tugas metode numerik pendidikan matematika umt

1.1 Barisan Fibonacci

METODE NUMERIK FIBONACCI

Amelia Noviasari 1384202071Denny Hardi 1384202110

Mona Yulinda Santika 1384202115Risti Apriani Dewi 1384202141Rudi Alviansyah 1384202100

March 11, 2016

Amelia Noviasari 1384202071 Denny Hardi 1384202110 Mona Yulinda Santika 1384202115 Risti Apriani Dewi 1384202141 Rudi Alviansyah 1384202100METODE NUMERIK FIBONACCI

Barisan Fibonacci

1 1.1 Barisan Fibonacci

Definisi

Barisan f0, f1, f2, f3, ..., fn − 2, fn − 1, fn disebut Fibonacci jika untukf0 = 1, f1 = 0 + f0, f2 = f0 + f1,f3 = f2 + f1,...,fn = fn − 2 + fn − 1

contoh barisan fibonacci1, 1, 2, 3, 5, 8, 13, 21, 34, 55, ...

Definisi

Barisan f0, f1, f2, f3, ..., fn − 2, fn − 1, fn disebut Fibonacci jika untukf0 = 1, f1 = 0 + f0, f2 = f0 + f1,f3 = f2 + f1,...,fn = fn − 2 + fn − 1

contoh barisan fibonacci1, 1, 2, 3, 5, 8, 13, 21, 34, 55, ...

Algoritma nilai optimal dengan Metode Fibonacci

Dicari nilai n terkecil

dibentukL0 = Fn+1Ln

dibentuk

λi = ai +F(n+1)−i−1

F(n+1)−i+1(bi − ai )

dibentukL0 = Fn+1Ln

dibentuk

λi = ai +F(n+1)−i−1

F(n+1)−i+1(bi − ai )

dibentukL0 = Fn+1Ln

dibentuk

λi = ai +F(n+1)−i−1

F(n+1)−i+1(bi − ai )

lanjutan

dicari

µi = ai +F(n+1)−i

F(n+1)−i+1(bi − ai )

jikaf (µi ) > f (λi )

ambil µi dan ai , masing-masing sebagai bi+1 dan ai+1

iterasi berhenti ketika bi − ai < 2δ

lanjutan

dicari

F(n+1)−i+1(bi − ai )

lanjutan

dicari

F(n+1)−i+1(bi − ai )

minimalkanf (x) = 2x3 − 3x2

dengan δ = 0, 1 pada selang −2 <= x <= 3

dengan cara analitik, diperoleh nilai x yang meminimalkanf (x) adalah x = 1

minimalkanf (x) = 2x3 − 3x2

dengan δ = 0, 1 pada selang −2 <= x <= 3

dengan cara analitik, diperoleh nilai x yang meminimalkanf (x) adalah x = 1

dibentukL0 = Fn+1Ln

Iterasi I

λ1 = a1 +F(7+1)−1−1

F(7+1)−1+1(b1 − a1)

λ1 = −2 +F(6)F(8)

(3 − (−2))

λ1 = −2 +13

λ1 = −0, 088

Iterasi I

λ1 = a1 +F(7+1)−1−1

F(7+1)−1+1(b1 − a1)

λ1 = −2 +F(6)F(8)

(3 − (−2))

λ1 = −2 +13

λ1 = −0, 088

Iterasi I

λ1 = a1 +F(7+1)−1−1

F(7+1)−1+1(b1 − a1)

λ1 = −2 +F(6)F(8)

(3 − (−2))

λ1 = −2 +13

λ1 = −0, 088

Iterasi I

λ1 = a1 +F(7+1)−1−1

F(7+1)−1+1(b1 − a1)

λ1 = −2 +F(6)F(8)

(3 − (−2))

λ1 = −2 +13

λ1 = −0, 088

lanjutan iterasi I

µ1 = a1 +F(7+1)−1

F(7+1)−1+1(b1 − a1)

