Lecture 3
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Homework 1 problems posted, due September
8.
Reminder: Wireshark Project 1 is due on
Tuesday.
In-class exercise from last time. Questions?
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CS 475 Networks - Lecture 3
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Outline
Chapter 1 - Foundation
1.1 Applications
1.2 Requirements
1.3 Network Architecture
1.4 Implementing Network Software
1.5 Performance
1.6 Summary
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Bandwidth
The primary network performance measures are
bandwidth (BW) and latency.
BW (or throughput) is the number of bits that can
be sent over the network in a period of time
(usually bits-per-second or bps).
The link BW is the maximum physical rate at
which data can be sent. The measured
performance (often called throughput) is the max
rate at which we can send data. The throughput
is a function of the link BW and other factors
(congestion, server load, etc)
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Bandwidth
Fig. 1.16: The bit width in a slower BW channel (a) is
wider than the bit width in a faster channel (b).
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Latency
Latency (or delay) is the time it takes a message
to travel across the network.
Latency = PropagationTime + TransmitTime
+ QueueTime
where
PropagationTime =
Distance/SpeedOfPropagation
TransmitTime = MessageSize/Bandwidth
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Latency
The PropagationTime is the time it takes for a
single pulse to be sent across the network. The
SpeedOfPropagation depends on the medium
(2x108 m/s for fiber, 2.3x108 m/s for cable and
3.0x108 m/s for wireless).
The TransmitTime is the time it takes to push the
entire message onto the medium. The
TransmitTime for a 1 MB file on a 10 Mbps link is
(1 MB x 8 b/B)/10 Mbps = 0.8 s.
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Latency
The QueueTime is the sum of the queuing delays
encountered at the switches and routers by the
message. The QueueTime is not included when
calculating the latency over a single link.
In some problems the round-trip-time or RTT is of
interest. The RTT includes the time it takes for a
signal to be sent plus the time it takes for an
acknowledgment to be received.
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Latency
For small messages
the latency is
dominated by the
propagation time,
for large messages
the transmit time
determines the
latency.
Fig. 1.17: Latency vs RTT for different
message sizes and link speeds.
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Delay-Bandwidth Product
The Delay-BW product represents the number of
bits in transit (on the network).
Alternatively, it can be thought of as the number
of bits transmitted before the first bit reaches the
receiver.
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High-Speed Networks
On a high-speed network, the RTT for an
acknowledgment can have a significant impact on
throughput:
Throughput = MessageSize/TransferTime
where
TransferTime = RTT + MessageSize/BW
Using 100 ms for a cross-country RTT, the
TransferTime for a 1 MB file on a 1 Gbps link is
108 ms for Throughput of only 74.1 Mbps.
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Application Requirements
Applications do not necessarily need “all the
bandwidth they can get”. Streaming video with a
resolution of 352 x 240 pixels and 24 bit color
would have a frame size of 247.5 KB. A frame
rate of 30 frames per second would require a
throughput of 75 Mbps. Much greater bandwidth
would be of no interest to this application.
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Application Requirements
The variation in latency or jitter can be important
in streaming applications. If a frame arrives late
quality will be affected.
The client can account for jitter via buffering. The
client must know the maximum jitter in order to
allocate a buffer of the appropriate size.
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Summary
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Chapter 1 lays the foundation for understanding
how to build a fully functional computer network
from the ground up.
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Identified the requirements
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Defined a layered architecture
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Discussed programmer interface between the
network protocols and applications
Identified performance metrics
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In-class Exercises
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1. Problem 1.5 on pages 61-62.
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2. Problem 1.19 on page 64.
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Turn these in at the end of class today.
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Work on Wireshark Project 1 and programming
problems for Homework 1.
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In-class Exercises
1. Problem 1.5 (page 61-62)
d = 4 km, s = 2x108 m/s
PropTime = d/s = 20 us
XmitTime = Size/BW => BW = Size/XmitTime
with Size = 100 B and XmitTime = 20 us
BW = (100 B x 8 b/B)/20 us = 40 Mbps
with Size = 512 B, BW = 204.8 Mbps
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In-class Exercises
2. Problem 1.19 (page 64)
a) delay x BW = 10 us x 100 Mbps = 1000 b
b) QTime = 12000 b/100 Mbps = 120 us
delay = 4 x 10 us + 3 x 120 us = 400 us
delay x BW = 40000 b
(12000 b in each switch, 1000 b on each link)
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In-class Exercises
2. Problem 1.19 (page 64), cont
c) delay x BW = 50 ms x 1.5 Mbps = 75 kb
d) delay = 2 x 35900 km/3x108 m/s = 239.3 ms
delay x BW = 239.3 ms x 1.5 Mbps
delay x BW = 359 kb
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