Transport Layer: TCP and UDP - Washington University in …jain/cse473-16/ftp/i_3tcp4.pdf · · 2016-10-03Lab 3A: UDP T Download the wireshark traces from ... 5. Examine the first

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http://ww

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Transport Layer: TC

P and UD

P

Raj Jain

Washington U

niversity in Saint LouisSaint Louis, M

O 63130

Jain@w

ustl.edu A

udio/Video recordings of this lecture are available on-line at:http://w

ww

.cse.wustl.edu/~jain/cse473-16/

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Overview

Transport Layer Design Issues:

Multiplexing/D

emultiplexing

Reliable Data Transfer

Flow control

Congestion controlU

DP

TCPHeader form

at, connection managem

ent, checksumCongestion Control

Note: This class lecture is based on Chapter 3 of the textbook

(Kurose and Ross) and the figures provided by the authors.

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Transport Layer Design Issues

1.Transport Layer Functions

2.M

ultiplexing and Dem

ultiplexing3.

Error Detection: Checksum

4.Flow

Control5.

Efficiency Principle6.

Error Control: Retransmissions

Overview

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Transport Layer

Transport = End-to-End Services Services required at source and destination system

sN

ot required on intermediate hops

Application

TransportN

etwork

Datalink

Physical

Netw

orkD

atalinkPhysical

Netw

orkD

atalinkPhysical

Application

TransportN

etwork

Datalink

Physical

End SystemRouter

RouterEnd System

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Transport Layer Functions1.

Multiplexing and dem

ultiplexing: Am

ong applications and processes at end system

s2.

Error detection: Bit errors3.

Loss detection: Lost packets due to buffer overflow

at intermediate system

s (Sequence numbers and

acks)4.

Error/loss recovery: Retransmissions

5.Flow

control: Ensuring receiver has buffers6.

Congestion C

ontrol: Ensuring network has capacity

Not all transports provide all functions

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Protocol LayersTop-D

own approach

Internetwork

Host to

Netw

ork

IP

Application

Transport

HTTP

TCP

FTPSM

TP

UD

P

Physical

EthernetW

i-FiPoint-to-Point

CoaxFiber

Wireless

P2PD

NS

Skype

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Multiplexing and D

emultiplexing

Transport Portsand Netw

ork addressesare used to separate flow

s

IP:128.3.4.1

Web

TCP

IP :209.3.1.1IP :125.5.1.1

User 1

ServerU

ser 2

DN

SA

pplication

Transport

Netw

ork

UD

P

Web

DN

SW

ebD

NS

TCPU

DP

TCPU

DP

Port #

Protocol Type

SP:3009D

P:80SA

: 128.3.4.1D

A: 209.3.1.1

SP:80D

P:3009SA

:209.3.1.1D

A:128.3.4.1

Ref: http://en.wikipedia.org/w

iki/List_of_TCP_and_UD

P_port_numbers

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User D

atagram Protocol (U

DP)

Connectionless end-to-end serviceProvides m

ultiplexing via portsError detection (Checksum

) optional. Applies to pseudo-

header(sam

e as TCP) and UD

P segment. If not used, it is set

to zero.N

o error recovery (no acks). No retransm

issions.U

sed by network m

anagement, D

NS, Stream

ed multim

edia(A

pplications that are loss tolerant, delay sensitive, or have their ow

n reliability mechanism

s)

SourcePort

Dest

PortCheck-

sumLength

16b16b

16bSize in bits

16b

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Error Detection: C

hecksumC

yclic Redundancy C

heck (CR

C): Pow

erful but generally requires hardw

areC

hecksum: W

eak but easily done in software

Example:1's com

plementof 1’s com

plement sum

of 16-bit w

ords with overflow

wrapped around

At receiver the sum

is all 1’s and the checksum is zero.

