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Transport Layer 3-1 Announcements Collect homework New homework: Ch3#3,4,7,10,12,16,18- 20,25,26,31,33,37 Due Wed Sep 24 Reminder: Project #1 due Monday First midterm on Monday Sep 29 In-class Closed book Chapters 1-3

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Page 1: Transport Layer3-1 Announcements r Collect homework r New homework: m Ch3#3,4,7,10,12,16,18-20,25,26,31,33,37 m Due Wed Sep 24 r Reminder: m Project #1

Transport Layer 3-1

Announcements

Collect homework New homework:

Ch3#3,4,7,10,12,16,18-20,25,26,31,33,37 Due Wed Sep 24

Reminder: Project #1 due Monday

First midterm on Monday Sep 29 In-class Closed book Chapters 1-3

Page 2: Transport Layer3-1 Announcements r Collect homework r New homework: m Ch3#3,4,7,10,12,16,18-20,25,26,31,33,37 m Due Wed Sep 24 r Reminder: m Project #1

Transport Layer 3-2

Chapter 3 outline

3.1 Transport-layer services

3.2 Multiplexing and demultiplexing

3.3 Connectionless transport: UDP

3.4 Principles of reliable data transfer

3.5 Connection-oriented transport: TCP segment structure reliable data transfer flow control connection

management

3.6 Principles of congestion control

3.7 TCP congestion control

Page 3: Transport Layer3-1 Announcements r Collect homework r New homework: m Ch3#3,4,7,10,12,16,18-20,25,26,31,33,37 m Due Wed Sep 24 r Reminder: m Project #1

Transport Layer 3-3

Fairness goal: if K TCP sessions share same bottleneck link of bandwidth R, each should have average rate of R/K

TCP connection 1

bottleneckrouter

capacity R

TCP connection 2

TCP Fairness

Page 4: Transport Layer3-1 Announcements r Collect homework r New homework: m Ch3#3,4,7,10,12,16,18-20,25,26,31,33,37 m Due Wed Sep 24 r Reminder: m Project #1

Transport Layer 3-4

Why is TCP fair?

Two competing sessions: Additive increase gives slope of 1, as throughout increases multiplicative decrease decreases throughput proportionally

R

R

equal bandwidth share

Connection 1 throughputConnect

ion 2

th

roughput

congestion avoidance: additive increaseloss: decrease window by factor of 2

congestion avoidance: additive increaseloss: decrease window by factor of 2

Page 5: Transport Layer3-1 Announcements r Collect homework r New homework: m Ch3#3,4,7,10,12,16,18-20,25,26,31,33,37 m Due Wed Sep 24 r Reminder: m Project #1

Transport Layer 3-5

Fairness (more)

Fairness and UDP Multimedia apps

often do not use TCP do not want rate

throttled by congestion control

Instead use UDP: pump audio/video at

constant rate, tolerate packet loss

Research area: TCP friendly datagram delivery (e.g., DCCP)

Fairness and parallel TCP connections

nothing prevents app from opening parallel connections between two hosts.

Web browsers do this Example: link of rate R

supporting 9 cnctions; new app asks for 1 TCP,

gets rate R*1/10 new app asks for 9 TCPs,

gets R*(9/18) = R/2 !

Page 6: Transport Layer3-1 Announcements r Collect homework r New homework: m Ch3#3,4,7,10,12,16,18-20,25,26,31,33,37 m Due Wed Sep 24 r Reminder: m Project #1

Transport Layer 3-6

Delay modeling

Q: How long does it take to receive an object from a Web server after sending a request?

Ignoring congestion, delay is influenced by:

TCP connection establishment

data transmission delay slow start

Notation, assumptions: Assume one link between

client and server of rate R S: MSS (bits) O: object size (bits) no retransmissions (no loss, no

corruption)

Window size: First assume: fixed congestion

window, W segments Then dynamic window,

modeling slow start

Page 7: Transport Layer3-1 Announcements r Collect homework r New homework: m Ch3#3,4,7,10,12,16,18-20,25,26,31,33,37 m Due Wed Sep 24 r Reminder: m Project #1

Transport Layer 3-7

Fixed congestion window (1)

First case:WS/R > RTT + S/R: ACK

for first segment in window returns before window’s worth of data sent

delay = 2RTT + O/R

Page 8: Transport Layer3-1 Announcements r Collect homework r New homework: m Ch3#3,4,7,10,12,16,18-20,25,26,31,33,37 m Due Wed Sep 24 r Reminder: m Project #1

Transport Layer 3-8

Fixed congestion window (2)

Second case: WS/R < RTT + S/R: wait

for ACK after sending window’s worth of data sent

delay = 2RTT + O/R+ (K-1)[S/R + RTT - WS/R]

(where K is the number of windows that cover the object.)

Page 9: Transport Layer3-1 Announcements r Collect homework r New homework: m Ch3#3,4,7,10,12,16,18-20,25,26,31,33,37 m Due Wed Sep 24 r Reminder: m Project #1

Transport Layer 3-9

TCP Delay Modeling: Slow Start (1)Now suppose window grows according to slow

start

Will show that the delay for one object is:

R

S

R

SRTTP

R

ORTTLatency P )12(2

where P is the number of times TCP idles at server:

}1,{min KQP

- where Q is the number of times the server idles if the object were of infinite size.

