1. nyatakan
bilangan-bilangan decimal berikut dalam system bilangan :
a) biner , b) octal , c) hexadecimal
a) biner , b) octal , c) hexadecimal
21 137 320 161 2400 9101
2.
binerkan kata berikut ini Fakultas-TIKOM
3. sebuah
hard diks 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 hard
diks tersebut telah berisikan system 190.199.199 bit, Berapakah sisa kapasitas
dari hard diks tersebut?
1.
A. biner
21= 11101 137= 10001001 320 =
101000000 161 = 10100001
21 : 2 = 10 => 1 137
: 2 = 68 => 1 320
: 2 = 160 => 0 161 : 2 =
80 => 1
10 : 2 = 5 => 0 68 : 2 = 34 => 0 160 : 2 = 80
=> 0 80 : 2 = 40 => 0
5 : 2 = 2 => 1 34
: 2 = 17 => 0 80
: 2 = 40 => 0 40 : 2
= 20 => 0
2 : 2 = 1 => 1 17
: 2 = 8 => 1 40
: 2 = 20 => 0 20 : 2
= 10 => 0
1 : 2 = 0 =>1 8
: 2 = 4 => 0 20
: 2 = 10 => 0 10 : 2
= 5 => 0
4
: 2 = 2 => 0 10
: 2 = 5 => 0 5 : 2
= 2 => 1
2
: 2 = 1 => 0 5
: 2 = 2 => 1 2 :
2 = 1 => 0
1
: 2 = 0 => 1 2
: 2 = 1 => 0 1 :
2 = 0 => 1
1
: 2 = 0 => 1
2400 = 10010110000 9101
= 10001100011101
2400 : 2 = 1200 => 0 9101 : 2
= 4550 => 1
1200 : 2 = 600 => 0 4550
: 2 = 2275 => 0
600 : 2 = 300 => 0 2275
: 2 = 1137 => 1
150 : 2 = 75 => 0 1137
: 2 = 568 => 1
75 : 2 = 37 => 1 568
: 2 = 284 => 0
37 : 2 = 18 => 1 284
: 2 = 142 => 0
18 : 2 = 9 => 0 142 : 2 = 71 => 0
9 : 2 = 4 => 1 71
: 2 = 35 => 1
4 : 2 = 2 => 0 35
: 2 = 17 => 1
2 : 2 = 1 => 0 17 : 2 = 8 => 1
1 : 2 = 0 => 1 8
: 2 = 4 => 0
4 : 2 = 2 => 0
2
: 2 = 1 => 0
1
: 2 = 0 => 1
1 B. octaL
21 : 8 = 2 sisa 5
2 : 8 = 0 sisa 2
Hasil dari 21(8) adalah 25
137 : 8 = 17 sisa 1
17 : 8 = 2 sisa
1
2 : 8 = 0
sisa 2
Hasil dari 137(8) adalah 211
320 : 8 = 40 sisa 0
40 : 8 = 5 sisa
0
5 : 8 = 0
sisa 5
Hasil dari 320(8) adalah 500
161 : 8 = 20 sisa 1
20 : 8 = 2 sisa
4
2 : 8 = 0
sisa 2
Hasil dari 161(8) adalah 241
2400 : 8 = 300 sisa 0
300 : 8 = 37
sisa 4
37 : 8 = 4
sisa 5
4 : 8 = 0
sisa 4
Hasil dari 2400(8) adalah 4540
9101 : 8 = 1137 sisa 5
1137 : 8 = 142
sisa 1
142 : 8 = 17
sisa 6
17 : 8 = 2
sisa 1
2 : 8 = 0
sisa 2
Hasil dari 9101(8) adalah 21615
1
c. Hexadesimal
21 : 16 = 1 sisa 5
1 : 16 = 0 sisa 1
Hasil dari 21(16) adalah 15
137 : 16 = 8 sisa 9
8 : 16 = 0
sisa 8
Hasil dari 137(16) adalah 89
320 : 16 = 20 sisa 0
20 : 16 = 1
sisa 4
1 : 16 =
0 sisa 1
Hasil dari 320(16) adalah 140
161 : 16 = 10 sisa 1
10 : 16 = 0 sisa
10 (10=A)
Hasil dari 161(16) adalah A1
2400 : 16 = 150 sisa 0
150 : 16 = 9
sisa 6
9 : 16 = 0
sisa 9
Hasil dari 2400(16) adalah 960
9101 : 16 = 568 sisa 13
(13=D)
568 : 16 = 35
sisa 8
35 : 16 = 2
sisa 3
2 : 16 = 0
sisa 2
Hasil dari 9101(16) adalah 238D
2.
