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CSE 461: Computer Networks Ratul Mahajan – ratul@cs Anirban Biswas – anirban@uw Michael Chenyang Fang – chenyf@uw Qian Yan – qiany7@uw

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Page 1: CSE 461: Computer Networks - courses.cs.washington.edu · Fundamentals – Channel throughput 1 Gbps 20ft container = 2,350,080 in3 (240 x 96 x 102) 3.5in SSD = 23 in3 (4 x 5.75 x

CSE461:ComputerNetworks

RatulMahajan– ratul@csAnirban Biswas– anirban@uw

MichaelChenyang Fang– chenyf@uwQianYan– qiany7@uw

Page 2: CSE 461: Computer Networks - courses.cs.washington.edu · Fundamentals – Channel throughput 1 Gbps 20ft container = 2,350,080 in3 (240 x 96 x 102) 3.5in SSD = 23 in3 (4 x 5.75 x

Whoweare

Page 3: CSE 461: Computer Networks - courses.cs.washington.edu · Fundamentals – Channel throughput 1 Gbps 20ft container = 2,350,080 in3 (240 x 96 x 102) 3.5in SSD = 23 in3 (4 x 5.75 x

Ratul

Page 4: CSE 461: Computer Networks - courses.cs.washington.edu · Fundamentals – Channel throughput 1 Gbps 20ft container = 2,350,080 in3 (240 x 96 x 102) 3.5in SSD = 23 in3 (4 x 5.75 x

Hieveryone,I’mAnir!(Anirban)

• I’minthe5th yearmasterprogram

• Ilovelowlevelsystemsprogramminganddistributedsystems

• Thisismy2nd quarterasaTA

• InternedatCDKGlobal,Facbeook andVMwareinthepast

• CurrentlydoingresearchintheSystemsLab

Page 5: CSE 461: Computer Networks - courses.cs.washington.edu · Fundamentals – Channel throughput 1 Gbps 20ft container = 2,350,080 in3 (240 x 96 x 102) 3.5in SSD = 23 in3 (4 x 5.75 x

ThisisMichael(Chenyang)

Q:WhatdoIlike?A:ScubaDiving+music

Q:Whatkindofmusicdoyoulike?A:Britishones,likethosemadebyTheStoneRoses

Q:Whatsongdoyoulistentorecently?A:StepintoMyWorldbyHurricane#1

Q:Anyadviceforthiscourse?A:Thiscourseissomuchfun.Justgotolectureandsectionanddon’thesitateaskingquestions.Therearealotofthingsgoingtobecoveredinthiscourseandalotnotgoingtobecoveredinthiscourse.IfyouareveryinterestedinNetworking,thebook(verygood)isyourbestfriend.

Myfavoritecartoon

MyFavoriteGuitarist

JohnSquire

WhatIamreading:

FarFromtheMaddingCrowd

Page 6: CSE 461: Computer Networks - courses.cs.washington.edu · Fundamentals – Channel throughput 1 Gbps 20ft container = 2,350,080 in3 (240 x 96 x 102) 3.5in SSD = 23 in3 (4 x 5.75 x

Qian(Will)Yan

• 5th YearMaster

• IskiALOT.

ComputerNetworks 6

Page 7: CSE 461: Computer Networks - courses.cs.washington.edu · Fundamentals – Channel throughput 1 Gbps 20ft container = 2,350,080 in3 (240 x 96 x 102) 3.5in SSD = 23 in3 (4 x 5.75 x

ClassStructure

Page 8: CSE 461: Computer Networks - courses.cs.washington.edu · Fundamentals – Channel throughput 1 Gbps 20ft container = 2,350,080 in3 (240 x 96 x 102) 3.5in SSD = 23 in3 (4 x 5.75 x

Grading

Assignments:10%• Readingandhomeworkfromthebook

Page 9: CSE 461: Computer Networks - courses.cs.washington.edu · Fundamentals – Channel throughput 1 Gbps 20ft container = 2,350,080 in3 (240 x 96 x 102) 3.5in SSD = 23 in3 (4 x 5.75 x

Grading

Assignments:10%SurpriseQuizzes:5%• Shortunannouncedquizzesduringthequarter• Droplowest

