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7 2013 5 29

6 (5/15) : 2 / 41

7 : 3 / 41

(univariate analysis) (multivariate analysis) ( ) 4 / 41

: 5 / 41

: WIDE 2001 1570 6 / 41

ITS 3 (, ) source: google crisis response 7 / 41

( ) 8 / 41

:.Locky WiFi iphone/android App GPS/WiFi WiFi 9 / 41

GPS (Global Positioning System) 30 GPS 100m 2000 10m : 3 GPS GPS GPS 4 : 10 / 41

GPS 11 / 41

: 12 / 41

: ping : 2 TCP 13 / 41

= + + + RTT(round trip time) : : : 400ms : : 14 / 41

( ) 1500 bytes at 10Mbps: 1.2 msec 1500 bytes at 100Mbps: 120 usec 1500 bytes at 1Gbps: 12 usec : 200,000 km/s 100km round-trip: 1 msec 20,000km round-trip: 200 msec RTT LEO (Low-Earth Orbit): 200 msec GEO (Geostationary Orbit): 600 msec 15 / 41

: : 16 / 41

pinger project the Internet End-to-end Performance Measurement (IEPM) project by SLAC using ping to measure rtt and packet loss around the world http://www-iepm.slac.stanford.edu/pinger/ started in 1995 over 600 sites in over 125 countries 17 / 41

pinger project monitoring sites monitoring (red), beacon (blue), remote (green) sites beacon sites are monitored by all monitors from pinger web site 18 / 41

pinger project monitoring sites in east asia monitoring (red) and remote (green) sites from pinger web site 19 / 41

pinger packet loss packet loss observed from N. Ameria exponential improvement in 10 years from pinger web site 20 / 41

pinger minimum rtt minimum rtts observed from N. America gradual shift from satellite to fiber in S. Asia and Africa from pinger web site 21 / 41

(linear regression) (least square method): y 500 400 IPv6 response time (msec) 300 200 100 x v4/v6 rtts 9.28 + 1.03 * x 0 0 100 200 300 400 500 IPv4 response time (msec) 22 / 41

(least square method) f(x) = b 0 + b 1 x b 1 = xy n xȳ x 2 n( x) 2 b 0 = ȳ b 1 x x = 1 n n i=1 x i ȳ = 1 n n i=1 n n xy = x i y i x 2 = (x i ) 2 y i i=1 i=1 23 / 41

i e i = y i (b 0 + b 1 x i ) n ē = 1 e i = 1 (y i (b 0 + b 1 x i )) = ȳ b 0 b 1 x n n i i 0 b 0 = ȳ b 1 x b 0 b 1 e i = y i ȳ + b 1 x b 1 x i = (y i ȳ) b 1 (x i x) SSE n n SSE = e 2 i = [(y i ȳ) 2 2b 1 (y i ȳ)(x i x) + b 2 1(x i x) 2 ] i=1 i=1 SSE n = 1 n (y i ȳ) 2 1 n 2b 1 (y i ȳ)(x i x) + b 2 1 n 1 (x i x) 2 n n n i=1 i=1 i=1 = σ 2 y 2b 1 σ 2 xy + b 2 1σ 2 x SSE b 1 b 1 2 b 1 0 1 d(sse) = 2σxy 2 n db + 2b 1σx 2 = 0 1 b 1 = σ2 xy σ 2 x = xy n xȳ x 2 n( x) 2 24 / 41

(principal component analysis; PCA) ( ) 25 / 41

2 ( 1 ) 1 2 3 x2 y2 y1 x1 26 / 41

( ) X d Y d d P Y = P X cov(y ) ( ) P P 1 = P ) Y cov(y ) = E[Y Y ] = E[(P X)(P X) ] = E[(P X)(X P )] = P E[XX ]P = P cov(x)p P cov(y ) = P P cov(x)p = cov(x)p P d 1 P = [P 1, P 2,..., P d ] cov(y ) ( ) cov(y ) = λ 1 0....... 0 λ d [λ 1 P 1, λ 2 P 2,..., λ d P d ] = [cov(x)p 1, cov(x)p 2,..., cov(x)p d ] λ i P i = cov(x)p i P i X P 27 / 41

1 : : : 2012 http://results.sportstats.ca/res2012/honolulumarathon m.htm 24,070 1. 2. 3 10 3 3. CDF 3 4. CDF 5. : PDF SFC-SFS : 2013 5 16 28 / 41

---- -------- ---- ----- ------------------- ------------ -- --- -------- ------- ------ ----- ------- --- Chip Pace Gender Category @10km @21.1 @30 Place Time /mi # Name City ST CNT Plce/Tot Plc/Tot Category Split1 Split2 1 02:12:31 5:04 6 Kipsang, Wilson Iten KEN 1/12690 1/16 MElite 31:40 1:07:07 1:3 2 02:13:08 5:05 7 Geneti, Markos Addis Ababa ETH 2/12690 2/16 MElite 31:39 1:07:02 1:3 3 02:14:15 5:08 11 Kimutai, Kiplimo Eldoret KEN 3/12690 3/16 MElite 31:40 1:07:02 1:3 4 02:14:55 5:09 2 Ivuti, Patrick Kangundo KEN 4/12690 4/16 MElite 31:40 1:07:02 1:3 5 02:15:17 5:10 12 Arile, Julius Kepenguria KEN 5/12690 5/16 MElite 31:39 1:07:02 1:3 6 02:15:53 5:11 9 Bouramdane, Abderr Champs De Cou MAR 6/12690 6/16 MElite 31:40 1:07:01 1:3 7 02:18:27 5:17 8 Manza, Nicholas Ngong Hills KEN 7/12690 7/16 MElite 31:39 1:07:01 1:3 8 02:19:46 5:20 1 Chelimo, Nicholas Ngong Hills KEN 8/12690 8/16 MElite 31:40 1:07:02 1:3 9 02:25:23 5:33 20850 Harada, Taku Nagoya-Shi AI JPN 9/12690 1/1238 M25-29 31:54 1:09:52 1:4 10 02:27:12 5:37 25474 Hagawa, Eiichi Matsumoto NA JPN 10/12690 1/1501 M30-34 32:46 1:12:21 1:4... Chip Time: Category: MElite, WElite, M15-19, M20-24,..., W15-29, W20-24,... No Age Country: 3-letter country code: e.g., JPN, USA UK 29 / 41

