Metode Numerik BISECTION

Metode Numerik BISECTION METHOD

Yoga Situmorang(4314100066)Sinar Baskoro Maulana Aji(4314100009)Sela Ferdiananta Ayun(4314100075)Vicko Ryangga Putra(4314100078)Abdul Holil Situmorang(4314100081)Fikri Imannudin(4314100087)Muhammad Iqbal Febriansyah Putra(4314100091)Nabielah Khiyaarul A(4314100095)

INSTITUT TEKNOLOGI SEPULUH NOPEMBERSURABAYA2015

BISECTION METHOD (Metode Setengah Interval)akar terkecil=xe akar terbesar=xu f ( xe ) . f ( xu ) < 0

akar kemudian adalah :

elevasi :

xrxexu III

SOAL PERTAMAMencari akar :- pilih x1 = 1.6f ( x1 ) = 0.9247393 x2 = 2.0 f ( x2 ) = -0.5656258

= 1.8

f ( ) = 0.2939756 akar berada diantara x = 1.8 dan x = 2.0x1 = 1.8f ( x1 ) = 0.2939756x2 = 2.0f ( x2 ) = -0.5656258

= 1.9 akar berada diantara x = 1.8 dan x = 2.0x1 = 1.8f ( x1 ) = 0.2939756x2 = 2.0f ( x2 ) = -0.5656258

= 1.9f ( ) = -0.1049169 akar berada diantara x = 1.8 dan x = 1.9x1 = 1.8f ( x1 ) = 0.2939756x2 = 1.9f ( x2 ) = -0.1049169

= 1.85f ( ) = 0.1019814 akar berada diantara x = 1.85 dan x = 1.9x1 = 1.85f ( x1 ) = 0.1019814x2 = 1.9f ( x2 ) = -0.1049169

= 1.875f ( ) = 0.0004306Seperti prosedur sebelumnya :x1 = 1.875f ( x1 ) = 0.0004306x2 = 1.9f ( x2 ) = -0.1049169

= 1.8875f ( ) = -0.051764kemudian :x1 = 1.875f ( x1 ) = 0.0004306x2 = 1.88125f ( x2 ) = -0.0255476

= 1.878125f ( ) = -0.0125288kemudian :x1 = 1.875f ( x1 ) = 0.0004306x2 = 1. 878125f ( x2 ) = -0.0125288

= 1.8765625f ( ) = -0.0060416kemudian :x1 = 1.875f ( x1 ) = 0.0004306x2 = 1. 8765625f ( x2 ) = -0.0060416




= 1.8750977f ( ) = -0.0000264kemudian :x1 = 1.8750977f ( x1 ) = -0.0000264x2 = 1.8751954f ( x2 ) = -0.0003778





= 1.875107f ( ) = -0.0000119

akar pertama adalah x = 1.875107

- Akar kedua :Ditentukan : x1 = 4.6f ( x1 ) = -4.5792724x2 = 4.8f ( x2 ) = 6.3136791

= 4.7f ( ) = 0.3188944kemudian :x1 = 4.6f ( x1 ) = -4.5792724x2 = 4.7f ( x2 ) = 0.3188944

= 4.65f ( ) = -2.2606573kemudian :x1 = 4.65f ( x1 ) = -2.2606573x2 = 4.7f ( x2 ) = 0.3188944


= 4.6875f ( ) = -0.3512151 kemudian :x1 = 4.6875f ( x1 ) = -0.3512151x2 = 4.7f ( x2 ) = 0.3188944



= 4.6953125f ( ) = 0.0656021

kemudian :x1 = 4.69375f ( x1 ) = -0.0182938x2 = 4.6953125f ( x2 ) = 0.0656021

= 4.6945313f ( ) = 0.0236208 kemudian :x1 = 4.69375f ( x1 ) = -0.0182938x2 = 4.6945313f ( x2 ) = 0.0236208

= 4.6941407f ( ) = 0.0026565 kemudian :x1 = 4.69375f ( x1 ) = -0.0182938x2 = 4.6941407f ( x2 ) = 0.0026565





= 4.6940797f ( ) = -0.006133

kemudian :x1 = 4.6940797f ( x1 ) = -0.006133x2 = 4.6940919f ( x2 ) = 0.0000412

= 4.690858f ( ) = -0.1728719

Akar kedua adalah x = 4.6940797

- Akar ketiga :Ditentukan : x1 = 7.8f ( x1 ) = 66.841864x2 = 7.9f ( x2 ) = -61.040369