µ1 = −2 +F(7)F(8)

(3 − (−2))

µ1 = −2 +21

µ1 = 1, 088

lanjutan iterasi I

µ1 = a1 +F(7+1)−1

F(7+1)−1+1(b1 − a1)

µ1 = −2 +F(7)F(8)

(3 − (−2))

µ1 = −2 +21

µ1 = 1, 088

lanjutan iterasi I

µ1 = a1 +F(7+1)−1

F(7+1)−1+1(b1 − a1)

µ1 = −2 +F(7)F(8)

(3 − (−2))

µ1 = −2 +21

µ1 = 1, 088

lanjutan iterasi I

µ1 = a1 +F(7+1)−1

F(7+1)−1+1(b1 − a1)

µ1 = −2 +F(7)F(8)

(3 − (−2))

µ1 = −2 +21

µ1 = 1, 088

lanjutan iterasi I

f (λ1) = 2λ13 − 3λ1

f (λ1) = 2(−0, 088)3 − 3(−0, 088)2

f (λ1) = −0, 001 − 0, 023

f (λ1) = −0, 024

lanjutan iterasi I

f (λ1) = 2λ13 − 3λ1

f (λ1) = 2(−0, 088)3 − 3(−0, 088)2

f (λ1) = −0, 001 − 0, 023

f (λ1) = −0, 024

lanjutan iterasi I

f (λ1) = 2λ13 − 3λ1

f (λ1) = 2(−0, 088)3 − 3(−0, 088)2

f (λ1) = −0, 001 − 0, 023

f (λ1) = −0, 024

lanjutan iterasi I

f (λ1) = 2λ13 − 3λ1

f (λ1) = 2(−0, 088)3 − 3(−0, 088)2

f (λ1) = −0, 001 − 0, 023

f (λ1) = −0, 024

lanjutan iterasi I

f (µ1) = 2µ13 − 3µ1

f (µ1) = 2(1, 088)3 − 3(1, 088)2

f (µ1) = 2, 576 − 3, 551

f (µ1) = −0, 975

f (λ1) > f (µ1)

λ1 = −0, 088(a2)

b1 = 3(b2)

lanjutan iterasi I

f (µ1) = 2µ13 − 3µ1

f (µ1) = 2(1, 088)3 − 3(1, 088)2

f (µ1) = 2, 576 − 3, 551

f (µ1) = −0, 975

f (λ1) > f (µ1)

λ1 = −0, 088(a2)

b1 = 3(b2)

lanjutan iterasi I

f (µ1) = 2µ13 − 3µ1

f (µ1) = 2(1, 088)3 − 3(1, 088)2

f (µ1) = 2, 576 − 3, 551

f (µ1) = −0, 975

f (λ1) > f (µ1)

λ1 = −0, 088(a2)

b1 = 3(b2)

lanjutan iterasi I

f (µ1) = 2µ13 − 3µ1

f (µ1) = 2(1, 088)3 − 3(1, 088)2

f (µ1) = 2, 576 − 3, 551

f (µ1) = −0, 975

f (λ1) > f (µ1)

λ1 = −0, 088(a2)

b1 = 3(b2)

lanjutan iterasi I

f (µ1) = 2µ13 − 3µ1

f (µ1) = 2(1, 088)3 − 3(1, 088)2

f (µ1) = 2, 576 − 3, 551

f (µ1) = −0, 975

f (λ1) > f (µ1)

λ1 = −0, 088(a2)

b1 = 3(b2)

lanjutan iterasi I

f (µ1) = 2µ13 − 3µ1

f (µ1) = 2(1, 088)3 − 3(1, 088)2

f (µ1) = 2, 576 − 3, 551

f (µ1) = −0, 975

f (λ1) > f (µ1)

