP/IP
Lecture1, TCP/IP
Lecture1, TCP/IP
Lecture1, TCP/IP
Lecture1, TCP/IP
Lecture1, TCP/IP
Lecture1, TCP/IP
Lecture1, TCP/IP
Lecture1, TCP/IP
Lecture1, TCP/IP
Lecture1, TCP/IP
Lecture1, TCP/IP
Lecture1, TCP/IP
Lecture1, TCP/IP
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Lecture1, TCP/IP
Lecture1, TCP/IP
Lecture1, TCP/IP
Lecture1, TCP/IP
Lecture1, TCP/IP
Lecture1, TCP/IP
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Lecture1, TCP/IP
Lecture1, TCP/IP
Lecture1, TCP/IP
Lecture1, TCP/IP
Lecture1, TCP/IP
Lecture1, TCP/IP
Lecture1, TCP/IP
Lecture1, TCP/IP
Lecture1, TCP/IP
Lecture1, TCP/IP
Lecture1, TCP/IP
Lecture1, TCP/IP
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Lecture1, TCP/IP

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  • 1. Lecture 1 1 Lecture 1: TCP/IP TCP/IP Layer Structure IP TCP UDP
  • 2. Lecture 1 2 Introduction To achieve the global connectivity, ideally we have a virtual global network to which every machine can connect to. However, in the real world, we do not have such global network. Many different networks with different technologies and protocols are existed all over the world. We called they are heterogeneous networks.
  • 3. Lecture 1 3 Introduction Alternative approach We establish physical links and routers to connect them together; we apply the same higher level communication protocol for each machine such that receivers can understand the content of packets sent from senders. Solution: TCP/IP
  • 4. Lecture 1 4 TCP/IP Model TCP/IP = Transmission Control Protocol/Internet Protocol. Developed in 1970s by the US Department of Defense. Application Transport (TCP) Internet (IP) Link Physical
  • 5. Lecture 1 5 TCP/IP Model Physical (Chapter 4) Link (Chapter 5) Internet (IP = Internet Protocol) specify the format of the packets sent across the Internet and the mechanisms used to forward packets from a station through one or more routers to the destination station.
  • 6. Lecture 1 6 TCP/IP Model Transport (TCP = Transmission Control Protocol) specify an end-to-end protocol for the reliable transfer of data between two programs. Application specify how one application uses an Internet.
  • 7. Lecture 1 7 TCP/IP Model TH IH HH Data TCP Data IP Data H-to-N Data HT Source machine Application Transport (TCP) Internet (IP) Host-to-Network Destination machine Application Transport (TCP) Internet (IP) Host-to-Network
  • 8. Lecture 1 8 IP Internet Protocol Main characteristics Hierarchical addressing: IP address are 32 bits in length and are used in the source and destination address fields of the IP datagram. Connectionless routing: each data packet is an individual datagram to do the routing.
  • 9. Lecture 1 9 IP Address Dotted Decimal Notation It is difficult for human being to read 32-bit IP addresses in technical documents or through application programs. Thus IP addresses are written as four decimal integers separated by decimal points, where each integer gives the value of one octet of the IP address (1 octet = 8 bits).
  • 10. Lecture 1 10 IP Address Example 1 10000000 00001010 00000010 00011110 is written as 128.10.2.30 Example 2 149.8.12.40 is written as 10010101 00001000 00001100 00101000
  • 11. Lecture 1 11 Two components: network id. and host id. Network id.: identifies the network; Host id.: identifies the station (or host computer) on that network (as identified by the network id.). IP Address IP address Network id. Host id.
  • 12. Lecture 1 12 IP Address Trade off between the size of the network id. field and that of the host id. field. Larger network id. ⇒ larger possible number of networks (in the Internet) with each network having smaller number of hosts. Larger host id. ⇒ larger number of hosts in a network but the possible number of networks is smaller. To accommodate networks of different sizes, we classified networks into 5 categories: A, B, C, D and E.
  • 13. Lecture 1 13 IP Address 0 Network Host 10 Network Host 110 Network Host 1110 Multicast address 11110 Reserved for future use Class A B C D E 32 bits
  • 14. Lecture 1 14 IP Address Class A 126 networks (7-bit network id. − 2 special cases); All 0s (0.0.0.0): allow only at system startup and never be a valid destination address. Once a machine learns its correct IP address, all 0s will not be used. All 1s (127.xx.yy.zz): reserved for loop back test (testing the TCP/IP on a local machine, send a packet from its output port and receive it from its own input port). 16.8 million hosts (24-bit host id. − 2 special cases); All 0s (xx.0.0.0): network. All 1s (xx.255.255.255): broadcast within the network.
