UTS Organisasi dan Arsitektur Komputer, Kelas D’11 FTIKOM UNMUL

1. Nyatakan bilangan-bilangan decimal berikut dalam sistem bilangan : a. biner, b. octal, c. hexadecimal.

21, 137, 320, 161, 2400, 9101.
Jawab :
a. Biner

• 21

21 : 2 = 10 sisa 1
10 : 2 = 5 sisa 0
5 : 2 = 2 sisa 1
2 : 2 = 1 sisa 0
1 : 2 = 1 sisa 1
21 = 00010101

• 137

137 : 2 = 68 sisa 1
68 : 2 = 34 sisa 0
34 : 2 = 17 sisa 0
17 : 2 = 8 sisa 1
8 : 2 = 4 sisa 0
4 : 2 = 2 sisa 0
2 : 2 = 1 sisa 0
1 : 2 = 1 sisa 1
137 = 10001001

• 320

320 : 2 = 160 sisa 0
160 : 2 = 80 sisa 0
80 : 2 = 40 sisa 0
40 : 2 = 20 sisa 0
20 : 2 = 10 sisa 0
10 : 2 = 5 sisa 0
5 : 2 = 2 sisa 1
2 : 2 = 1 sisa 0
1 : 2 = 1 sisa 1
320 = 101000000

• 161

161 : 2 = 80 sisa 1
80 : 2 = 40 sisa 0
40 : 2 = 20 sisa 0
20 : 2 = 10 sisa 0
10 : 2 = 5 sisa 0
5 : 2 = 2 sisa 1
2: 2 = 1 sisa 0
1 : 2 = 1 sisa 1
161 = 10100001

• 2400

2400 : 2 = 1200 sisa 0
1200 : 2 = 600 sisa 0
600 : 2 = 300 sisa 0
300 : 2 = 150 sisa 0
150 : 2 = 75 sisa 0
75 : 2 = 37 sisa 1
37 : 2 = 18 sisa 1
18 : 2 = 9 sisa 0
9 : 2 = 4 sisa 1
4 : 2 = 2 sisa 0
2 : 2 = 1 sisa 0
1 : 2 = 1 sisa 1
2400 = 100101100000

• 9101

9101 : 2 = 4550 sisa 1
4550 : 2 = 2275 sisa 0
2275 : 2 = 1137 sisa 1
1137 : 2 = 568 sisa 1
568 : 2 = 284 sisa 0
284 : 2 = 142 sisa 0
142 : 2 = 71 sisa 0
71 : 2 = 35 sisa 1
35 : 2 = 17 sisa 1
17 : 2 = 8 sisa 1
8 : 2 = 4 sisa 0
4 : 2 = 2 sisa 0
2 : 2 = 1 sisa 0
1 : 2 = 1 sisa 1
9101 = 10001110001101
b. Octal

• 21

21 : 8 = 2 sisa 5
2 : 8 = 1 sisa 2
21 = 25

• 137

137 : 8 = 17 sisa 1
17 : 8 = 2 sisa 1
2 : 8 = 1 sisa 2
137 = 211

• 320

320 : 8 = 40 sisa 0
40 : 8 = 5 sisa 0
5 : 8 = 1 sisa 5
320 = 500

• 161

161 : 8 = 20 sisa 1
20 : 8 = 2 sisa 4
2 : 8 = 1 sisa 2
161 = 241

• 2400

2400 : 8 = 300 sisa 0
300 : 8 = 37 sisa 4
37 : 8 = 4 sisa 5
4 : 8 = 1 sisa 4
2400 = 4540

• 9101

9101 : 8 = 1137 sisa 5
1137 : 8 = 142 sisa 1
142 : 8 = 17 sisa 6
17 : 8 = 2 sisa 1
2 : 8 = 1 sisa 2
9101 = 21615
c. Hexadesima

