EE 122: Midterm Review

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EE 122: Midterm ReviewIon Stoica

TAs: Junda Liu, DK Moon, David Zatshttp://inst.eecs.berkeley.edu/~ee122/fa09

(Materials with thanks to Vern Paxson, Jennifer Rexford,and colleagues at UC Berkeley)

http://inst.eecs.berkeley.edu/~ee122/fa09

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Announcements Midterm Information

Date: 19 October 2008 Time: 4:00 PM to 5:30 PM Closed book, open 8.5” x 11” crib sheet (both sides) No Blue Books; all answers on exam sheets we hand

out No calculators, PDAs, cell phones with cameras, etc. Please use PENCIL and bring ERASER

Ion, one additional office hour on Monday: 1-3pm

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Overview

Layering and e2e Argument Little Theorem Packet delays IP Forwarding and Addressing Stop-and-Wait and Sliding Window Bit encoding CSMA/CD & Ethernet & Ethernet

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Layering: The Problem

Re-implement every application for every technology?

No! But how does the Internet architecture avoid this?

Telnet FTP NFS

Packetradio

Coaxial cable

Fiberoptic

Application

TransmissionMedia

HTTP

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Layering: Solution Introduce an intermediate layer that provides a single

abstraction for various network technologies New application just need to be written for intermediate layer New transmission media just need to provide abstraction of

intermediate layer

SMTP SSH NFS

Packetradio

Coaxial cable

Fiberoptic

Application

TransmissionMedia

HTTP

Intermediate layer

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LayeringLayering is a particular form of modularization

System is broken into a vertical hierarchy of logically distinct entities (layers)

Service provided by one layer is based solely on the service provided by layer below

Rigid structure: easy reuse, performance suffers

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Layering: InternetUniversal Internet layer: Internet has only IP at the Internet layerMany options for modules above IPMany options for modules below IP

Internet

Net access/Physical

Transport

Application

IP

LAN Packetradio

TCP UDP

Telnet FTP DNS

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Hourglass

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Implications of Hourglass

Single Internet layer module:Allows networks to interoperate

Any network technology that supports IP can exchange packets

Allows applications to function on all networksApplications that can run on IP can use any network

Simultaneous developments above and below IP

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E2E Arguments: Where to Place Functionality?

Most influential paper about placing functionality is “End-to-End Arguments in System Design” by Saltzer, Reed, and Clark

“Sacred Text” of the InternetEndless disputes about what it meansEveryone cites it as supporting their position

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E2E Arguments: Moderate Interpretation

Think twice before implementing functionality in the network

If hosts can implement functionality correctly, implement it a lower layer only as a performance enhancement

But do so only if it does not impose burden on applications that do not require that functionality

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Overview

Layering and e2e Argument Little Theorem Packet delays IP Forwarding and Addressing Stop-and-Wait and Sliding Window Bit encoding CSMA/CD & Ethernet

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Little’s Theorem Assume a system (e.g., router, network, checkout line

in a supermarket) at which packets arrive at rate a(t) Let d(i) be the delay or service time of packet i , i.e.,

time packet i spends in the system What is the average number of packets in the

system?

systema(t) – arrival rate

d(i) = delay of packet i

Intuition: Assume arrival rate is a = 1 packet per second and the delay of

each packet is s = 4 seconds What is the average number of packets in the system?

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Example

Arrival rate = 1; delay = 4

Time = 0

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Example


Time = 1

delay = 1

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Example


Time = 2

delay = 1

delay = 2

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Example


Time = 3

delay = 2

delay = 3

delay = 1

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Example


Time = 4

delay = 3

delay = 4

delay = 2

delay = 1

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Example


Time = 4

delay = 3

delay = 2

delay = 1

Q: What is the average number of packets in system?

A: number_of_packets_in_system = avg_arrival_rate x avg_delay

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Overview

Layering and e2e Argument Little Theorem Packet Delays IP Forwarding and Addressing Stop-and-Wait and Sliding Window Bit encoding CSMA/CD & Ethernet

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Definitions Link bandwidth (capacity): maximum rate (in bps) at

which the sender can send data along the link Propagation delay: time it takes the signal to travel from

source to destination Packet transmission time: time it takes the sender to

transmit all bits of the packet Queuing delay: time the packet need to wait before being

transmitted because the queue was not empty when it arrived

Processing Time: time it takes a router/switch to process the packet header, manage memory, etc

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Sending One PacketR bits per second (bps)