1 1 1 1 0 0 1 1 0 0 1 1 0 0 1 1 0 1 1 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 1 1 0 1 1 1 0 1 1 1 0 1 1 1 0 1 1 1 1 0 1 1 1 0 1 1 1 0 1 1 1 1 0 0 1 0 1 0 0 0 1 0 0 0 1 0 0 0 0 1 1

wraparound

sum

checksum

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1’s Com

plement

2’s Com

plement:-ve of a num

ber is complem

ent+11 = 0001

-1 = 11112 = 0010

-2 = 11100 = 0000

-0 = 00001’s com

plement:-ve of a num

ber is it’s complem

ent1 = 0001

-1 = 11102 = 0010

-2 = 11010 = 0000

-0 = 11112’s C

omplem

ent sum: A

dd with carry. D

rop the final carry, if any.6-7 = 0110 + (-0111) = 0110 + 1001 = 1111 => -11’s com

plement sum

: Add w

ith carry. Add end-around carry back to sum

6-7 = 0110 + (-0111) = 0110+1000 = 1110 => -1C

omplem

ent of 1’s complem

ent sum: 0001

Checksum

: At the transm

itter: 0110 1000, append 0001A

t the receiver: 0110 1000 0001 compute checksum

of the full packet = com

plement of sum

= complem

ent of 1111 = 0000Ref: https://en.w

ikipedia.org/wiki/O

nes%27_com

plement

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Hom

ework 3A

Consider the following tw

o 16-bit words: A

BCD 1234

A.

What is the checksum

as computed by the sender

B.A

dd your answer of Part A

to the end of the packet and show

how the receiver w

ill compute the

checksum of the received three 16-bit w

ords and confirm

that there are no errors.C.

Now

assume that the first bit of the packet is flipped

due to an error. Repeat Part B at the receiver. Is the error detected?

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UD

P: Summ

ary1.

UD

P provides flow m

ultiplexing using port #s2.

UD

P optionally provides error detection using the checksum

3.U

DP does not have error or loss recovery

mechanism

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Lab 3A: U

DP

Dow

nload the wireshark traces from

http://gaia.cs.umass.edu/w

ireshark-labs/wireshark-traces.zip

Open the trace file http-ethereal-trace-5 in W

ireshark, View

Expand A

ll and answ

er the following questions?

1. Examine the first packet. In the printout for this packet, select the line

with U

DP. The U

PD header is highlighted in the trace. H

ow m

any fields are there in the U

DP header? N

ame these fields, their lengths and their

values.2. The value in the Length field is the length of w

hat? Verify your claim

w

ith your captured UD

P packet. 3. W

hat is the maxim

um num

ber of bytes that can be included in a UD

P payload?

4. What is the largest possible source port num

ber? 5. Exam

ine the first two U

DP packets. D

escribe the relationship between

the port numbers in the tw

o packets.

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Flow C

ontrolFlow

Control Goals:

