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UMD ENEE 426 - Introduction Rate and Latency

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Introduction,Rate and LatencyENEE 426 | Communication Networks | Spring 2008 Lecture 1Communication Networks• Why communicate? Necessary to support some application.• Example Applications– Audio communication• Radio, Telephone– Text communication• Email, SMS (text message), IM– Video and multimedia• Television, videoconferencing, streaming video– Information access• Web browsingENEE 426 | Communication Networks | Spring 2008 Lecture 1Application Requirements• Each application needs something different from a communications network• Two major performance properties:– Latency: length of time it takes to transmit and receive a message– Rate: amount of data that can be sent per unit of time (aka bandwidth or capacity)• Bandwidth ∝ Capacity = max(Rate)ENEE 426 | Communication Networks | Spring 2008 Lecture 1Application Performance RequirementsApplicationLatency (s)Rate (kbps)Email6010IM / SMS110iTunes Video10700 (iPod)Netflix Media303000 (DVD)Telephone0.18GSM Voice0.112.2Web Browsing2~ 1000Interactivity Amount of DataDifferent communicationsnetworks offer differentlatencies and rates.InteractiveHighContentENEE 426 | Communication Networks | Spring 2008 Lecture 1Communication Networks• Networks are graphs– Edges = communication links– Vertices = computers, switches, routers, etc• Each vertex has a rate and latency• Each edge has a rate and latencyENEE 426 | Communication Networks | Spring 2008 Lecture 1Processing and Queuing Delay• Graph vertices responsible for switching traffic across different links• Certain overhead (delay) involved in receiving, processing, and retransmitting data• Links may be saturated, packets queued• Packets may be buffered, introducing latencyABCA+B>C ???ENEE 426 | Communication Networks | Spring 2008 Lecture 1Communication Link Latency• Air, fiber optics– Data moves at the speed of light (3x108m/s)• Copper wires– Data moves at 2/3 speed of light (2x108m/s)• Distance affects latency– Satellite: 35e6 / 3e8 = 120 ms– Transcon: 5e6 / 3e8 = 17 ms– Cable: 20 / 2e8 = 100 ns35,000 km5,000 kmENEE 426 | Communication Networks | Spring 2008 Lecture 1Communication Link Rate• Rate measured in bits per second– bps = bits per second– Bps = bytes per second (8 bps)– kbps = kilobits per second (1000 bps)– kBps = kilobytes per second (8000 bps)• Terminology note: distinct from storage– Comms: kb = 103bits = 1000 bits– Storage: kb = 210bits = 1024 bits• Baud = symbols per second– Symbol could contain multiple bits– Example: Use 4 different voltages to convey 2 bits at a timeENEE 426 | Communication Networks | Spring 2008 Lecture 1Communication Link Rate• Rate affected by various components– Frequency with which you send data blocks– Number of bits per data block– Rate = bits-per-symbol x baud-rate• Examples:– Copper cable with two voltages changing voltage every 1 ms = 1 Mbps– WDM optical system using 16 different colors of light, pulsing every 1 ns = 16 Gbps– 256-QAM RF link with symbol time 10 ms = 800 kbpsENEE 426 | Communication Networks | Spring 2008 Lecture 10 5 10 15 20 25 30 35 40 45 50-8-6-4-2024681012Bounds on Rate• Transmit as fast as possible• Problem: Noise• Send too much data too quickly, and you can’t decode it in the presence of noise0 5 10 15 20 25 30 35 40 45 50-15-10-5051015Too many bits per baudENEE 426 | Communication Networks | Spring 2008 Lecture 1Bounds on Rate• The faster you transmit, the more error correction necessary• Error correction causes overhead, reducing rate• Information Theory provides a fundamental bound on communication performance• C = Capacity = theoretical maximum rate• B = Signal bandwidth = ½ x baud-rate• SNR = Signal to Noise Ratio– Signal power divided by noise power SNRBC  1log2ENEE 426 | Communication Networks | Spring 2008 Lecture 1Applications and Networks• Real-time applications require real-time network access– Satellite performs poorly for phone service• High-rate applications need high-rate network access– Streaming video requires broadband Internet– Mobile phones just now getting fast enough to handle multimedia (i.e. Verizon VCast)ENEE 426 | Communication Networks | Spring 2008 Lecture 1Bandwidth-Delay Product• Measure of the number of bits that can be “contained” by a communications channelSource DestinationTransmissionTimeTransmissionTimePropagationDelayRound-TripTime (RTT)Function ofLatencyFunction ofRateArea represents the bandwidth-delayproductENEE 426 | Communication Networks | Spring 2008 Lecture 1Bandwidth-Delay Product• Bandwidth-Delay product =Latency x Rate• seconds x bits/second = bits• Number of bits contained within the channel– Transmitted but not yet received• “Pipe” interpretation of a linkBandwidthDelayENEE 426 | Communication Networks | Spring 2008 Lecture 1Circuit vs Packet Switched• Original telephone network– Wires connecting every source-destination pair– Complexity O(n2)• Switching required– Single line from everycustomer connects to anoperator– O(n)ENEE 426 | Communication Networks | Spring 2008 Lecture 1Circuit Switched Networks• At time of call establishment, operator physically plugged a jumper cable between two lines to create an electrical connection• Fundamental concept behind phone networks• For two parties (src, dst) to communicate:– Src indicates to the network a desire to establish a connection to dst at rate R– Network examines all possible routes between src and dst to determine if a path exists supporting rate R– If such a path exists, the network allocates those resources to that connection• Still fundamental basis of the modern phone networkENEE 426 | Communication Networks | Spring 2008 Lecture 1Packet Switched Networks• Break data into packets• Rather than allocate resources with granularity of circuits, allocate per-packetData StreamCircuit Switched Packet SwitchedENEE 426 | Communication Networks | Spring 2008 Lecture 1Circuit vs Packet• Circuit switched networks have fixed allocations for guaranteed service• If someone isn’t fully utilizing their connection, others can’t take advantage of the available capacity• Example:– Nodes A and B are both fairly allocated 1Mbps circuits– If node A is not transmitting, B cannot take advantage of available capacity1 Mbps link2 Mbps link, 1 Mbps


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