T seconds

P bits

Bandwidth: R bpsPropagation delay: T sec

time

Transmission time = P/RT

Propagation delay =T = Length/speed

1m/speed = 3.3 usec in free space 4 usec in copper 5 usec in fiber

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Queueing The queue has Q bits when packet arrives packet

has to wait for the queue to drain before being transmitted

P bits

time

P/RT

Q bits

Queueing delay = Q/R

Capacity = R bpsPropagation delay = T sec

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Packet 1

Packet 1

Store & Forward

Packet 1

Queuing & processing

delay of Packet 1 at Node 2

Host 1 Host 2Node 1 Node 2

propagationdelay betweenHost 1 and Node 1

transmission time of Packet 1at Host 1

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Store & Forward: Various Capacities Example

A packet is stored (enqueued) before being forwarded (sent)

Sender Receiver

10 Mbps 5 Mbps 100 Mbps 10 Mbps

time

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Store & Forward: Multiple Packet Example

Sender Receiver

10 Mbps 5 Mbps 100 Mbps 10 Mbps

time

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Overview

Layering and e2e Argument Little Theorem Packet Delays IP Forwarding and Addressing Stop-and-Wait and Sliding Window Bit encoding

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Packet Forwarding Store a mapping between IP addresses and output

interfaces Forward an incoming packet based on its destination address

……

31.2.3.6 11.2.3.5

1

21.2.3.5

1.2.3.4

1.2.3.4 2

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Scalability Challenge Suppose hosts had arbitrary addresses

Then every router would need a lot of information …to know how to direct packets toward the host

host host host

LAN 1

... host host host

LAN 2

...

router router routerWAN WAN

1.2.3.4 5.6.7.8 2.4.6.8 1.2.3.5 5.6.7.9 2.4.6.9

1.2.3.41.2.3.5

forwarding table

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Solution: Hierarchical Addressing (IP Prefixes) Divided into network (left) & host portions (right) 12.34.158.0/24 is a 24-bit prefix with 29 addresses

Terminology: “Slash 24”

00001100 00100010 10011110 00000101

Network (24 bits) Host (8 bits)

12 34 158 5

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Scalability Improved Number related hosts from a common subnet

1.2.3.0/24 on the left LAN 5.6.7.0/24 on the right LAN

host host host

LAN 1

... host host host

LAN 2

...


1.2.3.4 1.2.3.7 1.2.3.156 5.6.7.8 5.6.7.9 5.6.7.212

1.2.3.0/245.6.7.0/24

forwarding table

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Easy to Add New Hosts No need to update the routers

E.g., adding a new host 5.6.7.213 on the right Doesn’t require adding a new forwarding entry

host host host

LAN 1

... host host host

LAN 2

...


1.2.3.4 1.2.3.7 1.2.3.156 5.6.7.8 5.6.7.9 5.6.7.212

1.2.3.0/245.6.7.0/24

forwarding table

host

5.6.7.213

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Classful Addressing Class A: if first byte in [0..127], assume /8 (top bit = 0)

Very large blocks (e.g., MIT has 18.0.0.0/8) Class B: first byte in [128..191] assume /16 (top bits = 10)

Large blocks (e.g,. UCB has* 128.32.0.0/16) Class C: [192..223] assume /24 (top bits = 110)

Small blocks (e.g., ICIR has 192.150.187.0/24) The “swamp” (many European networks, due to history)

0******* ******** ******** ********

10****** ******** ******** ********

110***** ******** ******** ********

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Classful Addressing (cont’d) Class D: [224..239] (top bits 1110)

Multicast groups Class E: [240..255] (top bits 11110)

Reserved for future use

What problems can classful addressing lead to? Only comes in 3 sizes Routers can end up knowing about a lot of class C’s

1110**** ******** ******** ********

11110*** ******** ******** ********

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Classless Inter-Domain Routing (CIDR)

IP Address : 12.4.0.0 IP Mask: 255.254.0.0

00001100 00000100 00000000 00000000

11111111 11111110 00000000 00000000

Address

Mask

for hosts Network Prefix

Use arbitrary length prefixesUse two 32-bit numbers to represent a network.

Network number = IP address + Mask

Written as 12.4.0.0/15 or 12.4/15

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Scalability: Address Aggregation

Provider is given 201.10.0.0/21 (201.10.0.x .. 201.10.7.x)

201.10.0.0/22 201.10.4.0/24 201.10.5.0/24 201.10.6.0/23

Provider

Routers in the rest of the Internet just need to know how to reach 201.10.0.0/21. The provider can direct the

IP packets to the appropriate customer.

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5 Minute Break

Questions Before We Proceed?

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Overview


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Automatic Repeat reQuest (ARQ)

Time

Packet

ACKTim

eout

Automatic Repeat Request Receiver sends

acknowledgment (ACK) when it receives packet

Sender waits for ACK and times out if does not arrive within some time period

Simplest ARQ protocol Stop and Wait Send a packet, stop and wait

until ACK arrives

Sender Receiver

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How Fast Can Stop-and-Wait Go? Suppose we’re sending from UCB to New York:

Bandwidth = 1 Mbps (megabits/sec) RTT = 100 msec Maximum Transmission Unit (MTU) = 1500 B = 12,000 b No other load on the path and no packet loss

What (approximately) is the fastest we can transmit using Stop-and-Wait? Answer: 12,000b/0.1s = 120 kbps

How about if Bandwidth = 1 Gbps?