1.Sender does not flood the receiver, 2.M

aximize throughput

Ack Pkt 1

Ack Pkt 2

Ack Pkt 3

SenderReceiver

SenderReceiver

Stop and Wait Flow

Control

Window

Flow C

ontrol

Throughput =Throughput =

L/RRTT+L/R

W L/R

RTT+L/R

Large RTTLow

Thruput

L/RRTT

L= Packet LengthR= Link bit Rate

University

inSSt.LSt.L

Ref: Textbook Section 3.4.2

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Sliding Window

Diagram

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Stop and Wait Flow

Control

first bit transmitted, t = 0 Sender

Receiver

RTT

last bit transmitted, t = L / R

First bit arrivesLast bit arrives, send A

CK

ACK

arrives, send next packet, t = RTT + L / R

U=

L / RRTT + L / R

=1

2+ 1

2tprop +tframe

tframe

=

Here,

= tprop /tframe

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Sliding Window

Protocol Efficiency

Data

Ack

tframe

tprop

U=

W tfram

e

2tprop +tframe

=

W2+1

1 if W>2

+1

Here,

= tprop /tframe

W=1

Stop and Wait

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Utilization: Exam

plesSatellite Link: O

ne-way Propagation D

elay = 270 ms

RTT=540 ms

Frame Size L = 500 Bytes = 4 kb

Data rate R = 56 kbps

tframe = L/R= 4kb/56kbps =0.071s =

71 ms

= tprop /tframe = 270/71 = 3.8

U = 1/(2

+1) = 0.12

Short Link: 1 km = 5

s (Assum

ing Fiber 200 m/

s), Rate=10 M

bps, Fram

e=500 bytes tfram

e = 4k/10M= 400

s=tprop /tfram

e =5/400=0.012 U

=1/(2+1)=0.98

Note: The textbook uses RTT in place of tprop and L/R for tfram

e

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Effect of Window

Size

ULarger window

is better for larger

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Efficiency PrincipleForallprotocols, the m

aximum

utilization (efficiency) is a non-increasing

function of

Max

Utilization

Not Possible

Protocol 1Protocol 2

Best possible

=tproptfram

e

=D

istance/Speed of SignalBits Transm

itted /Bit rate

=D

istance Bit rate

Bits Transmitted

Speed of Signal

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Error Control: R

etransmissions

Retransmit lost packets

Autom

atic Repeat reQ

uest (ARQ

)

Ack Pkt 1

Pkt 2

AckPkt 3

SenderReceiver

Stop and Wait A

RQ

Timeout

Timeout

Pkt 2

Pkt 3

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Go-Back-N

AR

Q

Receiver does not cache out-of-order frames

Sender has to go backand retransm

it all frames after

the lost frame

RcvrSender

1

Nack 1

1

0122 33

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Selective Repeat A

RQ

Receiver caches out-of-order frames

Sender retransmits only the lost fram

eA

lso known as selective rejectA

RQ RcvrSender

1

Nack 1

1

0124 35

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Selective Repeat: W

indow Size

012345670A

ck

Sequence number space >

2 window

sizeW

indow size <

2n-1

Timeout

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Performance: M

aximum

Utilization

Stop and Wait Flow

Control: U

= 1/(1+2)

Window

Flow C

ontrol:

Stop and Wait A

RQ

: U = (1-P)/(1+2

)G

o-back-N A

RQ

:

Selective Repeat A

RQ

:

U =

1W

>2

+1 W

/(2+1)

W< 2

+1{U

= (1-P)/(1+2

P)W

>2

+1 W

(1-P)/[(2+1)(1-P+W

P)] W

< 2+1

{U =

(1-P)W

>2

+1 W

(1-P)/(2+1)

W< 2

+1{

P = Probability of Loss

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Performance C

omparison

1.0

0.80.6

0.4

0.2

0.00.1

110

1001000

Utilization

Stop-and-wait

W= 127 G

o-back-N

W=7 G

o-back-N &

W= 7 Selective-repeat

W= 127 Selective-repeat

More bps or longer distance

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Transport Layer Design Issues

1.M

ultiplexing/demultiplexing by a com

bination of source and destination IP addresses and port num

bers.2.

Window

flow control is better for long-distance or high-speed

networks

3.Longer distance or higher speed

Larger Larger w

indow is better

4.Stop and and w

ait flow control is ok for short distance or low

-speed netw

orks5.

Selective repeat is better than stop and wait A

RQO

nly slightly better than go-back-N

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Hom

ework 3B

Problem 19 on page 302 of the textbook:

Consider the GBN

protocol with a sender w

indow size of 3 and a sequence num

ber range of 1,024. Suppose that at tim

e t, the next in-order packet that the receiver is expecting has a sequence num

ber of k. Assum

e that the medium

does not reorder m

essages. Answ

er the following questions:

A. W

hat are the possible sets of sequence numbers insdie the sender’s w

indow at tim

e t? Justify your answ

er.B. W

hat are all possible values of the ACK

field in all possible messages currently

propagating back to the sender at time t? Justify your answ

er.W

indow Flow

Control:

C. How

big window

(in number of packets) is required for the channel utilization to be

greater than 60% on a cross-country link of 4000 km

running at 20 Mbps using 1

kByte packets?Efficiency Principle:D

. Ethernet V1 access protocol w

as designed to run at 10 Mbps over 2.5 K

m using 1500

byte packets. This same protocol needs to be used at 100 M

bps at the same

efficiency. What distance can it cover if the fram

e size is not changed?