Page 10: CSE 461: Computer Networks - courses.cs.washington.edu · Fundamentals – Channel throughput 1 Gbps 20ft container = 2,350,080 in3 (240 x 96 x 102) 3.5in SSD = 23 in3 (4 x 5.75 x

Grading

Assignments:10%SurpriseQuizzes:10%3Projects:(15+15+15)%• 3codingexercises:

• Socketprogramming• LinkandNetworklayerbehavior• HTTPProxy

Page 11: CSE 461: Computer Networks - courses.cs.washington.edu · Fundamentals – Channel throughput 1 Gbps 20ft container = 2,350,080 in3 (240 x 96 x 102) 3.5in SSD = 23 in3 (4 x 5.75 x

Grading

Assignments:10%SurpriseQuizzes:5%3Projects:(15+15+15)%Midterm:15%Final:25%

Page 12: CSE 461: Computer Networks - courses.cs.washington.edu · Fundamentals – Channel throughput 1 Gbps 20ft container = 2,350,080 in3 (240 x 96 x 102) 3.5in SSD = 23 in3 (4 x 5.75 x

Grading

Assignments:10%SurpriseQuizzes:5%3Projects:(15+15+15)%Midterm:15%Final:25%

LatePolicy:• 10%penaltyforeachlateday• Eachperson getsthreelatedays

Page 13: CSE 461: Computer Networks - courses.cs.washington.edu · Fundamentals – Channel throughput 1 Gbps 20ft container = 2,350,080 in3 (240 x 96 x 102) 3.5in SSD = 23 in3 (4 x 5.75 x

Administrivia

• Officehours• OpportunitytohavemorepersonalinteractionswithbothmeandtheTAs.

• Tools• Mailinglist:primaryclasscommunications• CanvasAssignments:Homeworkandprojects• CanvasDiscussion:Backandforthdiscussionsonclasscontent• CanvasGradebook:Gradeswillbepostedhere

• Slides• AdaptedfromKurtisHeimerl,whoadaptedfromDavidWetherall• Iwillbepostingmyownslidesonline

Page 14: CSE 461: Computer Networks - courses.cs.washington.edu · Fundamentals – Channel throughput 1 Gbps 20ft container = 2,350,080 in3 (240 x 96 x 102) 3.5in SSD = 23 in3 (4 x 5.75 x

Laptoppolicy

• Laptopsarefine• IfyouaregoingtobeonFacebookdoitinthebackofclass

• Thisisdistractingtootherstudents• TAswillbeenforcingthispolicy

Page 15: CSE 461: Computer Networks - courses.cs.washington.edu · Fundamentals – Channel throughput 1 Gbps 20ft container = 2,350,080 in3 (240 x 96 x 102) 3.5in SSD = 23 in3 (4 x 5.75 x

Questions?

Page 16: CSE 461: Computer Networks - courses.cs.washington.edu · Fundamentals – Channel throughput 1 Gbps 20ft container = 2,350,080 in3 (240 x 96 x 102) 3.5in SSD = 23 in3 (4 x 5.75 x

CSE461:ComputerNetworks

Page 17: CSE 461: Computer Networks - courses.cs.washington.edu · Fundamentals – Channel throughput 1 Gbps 20ft container = 2,350,080 in3 (240 x 96 x 102) 3.5in SSD = 23 in3 (4 x 5.75 x

Focusofthecourse

Page 18: CSE 461: Computer Networks - courses.cs.washington.edu · Fundamentals – Channel throughput 1 Gbps 20ft container = 2,350,080 in3 (240 x 96 x 102) 3.5in SSD = 23 in3 (4 x 5.75 x

Focusofthecourse(2)

Three“networking”topics:

Distributedsystems

Networking

Communications

CSE452

CSE461

EE417

Page 19: CSE 461: Computer Networks - courses.cs.washington.edu · Fundamentals – Channel throughput 1 Gbps 20ft container = 2,350,080 in3 (240 x 96 x 102) 3.5in SSD = 23 in3 (4 x 5.75 x

Maingoals

1. Learnthefundamentalsofcomputernetworks2. LearnhowtheInternetworks• Whatreallyhappenswhenyou“browsetheweb”?• TCP/IP,DNS,HTTP,NAT,VPNs,802.11etc.

