<< Back to Networking Main Page      W e e k 8  N e t w o r k  S e r v i c e s C h a p t e r 13 & 14 Connection Oriented Connection V.S. Connectionless ex. telephone ex. postal system Establish a switched connection until done Address the message then send ATM IP datagram TCP IP A network provides either connected or connection-less. Connected requires two computers to establish a connection before sending. Connection-less do not need to establish a connection to send and receive frames (although addressed). Examples Technology Connection Oriented Connection-less used for LAN used for WAN Ethernet X X Token Ring X X FDDI X X Frame Relay X X SMDS X X ATM X X X LocalTalk X X   Addresses and Connection Identifiers (ATM Example) 12 bits of header (Virtual Path Identifier) is used for routing Virtual Path Identifier in in-port Virtual Path Identifier out out-port A 13 1 7 4 B 7 1 10 2 C 10 2 18 3 D 18 1 - 3 Whatever itself has identified with (integer)   Whatever destination has identified port with   Using connection identifiers reduces overhead because only 12 bits is used instead of a regular ATM address can be 160 bits. Delay and Throughput Throughput Throughput is the measure of the rate at which data can be sent through the network. It is measured (usually) in bps. Most have Mbps. Some even Gbps. Throughput is not bandwidth, as bandwidth is the hardware ability to send data and throughput is the networks ability to send data. Network delay which is measured in seconds, specifies how long a single bit remains in transit (in a network). Delay and Throughput, the Relationship As traffic increases so does delay. Let D = delay when network is idle Let U be a value that is between 1 and 0 and denoted utilization (of said network). Then D = D / (1-U) Delay-Throughput Product Delay-throughput product measures the volume of data that can be present on a network. Let T be throughput Let D be the delay Then the bits in transit at any time = T*D Protocols and Layers Protocol for LAYERS {stack(ISO Model)} Application (LAYER 7) Layer 7 : Application Application that transfers the files (FTP) Specified the details of how an application on one machine makes a request and how the application might respond on another Application (LAYER 7) <== identical message ==> Presentation (LAYER 6) Layer 6 : Presentation How to represent data. Needed to interpret one representation of one computer to the representation on a deferent computer Presentation (LAYER 6) <== identical message ==> Session (LAYER 5) Layer 5 : Session How to establish a communication session with a remote system Specification for security details such as authentication Session (LAYER 5) <== identical message ==> Transport (LAYER 4) Layer 4 : Transport How to handle details of reliable transfer Very complex layer This section discusses this layer Transport (LAYER 4) <== identical message ==> Network (LAYER 3) Layer 3 : Network How addresses are assigned How packets are forwarded from one end of a network to another Network (LAYER 3) <== identical message ==> Data Link (LAYER 2) Layer 2 : Data Link How to organize data into frames How to transmit frames over a network examples include byte stuffing and checksum Data Link (LAYER 2) <== identical message ==> Physical (LAYER 1) Layer 1 : Physical specifies the basic network hardware. Specification for RS-232 belongs in Layer 1 Physical (LAYER 1)   But......TCP IP protocal sweet is most comminly used   Applications or or   TCP (reliability, sequencing) sliding window UDP (like IP)     IP (Getting packets to and fro with addressing)   Network interface level   Sliding Window and Other Techniques Computers do not operate at the same speed. Data overrun can occur when a destination computer cannot absorb the data being sent to it because of the speed of the network. Techniques to handle Data Overrun Stop and go - A sender waits after sending each packet for an acknowledgment packet to come back from the recipient. Stop and go is externally inefficient use of network bandwidth. Sliding window - Sender and receiver agree on a fixed window size (maximum amount of data that can be sent before an acknowledgment arrives). EXAMPLE: Sender and receiver agree on a window size of 4 packets. Sender begins sending data (copies) by filling the first window Receiver must have buffer space to receive entire window When packets arrive in exact sequence it then gives them to the application Receiver then transits an acknowledgment When sender gets acknowledgment, it discards the copy and starts over Comparison Stop and go V.S. Sliding Window Techniques Attach a sequence # to each packet Time out tecqunique...if ack dosent return resend...ornacks if receiver missed a packet Have the receiver designate a sending speed according to the network congestion Two Sliding Windows go back: receiver has no buffer sends next one right away but only as many as the size of the window else stop application when no ack it sends the entire window (go back n) if time out (round trip time) - resend and clear acks and reset next packet to send if next packet to send is in window and transmitter is free - send and start timmer Selective Repeat Reciever will need a buffer as above but do not clear acks and do not reset packets to send packet numbering size = 2W-1 Class Examples: An issue is that I.D. Field numbers get too big! Example 1 optimal link 50 km or 50000 meters bandwidth 2.4 Gbs packet size of 424 b (both ways) speed of light in fiber 2E8 m/sec. propigation time = 50000/200000000 = 5/2000b = 2.5E-4 sec. transition time = 424b/2.4E9(b/sec) = 1.8E-7 sec. W = [2(propigation time) + 2(transition time)]/1.8E-7 = (5E-4 + 3.6E-7 / 1.8E-7) = 2.77E3 or 2778 Example 2 wireless link 20 km bandwidth 13 kbs packet size of 1000 b and an ack of 100 bits speed of light in air 3E8 m/sec. propigation time = 2E4/3.1E4(m/s) = .6E-4 sec. transition time (packet) = 10^3b/1.3E4 (b/sec) = .077 sec. transition time (ack) = .0077 sec. W = [2(propigation time) + (transition time for packet)(transition time for ack)]/transition time for packet = (.00012 + .077 + .0077) / .077 = .08482/.077 or 2