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TCP

1.TCP H

eader Format, O

ptions, Checksum2.

TCP Connection Managem

ent3.

Round Trip Time Estim

ation4.

Principles of Congestion Control5.

Slow Start Congestion Control

Overview

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Key Features of TC

PPoint-to-Point: O

ne sender, one receiverByte Stream

: No m

essage boundaries. TCP m

akes “segments”

Maxim

um segm

ent size(M

SS)C

onnection Oriented: H

andshake to initialize states before data exchangeFull D

uplex: Bidirectional data flow in one connection

Reliable: In-order byte delivery

Flow control: To avoid receiver buffer overflow

Congestion control: To avoid netw

ork router buffer overflow

BytesBytes

Segments

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TCP

Transmission Control Protocol

Key Services:Send: Please send w

hen convenient D

ata stream push: D

estination TCP, please deliver it im

mediately to the receiving application.

Source TCP, please send it now.

Set on last packet of an application message.

Urgent data signaling: D

estination TCP, please give this urgent data to the user out-of-band. G

enerally used for CTRL-C.

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TCP Segm

ent Format (C

ont)

Source PortD

est PortSeq N

oA

ck No

Data

Offset

Window

Resvd

ChecksumU

rgentO

ptions

Data

UA

PR

SF

16b16bPad

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TCP H

eader Fields

Source Port (16 bits): Identifies source user processD

estination Port(16 bits)21 = FTP, 23 = Telnet, 53 = D

NS, 80 = H

TTP, ...Sequence N

umber

(32 bits): Sequence number of the

first byte in the segment. If SY

N is present, this is

the initial sequence number (ISN

) and the first data byte is ISN

+1.A

ck number

(32 bits): Next byte expected

Data offset(4 bits): N

umber of 32-bit w

ords in the headerR

eserved(6 bits)

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TCP H

eader (Cont)

Control(6 bits):U

rgent pointer field significant, A

ck field significant, Push function,

Reset the connection, Synchronize the sequence num

bers, N

o more data from

sender

Window

(16 bits): W

ill accept [Ack] to [A

ck]+[window

]-1

ACK

URG

PSHRST

SYN

FIN

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TCP H

eader (Cont)

Checksum

(16 bits): covers the segment plus a

pseudo header. Includes the following fields from

IP header: source and dest adr, protocol, segm

ent length. Protects from

IP misdelivery.

Urgent pointer

(16 bits): Points to the byte following

urgent data. Lets receiver know how

much data it

should deliver right away out-of-band.

Options(variable):

Max segm

ent size (does not include TCP header, default 536 bytes), W

indow scale factor, Selective

Ack perm

itted, Timestam

p, No-O

p, End-of-options

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TCP O

ptions

End of Options: Stop looking for further option

No-op: Ignore this byte. U

sed to align the next option on a 4-byte w

ord boundaryM

ax Segment Size (M

SS):Does notinclude TCP header

Kind

LengthM

eaning0

1End of Valid options in header

11

No-op

24

Maxim

um Segm

ent Size3

3W

indow Scale Factor

810

Timestam

p

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TCP C

hecksumChecksum

is the 16-bit one's complem

ent of the one's com

plement sum

of a pseudo header of inform

ation from the IP header, the TCP header, and

the data, padded with zero octets at the end (if

necessary) to make a m

ultiple of two octets.

Checksum field is filled w

ith zeros initiallyTCP length (in octet) is not transm

itted but used in calculations. Efficient im

plementation in RFC1071.