λ1 = −0, 088(a2)

b1 = 3(b2)

lanjutan iterasi I

f (µ1) = 2µ13 − 3µ1

f (µ1) = 2(1, 088)3 − 3(1, 088)2

f (µ1) = 2, 576 − 3, 551

f (µ1) = −0, 975

f (λ1) > f (µ1)

λ1 = −0, 088(a2)

b1 = 3(b2)

Iterasi II

λ2 = a2 +F(7+1)−2−1

F(7+1)−2+1(b2 − a2)

λ2 = −0, 088 +F(5)F(7)

(3 − (−0, 088))

λ2 = −0, 088 +8

21(3, 088)

λ2 = 1, 088

Iterasi II

λ2 = a2 +F(7+1)−2−1

F(7+1)−2+1(b2 − a2)

λ2 = −0, 088 +F(5)F(7)

(3 − (−0, 088))

λ2 = −0, 088 +8

21(3, 088)

λ2 = 1, 088

Iterasi II

λ2 = a2 +F(7+1)−2−1

F(7+1)−2+1(b2 − a2)

λ2 = −0, 088 +F(5)F(7)

(3 − (−0, 088))

λ2 = −0, 088 +8

21(3, 088)

λ2 = 1, 088

Iterasi II

λ2 = a2 +F(7+1)−2−1

F(7+1)−2+1(b2 − a2)

λ2 = −0, 088 +F(5)F(7)

(3 − (−0, 088))

λ2 = −0, 088 +8

21(3, 088)

λ2 = 1, 088

lanjutan iterasi II

µ2 = a2 +F(7+1)−2

F(7+1)−2+1(b2 − a2)

µ2 = −0, 088 +F(6)F(7)

(3 − (−0, 088))

µ2 = −0, 088 +13

21(3, 088)

µ2 = 1, 744

lanjutan iterasi II

µ2 = a2 +F(7+1)−2

F(7+1)−2+1(b2 − a2)

µ2 = −0, 088 +F(6)F(7)

(3 − (−0, 088))

µ2 = −0, 088 +13

21(3, 088)

µ2 = 1, 744

lanjutan iterasi II

µ2 = a2 +F(7+1)−2

F(7+1)−2+1(b2 − a2)

µ2 = −0, 088 +F(6)F(7)

(3 − (−0, 088))

µ2 = −0, 088 +13

21(3, 088)

µ2 = 1, 744

lanjutan iterasi II

µ2 = a2 +F(7+1)−2

F(7+1)−2+1(b2 − a2)

µ2 = −0, 088 +F(6)F(7)

(3 − (−0, 088))

µ2 = −0, 088 +13

21(3, 088)

µ2 = 1, 744

lanjutan iterasi II

f (λ2) = 2λ23 − 3λ2

f (λ2) = 2(1, 088)3 − 3(1, 088)2

f (λ2) = 2, 576 − 3, 551

f (λ2) = −0, 975

lanjutan iterasi II

f (λ2) = 2λ23 − 3λ2

f (λ2) = 2(1, 088)3 − 3(1, 088)2

f (λ2) = 2, 576 − 3, 551

f (λ2) = −0, 975

lanjutan iterasi II

f (λ2) = 2λ23 − 3λ2

f (λ2) = 2(1, 088)3 − 3(1, 088)2

f (λ2) = 2, 576 − 3, 551

f (λ2) = −0, 975

lanjutan iterasi II

f (λ2) = 2λ23 − 3λ2

f (λ2) = 2(1, 088)3 − 3(1, 088)2

f (λ2) = 2, 576 − 3, 551

f (λ2) = −0, 975

lanjutan iterasi II

f (µ2) = 2µ23 − 3µ2

f (µ2) = 2(1, 744)3 − 3(1, 744)2

f (µ2) = 10, 609 − 9, 125

f (µ2) = 1, 484

f (µ2) > f (λ2)

