Análise de Espaço em Arquivos Indexados

Prévia do material em texto

Nome = Leonardo Rothier Soares Cardoso 
 
t= 2KB ( tamanho de bloco de disco ) = 2 x 1024B 
 
(Funcionarios) 
 
n = 3.500 registros 
r= 11B + 160B + 11B = 182B 
 
 
Chave Primária: CPF -> 11B 
Chave Secundária: CPF_Supervisor -> 11B 
 
Arquivo Indexado 
F = ⎣ t/r ⎦ = ⎣ (2 x 1024B)/182 ⎦ ≈ ⎣11,25⎦ = 11(piso) 
B = ⎡ n/F ⎤ = (3500 /11) ≈ 318,18 = 319 
U = t - ( F x r ) = 2 x 1024B - ( 11 x 182 B) = 2048 - 2002 = 46B 
 Consumo de espaço -> B x t = 319 x 2KB = 319 x 2 KB = 638KB ≈ 0,62 MB 
 
Índice Primário 
Fp = ⎣ (2 x 1024B)/(11B+16B) ⎦ = ⎣75,85⎦ = 75 
Bp = ⎡ B/Fp ⎤ = (319 /75) ≈ 4,25 = 5 
Up = t – (Fp x Rp) = 2048B – 75 x (11B+16B) = 23B 
Sp = t x Bp = 2KB x 5 = 10KB 
Ap = ⎡ log Bp ⎤ = ⎡ log 5 ⎤ = ⎡2,32⎤ = 3 + 1 = 4 
 
Índice Secundário 
Fp = ⎣ (2 x 1024B)/(11B+16B) ⎦ = ⎣75,85⎦ = 75 
Bp = ⎡ B/Fp ⎤ = ⎡ (3.500 /75) ⎤ ≈ ⎡46,66⎤ = 47 
Up = t – (Fp x Rp) = 2048B – 75 x (11B+16B) = 23B 
Sp = t x Bp = 2KB x 47 = 94KB 
Ap = ⎡ log Bp ⎤ = ⎡ log 47 ⎤ = ⎡5,55⎤ = 6 + 1 = 7 
 
(Clientes) 
 
n = 100.000 registros 
r = 11B + 160B + 200B + 16B + 12B + 1B = 400B 
 
Chave Primária: CPF -> 11B 
 
Arquivo Indexado 
F = ⎣ t/r ⎦ = ⎣ ( 2 x 1024B)/400B ⎦ = ⎣ 5,12 ⎦ = 5 
B = ⎡ n/F ⎤ = ⎡ 100.000 / 5 ⎤ = ⎡ 20.000 ⎤ = 20.000 
U = t - ( F x r ) = 2 x 1024B - ( 5 x 400B ) = 48B 
Consumo de espaço -> B x t = 20.000 x 2KB = 40.000 KB ≈ 39,06 MB 
 
Índice Primário 
Fp = ⎣ (2 x 1024B)/(11B+16B) ⎦ = ⎣75,85⎦ = 75 
Bp = ⎡ B/Fp ⎤ = ⎡ (20.000 /75) ⎤ ≈ ⎡266,66⎤ = 267 
Up = t – (Fp x Rp) = 2048B – 75 x (11B+16B) = 23B 
Sp = t x Bp = 2KB x 267 = 534KB 
Ap = ⎡ log Bp ⎤ = ⎡ log 267 ⎤ = ⎡8.06⎤ = 9 + 1 = 10 
 
(Aluguel) 
 
n = 20.000.000 registros 
r = 11B + 24B + 12B + 10B + 24B + 11B = 92B 
 
Chave Primária: DataLocacao -> 12B 
Chave Secundária: CPF_Cliente -> 11B, ID_Midia -> 24B, CPF_Funcionario - > 11B 
 