3. UnderstandhowandwhyofInternetdesign• SDN,LoadBalancers,Architectures

Page 20: CSE 461: Computer Networks - courses.cs.washington.edu · Fundamentals – Channel throughput 1 Gbps 20ft container = 2,350,080 in3 (240 x 96 x 102) 3.5in SSD = 23 in3 (4 x 5.75 x

Whylearnthefundamentals?

Intellectualinterest

Reinvention,broadapplicability• Non-Internetnetworks• ChangingInternet

Page 21: CSE 461: Computer Networks - courses.cs.washington.edu · Fundamentals – Channel throughput 1 Gbps 20ft container = 2,350,080 in3 (240 x 96 x 102) 3.5in SSD = 23 in3 (4 x 5.75 x

Fundamentals- Reliablecommunication

Page 22: CSE 461: Computer Networks - courses.cs.washington.edu · Fundamentals – Channel throughput 1 Gbps 20ft container = 2,350,080 in3 (240 x 96 x 102) 3.5in SSD = 23 in3 (4 x 5.75 x

Fundamentals– Channelthroughput

1Gbps20ftcontainer=2,350,080in3(240x96x102)3.5inSSD=23in3 (4x5.75x1)SSDs/container=50K(50%packingefficiency)Containercapacity=25PB(512GBperSSD)Containerspeed=100mphSEA<>NYCthroughput=~2000Gbps

Page 23: CSE 461: Computer Networks - courses.cs.washington.edu · Fundamentals – Channel throughput 1 Gbps 20ft container = 2,350,080 in3 (240 x 96 x 102) 3.5in SSD = 23 in3 (4 x 5.75 x

Fundamentals– Reinvention

• TheInternetisconstantlybeingre-invented!• GrowthovertimeandtechnologytrendsdriveupheavalsinInternetdesign

• Today’sInternetisdifferentfromyesterday’s• Andtomorrow’swillbedifferentagain• Butthefundamentalsremainthesame

Page 24: CSE 461: Computer Networks - courses.cs.washington.edu · Fundamentals – Channel throughput 1 Gbps 20ft container = 2,350,080 in3 (240 x 96 x 102) 3.5in SSD = 23 in3 (4 x 5.75 x

Internetgrowth

Page 25: CSE 461: Computer Networks - courses.cs.washington.edu · Fundamentals – Channel throughput 1 Gbps 20ft container = 2,350,080 in3 (240 x 96 x 102) 3.5in SSD = 23 in3 (4 x 5.75 x

Exampleupheavals

Change EnablingTechnologyEmergenceofWeb ContentDistributionNetworksPiracy Peer-to-peer filesharingInternetofThings IPv6Mobile Devices Wireless,HighbandwidthcellularCloudcomputing VirtualizationCryptocurrencies Blockchains…. ….

Page 26: CSE 461: Computer Networks - courses.cs.washington.edu · Fundamentals – Channel throughput 1 Gbps 20ft container = 2,350,080 in3 (240 x 96 x 102) 3.5in SSD = 23 in3 (4 x 5.75 x

Maingoals

1. Tolearnthefundamentalsofcomputernetworks2. LearnhowtheInternetworks• Whatreallyhappenswhenyou“browsetheweb”?• TCP/IP,DNS,HTTP,NAT,VPNs,802.11etc.

3. UnderstandhowandwhyofInternetdesign• SDN,LoadBalancers,Architectures

Page 27: CSE 461: Computer Networks - courses.cs.washington.edu · Fundamentals – Channel throughput 1 Gbps 20ft container = 2,350,080 in3 (240 x 96 x 102) 3.5in SSD = 23 in3 (4 x 5.75 x

Whocaresabouttheinternet?

1. Curiosity2. Impactonourworld3. Jobprospects!

Page 28: CSE 461: Computer Networks - courses.cs.washington.edu · Fundamentals – Channel throughput 1 Gbps 20ft container = 2,350,080 in3 (240 x 96 x 102) 3.5in SSD = 23 in3 (4 x 5.75 x

Fromthisexperimentalnetwork(~1970)…

(a)Dec.1969. (b)July1970. (c)March1971.

