Rabu, 11 April 2012

SOAL DAN JAWABAN KUIS 1 ORKOM

1. nyatakan bilangan-bilangan decimal berikut dalam system bilangan :
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

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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