Arquivo Indexado 
F = ⎣ t/r ⎦ = ⎣ (2 x 1024B)/92B ⎦ = ⎣ 22,26 ⎦ = 22 
B = ⎡ n/F ⎤ = ⎡ 20.000.000 / 22 ⎤ = ⎡ 909.090,90 ⎤ = 909.091 
U = t - ( F x r ) = 2 x 1024B - ( 22 x 92B) = 24B 
Consumo de espaço -> B x t = 20.000 x 2KB = 40.000 KB ≈ 39,06 MB 
 
Índice Primário 
Fp = ⎣ (2 x 1024B)/(12B+16B) ⎦ = ⎣73,14⎦ = 73 
Bp = ⎡ B/Fp ⎤ = ⎡ (909.091/73) ⎤ = ⎡12.453,30⎤ = 12.454 
Up = t – (Fp x Rp) = 2048B – 73 x (12B+16B) = 4B 
Sp = t x Bp = 2KB x 12.454 = 24,32MB 
Ap = ⎡ log Bp ⎤ = ⎡ log 12.454 ⎤ = ⎡13.60⎤ = 14 + 1 = 15 
 
Índice Secundário - CPF_Cliente 
Fs = ⎣ (2 x 1024B)/(11B+16B) ⎦ = ⎣75,85⎦ = 75 
Bs = ⎡ n/Fs ⎤ = ⎡ (20.000.000/75) ⎤ = ⎡266.666,66⎤ = 266.667 
Us = t – (Fs x Rs) = 2048B – 75 x (11B+16B) = 2048B – 2.025B = 23B 
Ss = t x Bs = 2KB x 266.667 = 520,83MB 
As = ⎡ log Bs ⎤ = ⎡ log 266.667 ⎤ = ⎡18.02⎤ = 19 + 1 = 20 
 
Índice Secundário - ID_Midia 
Fs = ⎣ (2 x 1024B)/(24B+16B) ⎦ = ⎣51,2⎦ = 51 
Bs = ⎡ n/Fs ⎤ = ⎡ (20.000.000/51) ⎤ = ⎡392.156,86⎤ = 392.157 
Us = t – (Fs x Rs) = 2048B – 51 x (24B+16B) = 2048B – 2040B = 8B 
Ss = t x Bs = 2KB x 392.157 = 765,93MB 
As = ⎡ log Bs ⎤ = ⎡ log 392.157 ⎤ = ⎡18.58⎤ = 19 + 1 = 20 
 
Índice Secundário - CPF_Funcionario 
Fs = ⎣ (2 x 1024B)/(11B+16B) ⎦ = ⎣75,85⎦ = 75 
Bs = ⎡ n/Fs ⎤ = ⎡ (20.000.000/75) ⎤ = ⎡266.666,66⎤ = 266.667 
Us = t – (Fs x Rs) = 2048B – 75 x (11B+16B) = 2048B – 2.025B = 23B 
Ss = t x Bs = 2KB x 266.667 = 520,83MB 
As = ⎡ log Bs ⎤ = ⎡ log 266.667 ⎤ = ⎡18.02⎤ = 19 + 1 = 20 
 
(Midias) 
 
n = 10.000.000 registros 
r = 24B + 8B + 24B + 16B = 72B 
 
Chave Primária: Identificador -> 24B 
Chave Secundária: CodFilme -> 16B 
 
Arquivo Indexado 
F = ⎣ t/r ⎦ = ⎣ (2 x 1024B)/72B ⎦ ≈ ⎣ 28,44 ⎦ = 28 
B = ⎡ n/F ⎤ = ⎡ 10.000.000 / 28 ⎤ ≈ ⎡ 357.142,85 ⎤ = 357.143 
U = t - ( F x r ) = 2 x 1024B - ( 28 x 72B) = 32B 
Consumo de espaço -> B x t = 357.143 x 2KB = 714286 KB ≈ 697,54 MB 
 