  • 15. Lecture 1 15 IP Address Class B 16,382 networks (14-bit network id. − 2); 65534 hosts (16-bit host id. − 2); Class C 2 million networks (21-bit network id. − 2); 254 hosts (8-bit host id. − 2); Class D 28 bits to specify a multicast group; can be used only as destination address;
  • 16. Lecture 1 16 IP Header Version IHL Type of service Total length Identification D F M F Fragment offset Time to live Protocol Header checksum Source address Destination address Options (0 or more words) 32 bits
  • 17. Lecture 1 17 IP Header Total length: the total length of the datagram (including header); Datagram = IP-PDU, packet routed in IP layer; Maximum: 65,535 bytes (1 byte = 8 bits); How to handle a message if its size is larger than the maximum packet size of a physical network? Solution: fragmentation and reassembling.
  • 18. Lecture 1 18 IP Header: Fragmentation Fragmentation of IP datagram is allowed because This feature facilitates building an Internet with components networks accommodating different maximum packet sizes. IP datagram Datagram header Datagram Data Area Frame Header Frame Data Area Physical Network frame
  • 19. Lecture 1 19 IP Header: Fragmentation When an IP datagram is encapsulated by a physical network frame (e.g. Ethernet frame), since the size of a frame is limited (e.g. for fairness), the size of an IP datagram is also limited. Thus we need to apply fragmentation for the transmitted message longer than the limited size. Packet too long may suffer long transmission delay and even cause network congestion. Thus short packet is preferred.
  • 20. Lecture 1 20 IP Header: Fragmentation When the size of a datagram is larger than the maximum one, a router breaks the datagram up into a number of small fragments. The destination host's IP layer can then reassemble the fragments back to the complete datagram before passing it up to upper layer protocol (say TCP) entity.
  • 21. Lecture 1 21 IP Header: Fragmentation Identifier: When a large datagram needs to be fragmented, all its fragments carry the same value in the identifier field. The destination host can determine which datagram the current fragment belongs to and reassemble the original datagram. DF: when it sets to 1, it tells the Internet (router) not to fragment the datagram.
  • 22. Lecture 1 22 IP Header: Fragmentation MF: when it sets to 1, it stands for 'more fragment'. All fragments of a datagram except the last one have this bit set. Fragment offset: it tells where in the containing datagram this fragment belongs. To reassemble, the destination host must obtain all fragments starting with the fragment that has offset 0 through the fragment with the highest offset.
  • 23. Lecture 1 23 IP Header Time to live: it specifies how long, in seconds, the datagram is allowed to remain in the internet system. Protocol: it tells the network access layer in the destination host, which upper protocol process to give the datagram to. Usually it will be TCP or UDP.
  • 24. Lecture 1 24 IP Header Header checksum: a checksum verifying the header only; Source address; Destination address; Options
  • 25. Lecture 1 25 IP Routing hosts messag e NetA NetC NetB NetD R(AC) H1 H2 H4 H3 R(ABD)
  • 26. Lecture 1 26 IP Routing Consider H1 would like to send a packet to H3. H1 is the end station of the network NetA, and H2, H3 and H4 are the end stations of the network NetD. H1 communicates with other stations by using the native protocol of the network NetA (say PrA, e.g. Ethernet). Similarly, H2, H3 and H4 communicate with each other with the native protocol of NetD (say PrD, e.g. Token Ring).
  • 27. Lecture 1 27 IP Routing It is possible that PrA, PrB , PrC and PrD are not the same. H1 does so by using IP protocol which H1, R(ABD) and H3 all understand and agree upon. H1 puts H3's IP address in the destination address and its own IP address in the source address. H1 also puts the destination address of R(ABD) in the destination address field of the header of PrA-PDU.
  • 28. Lecture 1 28 IP Routing When the PrA-PDU is routed by NetA to the destination R(ABD), R(ABD) will extract the IP datagram from the PrA-PDU and look at the destination address and decide that the destination is on H3. So R(ABD) sends the IP datagram to station H3, this time embedding the datagram in a PrD-PDU. When H3 receives the PrD-PDU, it will extract the IP datagram and obtain the data.