- and K is the number of windows that cover the object.

Page 10: Transport Layer3-1 Announcements r Collect homework r New homework: m Ch3#3,4,7,10,12,16,18-20,25,26,31,33,37 m Due Wed Sep 24 r Reminder: m Project #1

Transport Layer 3-10

TCP Delay Modeling: Slow Start (2)

RTT

initia te TCPconnection

requestobject

first w indow= S /R

second w indow= 2S /R

third w indow= 4S /R

fourth w indow= 8S /R

com pletetransm issionobject

delivered

tim e atc lient

tim e atserver

Example:• O/S = 15 segments• K = 4 windows• Q = 2• P = min{K-1,Q} = 2

Server idles P=2 times

Delay components:• 2 RTT for connection establ. and request• O/R to transmit object• time server idles due to slow start

Server idles: P = min{K-1,Q} times

Page 11: Transport Layer3-1 Announcements r Collect homework r New homework: m Ch3#3,4,7,10,12,16,18-20,25,26,31,33,37 m Due Wed Sep 24 r Reminder: m Project #1

Transport Layer 3-11

TCP Delay Modeling (3)

R

S

R

SRTTPRTT

R

O

R

SRTT

R

SRTT

R

O

idleTimeRTTR

O

P

kP

k

P

pp

)12(][2

]2[2

2delay

1

1

1

windowth after the timeidle 2 1 kR

SRTT

R

S k

ementacknowledg receivesserver until

segment send tostartsserver whenfrom time RTTR

S

window kth the transmit totime2 1

R

Sk

RTT

initia te TCPconnection

requestobject

first w indow= S /R

second w indow= 2S /R

third w indow= 4S /R

fourth w indow= 8S /R

com pletetransm issionobject

delivered

tim e atc lient

tim e atserver

Page 12: Transport Layer3-1 Announcements r Collect homework r New homework: m Ch3#3,4,7,10,12,16,18-20,25,26,31,33,37 m Due Wed Sep 24 r Reminder: m Project #1

Transport Layer 3-12

TCP Delay Modeling (4)

)1(log

)}1(log:{min

}12:{min

}/222:{min

}222:{min

2

2

110

110

S

OS

Okk

S

Ok

SOk

OSSSkK

k

k

k

Calculation of Q, number of idles for infinite-size object,is similar (see HW).

Recall K = number of windows that cover object

How do we calculate K ?

Page 13: Transport Layer3-1 Announcements r Collect homework r New homework: m Ch3#3,4,7,10,12,16,18-20,25,26,31,33,37 m Due Wed Sep 24 r Reminder: m Project #1

Transport Layer 3-13

HTTP Modeling Assume Web page consists of:

1 base HTML page (of size O bits) M images (each of size O bits)

Non-persistent HTTP with no parallel connections: M+1 TCP connections in series Response time = (M+1)O/R + (M+1)2RTT + sum of idle

times Persistent HTTP with no parallel connections:

2 RTT to request and receive base HTML file 1 RTT to request and receive M images Response time = (M+1)O/R + 3RTT + sum of idle times

Non-persistent HTTP with X parallel connections Suppose M/X integer. 1 TCP connection for base file M/X sets of parallel connections for images. Response time = (M+1)O/R + (M/X + 1)2RTT + sum of

idle times

Page 14: Transport Layer3-1 Announcements r Collect homework r New homework: m Ch3#3,4,7,10,12,16,18-20,25,26,31,33,37 m Due Wed Sep 24 r Reminder: m Project #1

Transport Layer 3-14

02468

101214161820

28Kbps

100Kbps

1Mbps

10Mbps

non-persistent

persistent

parallel non-persistent

HTTP Response time (in seconds)RTT = 100 msec, O = 5 Kbytes, M=10 and X=5

For low bandwidth, connection & response time dominated by transmission time.

Persistent connections only give minor improvement over parallel connections.

Page 15: Transport Layer3-1 Announcements r Collect homework r New homework: m Ch3#3,4,7,10,12,16,18-20,25,26,31,33,37 m Due Wed Sep 24 r Reminder: m Project #1

Transport Layer 3-15

0

10

20

30

40

50

60

70

28Kbps

100Kbps

1Mbps

10Mbps

non-persistent

persistent

parallel non-persistent

HTTP Response time (in seconds)

RTT =1 sec, O = 5 Kbytes, M=10 and X=5

For larger RTT, response time dominated by TCP establishment & slow start delays. Persistent connections now give important improvement: particularly in high delaybandwidth networks.

Page 16: Transport Layer3-1 Announcements r Collect homework r New homework: m Ch3#3,4,7,10,12,16,18-20,25,26,31,33,37 m Due Wed Sep 24 r Reminder: m Project #1

Transport Layer 3-16

Chapter 3: Summary principles behind transport

layer services: multiplexing,

demultiplexing reliable data transfer flow control congestion control

instantiation and implementation in the Internet UDP TCP

Next: leaving the network

“edge” (application, transport layers)

into the network “core”