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Understanding Network Performance: Circuit vs Packet Switching

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20 diapositives · 374 Views · Sep 11, 2014

This lecture covers key concepts in network performance, focusing on the differences between circuit switching and packet switching. It explains how dedicated links function in circuit-switched networks, offering guaranteed performance through call setups, while packet-switched networks transmit data in small packets, allowing for more flexible bandwidth utilization. The discussion includes statistical multiplexing, queuing delays, and examples of bandwidth sharing. Also covered are practical scenarios showing throughput calculations and delays in transmission, highlighting their impact on network performance.

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  1. CS 457 - Lecture 2Network Performance Fall 2011

  2. Circuit switching dedicate link bandwidth & switch capacity to each “call” Requires call setup Guaranteed performance Packet switching Packet: small chunks of data Send packets as soon as link available Switch receives a full packet then forwards it towards the destination How Is the Link Shared?Circuit Switched or Packet Switched

  3. FDM • frequency • time • TDM • frequency • time Circuit Switching: FDM and TDM • 2 users • Example:

  4. Packet Switching: Statistical Multiplexing • Store-and-forward • Packet switch can temporarily buffer up packets • Introduces queueing delay • Packets get dropped when the queue is full

  5. Given 1 megabits/sec (1 Mbps) link Assume each user: User send 100,000 bits/sec when “active” User is active 10% of time 100% of capacity used if 10 active users. Circuit-switching: Link can support 10 users Packet Switching: Link can support 35 users Prob.(n > 10)  0.0004 How Many Users Can Share? • 1 Mbps link

  6. Given 1 megabits/sec (1 Mbps) link Divided into 10 distinct slices (TDM or FDM) No Interaction between users Potential bandwidth for my connection is 1 Mbps/10 = 100,000 bps How long to send 4 Mb file? 4 Mb / 100,000 bps = 40 sec Plus some delay to setup connection Assume 1 seconds Total of 41 seconds Throughput = 97,560 bps How long to send 100,000b file? 100,000b / 100,000bps = 1 sec Total of 2 seconds Throughput = 50,000 bps Circuit Switching Performance • 1 Mbps link

  7. Packet Switching Performance = L / R • Transmission • R = link bandwidth (bps) • L = packet length (bits) • Propagation • d = length of physical link • s = propagation speed in medium (~2x108 m/sec) Queueing = #packets in queue X transmission time of each packet = d/s • transmission • A • propagation • C • B • queueing

  8. Delay on A Single Link • Relevant Specifications • Bandwidth: R = 1 Mbps • Packet Size: L = 1000 bits • Link length: d = 100 km • Propagation Speed: s = 2.0 x 10^8 m/sec (typical fiber) • Assumptions • qlength = 2 Packets in queue when our packet arrives • Total Delay = transmit + prop + queue = (L/R) + (d/s) + (qlength * L/R) = (1000/1000000) + (100000/2 x 10^8) + (2*1000/1000000) = 3.5 ms

  9. Some Comments about Units • Bits or Bytes • Bits denoted by “b” • Bytes denoted by “B” • Mb = megabits while MB = megabytes • Kb = kilobits while KB = kilobytes • How big is K and M? It depends…. • Mega = 2^20, Kilo = 2^10 • Mega = 10^6, Kilo = 10^3 • Bandwidth uses powers of 10 • Tied to MHz which is 10^6 hertz • So bandwidth of 1 Mbps = 10^ bits per second • Messages use powers of 2 • Tied to computer memory measures in powers of 2 • So packet/file/message of 1 Mb = 2^20 bits

  10. Example Problem • Find total time to transfer a file assuming • File size is 1.5 MB, RTT is 80 ms, Packet size is 1KB • Initial 2*RTT “handshake” before sending data • Bandwidth is 10 Mbps and packets sent continuously • Total Time to get all bytes to receiver is:Handshake + transmit + prop2*RTT + FileSize/R + ½ RTT2 * 80 ms + 1.5 MB/ 10 Mbps + ½ 80 ms160 ms + 12,582.912 b/ 10,000,000 b/s + 40 ms approx 1.458 seconds

  11. Network latency • 2 • 1 • 3 • A: my laptop • B: www.cnn.com • Time to send a packet from point A to point B • sum of delays across each hop along the path • RTT: round-trip-time

  12. Bandwidth and Latency Bandwidth matters L large => L/R dominates Increasing R reduces latency Prop delay matters L small => L/R also small Increasing R has no impact

  13. Examples

  14. Store and Forward: Entire packet must arrive at router before it can be transmitted on next link: Let dqueue = dprop = 0 Takes dtrans = L/R seconds to transmit (push out) packet of L bits on to link of R bps Example: L = 8000 bits (1000bytes) R = 2 Mbps Delay(A-B) = 3L/R = 12 msec Store and Forward (1/2) • L • R • R • R

  15. Application Performance • Application input to network • Traffic data rate • Traffic pattern (bursty or constant bit rate) • Traffic target (multipoint or single destination, mobile or fixed) • Network service delivered to application • Delay sensitivity • Loss sensitivity

  16. Email, Reliable File Transfer • Loss sensitive • Not delay sensitive relative to round trip times • Point-to-point • Bursty

  17. Remote Login • Loss sensitive • Delay sensitive • Subject to interactive constraints • Can tolerate up to several hundreds of milliseconds • Bursty • Point to point

  18. Network Audio • Relatively low bandwidth • Digitized samples, packetized • Delay variance sensitive • Loss tolerant • Possibly multipoint, long duration sessions • Natural limit to number of simultaneous senders

  19. Network Video • High bandwidth • Compressed video, bursty • Loss tolerance function of compression • Delay tolerance a function of interactivity • Possibly multipoint • Larger number of simultaneous sources

  20. Web • Transactional traffic • Short requests, possibly large responses • Loss (bug?) tolerant • Delay sensitive • Human interactivity • Point-to-point (multipoint is asynchronous)