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Allowing Multiple Packets in Flight

“In Flight” = “Unacknowledged” Sender-side issue: how many packets (bytes)? Receiver-side issue: how much buffer for data

that’s “above a sequence hole”? I.e., data that can’t be delivered since previous data is

missing

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Sliding Window Example (This is NOT TCP !) Sender

Sending rate = 1 pkt/s Receiver:

Delivering rate = 0.5 pkt/s Delivers packets in sequence to application Acknowledges (acks) each delivered pkt Send negative ack. (nack) if packet lost

Round-trip time = 4 sec Receiver Window = 4 packets Note: max. achievable throughput = 0.5pkt/s

Sliding Window ExampleSender Receiver1 1 1s

2s3s

4s5s6s7s8s

9s10s11s12s13s14s

Sender, at 1s Send 1st pkt

Sliding Window ExampleSender Receiver1

1

1 1s2s3s

4s5s6s7s8s

9s10s11s12s13s14s

Sender, at 1s Send 1st pkt

Receiver, at 3s Get 1st pkt Deliver 1st pkt

to appl. Send ack=1 to

sender

ack=1

Sliding Window Example

Sender, at 2s Send 2nd pkt,

which is lost

Sender Receiver1 1

ack=1

1s2s3s

4s5s6s7s8s

9s10s11s12s13s14s

2 1 2

Sliding Window Example1s2s3s

4s5s6s7s8s

9s10s11s12s13s14s

Sender Receiver1

2 13 2 1

3 2

123 ack=1

nack=2

Sender, at 3s Send 3nd pkt

Receiver, at 5s: Get 3rd pkt;

doesn’t deliver it since out of seq.

Send nack=2 (request 2nd pkt)


4s5s6s7s8s

9s10s11s12s13s14s

Sender Receiver1

2 13 2 1

3 2

123 ack=1

nack=2

Sender, at 4s Send 4th pkt Receiver

window full!

Receiver, at 6s Get 4th packet

4 3 2 1 4

4 3 2


4s5s6s7s8s

9s10s11s12s13s14s

Sender Receiver1

2 13 2 1

3 2

123 ack=1

nack=2

Sender, at 5s Get ack=1 Remove 1st pkt

from buffer Send 5th pkt;

now 2, 3, 4, 5 are in flight (window full!)

Receiver, at 7s Get 5th pkt

4 3 2 1 4

4 3 25 4 3 2 5

4 35 2


4s5s6s7s8s

9s10s11s12s13s14s

Sender Receiver1

2 13 2 1

3 2

123 ack=1

nack=2

Sender, at 7s Get nack=2 Resend pkt 2

Receiver, at 9s Get 2nd pkt Deliver it to

appl. Send ack=2

4 3 2 1 4

4 3 25 4 3 2 5

4 35

ack=2

25 4 3 2 2

4 35 2


4s5s6s7s8s

9s10s11s12s13s14s

Sender Receiver1

2 13 2 1

3 2

123 ack=1

nack=2

Sender, at 11s Get ack=2 Send pkt 6;

pkts 3, 4, 5, 6 are in-flight

Receiver, at 11s Deliver 3d pkt

to appl. (recall, delivery rate is 1pkt every 2s)

Send ack=3

4 3 2 1 4

4 3 25 4 3 2 5

4 35

ack=2

25 4 3 2 2

4 35 2

6 5 4 3 4 35ack=36

Sliding Window Example

16s17s18s19s20s

If no more losses, throughput = 0.5pkt/sec

This is max throughput as receiver cannot deliver more than 0.5pkt/sec

6 5 4 3 4 35ack=3

7 6 5 4

11s12s13s14s15s

5 46

8 7 6 5 6 57

ack=4

9 8 7 6 5 46

4 35

ack=5

ack=6

6

7

8

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What If Sending & Delivey Rates are 2 pkt/s?