Source Adr

Dest. A

drZeros

ProtocolTCP Length

TCP Header

TCP data32

328

816

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TCP C

onnection Managem

entConnection Establishm

entThree w

ay handshakeSY

N flag set

Request for connection

SYN

, ISN = 100

SYN

, ISN = 350, A

ck 101

Ack 351

Connection Termination

Close with FIN

flag setA

bort

FINA

ck

Ack

FIN

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Example R

TT estimation:

RTT: gaia.cs.umass.edu to fantasia.eurecom

.fr

100

150

200

250

300

350

18

1522

2936

4350

5764

7178

8592

99106

time (seconnds)

RTT (milliseconds)

SampleR

TTEstim

ated RTT

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Round Trip Tim

e Estimation

Measured round trip tim

e (SampleRTT) is very random

. Estim

atedRTT=(1-)Estim

atedRTT+Sam

pleRTTD

evRTT = (1-)D

evRTT+ |Sam

pleRTT-EstmatedRTT|

TimeoutInterval=Estim

atedRTT+4D

evRTT

Value

ProbabilityVery low

probability of false tim

eout

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Our R

esearch on Congestion C

ontrol

Early 1980s Digital Equipm

ent Corporation (DEC) introduced Ethernet

productsN

oticed that throughput goes down w

ith a higher-speed link in middle

(because no congestion mechanism

s in TCP)Results:1.

Timeout

Congestion Reduce the TCP w

indow to one on a tim

eout [Jain 1986]2.

Routers should set a bit when congested (D

ECbit). [Jain, Ram

akrishnan, Chiu 1988]3.

Introduced the term “Congestion A

voidance”4.

Additive increase and m

ultiplicative decrease (AIM

D principle)

[Chiu and Jain 1989]There w

ere presented to IETF in 1986.Slow

-start based on Timeout and A

IMD

[Van Jacobson 1988]

1Mbps

1Mbps

1Mbps

Time=6 m

inutes1M

bps10M

bps1M

bpsTim

e=6 hoursBit in header

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Slow Start C

ongestion Control

Window

= Flow control avoids receiver overrun

Need congestion control to avoid netw

ork overrunThe sender m

aintains two w

indows:

Credits from the receiver

Congestion window

from the netw

orkCongestion w

indow is alw

ays less than the receiver window

Starts with a congestion w

indow (CW

ND

) of 1 max segm

ent size (M

SS) D

o not disturb existing connections too much.

Increase CWN

D by 1 M

SS every time an ack is received

Assum

e CWN

D is in bytes

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Slow Start (C

ont)If segm

ents lost, remem

ber slow start threshold (SSThresh) to

CWN

D/2

Set CWN

D to 1 M

SSIncrem

ent by 1MSS per ack until SSThresh

Increment by 1 M

SS*MSS/CW

ND

per ack afterwards

Time

CongestionW

indowCW

ND

Receiver Window

IdleInterval

Timeout

1MSS

SSThresh

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Slow Start (C

ont)A

t the beginning, SSThresh = Receiver window

After a long idle period (exceeding one round-trip tim

e), reset the congestion w

indow to one.

If CWN

D is W

MSS, W

acks are received in one round trip.Below

SSThresh, CWN

D is increased by 1M

SS on every ackCW

ND

increases to 2W M

SS in one round tripCW

ND

increases exponentially with tim

eExponential grow

th phase is also known as “Slow start” phase

Above SSThresh, CW

ND

is increased by MSS/CW

ND

on every ack

CWN

D increases by 1 M

SS in one round tripCW

ND

increases linearly with tim

eThe linear grow

th phase is known as “congestion avoidance”

phase

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AIM

D Principle

Additive Increase, M

ultiplicative D

ecreaseW

1+W2 = Capacity

Efficiency, W

1=W2

Fairness(W

1,W2) to (W

1+W

,W2+

W)

Linear increase (45line)