  • 29. Lecture 1 29 IP Routing ET = Ethernet Tail TT = Token Ring Tail Px(y) = Physical address of y in x H1 App. TCP IP H-to-N H3 Datagram IP Header Data S=H1 D=H3 ... IP H-to-N App. TCP IP H-to-N Router R(ABD) D=PrA(R(ABD)) … ET Ethernet Header D=PrD(H3) … TTDatagram Token Ring Header Network NetA Network NetD Datagram
  • 30. Lecture 1 30 IP Routing How does the router make a suitable decision to route the packet to H3? Solution: Routing table. A network on the Internet is usually designated by the network prefix of its IP address followed by appending 0's to the suffix. Example: the network 144.214 in the next slide is usually designated 144.214.0.0.
  • 31. Lecture 1 31 IP Routing A router is connected to more than one network. Hence it has multiple IP addresses. Example: the router below has two IP addresses: 144.214.0.15 and 144.120.12.9. router 144.214.0.0 144.120.0.0 144.214.0.15 144.120.12.9
  • 32. Lecture 1 32 IP Routing R1 R2 R3 NetA NetB NetC NetD NetE Routing table at R2 Destination Next hop NetA R1 NetB R1 NetC direct deliver NetD direct deliver NetE R3
  • 33. Lecture 1 33 IP Routing In practice, the networks are identified by its IP address. The router uses a network-id mask (also called a subnet mask) to extract the network id from a (destination station) IP address and search the routing table for a match.
  • 34. Lecture 1 34 IP Routing R1 R2 R3 20.0.0.0 40.0.0.0 128.1.0.0 192.4.10.0 144.214.0.0 40.0.0.7 128.1.0.8 144.214.0.5128.1.0.9 192.4.10.9 20.0.0.7 192.4.10.8
  • 35. Lecture 1 35 IP Routing Routing table at R2 would look like Destination Mask Next hop 20.0.0.0 255.0.0.0 128.1.0.8 40.0.0.0 255.0.0.0 128.1.0.8 128.1.0.0 255.255.0.0 direct deliver 192.4.10.0 255.255.255.0 direct deliver 144.214.0.0 255.255.0.0 192.4.10.8
  • 36. Lecture 1 36 IP Routing Example Datagram P arrive at R2 with destination address 144.214.10.18. For each entry in the routing table, the corresponding mask is “anded” with the destination address and the result (144.214.0.0) is compared to the destination (network) field. If a match is found, it will be sent to the address at the next hop field (192.4.10.8).
  • 37. Lecture 1 37 Companion IP Protocols The core IP protocol is for the sending of datagrams between stations across the Internet. There are a number of companion protocols to handle other functions. Two important protocols will be described: ICMP (Internet Control Message Protocol) and ARP (Address Resolution Protocol).
  • 38. Lecture 1 38 ICMP It is used to communicate control messages between host and router, among routers and between hosts. ICMP messages are embedded in the data field of a datagram and the protocol type is set to 1. IP header IP data field IP datagramICMP message Protocol = 1
  • 39. Lecture 1 39 ICMP Most ICMP messages are for signaling error or unusual situations. Messages between routers and hosts: 'can't reach destination' 'Time-to-live expired' 'illegal parameter' 'slow down - congestion' 'there is a better route to send data', …, etc.
  • 40. Lecture 1 40 ICMP Messages between hosts: 'can't read application' 'reassembly time expired' 'strange parameter' 'slow down - congestion' 'echo request' 'echo reply', …, etc.
  • 41. Lecture 1 41 ARP When an IP datagram arrives at a destination router, the router will send the datagram to the destination host over the destination network. Since the format of the physical network address (e.g. Ethernet address) is different from that of the IP address, usually the router has a table to map the destination IP address to its corresponding physical network address.
  • 42. Lecture 1 42 ARP The router then sends the datagram to the destination by encapsulating the datagram in the corresponding physical network address. However, if the router does not know it, how does the router send the datagram? The router may not know the mapping if the configuration of the physical network is changed, or the station is just joined the network. Solution: ARP
  • 43. Lecture 1 43 ARP R 144.214.0.0 144.120.0.0 144.214.01.5 144.120.12.9 Datagram with destination address Ap
  • 44. Lecture 1 44 ARP An IP datagram with destination address Ap (e.g. 144.120.60.8) arrives at the (destination) router R. The router wants to know the Ethernet address of the station with IP address Ap. R broadcasts a request: “Who owns IP address Ap?” on the destination LAN (e.g. 144.120.0.0).