Throughput limited by window size = Window/RTT = 4pkt/4 = 1 pkt/s

Sender Receiver1 1,2 1s

2s3s

4s5s6s7s8s

9s10s

2

121234 3,4

34ack=1, ack=2

3456 ack=3, ack=4

5,6

5678

5678

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Performance with Sliding Window

Given previous UCB New York 1 Mbps path with 100 msec RTTand Sender (and Receiver) window = 100 Kb = 12.5 KB

• How fast can we transmit?• Answer: min(100Kb/0.1s, 1Mbps) = 1 Mbps

• What about with 12.5 KB window & 1 Gbps path?• Window required to fully utilize path:

• Bandwidth-delay product (or “delay-bandwidth product”)• 1 Gbps * 100 msec = 100 Mb = 12.5 MB• Note: large window = many packets in flight

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Overview


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Encoding Signals propagate over physical links

How do we represent the bits? Physical layer issue

Simplify some electrical engineering details Assume two discrete signals, high and low E.g., could correspond to two different voltages

Basic approach High for a 1, low for a 0 How hard can it be?

Sender & receiver agree: what’s “high”, what’s “low” And: when to read the signal

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Non-Return to Zero (NRZ) 1 high signal; 0 low signal

(Actual signals are of course not so sharp) How does receiver know where one bit stops and another

begins? 0 0 1 0 1 0 1 1 0

NRZ(non-return to zero)

Clock

Receiver reads the signal on the clock’s leading edge

Sender begins to transmit signal on clock’s falling edge

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Clock Coordination How do the sender and receiver agree on the running

of the clock? Problem: without explicit synchronization, receiver’s

clock can drift with respect to sender’s0 0 1 0 1 0 1 1 0

NRZ

Clock

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Clock Recovery

To avoid clock drift, we use the signal itself to coordinate

Whenever see a transition (0 1 or 1 0) we know that corresponds to sender clock’s trailing edge So pull our clock in phase towards it

Problem with NRZ: long strings of 0s or 1s Quite common No transitions from low-to-high, or high-to-low Clock recovery fails and receiver’s clock begins to drift

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Non-Return to Zero Inverted (NRZI) 1 make transition; 0 stay at same level Fixes previous problem for long sequences of 1’s But not for 0’s

0 0 1 0 1 0 1 1 0

Clock(read on

falling edge)

NRZI(non-return to zero

inverted)

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Manchester Encoding 1 high-to-low transition; 0 low-to-high

transition Addresses clock recovery problems But: physical signaling must be twice as fast

To support 2 transitions/cycle Efficiency of 50%

0 0 1 0 1 0 1 1 0

Clock(read on each

rising edge)

Manchester

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4-bit/5-bit (100Mb/s Ethernet) Goal: address inefficiency of Manchester encoding, while

avoiding long periods of no transition Solution:

Use 5 bits to encode every sequence of four bits such that No 5 bit code has more than one leading 0 or two trailing 0’s

Use NRZI to then encode the 5 bit codes Efficiency is 80%

0000 111100001 010010010 101000011 101010100 010100101 010110110 011100111 01111

1000 100101001 100111010 101101011 101111100 110101101 110111110 111001111 11101

4-bit 5-bit 4-bit 5-bit

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Overview


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Ethernet: CSMA/CD Protocol Carrier sense: wait for link to be idle Collision detection: listen while transmitting

No collision: transmission is complete Collision: abort transmission & send jam signal

Random access: exponential back-off After collision, wait a random time before trying again After mth collision, choose K randomly from {0, …, 2m-

1} … and wait for K*512 bit times before trying again

The wired LAN technology Hugely successful: 3/10/100/1000/10000 Mbps

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Minimum Packet Sizepropagation delay (d)a) Time = t; Host 1

starts to send frame

Host 1 Host 2

propagation delay (d)Host 1 Host 2

b) Time = t + d; Host 2 starts to send a frame, just before it hears fromhost 1’s frame

propagation delay (d)Host 1 Host 2c) Time = t + 2*d; Host 1

hears Host 2’s frame detects collision

2*d < min_frame_size/bandwidth 2*(max_length/light_speed) < min_frame_size/bandwidth max_length < min_frame_size*light_speed/(2*bandwidth)

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Minimum Packet Sizepropagation delay (d)a) Time = t; Host 1

starts to send frame

Host 1 Host 2

propagation delay (d)Host 1 Host 2

b) Time = t + d; Host 2 starts to send a frame, just before it hears fromhost 1’s frame

propagation delay (d)Host 1 Host 2c) Time = t + 2*d; Host 1

hears Host 2’s frame detects collision

min_frame_size = 512b; light_speed = 2.5*108mps; bandwidth = 10Mbps max_length = (min_frame_size)*(light_speed)/(2*bandwidth) = = (8*64b)*(2.5*108mps)/(2*107 bps) = 6400m approx

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Midterm Information Date: 19 October 2008 Time: 4:00 PM to 5:30 PM Closed book, open 8.5” x 11” crib sheet (both sides) No Blue Books; all answers on exam sheets we hand out No calculators, PDAs, cell phones with cameras, etc. Please use PENCIL and bring ERASER

Ion, one additional office hour on Monday: 1-3pm

Documents

EE 122: Midterm Review