Metode Numerik Ayu [PDF]

Citation preview

Soal 11.Tentukan fungsi lanjar yang mencocokan titik-titik data berikut dengan metode regresi. x



1.0



1.5



2.0



2.5



3.0



Y



2.0



3.2



4.1



4.9



5.9



12.Diberikan titik-titik (x,y) sebagai berikut X



1



2



3



4



5



Y



0.6



0.9



4.3



7.6



12.6



a. Cocokkan titik di tabel masing-masing dengan fungsi f(x) = Ce^bx dan f(x)^b b. Hitung deviasi = yi - f(xi),kemudian tentukan galat RMSnya berdasarkan galat RMS,fungsi hampiran mana yang terbaik?



Penyelesaian 11. i



xi 1 2 3 4 5



Total



𝑛 𝑥𝑖



(



yi 1 1.5 2 2.5 3 10



𝑥𝑖 ) 𝑋𝑖 2



xi^2 2 3.2 4.1 4.9 5.9 20.1



1 2.25 4 6.25 9 22.5



𝑦𝑖 𝑎 [ ] = [ ] 𝑏 𝑥𝑖𝑦𝑖



xiyi 2 4.8 8.2 12.25 17.7 44.95



f(xi)=a+bx deviasi (deviasi)^2 2.12 -0.12 0.0144 3.07 0.13 0.0169 4.02 0.08 0.0064 4.97 -0.07 0.0049 5.92 -0.02 0.0004 0.043



(



5 10



𝑎 20.1 [ ] = [ ] 𝑏 44.95



10 ) 22.5



𝑎 [ ] = 𝑏



22.5 ( −10



1 (5)(22.5)−(10)(10)



20.1 − 10 ) ( ) 44.95 5



𝑎 0.22 [ ] =( ) 𝑏 1.9 a = 0.22



b = 1.9



maka, persamaan regresinya : f(x) = 0.22 + 1.9x Galat RMS = ERMS = [



0.043 1/2 5



]



= 0.09274



12. a. xi



yi 1 2 3 4 5



0.6 0.9 4.3 7.6 12.6



Xi=ln(xi) Yi=ln(yi) 0 -0.510826 0.693147 -0.105361 1.098612 1.458615 1.386294 2.028148 1.609438 2.533697 4.787492 5.404274



y = Cxb ln(y) = ln(C) + b ln(x) defenisikan : Y = ln(y) a = ln(C) X = ln(x)



Xi^2



XiYi



0 0 0.480453 -0.07303 1.206949 1.602452 1.921812 2.81161 2.59029 4.077828 6.199504 8.41886



y=Cx^b 2.3225 9.342953 21.09165 37.58483 58.83385



Persamaan regresi lanjar : Y = a + bX (



(



5 4.7875



𝑎 [ ] = 𝑏



𝑥𝑖 ) 𝑋𝑖 2



𝑛



𝑥𝑖



𝑦𝑖 𝑎 [ ] = [ ] 𝑏 𝑥𝑖𝑦𝑖



𝑎 5.4043 [ ] = [ ] 𝑏 8.4189



4.7875 ) 6.1995



1 (5)(6.1995)−(4.7875)(4.7875)



6.1995 − 4.7875 5.4043 ) ( ) −4.7875 5 8.4189



(



𝑎 0.8421 [ ] =( ) 𝑏 2.0082 a = 0.8421



b = 2.0082



C = ea = 2.720.8421 = 2.3225 Maka, y = Cxb = 2.3225x2.0082 Galat RMS = ERMS = [



4916.74 1/2 5



]



= 31.3584



b. xi



yi 1 2 3 4 5 15



xi^2 0.6 0.9 4.3 7.6 12.6



1 4 9 16 25 55



Yi=ln(yi) -0.510826 -0.105361 1.458615 2.028148 2.533697 5.404274



xiYi -0.51083 -0.21072 4.375845 8.112593 12.66848 24.43538



y=Ce^(bx) 0.10976495 0.04821267 0.02117672 0.00930157 0.00408558



y = Cebx ln(y) = ln(C) + bx ln(e)



ln(e) = 1



ln(y) = ln(C) + bx defenisikan : Y = ln(y) a = ln(C) X=x Persamaan regresi lanjar : Y = a + bX (



(



𝑥𝑖 ) 𝑋𝑖 2



𝑦𝑖 𝑎 [ ] = [ ] 𝑏 𝑥𝑖𝑦𝑖



𝑎 15 5.4043 ) [ ] = [ ] 𝑏 55 24.4354



5 15



𝑎 [ ] = 𝑏



𝑛 𝑥𝑖



1 (5)(55)−(15)(15)



𝑎 −1.3859 [ ] =( ) 𝑏 −0.8222 a = -1.3859



b = -0.8222



C = ea = 2.72-1.3859 = 0.2499 Maka, y = Cebx = 0.2499e-0.8222x



55 − 15 5.4043 ) ( ) −15 5 24.4354



(



deviasi 1 deviasi 2 (deviasi 1)^2 (deviasi 2)^2 -2.833326 -0.620591 8.02773409 0.385132666 -9.448313 -0.153573 89.2706266 0.023584722 -19.63304 1.4374383 385.4561968 2.066228888 -35.55668 2.0188467 1264.27766 4.075741925 -56.30016 2.5296112 3169.7075 6.398933006 4916.739718 12.94962121



Galat RMS = ERMS = [



12.9496 1/2 5



]



= 1.6093



Maka, berdasarkan Galat RMSnya fungsi hampiran yang terbaik adalah fungsi y = Cebx.