Page 29: CSE 461: Computer Networks - courses.cs.washington.edu · Fundamentals – Channel throughput 1 Gbps 20ft container = 2,350,080 in3 (240 x 96 x 102) 3.5in SSD = 23 in3 (4 x 5.75 x

Tothis…

Page 30: CSE 461: Computer Networks - courses.cs.washington.edu · Fundamentals – Channel throughput 1 Gbps 20ft container = 2,350,080 in3 (240 x 96 x 102) 3.5in SSD = 23 in3 (4 x 5.75 x

Tothis!(2011)

Page 31: CSE 461: Computer Networks - courses.cs.washington.edu · Fundamentals – Channel throughput 1 Gbps 20ft container = 2,350,080 in3 (240 x 96 x 102) 3.5in SSD = 23 in3 (4 x 5.75 x

Andthis(2015)!

• Aneverydayinstitutionusedatwork,home,andon-the-go• Visualizationcontainsmillionsofservers• Red=.com,Yellow=.org

• Networknowcontainsliterally3billionpeople!

Page 32: CSE 461: Computer Networks - courses.cs.washington.edu · Fundamentals – Channel throughput 1 Gbps 20ft container = 2,350,080 in3 (240 x 96 x 102) 3.5in SSD = 23 in3 (4 x 5.75 x

Internet– SocietalImpact

• Anenablerofsocietalchange• Easyaccesstoknowledge• Electroniccommerce• Personalrelationships• Privatecommunications

Page 33: CSE 461: Computer Networks - courses.cs.washington.edu · Fundamentals – Channel throughput 1 Gbps 20ft container = 2,350,080 in3 (240 x 96 x 102) 3.5in SSD = 23 in3 (4 x 5.75 x

Internet– Economicimpact

• Anengineofeconomicgrowth• Informationsources

• Andlotsofethicalquestions!• Onlinemarketplaces• Socialmedia/Crowdsourcing

Page 34: CSE 461: Computer Networks - courses.cs.washington.edu · Fundamentals – Channel throughput 1 Gbps 20ft container = 2,350,080 in3 (240 x 96 x 102) 3.5in SSD = 23 in3 (4 x 5.75 x

Maingoals

1. Tolearnthefundamentalsofcomputernetworks2. LearnhowtheInternetworks• Whatreallyhappenswhenyou“browsetheweb”?• TCP/IP,DNS,HTTP,NAT,VPNs,802.11etc.

3. UnderstandhowandwhyofInternetdesign• SDN,LoadBalancers,Architectures

Page 35: CSE 461: Computer Networks - courses.cs.washington.edu · Fundamentals – Channel throughput 1 Gbps 20ft container = 2,350,080 in3 (240 x 96 x 102) 3.5in SSD = 23 in3 (4 x 5.75 x

Architectures

Lotsofwaystobuildnetworkswithdifferenttradeoffs• Internet-- openaccess

• Flexibility++,Privacy++,Security--• Cellular-- identityfirst

• Flexibility--,Privacy--,Security++,

Page 36: CSE 461: Computer Networks - courses.cs.washington.edu · Fundamentals – Channel throughput 1 Gbps 20ft container = 2,350,080 in3 (240 x 96 x 102) 3.5in SSD = 23 in3 (4 x 5.75 x

Not aCourseGoal

TolearnITjobskills• Howtoconfigurespecificequipmentortechnologies

• e.g.,Ciscocertifications,• Technicalwhack-a-mole

• Butcoursematerialisrelevant,andweusehands-ontools• Hopefullyyou’llbeabletousethesetoolstobuildstuffattheendofclass

Page 37: CSE 461: Computer Networks - courses.cs.washington.edu · Fundamentals – Channel throughput 1 Gbps 20ft container = 2,350,080 in3 (240 x 96 x 102) 3.5in SSD = 23 in3 (4 x 5.75 x

Maingoals

1. Tolearnthefundamentalsofcomputernetworks2. LearnhowtheInternetworks• Whatreallyhappenswhenyou“browsetheweb”?• TCP/IP,DNS,HTTP,NAT,VPNs,802.11etc.

3. UnderstandhowandwhyofInternetdesign• SDN,LoadBalancers,Architectures