(W1,W

2) to (kW1,kW

2)M

ultiplicative decrease (line through origin)

CapacityW

1

C

FairEfficient

Ref: D. Chiu and Raj Jain, "A

nalysis of the Increase/Decrease A

lgorithms for

Congestion Avoidance in C

omputer N

etworks," Journal of Com

puter N

etworks and ISD

N, Vol. 17, N

o. 1, June 1989, pp. 1-14, http://w

ww.cse.w

ustl.edu/~jain/papers/cong_av.htm

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Fast Retransm

itO

ptional –im

plemented in TCP Reno

(Earlier version was TCP Tahoe)

Duplicate A

ck indicates a lost/out-of-order segment

On receiving 3 duplicate acks (4

thack for the same segm

ent): Enter Fast Recovery m

odeRetransm

it missing segm

entSet SSThresh=CW

ND

/2Set CW

ND

=SSThresh+3 MSS

Every subsequent duplicate ack: CWN

D=CW

ND

+1MSS

When a new

ack (not a duplicate ack) is receivedExit fast recoverySet CW

ND

=SSTHRESH

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TCP C

ongestion ControlState D

iagram

SlowStart

CongestionAvoidance

FastRecovery

Dup A

ckD

upAckCount++

CWN

D<SSThresh, N

ew A

ckCW

ND

=CWN

D+M

SSD

upAckCount=0

Transmit new

segment as allow

ed

CWN

D=1M

SSSSThresh=RcvrW

in/2D

upAckCount=0

Transmit one segm

entD

upAckCount==3

SSThresh=CWN

D/2

Cwnd=ssthresh+3M

SSRetransm

it missing

segment

Timeout

SSTthresh=CWN

D/2

CWN

D=1M

SSD

upAckCount=0

Retransmit m

issing segm

ent

Dup A

ckD

upAckCount++

Dup A

ckCW

ND

=CWN

D+1M

SSTransm

it new segm

ents as allowed

DupA

ckCount==3SSThresh=CW

ND

/2Cw

nd=ssthresh+3MSS

Retransmit m

issing segm

ent

New

Ack

CWN

D=CW

ND

+MSS*M

SS/CWN

DD

upAckCount=0

Transmit new

segment as allow

ed

IdleIdle

Idle

Note: The state transition

diagram in the textbook

does not show Idle state

CWN

D>SSThresh

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TCP A

verage ThroughputA

verage Throughput =

Here, P = Probability of Packet loss

Note 1: The form

ula is an approximation w

hich does not apply at P=0 or P=1. A

t P=1, the throughput is zero. At P=0, the

throughput is min{1, (Receiver W

indow/RTT)}

Note 1: The textbook uses L for probability of packet loss but it

was used earlier for length of packets.

1.22 MSS

RTT P

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Explicit Congestion N

otification (ECN

)Explicit congestion notification (ECN

) is based on our DECbit

research. Two bits in IP H

eader:00: Transport is not capable of ECN

(e.g., UD

P)01: Transport is capable of ECN10: Transport is capable of ECN11: Congestion experienced (CE)

When a router encounters congestion, instead of dropping the

datagram, it m

arks the two bits as “11” congestion experienced

Transport

Netw

orkD

atalink

Transport

Netw

orkD

atalinkECN

11

ECE1

CWR

1

Application

Application

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ECN

(Cont)

On receiving “CE” code point, the receiver sends “ECN

Echo (ECE)” flag in the TCP headerO

n seeing the ECE flag, the source reduces its congestion w

indow, and sets “Congestion W

indow Reduced (CW

R) flag in outgoing segm

entO

n receiving “CWR” flag, the receiver, stops setting ECE bit

Transport

Netw

orkD

atalink

Transport

Netw

orkD

atalinkECN

11

ECE1

CWR

1

Application

Application

Ref: https://en.wikipedia.org/w

iki/Explicit_Congestion_Notification

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TCP: Sum

mary

1.TCP uses port num

bers for multiplexing

2.TCP provides reliable full-duplex connections.

