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Fractional-number Algorithm
11
ALGORITMA MATEMATIKA INFORMASI: Fractional-Number Ir. Sihar, MT. T. Informatika –Fak. Teknologi Informasi Bandung –2009 FTI305 Organisasi dan Arsitektur Komputer (3 sks)
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Transcript of Fractional-number Algorithm
ALGORITMA MATEMATIKA
INFORMASI: Fractional-Number
Ir. Sihar, MT.
T. Informatika – Fak. Teknologi Informasi
Bandung – 2009
FTI305 Organisasi dan Arsitektur Komputer (3 sks)
Daftar Pustaka1) Bell, C. Gordon, Newell, A. Computer Structures: Readings and
Examples. McGraw-Hill computer science series. 1971.
2) Hennessy, J.L., Patterson, D.A. Computer Architecture: A Quantitative
Approach. Morgan-Kaufmann. 2004.
3) Larry L. Wear, L.L., Pinkert, J.R., Lane, W.G. Computers, An Introduction
to Hardware and Software Design. McGraw-Hill. 1999.
4) Scott, N.R. Computer Number Systems and Arithmetic. Prentice-Hall.
1985.
5) Simamora, S.N.M.P. “Diktat Kuliah Algoritma dan Pemrograman-I”,
Program studi Teknik Informatika. Fak. Teknik. UTAMA. Bandung. 2009.
6) Simamora, S.N.M.P. “Diktat Kuliah IF101 Pengantar Teknik Informatika”.
Departemen Teknik Informatika. Fak. Teknik. ITHB. Bandung. 2002.
7) Simamora, S.N.M.P. “Diktat Kuliah CE113 Sistem Komputer”. Program
studi Teknik Komputer. Politeknik TELKOM. Bandung. 2007.
8) Simamora, S.N.M.P. “Diktat Sistem Mikroprosesor dan Antar-muka”.
Dept. Sistem Komputer, Fak. Teknik. ITHB. Bandung. 2002.
Konsep dan Pola A-1 =A
1
A-3 =)(
1
AAA ∗∗
A ÷ B = C sisa D; maka A%B = D
jika: X ÷ B = TB
z; maka: X←T∗B+z;
2-1 =2
1
2-3 =8
1
2-5 =32
1
8-1 =8
1
8-2 =64
1
8-3 =512
1
16-1 =16
1
16-2 =256
1
16-3 =4096
1
(2.5)10 = (2.0)10 + (0.5)10
14= 0∗17+14
17
Jika maka: bac = bc
a=)log(
Hukum Komutatif: A ∗ B = B ∗ A;
Hukum Distributif: A ∗ (B+C) = (A ∗ B) + (A ∗ C);
Hukum Asosiatif: (A ∗ B) ∗ C = A ∗ (B ∗ C);
11= 3∗19+11
193
(2.5)16 = (2.0)16 + (0.5)16
(5)10 = (5)16 = (5)8
“Suatu bilangan jika
dikalikan dengan 0, maka
hasilnya selalu 0”
“Suatu bilangan jika
dipangkatkan dengan 0,
maka hasilnya selalu 1”
“Suatu bilangan jika
dipangkatkan dengan 1, maka
hasilnya bilangan itu sendiri”
= 0.5
= 0.125
= 0.03125
Kasus: Transformasikan expression berikut dalam operator ADD.
48 ∗ 4 = ...?...
Solusi: 048
+4848
+9648
+144
48+
192
maka: 48∗4 = 192
(1)
(2)
(3)
(4)
Tampilan jalannya program:
Alternatif-1:
Alternatif-2:
Alternatif-3:
Kasus: Transformasikan expression berikut dalam operator SUB.
18 ÷ 3 = ...?...
Solusi:
18
selesai
Lakukan proses pengurangan oleh pembagi terhadap nilai yang dibagi sampai hasil
pengurangan = 0.
3-
0
153
-12
3-
93
-63
-33
-
(1)
(2)
(3)
(4)
(5)
(6)
maka: 18 ÷ 3 = 6
Alternatif-1:
Alternatif-2:
Tampilan jalannya program:
Kasus:BIN⇒DEC (0.001)2 = ( ... )10
Solusi: = (0)(2-1) + (0)(2-2) + (1)(2-3)
= 0 + 0 +
= 8
1
108
1
(0.001)2
Kasus: (101.101)2 = ( ... )10
Solusi:
= (1)(2-1) + (0)(2-2) + (1)(2-3)
= + 0 +
=
=
8
1
108
5
(0.101)2
(101.101)2 = (101.0)2 + (0.101)2
Bagian-1 Bagian-2
= (1)(22) + (0)(21) + (1)(20)
= 4 + 0 + 1
= (5)10
(101.0)2
Bagian-1 Bagian-2
2
1
8
)14( +
Bagian-1 + Bagian-2 = (5)10 + 108
5
maka: (101.101)2 =108
55
DEC⇒BIN Kasus: (0.8125)10 = ( ... )2
Solusi:
0.8125 ∗ 2 = 1.625
2-4 =16
1
8-3 =512
1
= 0.0625
= 0.001953
0.625 ∗ 2 = 1.25
1
1
0.25 ∗ 2 = 0.5
0
0.5 ∗ 2 = 1.0
stop!
