02 – Performance Basics 1CS 332 - Computer Networks.

02 – Performance Basics

1CS 332 - Computer Networks

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.


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 μsec (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

Delay x bandwidth product

• For a 10 Mbps channel with latency of 2 msec:d x b = (10 x 106 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.


Latency and Distance

• How long is a 2 msec latency channel?

• Speed of light in copper = 2.3 x 108 m/sec

2 msec x 2.3 x 108 m/sec =

4.6 x 105 meters =

460 kilometers = 286 miles


Maximum Bandwidth

• Nyquist: In a noise-free channel of bandwidth H, where signals consist of V discrete levels,

max. data rate = 2H log2 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


Maximum Bandwidth

• Shannon: extended to noisy channels• Signal to noise ratio S/N, expressed in dB• dB = 10 x log10(S/N)

max. rate = H log2 (1 + S/N)• Example: 56 Khz channel, S/N = 10,000

(40 dB)max. rate = 730 Kbps

NOTE: regardless of value of V!


Calculating transfer time

Example: How long to transmit 200 KB file, if:

• 3 Mbps channel (3 x 106 bits per sec)• RTT = 50 msec (→ prop. delay = 25 msec)• Handshake = 3 RTTs• Packet size = 4 KB = 4 x 210 bytes x 8

bits/byte = 32768 bits• We must insert 1 RTT delay between

packets8CS 332 - Computer Networks

Calculating transfer time

Time to send 1 packet =(32768 bits / packet) / (3 x 106 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


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)


Multiplexing styles

• Choose some characteristic of the transmission process that we can subdivide reliably. – Frequency bands (radio, old phone system)– Time– Wavelength (optical)


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


• Bandpass filters separate the original signals, which are then demodulated to recover the data pattern



• 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.)


Time Division Multiplexing

• Transmission time divided into "frames" with some number of slots.

• Each conversation is assigned to a particular slot



• If there is no data for a given conversation, that slot in the frame is empty



• 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.


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.


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


02 – Performance Basics 1CS 332 - Computer Networks.

Documents

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