3.TCP is stream

based and has window

flow control

4.Slow

-start congestion control works on tim

eout5.

Explicit congestion notification works using ECN

bits

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Hom

ework 3C

Consider Figure below. A

ssuming TCP Reno is the

protocol experiencing the behavior shown above, answ

er the follow

ing questions. In all cases, you should provide a short discussion justifying your answ

er.

TCP W

indow Size

0 5 10 15 20 25 30 35 40 4505

1015

2025

30Transm

ission Round

CWND

Round

CW

ND

11

22

34

48

516

632

733

834

935

1036

1137

1238

1339

1440

1541

1642

1721

1822

1923

2024

2125

2226

231

242

254

268

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Hom

ework 3C

(Cont)

A. Identify the interval of tim

e when TCP slow

start is operating.B. Identify the intervals of tim

e when TCP congestion

avoidance is operating.C. A

fter the 16thtransm

ission round, is segment loss detected

by a triple duplicate ACK

or by a timeout?

D. A

fter the 22ndtransm

ission round, is segment loss detected

by a triple duplicate ACK

or by a timeout?

E. What is the initial value of ssthresh at the first transm

ission round?F .W

hat is the value of ssthresh at the 18thtransm

ission round?G

. What is the value of ssthresh at the 24

thtransmission round?

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Hom

ework 3C

(Cont)

H. D

uring what transm

ission round is the 70thsegm

ent sent?I. A

ssuming a packet loss is detected after the 26

thround by the receipt of a triple duplicate A

CK, w

hat will be the values of the

congestion window

size and of ssthresh?J. Suppose TCP Tahoe is used (instead of TCP Reno), and assum

e that triple duplicate ACK

s are received at the 16th

round. What are the ssthresh and the congestion w

indow size at

the 19thround?

K. A

gain suppose TCP Tahoe is used, and there is a timeout

event at the end of 22ndround. H

ow m

any packets have been sent out from

17thround till 22

ndround, inclusive?

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Summ

ary

1.M

ultiplexing/demultiplexing by a com

bination of source and destination IP addresses and port num

bers.2.

Longer distance or higher speed Larger

Larger window

is better3.

Window

flow control is better for long-distance or high-speed

networks

4.U

DP is connectionless and sim

ple. N

o flow/error control. H

as error detection.5.

TCP provides full-duplex connections with

flow/error/congestion control.

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Lab 3B: TCP

In Wireshark, open the trace file tcp-ethereal-trace-1 in

http://gaia.cs.umass.edu/w

ireshark-labs/wireshark-traces.zip

and answ

er the following:

1. What is the IP address and TCP port num

ber used by the client com

puter (source) to transfer the file to gaia.cs.umass.edu?

2. What is the IP address and port num

ber used by gaia.cs.umass.edu to

receive the file.3. W

hat is the sequence number of the TCP SY

N segm

ent that is used to initiate the TCP connection betw

een the client computer and

gaia.cs.umass.edu? W

hat is it in the segment that identifies the

segment as a SY

N segm

ent?4. W

hat is the sequence number of the SY

N A

ck segment sent by

gaia.cs.umass.edu to the client com

puter in reply to the SYN

? What

is the value of the ACK

nowledgem

ent field in the SYN

Ack

segment? H

ow did gaia.cs.um

ass.edu determine that value? W

hat is it in the segm

ent that identifies the segment as a SY

N A

ck segment?

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Lab 3B: TCP (C

ont)5. W

hat is the sequence number of the TCP segm

ent containing the H

TTP POST com

mand? N

ote that in order to find the POST

comm

and, you’ll need to dig into the packet content field at the bottom

of the Wireshark w

indow, looking for a segm

ent with a

“POST” w

ithin its DA

TA field.