  • 45. Lecture 1 45 ARP Only the destination with the IP address Ap will response, giving its physical network address to R (e.g. Ethernet address E). Then R updates its table and send the datagram to the destination. Note that a source station can use ARP to find the local network address of the router if necessary.
  • 46. Lecture 1 46 TCP Transmission Control Protocol Functions: To provide a point-to-point reliable connection oriented service for upper (application) layer entities. To provide for multiplexing of multiple transport connections over a single network.
  • 47. Lecture 1 47 TCP Segment TCP PDUs are called segments. Fixed size header (20 bytes); The data field can be up to 216 − 40 bytes (TCP and IP headers), i.e. 65,495 bytes.
  • 48. Lecture 1 48 TCP Segment Data (optional) Destination port Acknowledgement number Options (0 or more 32-bit words) Source port Sequence number TCP header length F I N R S T P S H A C K U R G S Y N Window size Urgent pointerChecksum 32 bits
  • 49. Lecture 1 49 TCP Segment Source port and destination port: TCP port numbers that identify the application programs at the ends of the connection. A port number plus an IP address form an unique transport service access point (TSAP). Sequence number (SEQ): identify the position in the sender’s byte stream of the data in the segment.
  • 50. Lecture 1 50 TCP Segment Acknowledgement number (ACKN): identifies the number of the octet that the source expects to receive next. TCP header length: TCP segment’s header (in units of 32-bit words);
  • 51. Lecture 1 51 TCP Segment URG When it sets to 1, the urgent pointer is in use. The pointer is used to specify the position in the segment where urgent data ends. This is used to draw attention of the receiver. ACK When it sets to 1, the field of the acknowledgement number (ACKN) is valid.
  • 52. Lecture 1 52 TCP Segment PSH When it sets to 1, it indicates to the receiver that it should deliver the data (and any already buffered) to the application program. Otherwise, the receiver may buffer (and only deliver when buffer is full) for efficiency. RST When it sets to 1, reset the connection.
  • 53. Lecture 1 53 TCP Segment SYN Used for connection set-up; SYN = 1, ACK = 0 ⇒ connection set-up request; SYN = 1, ACK = 1 ⇒ connection set-up accept; FIN Used for connection release; When it sets to 1, the sender has reached end of its byte stream.
  • 54. Lecture 1 54 TCP Services Connection set-up three-way handshake Sender Events Receiver Events Network Message s SYN(SEQ = x) SYN(SEQ = y, ACKN = x + 1) SYN(SEQ = x + 1, ACKN = y + 1)
  • 55. Lecture 1 55 TCP Services The advantage of three-way handshake is that it still works even the TCP segment containing the connection-accept segment is lost. Note that a new set of starting sequence numbers is used on connection set-up. This is to avoid any segment from a previous connection session between the same processes from confusing the current connection.
  • 56. Lecture 1 56 TCP Services Connection release FIN FIN ACK Sender Events Receiver Events Network Message s
  • 57. Lecture 1 57 TCP Services Data transfer Damaged and lost segments are handled by a positive acknowledgement time-out retransmission mechanism. Duplicated and out of order segments are detected by use of the sequence number field.
  • 58. Lecture 1 58 TCP Services Flow control It is affected by a window mechanism. The send window size can be dynamically changed by the receiver (based on its buffer condition). Window advertisement (the window size field in the TCP header) specifies how many octets (1 octet = 8 bits) of data that the receiver is prepared to accept. Example : maximum segment size of the sender is 1000 octets and maximum window advertisement is 2000 octets.
  • 59. Lecture 1 59 TCP Services advertise window = 2500 send data octets 1 - 1000 send data octets 1001 - 2000 send data octets 2001 - 2500 ack up to 1000, window = 1500 ack up to 2000, window = 500 ack up to 2500, window = 0 Sender Events Receiver Events Network Message s receive ack for 1000 receive ack for 2000 receive ack for 2500 application reads 2000 octets ack up to 2500, window = 2000 ack up to 3500, window = 1000 ack up to 4500, window = 0 application reads 1000 octets ack up to 4500, window = 1000 send data octets 2501 - 3500 send data octets 3501 - 4500 receive ack for 4500 receive ack for 3500 receive ack for 4500
  • 60. Lecture 1 60 TCP Services Congestion control: slow-start algorithm TCP (sender) maintains two windows, a send window Ws which is set by receiver’s window advertisement, and a congestion window Wc. The sender uses the smaller of the two for actual transmission. A threshold, T, is an integer such that the congestion window will increase exponentially before reaching the threshold. Usually T will be initially set to 64k bytes.
