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COMT 625 2
Outline
• General protocol principles for flow control and error handling, including:
• Examples of protocols using these schemes• Principles of interconnecting networks• Bridges, Switches, Routers, Gateways
COMT 625 3
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
COMT 625 5
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
COMT 625 6
Flow Control Layers
Transport
Network
Data Link
Physical
Data Link
Physical
Transport
Network
Data Link
Physical
COMT 625 12
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)
COMT 625 13
Propagation Time
• 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
COMT 625 17
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
COMT 625 18
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
Beff =W
ti + 2 ∗tpEff =
BeffB
COMT 625 19
How do we get high efficiency?
• Note that ti is related to W and Bti =
WB orW =ti ∗B
Eff =BeffB
=1B(
Wti + 2∗tp
)
=ti
ti + 2 ∗tpEff(ti + 2∗tp) =ti
ti −Eff∗ti =Eff∗2∗tp
ti =Eff
1−Eff∗2∗tp orW=B∗ Eff
1−Eff∗2∗tp
COMT 625 20
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
COMT 625 21
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 referred to as the Round Trip Time (RTT), and may include queue delays as well as multiple transmission and propagation times
Beff =W
ti + 2 ∗tp
Beff =W
RTT
COMT 625 22
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