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Lecture 14: Transport Layer

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ECS 152B Computer Networks. How does actual TCP ... Two cases. Demet Aksoy. 5. ECS 152B Computer Networks. Dynamic Congestion Window. K = log (O/S 1) 2 ... – PowerPoint PPT presentation

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Title: Lecture 14: Transport Layer


1
Lecture 14 Transport Layer
  • Prev. summary
  • TCP flow control
  • TCP congestion control

Application
Transport
Network
Link
  • Todays lecture
  • TCP latencies

2
Overview
3
How does actual TCP latencies look like?
  • Assumptions
  • Assume fixed congestion window, W
  • throughput of R bps
  • S MSS (bits)
  • O object size (bits)
  • no retransmissions (no loss, no corruption)
  • Q latency until receipt of the object?
  • TCP connection establishment
  • data transfer delay at the rate arranged by the
    window size

4
Fixed Congestion Window
Two cases
1) ACK arrives before windows worth of data
sent 2) Wait for ACK after sending windows worth
of data sent
  • fixed congestion window, W
  • throughput of R bps
  • S MSS (bits)
  • O object size (bits)
  • no retransmissions
  • Case I
  • WS/R gt RTT S/R

Case 2 latency 2RTT O/R (K-1)S/R RTT -
WS/R
Case 1 latency 2RTT O/R
latency 2RTT O/R max0, (K-1)S/R RTT -
WS/R
5
Dynamic Congestion Window
K windows cover the object
6
Dynamic Congestion Window
Out of infinite windows, will stall Q times Q
log (1RTT/(S/R)) 1
2
Out of K windows, will stall P times P min
Q, K-1
p
latency 2RTT O/R P(RTTS/R) - (2 -1)S/R
7
Example Scenarios
O 100 Kbytes S 536 bytes K 8 RTT100 ms
a) Large Object Size
8
Large Object
O 100 Kbytes S 536 bytes K 8 RTT100 ms
a) Large Object Size
9
Large vs Small Object
O 5 Kbytes S 536 bytes K 4 RTT100 ms
O 100 Kbytes S 536 bytes K 8 RTT100 ms
10
Example Scenarios
O 5 Kbytes S 536 bytes K 4 RTT1 sec
c) Larger RTT
11
Larger RTT
O 5 Kbytes S 536 bytes K 4 RTT1 sec
O 5 Kbytes S 536 bytes K 4 RTT100 ms
12
In Practice, can be complicated...
Host A
Host B
initialize Congwin 1 for (each segment ACKed)
Congwin until (loss event OR
CongWin gt threshold)
one segment
two segments
two segments
  • In practice, slow start can result in a slower
    increase than exponential due to the position of
    lost packets.
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