1. 02 – Performance Basics 1CS 332 - Computer Networks

2. Physical Layer Characteristics latency – the time it takes for a signal to propagate from one end of a channel to the other. Determined by distance and speed of signal propagation in the medium Also called propagation delay or just delay Round Trip Time – 2 x latency (RTT) Latency includes buffering time at intermediate nodes in switched networks; may include message transmission time. 2CS 332 - Computer Networks

3. Bandwidth Technically, the width of a frequency band, measured in Hertz (cycles per second) In networking, we typically mean how many bits per second we can transmit Conversely, how long it takes to transmit a bit: –On 10Mbps channel, 1 bit takes 0.1 μ sec (1 x 10 -7 secs) Throughput – usually the measured performance of a channel. Usually less than the theoretical bandwidth. How do we make a channel faster? 3CS 332 - Computer Networks

4. Delay x bandwidth product For a 10 Mbps channel with latency of 2 msec: d x b = (10 x 10 6 bits / sec) x (2 x 10 -3 sec) = 20,000 bits This is the amount of data that is in transit in the channel at any given time Important in end-to-end protocol design. Sender has transmitted this many bits before receiver accepts the first one sent. 4CS 332 - Computer Networks

5. Latency and Distance How long is a 2 msec latency channel? Speed of light in copper = 2.3 x 10 8 m/sec 2 msec x 2.3 x 10 8 m/sec = 4.6 x 10 5 meters = 460 kilometers = 286 miles 5CS 332 - Computer Networks

6. Maximum Bandwidth Nyquist: In a noise-free channel of bandwidth H, where signals consist of V discrete levels, max. data rate = 2H log 2 V bits/sec So, in a system using binary signaling, we can transmit twice as many bits as the width of the band. We can increase the data rate by increasing V, the number of discrete levels 6CS 332 - Computer Networks

7. Maximum Bandwidth Shannon: extended to noisy channels Signal to noise ratio S/N, expressed in dB dB = 10 x log 10 (S/N) max. rate = H log 2 (1 + S/N) Example: 56 Khz channel, S/N = 10,000 (40 dB) max. rate = 730 Kbps NOTE: regardless of value of V! 7CS 332 - Computer Networks

8. Calculating transfer time Example: How long to transmit 200 KB file, if: 3 Mbps channel (3 x 10 6 bits per sec) RTT = 50 msec (→ prop. delay = 25 msec) Handshake = 3 RTTs Packet size = 4 KB = 4 x 2 10 bytes x 8 bits/byte = 32768 bits We must insert 1 RTT delay between packets 8CS 332 - Computer Networks

9. Calculating transfer time Time to send 1 packet = (32768 bits / packet) / (3 x 10 6 bits/sec) = 1.092267 x 10 -2 sec/packet (10.92267 ms) Total transfer time = 150 ms (handshake) + 25 ms (propagation time for 1st bit) + (10.92267 ms x 50 packets) + 50 ms x 49 (RTT delays between packets) = 3171.1335 ms 9CS 332 - Computer Networks

10. Multiplexing We want to support n simultaneous "conversations" at y bps between point A and point B. Should we: –String n wires supporting y bps between A and B –String one wire supporting n x y bps between A and B and share it multiplex – combine multiple conversations for transmission on a single channel (mux) demultiplex – split received signal into original conversations (demux) 10CS 332 - Computer Networks

11. Multiplexing styles Choose some characteristic of the transmission process that we can subdivide reliably. –Frequency bands (radio, old phone system) –Time –Wavelength (optical) 11CS 332 - Computer Networks

12. Frequency Division Multiplexing Each conversation is modulated at a different frequency Signals from different conversations are combined for transmission Multiplexing images by Ken Williams, NCA&T Univ. 12CS 332 - Computer Networks

13. Frequency Division Multiplexing Bandpass filters separate the original signals, which are then demodulated to recover the data pattern 13CS 332 - Computer Networks

14. Frequency Division Multiplexing For voice transmission, 3000hz channels are used, frequency shifted into 4000hz frequency bands to leave 1000hz guard bands. The radio spectrum is divided this way for various different applications (e.g. radio, TV, cellular, low-power apps, etc.) 14CS 332 - Computer Networks

15. Time Division Multiplexing Transmission time divided into "frames" with some number of slots. Each conversation is assigned to a particular slot 15CS 332 - Computer Networks

16. Time Division Multiplexing If there is no data for a given conversation, that slot in the frame is empty 16CS 332 - Computer Networks

17. Time Division Multiplexing Synchronous: slots are dedicated to conversations whether there is data to send or not. Fine for sources that continuously send Asynchronous: slots are used by whatever conversation has data. Must send an index to indicate which conversation each slot belongs to. Better for bursty data sources. During a "slot" (or time slice), sender uses the entire bandwidth of the channel. 17CS 332 - Computer Networks

18. TDM – Voice network T1 digital signalling standard transmits 24 conversations simultaneously Each channel has 7 bits of sample data plus 1 bit of control information per 193 bit frame (the extra bit is for frame synchronization). To avoid detectable gaps in voice conversations, each channel must transmit a sample 8000 times per second. This requires a data capacity for a T1 link of 1.544 Mbps. 18CS 332 - Computer Networks

19. Multiplexing In FDM and STDM, we handle only a fixed number of "conversations." If we have more, some must wait (busy signal) Statistical Multiplexing is like ATDM, except there are no slots. Each "conversation" is allowed to send at most one packet at a time. If there is only one active conversation, it can send continuously 19CS 332 - Computer Networks