EE 122: Midterm Review - University of California, Berkeleyinst.eecs.berkeley.edu/.../fa09/notes/14-MidtermReviewx4.pdf4 13 Little’s Theorem Assume a system (e.g., router, network,

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EE 122: Midterm Review Ion Stoica

TAs: Junda Liu, DK Moon, David Zats

http://inst.eecs.berkeley.edu/~ee122/fa09 (Materials with thanks to Vern Paxson, Jennifer Rexford,

and colleagues at UC Berkeley) 2

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

Packet radio

Coaxial cable

Fiber optic

Application

Transmission Media

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

Packet radio

Coaxial cable

Fiber optic

Application

Transmission Media

HTTP

Intermediate layer

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Layering

  Layering 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: Internet Universal Internet layer:   Internet has only IP at the Internet layer   Many options for modules above IP   Many options for modules below IP

Internet

Net access/ Physical

Transport

Application

IP

LAN Packet radio

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 networks  Applications 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 Internet  Endless disputes about what it means  Everyone 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?

system a(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

5

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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 20

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 Packet R bits per second (bps)

T seconds

P bits

Bandwidth: R bps Propagation delay: T sec

time

Transmission time = P/R T

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/R T

Q bits

Queueing delay = Q/R

Capacity = R bps Propagation 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 2 Node 1 Node 2

propagation delay between Host 1 and Node 1

transmission time of Packet 1 at 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 26

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

… …

3 1.2.3.6 1 1.2.3.5

1

2 1.2.3.5

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! router!WAN! 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

1.2.3.4

1.2.3.5

forwarding table! 30

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!


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

1.2.3.0/24

5.6.7.0/24

forwarding table!32

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!


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

1.2.3.0/24

5.6.7.0/24

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 prefixes Use two 32-bit numbers to represent a network.

Network number = IP address + Mask

Written as 12.4.0.0/15 or 12.4/15 36

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

ACK Tim

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

Sliding Window Example (This is NOT TCP !)   Sender

  Sending rate = 1 pkt/s   Receiver:

  Delivering rate = 0.5 pkt/s   Delivers packets in 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

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Sliding Window Example Sender Receiver 1 1 1s

2s 3s

4s 5s 6s 7s 8s

9s 10s 11s 12s 13s 14s

  Sender, at 1s   Send 1st pkt

Sliding Window Example Sender Receiver 1

1

1 1s 2s 3s

4s 5s 6s 7s 8s

9s 10s 11s 12s 13s 14s

  Sender, at 1s   Send 1st pkt

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

to appl.   Send ack=1 to

sender

ack=1

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Sliding Window Example

  Sender, at 2s   Send 2nd pkt,

which is lost

Sender Receiver 1 1

ack=1

1s 2s 3s

4s 5s 6s 7s 8s

9s 10s 11s 12s 13s 14s

2 1 2

Sliding Window Example 1s 2s 3s

4s 5s 6s 7s 8s

9s 10s 11s 12s 13s 14s

Sender Receiver 1

2 1 3 2 1

3

1 2 3 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)


4s 5s 6s 7s 8s

9s 10s 11s 12s 13s 14s

Sender Receiver 1

2 1 3 2 1

3

1 2 3 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


4s 5s 6s 7s 8s

9s 10s 11s 12s 13s 14s

Sender Receiver 1

2 1 3 2 1

3

1 2 3 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 5 4 3 2 5

4 3 5

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4s 5s 6s 7s 8s

9s 10s 11s 12s 13s 14s

Sender Receiver 1

2 1 3 2 1

3

1 2 3 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 5 4 3 2 5

4 3 5

ack=2

5 4 3 2 2

4 3 5 2


4s 5s 6s 7s 8s

9s 10s 11s 12s 13s 14s

Sender Receiver 1

2 1 3 2 1

3

1 2 3 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 5 4 3 2 5

4 3 5

ack=2

5 4 3 2 2

4 3 5 2

6 5 4 3 4 3 5 ack=3 6

Sliding Window Example

16s 17s 18s 19s 20s

  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 3 5 ack=3

7 6 5 4

11s 12s 13s 14s 15s

5 4 6

8 7 6 5 6 5 7

ack=4

9 8 7 6 5 4 6

4 3 5

ack=5

ack=6

6

7

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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 Receiver 1 1,2 1s

2s 3s

4s 5s 6s 7s 8s

9s 10s

2

1 2 1 2 3 4 3,4

3 4 ack=1, ack=2

3 4 5 6 ack=3, ack=4

5,6

5 6 7 8

5 6 7 8

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Performance with Sliding Window

  Given previous UCB ↔ New York 1 Mbps path with 100 msec RTT and 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’s 0 0 1 0 1 0 1 1 0

NRZ

Clock 58

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 11110 0001 01001 0010 10100 0011 10101 0100 01010 0101 01011 0110 01110 0111 01111

1000 10010 1001 10011 1010 10110 1011 10111 1100 11010 1101 11011 1110 11100 1111 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 64

Minimum Packet Size propagation 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 from host 1’s frame

propagation delay (d) Host 1 Host 2 c) 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 Size propagation 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 from host 1’s frame

propagation delay (d) Host 1 Host 2 c) 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 66

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

EE 122: Midterm Review - University of California, Berkeleyinst.eecs.berkeley.edu/.../fa09/notes/14-MidtermReviewx4.pdf4 13 Little’s Theorem Assume a system (e.g., router, network,

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