1 1 ( ) No Age n mean stddev median all 24,070 369.1 94.2 357 men 12,532 350.5 93.2 338 women 11,537 389.3 91.0 381 30 / 41

# regular expression to read chiptime and category from honolulu marathon data re = /\s*\d+\s+(\d{2}:\d{2}:\d{2})\s+.*((?:[mw](?:elite \d{2}\-\d{2}) No Age))/ filename = ARGV[0] open(filename, r ) do io io.each_line do line if re.match(line) puts "#{$1}\t#{$2}" end end end 31 / 41

1 2 3 10 3 count count count 1200 1000 800 600 400 200 0 100 200 300 400 500 600 700 800 900 finish time (minutes) with 10-minute-bin 1200 1000 800 600 400 200 0 100 200 300 400 500 600 700 800 900 finish time (minutes) with 10-minute-bin 1200 1000 800 600 400 200 0 100 200 300 400 500 600 700 800 900 finish time (minutes) with 10-minute-bin ( ) ( ) ( ) 32 / 41

1 3 CDF 3 1 0.9 0.8 all men women 0.7 0.6 CDF 0.5 0.4 0.3 0.2 0.1 0 100 200 300 400 500 600 700 800 900 finish time (minutes) 33 / 41

y y : correlation-data-1.txt, correlation-data-2.txt 80 70 60 50 100 80 60 40 30 40 20 10 0 0 20 40 60 80 100 120 140 160 x 20 0 0 20 40 60 80 100 120 140 x data-1:r=0.87 (left), data-2:r=-0.60 (right) 34 / 41

: #!/usr/bin/env ruby # regular expression for matching 2 floating numbers re = /([-+]?\d+(?:\.\d+)?)\s+([-+]?\d+(?:\.\d+)?)/ sum_x = 0.0 # sum of x sum_y = 0.0 # sum of y sum_xx = 0.0 # sum of x^2 sum_yy = 0.0 # sum of y^2 sum_xy = 0.0 # sum of xy n = 0 # the number of data ARGF.each_line do line if re.match(line) x = $1.to_f y = $2.to_f sum_x += x sum_y += y sum_xx += x**2 sum_yy += y**2 sum_xy += x * y n += 1 end end r = (sum_xy - sum_x * sum_y / n) / Math.sqrt((sum_xx - sum_x**2 / n) * (sum_yy - sum_y**2 / n)) printf "n:%d r:%.3f\n", n, r 35 / 41

y y : correlation-data-1.txt, correlation-data-2.txt f(x) = b 0 + b 1 x b 1 = xy n xȳ x 2 n( x) 2 b 0 = ȳ b 1 x 80 5.75 + 0.45 * x 100 72.72-0.38 * x 70 60 80 50 60 40 30 40 20 10 20 0 0 0 20 40 60 80 100 120 140 160 0 20 40 60 80 100 120 140 x x data-1:r=0.87 (left), data-2:r=-0.60 (right) 36 / 41

: #!/usr/bin/env ruby # regular expression for matching 2 floating numbers re = /([-+]?\d+(?:\.\d+)?)\s+([-+]?\d+(?:\.\d+)?)/ sum_x = sum_y = sum_xx = sum_xy = 0.0 n = 0 ARGF.each_line do line if re.match(line) x = $1.to_f y = $2.to_f sum_x += x sum_y += y sum_xx += x**2 sum_xy += x * y n += 1 end end mean_x = Float(sum_x) / n mean_y = Float(sum_y) / n b1 = (sum_xy - n * mean_x * mean_y) / (sum_xx - n * mean_x**2) b0 = mean_y - b1 * mean_x printf "b0:%.3f b1:%.3f\n", b0, b1 37 / 41

: set xrange [0:160] set yrange [0:80] set xlabel "x" set ylabel "y" plot "correlation-data-1.txt" notitle with points, \ 5.75 + 0.45 * x lt 3 38 / 41

7 : 39 / 41

8 (6/5) : 2 6/19 ( ) 6 (18:10-19:40) λ13 7/17 ( ) 4 (14:45-16:15) ϵ12 40 / 41

[1] Ruby official site. http://www.ruby-lang.org/ [2] gnuplot official site. http://gnuplot.info/ [3] Mark Crovella and Balachander Krishnamurthy. Internet measurement: infrastructure, traffic, and applications. Wiley, 2006. [4] Pang-Ning Tan, Michael Steinbach and Vipin Kumar. Introduction to Data Mining. Addison Wesley, 2006. [5] Raj Jain. The art of computer systems performance analysis. Wiley, 1991. [6] Toby Segaran. ( )... 2008. [7] Chris Sanders. ( ). 2 Wireshark.. 2012. [8]... 2011. [9],.., 2010. [10],.., 2009. 41 / 41