comt 4291 performance of data communications protocols general protocol concepts
TRANSCRIPT
COMT 429
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Performance of Data Communications Protocols
General Protocol Concepts
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OSI Model for Networks
7 Application Layer
6 Presentation Layer
5 Session Layer
4 Transport Layer
3 Network Layer
2 Data Link Layer
1 Physical Layer
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Flow Control
• Synchronize devices or routines at the sender and the receiver
• Basis for error recovery• May exist in every protocol layer
– Layer 2 for device to device– Layer 4 for user to user
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Flow Control Layers
Transport
Network
Data Link
Physical
Data Link
Physical
Transport
Network
Data Link
Physical
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Stop-and-Wait
Sender Receiver
ABCD
HG
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F
J
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Transmit
Sender Receiver
ABCD
HG
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F
J
E
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Propagate
Sender Receiver
ABCD
HG
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Ack-Transmit
Sender Receiver
BCD
HG
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Ack
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Ack-Propagate
Sender Receiver
BCD
HG
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Ack
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Transmission Time
• Definition: the time required for the transmitter to create the electrical signals needed to completely represent the message
• Depends on the length (in bits) of the message and the circuit speed (bits/sec).
• (Transmission Time) = (Message Length)/(Circuit Speed)
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Propagation TimeData Link Layer
• Definition: the time required for the electrical signals to travel from sender to receiver
• Depends on the length of the circuit and the signal speed.
• (Propagation Time) = (Circuit Length)/(Signal Speed)
• Speed of Light– In Vacuum: 300,000 km/sec– In a typical wire: 200,000 km/sec
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Propagation TimeTransport Layer
• Definition: the total time required for the packet to travel from sender to receiver
• Depends on all intermediate Data Link Layer propagation times as well as all queue and buffer delays. Also includes all transmission times except for the first one (from the sender to the first network device).
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Short Message orHigh Bit Rate
Sender Receiver
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Long Message orLow Bit Rate
Sender Receiver
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Stop and Wait Time Line
TimePacket ready
length=W bits
Last bittransmitted
Transmission Time tI
Last bit received
Ack returned
Propagation Time tp
Propagation Time tp
Note:
We ignore the length of the acknowledgement
Next packet ready
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What is the transmission rate?
• In the diagram, “W” bits were transmitted (Message Length)
• The elapsed time was ti+2*tp
• Define Beff as the effective bit rate, B as the circuit speed, and Eff as the efficiency
pout tt 2
pioutiT ttttt 2
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How do we get high efficiency?
• Note that ti is related to W and B
ppI
pII
ipI
pI
i
pI
eff
II
tEff
EffBWt
Eff
Efft
tEfftEfft
tttEff
tt
t
tt
W
BB
BEff
BtWB
Wt
21
or 21
2
)2(
2
)2
(1
or
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Why are very large packets impractical?
• Intermediate routers may not be able to store the packet
• Sharing a circuit is impractical because long packets tie up the circuit for long times
• Error checks are performed on a per-packet basis; we may need to re-transmit the entire (large) packet after an error
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An aside …
• The time in the denominator is the total time elapsed from the start of transmission to the receipt of a “reaction” from the other side
• This time is more generally refered to as the Round Trip Time (RTT); for Transport Layer protocols this includes queue, buffer, and transmisson times as well.
Beff W
ti 2 tp
Beff W
RTT
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What is the answer?
• On fast circuits (or on circuits with high RTT) I need a large amount of data “in flight”
• We don’t want large packets• Therefore, we have to allow
transmission of additional packets before the first packet is acknowledged
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Sliding WindowSender Receiver
ABCD
HG
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F
J
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0
21
10
67
45
3
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Sliding WindowSender Receiver
ABCD
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Sliding WindowSender Receiver
AB
CD
HG
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0
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Sliding WindowSender Receiver
AB
CD
HG
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0
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Sliding WindowSender Receiver
BC
D
HG
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0
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3
Ack
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Sliding WindowSender Receiver
BC
D
HG
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0
21
10
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Ack
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Sliding WindowSender Receiver
CD
HG
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F
J
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0
21
10
67
45
3
AckAck