Fakultas-TIKOM
F =46 a
= 61 k
= 611
(4*161)+(6*160)=64+6=70 (6*161)+(1*160)= 96+1 = 97 (6*16)+(11*16)= 96 + 11 = 107
70 : 2 = 35 => 0 97
: 2 = 48 => 1 107
: 2 = 53 => 1
35 : 2 = 17 => 1 48
: 2 = 24 => 0 53
: 2 = 26 => 1
17 : 2 = 8 => 1 24
: 2 = 12 => 0 26
: 2 = 13 => 0
8 : 2 = 4 => 0 12
: 2 = 6 => 0 13
: 2 = 6 => 1
4 : 2 = 2 => 0 6
: 2 = 3 => 0 6
: 2 = 3 => 0
2 : 2 = 1 => 0 3
: 2 = 1 => 1 3
: 2 = 1 => 1
1 : 2 = 0 => 1 1
: 2 = 0 => 1 1
: 2 = 0 => 1
Biner : 01000110 Biner
: 01100001 Biner
: 01101011
u = 75 l=
612
(7*161)+(5*160)= 112 + 5 = 117 (6*161)+(12*160)=
96 + 12 =108
117 : 2 = 58 => 1 108
: 2 = 54 => 0
58 : 2 = 29 => 0 54
: 2 = 27 => 0
29 : 2 = 14 => 1 27
: 2 = 13 => 1
14 : 2 = 7 => 0 13
: 2 = 6 => 1
7 : 2 = 3 => 1 6
: 2 = 3 => 0
3 : 2 = 1 => 1 3
: 2 = 1 => 1
1 : 2 = 0 => 1 1
: 2 = 0 => 1
Biner : 01110101 biner
: 01101100
t : 74 a
= 61 s
= 73
(7*161)+(4*160)= 112 + 4 = 116 (6*161)+(1*160)=
96+1 = 97 (7*161)+(3*160)=
112 + 3 = 115
116 : 2 = 63 => 0 97
: 2 = 48 => 1 115
: 2 = 62 => 1
63 : 2 = 31 => 1 48
: 2 = 24 => 0 62
: 2 = 31 => 0
31 : 2 = 15 => 1 24
: 2 = 12 => 0 31
: 2 = 15 => 1
15 : 2 = 7 => 1 12
: 2 = 6 => 0 15
: 2 = 7 => 1
7 : 2 = 3 => 1 6
: 2 = 3 => 0 7
: 2 = 3 => 1
3 : 2 = 1 => 1 3
: 2 = 1 => 1 3
: 2 = 1 => 1
1 : 2 = 0 => 1 1
: 2 = 0 => 1 1
: 2 = 0 => 1
Biner : 01111110 Biner
: 01100001 Biner
: 01111101
-
= 213 T=
54
(2*161)+(13*160)= 32 + 13 = 45 (5*161)+(4*160)=
80 + 4 = 84
45 : 2 = 22 => 1 84
: 2 = 42 => 0
22 : 2 = 11 => 0 42
: 2 = 21 => 0
11 : 2 = 5 => 1 21
: 2 = 10 => 1
5 : 2 = 2 => 1 10
: 2 = 5 => 0
2 : 2 = 1 => 0 5
: 2 = 2 => 1
1 : 2 = 0 => 1 2
: 2 = 1 => 0
Biner : 0101101 1
: 2 = 0 => 1
Biner
: 01010100
I = 49 K
= 411
(4*161)+(9*160)= 64 + 9 = 73 (4*161)+(11*160)=
64 + 11 = 75
73 : 2 = 36 => 1 75
: 2 = 37 => 1
36 : 2 = 18 => 0 37
: 2 = 18 => 1
18 : 2 = 9 => 0 18
: 2 = 9 => 0
9 : 2 = 4 => 1 9
: 2 = 4 => 1
4 : 2 = 2 => 0 4
: 2 = 2 => 0
2 : 2 = 1 => 0 2
: 2 = 1 => 0
1 : 2 = 0 => 1 1
: 2 = 0 => 1
Biner : 01001001 Biner : 01001011
Biner : 01001001 Biner : 01001011
O = 415 M=
413
(4*161)+(15*160)= 64 + 15= 79 (4*161)+(13*160)=
64 + 13 = 77
79 : 2 = 39 => 1 77
: 2 = 38 => 1
39 : 2 = 19 => 1 38
: 2 = 19 => 0
19 : 2 = 9 => 1 19
: 2 = 9 => 1
9 : 2 = 4 => 1 9
: 2 = 4 => 1
4 : 2 = 2 => 0 4
: 2 = 2 => 0
4 : 2 = 2 => 0 4
: 2 = 2 => 0
1 : 2 = 0 => 1 1
: 2 = 0 => 1
Biner : 01001111 Biner
: 01001101
Dari hasil ilustrasi
di atas maka bilangan biner dari “ Fakultas-TIKOM” adalah “
01000110 01100001 01101011 01110101 01101100 01111110 01100001 01111101 0101101 01010100 01001001 01001011 01001111 01001101 ”
3.
Hitunglah sisa hard diks
Diketahui :
1
byte = 8 bit
1
kilobyte = 1024 byte
1
megabyte = 1024 kilobyte
1
gigabyte = 1024 megabyte
1
terabyte = 1024 gigabyte
2Tb = 1.099.511.627.776Byte * 2 = 2.199.023.255.552 byte * 8 =
17.529.186.044.416 bit
a). 160 Mb = 1.048. 576 * 160 = 167.772.160 Byte* 3.210.987 file =
538.714.224.721.920 byte* 8 =
4.309.713.797.775.360 bit
b). 190 Kb =1024 * 190 = 194.560Byte * 9456 File = 1.839.759.360 Byte *
8 = 14.706.674.880 bit
c). 19. 191.199 * 19.190 = 368.279.108.810 bit
hard diks telah berisi 190.199.199 bit
190.199.199bit + 4.309.713.797.775.360bit + 14.706.674.880bit +
368.279.108.810bit = 4.310.096.973.758.249 bit
17.529.186.044.416 bit - 4.310.096.973.758.249 bit = - 4292567787713833
bit
jadi hard diks tersisa - 4292567787713833 bit
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