1
direkonstruksi: (0.1101)2
maka: (0.8125)10 = (0.1101)2
Kasus: (3.125)10 = ( ... )2
Solusi:
0.125 ∗ 2 = 0.25
0.25 ∗ 2 = 0.5
0
0
0.5 ∗ 2 = 1.0
1
stop!
direkonstruksi: (0.001)2
maka: (0.125)10 = (0.001)2
(3.125)10 = (3.0)10 + (0.125)10
Bagian-1 Bagian-2
(3.0)10 = (011)2
Bagian-1
Bagian-2
Bagian-1 + Bagian-2 = (011.0)2 + (0.001)2 = (011.001)2
maka: (3.125)10 = (011.001)2
HEX⇒DEC Kasus: (0.A01)16 = ( ... )10
Solusi: = (A)(16-1) + (0)(16-2) + (1)(16-3)
= (10) + 0 +
=
4096
1
1010 4096
2561
4096
12560
=
+
(0.A01)16
16
1
Kasus: (1B1.C4)16 = ( ... )10
Solusi:
= (12)(16-1) + (4)(16-2)
= +
=
=
256
4
10256
196
(0.C4)16
(1B1.C4)16 = (1B1.0)16 + (0.C4)16
Bagian-1 Bagian-2
= (1)(162) + (11)(161) + (1)(160)
= 256 + 176 + 1
= (433)10
(1B1.0)16
Bagian-1 Bagian-2
16
12
256
)4192( +
Bagian-1 + Bagian-2 = (433)10 + 10256
196
maka: (1B1.C4)16 =10256
196433
Kasus:
DEC⇒HEX
(30.066406)10 = ( ... )16
Kasus: (0.757813)10 = ( ... )16
Solusi:
0.757813 ∗ 16 = 12.125
0.125 ∗ 16 = 2.0
12
2
stop!
direkonstruksi: (0.C2)16
maka: (0.757813)10 = (0.C2)16
⇒ C
16-3 =
16-4 =
4096
1= 0.000244
65536
1= 1.5288 x 10-5
(30.066406)10 = (30.0)10 + (0.066406)10
Bagian-1 Bagian-2
(30.0)10 = ( ... )16
Bagian-1
Solusi:
30 ÷ 16 = 1 sisa 14
1 ÷ 16 = 0 sisa 1E 0x1E
0.0625 ∗ 16 = 1.0
1
1
0.066406 ∗ 16 = 1.0625
stop!
direkonstruksi: (0.11)16
Bagian-2
maka: (0.0066406)10 = (0.11)16
Bagian-1 + Bagian-2 = (1E.0)16 + (0.11)16 = (1E.11)16
maka: (30.066406)10 = (1E.11)16
OCT⇒DEC
Bagian-1 Bagian-2
Kasus: (0.701)8 = ( ... )10
Solusi: = (7)(8-1) + (0)(8-2) + (1)(8-3)
= (7) + 0 +
=
512
1
1010 512
449
512
1448
=
+
(0.701)8
8
1
Kasus: (71.17)8 = ( ... )10
Solusi: (71.17)8 = (71.0)8 + (0.17)8
= (1)(8-1) + (7)(8-2)
= +
=
=
64
7
1064
15
(0.17)8= (7)(81) + (1)(80)
= 56 + 1
= (57)10
(71.0)8
Bagian-1 Bagian-2
8
1
64
)78( +
Bagian-1 + Bagian-2 = (57)10 + 1064
15
maka: (71.17)8 =1064
1557
Kasus:
DEC⇒OCT Kasus: (0.421875)10 = ( ... )8
Solusi:
0.421875 ∗ 8 = 3.375
0.375 ∗ 8 = 3.0
3
3direkonstruksi: (0.33)8
maka: (0.421875)10 = (0.33)8
(71.65625)10 = ( ... )8
stop!
(71.65625)10 = (71.0)10 + (0.65625)10
Bagian-1 Bagian-2
Solusi:
(71.0)10 = ( ... )8
Bagian-1
71 ÷ 8 = 8 sisa 78 ÷ 8 = 1 sisa 0 0107
0.25 ∗ 8 = 2.0
2
5
0.65625 ∗ 8 = 5.25
stop!
direkonstruksi: (0.52)8
Bagian-2
maka: (0.65625)10 = (0.52)8
Bagian-1 + Bagian-2 = (107.0)8 + (0.52)8 = (107.52)8
maka: (71.65625)10 = (107.52)8
1 ÷ 8 = 0 sisa 1