  • 61. Lecture 1 61 TCP Services Procedure: 1. Wc = 1. 2. When (i) a window is sent, (ii) there is no time-out, and (iii) Wc is smaller than the threshold, Wc = min(2 × Wc, threshold) (growth rate is exponential). 3. When (i) a window is sent, (ii) there is no time-out, and (iii) Wc is not smaller than the threshold, Wc = Wc + 1 (growth rate is linear). 4. When a time-out occurs, T = Wc / 2 and Wc = 1.
  • 62. Lecture 1 62 TCP Services Slow-Start 0 10 20 30 40 50 60 70 80 0 5 10 15 20 Transmission numbers Congestionwindow(kbytes) Timeout Threshold New Threshold
  • 63. Lecture 1 63 TCP Services Trans. No. Wc (kbytes) Trans. No. Wc (kbytes) 0 1 13 1 1 2 14 2 2 4 15 4 3 8 16 8 4 16 17 16 5 32 18 32 6 64 19 35 7 65 20 36 8 66 21 37 9 67 22 38 10 68 23 39 11 69 24 40 12 (Timeout) 70
  • 64. Lecture 1 64 TCP Multiplexing A host use an unique IP address to communicate through the Internet. Within that machine, there may be multiple application programs requiring remote communication services. The TCP layer implements multiple transport connections over a single network interface.
  • 65. Lecture 1 65 TCP Multiplexing Host A Host B Internet TCP IP X Y TCP IP M N 144.214.12.38 205.10.11.09 144.214.12.38:23144.214.12.38:290 205.10.11.09:2529 205.10.11.09:1326
  • 66. Lecture 1 66 UDP User Data Protocol Connectionless transport protocol suitable for applications requiring short communication exchanges; packet is up to 64 kbytes. 32 bits Destination portSource port ChecksumDatagram length User data
  • 67. Lecture 1 67 Tutorial 1 1. When an IP datagram is to be routed through a network whose maximum packet size is smaller than that of the datagram, it is fragmented into smaller datagrams. Where do you think is better to reassemble the datagram? At the next router or at the destination host (IP layer)? Explain.
  • 68. Lecture 1 68 Tutorial 1 2. Most IP datagram reassembly algorithms have a timer to avoid having a lost fragment tie up reassembly buffers forever. Suppose a datagram is fragmented into 4 fragments. The first 3 fragments arrive, but the last one is delayed. Eventually the timer goes off and the three fragments in the receiver’s memory are discarded. A little later, the last fragment stumbles in. What should be done with it?
  • 69. Lecture 1 69 Tutorial 1 3. How many responses a router expects to get when it broadcasts an ARP request? Why? 4. You have just explained the ARP protocol to a friend. When you are all done, he says: “I have got it. ARP provides a service to the network layer, so it is part of the data link layer.” What do you say to him?
  • 70. Lecture 1 70 Tutorial 1 5. Write out the following IP address in dotted decimal format: 10010000 11001000 00100101 01000001 6. Is the IP address space efficiently utilized? Explain. Suppose that instead of using 16 bits for the network part of a class B address, 20 bits has been used. How many class B networks would there have been?
  • 71. Lecture 1 71 Tutorial 1 7. What is the size of the port number space for TCP? What is the maximum size of a TCP segment? Under what condition is this maximum size achievable? 8. Consider a TCP connection over the Internet. When a time-out occurs on the sending of a segment, which is by far most likely the cause: (i) congestion, (ii) error: damaged or lost IP datagram (which encapsulates the TCP segment). Explain.
  • 72. Lecture 1 72 Tutorial 1 9. Consider the slow start flow control algorithm used in TCP. Suppose the maximum segment size is 1 Kbytes. Suppose the congestion window Wc just before a time- out was 32 Kbytes. What are the congestion window sizes for the first 8 transmissions after the time-out? Assume that there are no time-outs during these 8 transmissions.

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