• 21

21 : 16 = 1 sisa 5
1 : 16 = 1 sisa 1
21 = 15

• 137

137 : 16 = 8 sisa 9
8 : 16 = 1 sisa 8
137 = 89

• 320

320 : 16 = 20 sisa 0
20 : 16 = 1 sisa 4
1 : 16 = 1 sisa 1
320 = 140

• 161

161 : 16 = 10 sisa 1
10 : 16 = 1 sisa 10 = A
161 = A1

• 2400

2400 : 16 = 150 sisa 0
150 : 16 = 9 sisa 6
9 : 16 = 1 sisa 9
2400 = 960

• 9101

9101 : 16 = 568 sisa 13 = D
568 : 16 = 35 sisa 8
35 : 16 = 2 sisa 3
2 : 16 = 1 sisa 2
9101 = 238D

2. Binerkan kata berikut ini “Fakultas F-TIKOM”

• F = 46

(4 x 16¹) + (6 x 16⁰)
= 64 + 6 = 70
01000110

• a = 61

(6 x 16¹) + (1 x 16⁰)
= 96 + 1 = 97
01100001

• k = 6B

(6 x 16¹) + (11 x 16⁰)
= 96 + 11 = 107
01101011

• u = 75

(7 x 16¹) + (5 x 16⁰)
= 112 + 5 = 117
01110101

• l = 6C

(6 x 16¹) + (6 x 16⁰)
= 96 + 12 = 108
01101100

• t =74

(7 x 16¹) + (4 x 16⁰)
= 112 + 4 = 116
01110100

• a = 61

(6 x 16¹) + (1 x 16⁰)
= 96 + 1 = 97
01100001

• s = 73

(7 x 16¹) + (3 x 16⁰)
= 112 + 3 = 115
01110011

• F = 46

(4 x 16¹) + (6 x 16⁰)
= 64 + 6 = 70
01000110

• - = 2D

(2 x 16¹) + (13 x 16⁰)
= 32 + 13 = 45
00101101

• T = 54

(5 x 16¹) + (4 x 16⁰)
= 80 + 4 = 84
01010100

• I = 49

(4 x 16¹) + (9 x 16⁰)
= 64 + 9 = 73
01001001

• K = 4B

(4 x 16¹) + (11 x 16⁰)
= 64 + 11 = 75
01001011

• O = 4F

(4 x 16¹) + (15 x 16⁰)
= 64 + 15 = 79
01001111

• M = 4D

(4 x 16¹) + (13 x 16⁰)
= 64 + 13 = 77
01001101

Fakultas F-TIKOM = 01000110 01100001 01101011 01110101 01101100 01110100 01100001 01110011 01000110 00101101 01010100 01001001 01001011 01001111
3. Sebuah harddisk 2 Tb berisikan file-file sebagai berikut :
a. File Video 3.210.987, dengan per file 160 Mb dengan durasi 60 menit dengan penjelasan setengah dari file tersebut durasi 30 menit dan setengahnya 49 menit.
b. File Save Game 9.456, dengan per file 190 Kb
c. File Gambar 19.190, dengan per file 19.191.199 bit
Jika harddisk tersebut telah berisikan sistem 190.199.199 bit, berapakah sisa kapasitas dari harddisk tersebut?
Jawab :

• Harddisk 2 Tb = 2048 Gb = 2097152 Mb = 2147483648 Kb = 2199023255552 byte = 17592186044416 bit
• Sistem = 190.199.199 bit

a. 160 Mb = 163840 Kb = 167772160 byte = 171798691840 bit
File Video 3.210.987  x 171798691840 bit = 551643366115246080 bit
b. 190 Kb = 194560 byte = 199229440 bit
File Save Game 9.456 x 199229440 bit = 1883913584640 bit
c. File Gambar 19.190 x 19.191.199 bit = 368279108810 bit
551643366115246080 bit + 1883913584640 bit + 368279108810 bit + 190199199 bit
= 551645618498138729 bit
kapasitas hdd – 551645618498138729 bit
= 17592186044416 bit – 551645618498138729 bit = -551628026312094313 bit