µ2 = −1, 744(b3)

a2 = −0, 088(a3)

lanjutan iterasi II

f (µ2) = 2µ23 − 3µ2

f (µ2) = 2(1, 744)3 − 3(1, 744)2

f (µ2) = 10, 609 − 9, 125

f (µ2) = 1, 484

f (µ2) > f (λ2)

µ2 = −1, 744(b3)

a2 = −0, 088(a3)

lanjutan iterasi II

f (µ2) = 2µ23 − 3µ2

f (µ2) = 2(1, 744)3 − 3(1, 744)2

f (µ2) = 10, 609 − 9, 125

f (µ2) = 1, 484

f (µ2) > f (λ2)

µ2 = −1, 744(b3)

a2 = −0, 088(a3)

lanjutan iterasi II

f (µ2) = 2µ23 − 3µ2

f (µ2) = 2(1, 744)3 − 3(1, 744)2

f (µ2) = 10, 609 − 9, 125

f (µ2) = 1, 484

f (µ2) > f (λ2)

µ2 = −1, 744(b3)

a2 = −0, 088(a3)

lanjutan iterasi II

f (µ2) = 2µ23 − 3µ2

f (µ2) = 2(1, 744)3 − 3(1, 744)2

f (µ2) = 10, 609 − 9, 125

f (µ2) = 1, 484

f (µ2) > f (λ2)

µ2 = −1, 744(b3)

a2 = −0, 088(a3)

lanjutan iterasi II

f (µ2) = 2µ23 − 3µ2

f (µ2) = 2(1, 744)3 − 3(1, 744)2

f (µ2) = 10, 609 − 9, 125

f (µ2) = 1, 484

f (µ2) > f (λ2)

µ2 = −1, 744(b3)

a2 = −0, 088(a3)

lanjutan iterasi II

f (µ2) = 2µ23 − 3µ2

f (µ2) = 2(1, 744)3 − 3(1, 744)2

f (µ2) = 10, 609 − 9, 125

f (µ2) = 1, 484

f (µ2) > f (λ2)

µ2 = −1, 744(b3)

a2 = −0, 088(a3)

Iterasi III

λ3 = a3 +F(7+1)−3−1

F(7+1)−3+1(b3 − a3)

λ3 = −0, 088 +F(4)F(6)

(1, 744 − (−0, 088))

λ3 = −0, 088 +5

13(1, 832)

λ3 = −0, 617

Iterasi III

λ3 = a3 +F(7+1)−3−1

F(7+1)−3+1(b3 − a3)

λ3 = −0, 088 +F(4)F(6)

(1, 744 − (−0, 088))

λ3 = −0, 088 +5

13(1, 832)

λ3 = −0, 617

Iterasi III

λ3 = a3 +F(7+1)−3−1

F(7+1)−3+1(b3 − a3)

λ3 = −0, 088 +F(4)F(6)

(1, 744 − (−0, 088))

λ3 = −0, 088 +5

13(1, 832)

λ3 = −0, 617

Iterasi III

λ3 = a3 +F(7+1)−3−1

F(7+1)−3+1(b3 − a3)

λ3 = −0, 088 +F(4)F(6)

(1, 744 − (−0, 088))

λ3 = −0, 088 +5

13(1, 832)

λ3 = −0, 617

lanjutan iterasi III

µ3 = a3 +F(7+1)−3

F(7+1)−3+1(b3 − a3)

µ3 = −0, 088 +F(5)F(6)

(1, 744 − (−0, 088))

µ3 = −0, 088 +8

13(1, 832)

µ3 = 1, 039

µ3 = a3 +F(7+1)−3

F(7+1)−3+1(b3 − a3)

µ3 = −0, 088 +F(5)F(6)

(1, 744 − (−0, 088))

µ3 = −0, 088 +8

13(1, 832)

µ3 = 1, 039

µ3 = a3 +F(7+1)−3

F(7+1)−3+1(b3 − a3)

µ3 = −0, 088 +F(5)F(6)