= 7.85f ( ) = 6.1078946kemudian :x1 = 7.85f ( x1 ) = 6.1078946x2 = 7.9f ( x2 ) = -61.040369












= 7.8547486f ( ) = 0.0114011

Akar ketiga adalah x =7.8547608Jadi ketiga akar tersebut adalah :x = 1.875107x = 4.694797x = 7.8547608

SOAL KEDUAMencari akar pertama:- pilih x1 = 1800f ( x1 ) = 0.5 x2 = 2200f ( x2 ) = -0.1428

= 2000f ( ) = 0.1580 akar berada diantara x = 2200 dan x = 2000x1 =2200f ( x1 ) = -0.1428x2 = 2000f ( x2 ) = 0.1580

= 2100 f ( ) = 0.00000Mencari akar kedua: akar berada diantara x = 2700 dan x = 3600x1 = 2700f ( x1 ) = -0.5x2 = 3600f ( x2 ) = 0.5

=3150f ( ) = -0.2071akar berada diantara x = 3150 dan x = 3600x1 =3150f ( x1 ) = -0.2071x2 =3600f ( x2 ) = 0.5

=337.50f ( ) = 0.1173akar berada diantara x = 3150 dan x = 337.50x1 =3150f ( x1 ) = -0.2071x2 =337.50f ( x2 ) =0.1173

=326.250f ( ) = -0.0556akar berada diantara x = 326.250 dan x = 337.50x1 =326.250f ( x1 ) = -0.0556x2 =337.50f ( x2 ) =0.1173

=331.8750f ( ) = 0.0286akar berada diantara x = 331.8750 dan x = 326.250x1 =331.8750f ( x1 ) = 0.0286x2 =326.250f ( x2 ) =-0.0556

=329.06250f ( ) = -0.0141akar berada diantara x = 330.41880 dan x = 329.06250x1 =330.41880f ( x1 ) = 0.0071x2 =329.06250f ( x2 ) =-0.0141

=329.7656f ( ) = -0.0035

akar berada diantara x = 329.76560 dan x = 330.41880x1 =329.76560f ( x1 ) = -0.0035x2 =330.41880f ( x2 ) =0.0071


=329.94140f ( ) = -0.0008akar berada diantara x = 329.94140dan x = 330.11720x1 =329.94140f ( x1 ) = -0.0008x2 =330.11720f ( x2 ) =0.0017

=330.02930f ( ) = 0.0004akar berada diantara x = 330.02930dan x = 329.94140x1 =330.02930f ( x1 ) = 0.0004x2 =329.94140f ( x2 ) =-0.0008

=329.98540f ( ) = -0.0002

akar berada diantara x = 329.98540 dan x = 330.02930x1 =329.98540f ( x1 ) = -0.0002x2 =330.02930f ( x2 ) =0.0004


=329.99640f ( ) = 0.00005akar berada diantara x = 329.99640 dan x = 330.00740x1 =329.99640f ( x1 ) = 0.00005x2 =330.00740f ( x2 ) =0.0001

=330.00190f ( ) = 0.00003akar berada diantara x = 330.00190 dan x = 329.99640x1 =330.00190f ( x1 ) = 0.00003x2 =329.99640f ( x2 ) =0.00005

=329.99920f ( ) = -0.00001akar berada diantara x = 329.99920dan x = 330.00190x1 =329.99920f ( x1 ) = -0.00001x2 =330.00190f ( x2 ) =0.00003

=330.00060f ( ) = 0.00000Mencari akar ketiga:- pilih x1 = 5400f ( x1 ) = 0.5 x2 = 5800f ( x2 ) = -0.14

= 5600f ( ) = -0.15798akar berada diantara x = 5600 dan x = 5800x1 =5600f ( x1 ) = -0.15798x2 = 5800f ( x2 ) = -0.14

=5700f ( ) = 0

S T A R TWRITETHE ROOT ARE r DO 8 I = 1,38DC= 0,5XEN= XST+ DXXST= XST+ DX4FXST= SIN(XST) * COS(XST) + 1FXEN= SIN(XEN) * COS(XEN) + 1IS FXST * FXEN. LT. 0,0 XST = XST + DX XEN = XST - DX FXST = SIN(XST) * COS(XST) + 1YES3 SINh X = e - e ITER = L . 0.5/1.10L . 2 XST = -0.1X-XN6N2

Metode Numerik BISECTION

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