6. Consider the TCP segment containing the H

TTP POST as the first

segment in the TCP connection. W

hat are the sequence numbers of

the first six segments in the TCP connection (including the segm

ent containing the H

TTP POST)? A

t what tim

e was each segm

ent sent? W

hen was the A

CK for each segm

ent received? Given the difference

between w

hen each TCP segment w

as sent, and when its

acknowledgem

ent was received, w

hat is the RTT value for each of the six segm

ents?

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Lab 3B: TCP (C

ont)7. W

hat is the length of each of the first six TCP segments?

8. What is the m

inimum

amount of available buffer space advertised at

the received for the entire trace? Does the lack of receiver buffer

space ever throttle the sender?9. A

re there any retransmitted segm

ents in the trace file? What did you

check for (in the trace) in order to answer this question?

10. How

much data does the receiver typically acknow

ledge in an A

CK? Can you identify cases w

here the receiver is ACK

ing every other received segm

ent (see Table 3.2 on page 247 in the text).11. W

hat is the throughput (bytes transferred per unit time) for the TCP

connection? Explain how you calculated this value.

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Optional H

omew

ork 3DTry but do not subm

it.A

TCP entity opens a connection and uses slow start.

Approxim

ately how m

any round-trip times are

required before TCP can send N segm

ents.

CWN

D=1

CWN

D=2

CWN

D=4

Hint:

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Acronym

sA

CKA

CKnow

ledgement

AIM

DA

dditive increase and multiplicative decrease

ARQ

Autom

atic Repeat RequestCE

Congestion ExperiencedCRC

Cyclic Redundancy Check CW

ND

Congestion Window

CWR

Congestion Window

Reduced D

AD

estination Address

DEC

Digital Equipm

ent CorporationD

ECbitD

EC's bit based congestion scheme

DevRTT

Deviation of RTT

DN

SD

omain N

ame System

DP

Destination Port

ECEExplicit Congestion Experienced

ECNExplicit Congestion N

otificationFIN

Final

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Acronym

s (Cont)

FTPFile Transfer Protocol

GBN

Go-Back N

HTTP

Hyper-Text Transfer Protocol

IETFInternet Engineering Task Force

IPInternet Protocol

ISNInitial Sequence N

umber

kBK

ilo-ByteM

SSM

aximum

segment size

PBXPrivate Branch Exchange

PSHPush

RFCRequest for Com

ments

RMResource M

anagement

RSTReset

RTTRound-Trip Tim

eSA

Source Address

SACK

Selective Acknolow

ledgement

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Acronym

s (Cont)

SMTP

Simple M

ail Transfer ProtocolSP

Source PortSSThresh

Slow Start Threshold

SYN

SynchronizationSY

NA

CKSY

N A

cknowledgem

entTCP

Transmission Control Protocol

UD

PU

ser Datagram

ProtocolU

RGU

rgentV

CIV

irtual Circuit Identifiers

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Scan This to Dow

nload These Slides

Raj Jainhttp://rajjain.com

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Related M

odules

Audio/Video Recordings and Podcasts of

Professor Raj Jain's Lectures, https://w

ww.youtube.com

/channel/UCN

4-5wzN

P9-ruOzQ

Ms-8N

Uw

CSE473S: Introduction to Computer N

etworks (Fall 2016),

http://ww

w.cse.wustl.edu/~jain/cse473-16/index.htm

l

Wireless and M

obile Netw

orking (Spring 2016), http://w

ww.cse.w

ustl.edu/~jain/cse574-16/index.html

CSE571S: Netw

ork Security (Fall 2014), http://w

ww.cse.w

ustl.edu/~jain/cse571-14/index.html

CSE 473s: Introduction to Computer N

etworks

(Course Overview

), http://w

ww.cse.w

ustl.edu/~jain/cse473-16/ftp/i_0int.pdf

Documents

Transport Layer: TCP and UDP - Washington University in …jain/cse473-16/ftp/i_3tcp4.pdf · · 2016-10-03Lab 3A: UDP T Download the wireshark traces from ... 5. Examine the first