(1, 744 − (−0, 088))

µ3 = −0, 088 +8

13(1, 832)

µ3 = 1, 039

µ3 = a3 +F(7+1)−3

F(7+1)−3+1(b3 − a3)

µ3 = −0, 088 +F(5)F(6)

(1, 744 − (−0, 088))

µ3 = −0, 088 +8

13(1, 832)

µ3 = 1, 039

f (λ3) = 2λ33 − 3λ3

f (λ3) = 2(0, 617)3 − 3(0, 617)2

f (λ3) = 0, 470 − 1, 142

f (λ3) = −0, 672

f (λ3) = 2λ33 − 3λ3

f (λ3) = 2(0, 617)3 − 3(0, 617)2

f (λ3) = 0, 470 − 1, 142

f (λ3) = −0, 672

f (λ3) = 2λ33 − 3λ3

f (λ3) = 2(0, 617)3 − 3(0, 617)2

f (λ3) = 0, 470 − 1, 142

f (λ3) = −0, 672

f (λ3) = 2λ33 − 3λ3

f (λ3) = 2(0, 617)3 − 3(0, 617)2

f (λ3) = 0, 470 − 1, 142

f (λ3) = −0, 672

f (µ3) = 2µ33 − 3µ3

f (µ3) = 2(1, 039)3 − 3(1, 039)2

f (µ3) = 2, 243 − 3, 239

f (µ3) = −1, 086

f (λ3) > f (µ3)

λ3 = 0, 617(a4)

b3 = 1, 744(b4)

f (µ3) = 2µ33 − 3µ3

f (µ3) = 2(1, 039)3 − 3(1, 039)2

f (µ3) = 2, 243 − 3, 239

f (µ3) = −1, 086

f (λ3) > f (µ3)

λ3 = 0, 617(a4)

b3 = 1, 744(b4)

f (µ3) = 2µ33 − 3µ3

f (µ3) = 2(1, 039)3 − 3(1, 039)2

f (µ3) = 2, 243 − 3, 239

f (µ3) = −1, 086

f (λ3) > f (µ3)

λ3 = 0, 617(a4)

b3 = 1, 744(b4)

f (µ3) = 2µ33 − 3µ3

f (µ3) = 2(1, 039)3 − 3(1, 039)2

f (µ3) = 2, 243 − 3, 239

f (µ3) = −1, 086

f (λ3) > f (µ3)

λ3 = 0, 617(a4)

b3 = 1, 744(b4)

f (µ3) = 2µ33 − 3µ3

f (µ3) = 2(1, 039)3 − 3(1, 039)2

f (µ3) = 2, 243 − 3, 239

f (µ3) = −1, 086

f (λ3) > f (µ3)

λ3 = 0, 617(a4)

b3 = 1, 744(b4)

f (µ3) = 2µ33 − 3µ3

f (µ3) = 2(1, 039)3 − 3(1, 039)2

f (µ3) = 2, 243 − 3, 239

f (µ3) = −1, 086

f (λ3) > f (µ3)

λ3 = 0, 617(a4)

b3 = 1, 744(b4)

f (µ3) = 2µ33 − 3µ3

f (µ3) = 2(1, 039)3 − 3(1, 039)2

f (µ3) = 2, 243 − 3, 239

f (µ3) = −1, 086

f (λ3) > f (µ3)

λ3 = 0, 617(a4)

b3 = 1, 744(b4)

Tabel perhitungan dengan Metode Fibonacci

x∗ = a8 +

(b8 − a8

x∗ = 1, 039 +

(1, 180 − 1, 039

)x∗ = 1, 109

x∗ = a8 +

(b8 − a8

x∗ = 1, 039 +

(1, 180 − 1, 039

x∗ = 1, 109

x∗ = a8 +

(b8 − a8

x∗ = 1, 039 +

(1, 180 − 1, 039

)x∗ = 1, 109

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