Índice Primário 
Fp = ⎣ (2 x 1024B)/(24B+16B) ⎦ = ⎣51,20⎦ = 51 
Bp = ⎡ B/Fp ⎤ = ⎡ (357.143/51) ⎤ = ⎡7.002,80⎤ = 7003 
Up = t – (Fp x Rp) = 2048B – 51 x (24B+16B) = 8B 
Sp = t x Bp = 2KB x 7003 = 13,67MB 
Ap = ⎡ log Bp ⎤ = ⎡ log 7003 ⎤ = ⎡12.77⎤ = 13 + 1 = 14 
 
Índice Secundário - Cod_Filme 
Fs = ⎣ (2 x 1024B)/(16B+16B) ⎦ = ⎣64⎦ = 64 
Bs = ⎡ n/Fs ⎤ = ⎡ (10.000.000/64) ⎤ = ⎡156.250⎤ = 156.250 
Us = t – (Fs x Rs) = 2048B – 64 x (16B+16B) = 2048B – 2.048B = 0B 
Ss = t x Bs = 2KB x 156.250 = 305,17MB 
As = ⎡ log Bs ⎤ = ⎡ log 156.250 ⎤ = ⎡17.25⎤ = 18 + 1 = 19 
 
 
 
(Pagamentos) 
 
n = 50.000.000 registros 
r = 48B + 11B + 24B + 12B + 12B + 24B = 131B 
 
Chave Primária: Codigo -> 48B 
Chave Secundária: CPF_Cliente -> 11B, ID_Midia -> 24B, DataLocacao -> 12B 
 
Arquivo Indexado 
F = ⎣ t/r ⎦ = ⎣ (2 x 1024B)/131B ⎦ ≈ ⎣ 15,63 ⎦ = 15 
B = ⎡ n/F ⎤ = ⎡ 50.000.000 / 15 ⎤ ≈ ⎡ 3.333.333,3 ⎤ = 3.333.334 
U = t - ( F x r ) = 2 x 1024B - ( 15 x 131B) = 83B 
Consumo de espaço -> B x t = 3.333.334 x 2KB = 6.666.668 KB ≈ 6,35 GB 
 
Índice Primário 
Fp = ⎣ (2 x 1024B)/(48B+16B) ⎦ = ⎣32⎦ = 32 
Bp = ⎡ B/Fp ⎤ = ⎡ (3.333.334/73) ⎤ = ⎡104.166,68⎤ = 104.167 
Up = t – (Fp x Rp) = 2048B – 32 x (48B+16B) = 0B 
Sp = t x Bp = 2KB x 104.167 = 203,45MB 
Ap = ⎡ log Bp ⎤ = ⎡ log 104.167 ⎤ = ⎡16.66⎤ = 17 + 1 = 18 
 
Índice Secundário - ID_Midia 
Fs = ⎣ (2 x 1024B)/(24B+16B) ⎦ = ⎣51,2⎦ = 51 
Bs = ⎡ n/Fs ⎤ = ⎡ (50.000.000/51) ⎤ = ⎡980.392,15⎤ = 980.393 
Us = t – (Fs x Rs) = 2048B – 51 x (24B+16B) = 2048B – 2.040B = 8B 
Ss = t x Bs = 2KB x 980.393 = 1,86GB 
As = ⎡ log Bs ⎤ = ⎡ log 980.393 ⎤ = ⎡19.90⎤ = 20 + 1 = 21 
 
Índice Secundário - CPF_Cliente 
Fs = ⎣ (2 x 1024B)/(11B+16B) ⎦ = ⎣75,85⎦ = 75 
Bs = ⎡ n/Fs ⎤ = ⎡ (50.000.000/75) ⎤ = ⎡666.666,66⎤ = 666.667 
Us = t – (Fs x Rs) = 2048B – 75x (11B+16B) = 2048B – 2025B =23B 
Ss = t x Bs = 2KB x 666.667 = 1.333.334KB ~ 1,27GB 
As = ⎡ log Bs ⎤ = ⎡ log 666.667 ⎤ = ⎡19.34⎤ = 20 + 1 = 21 
 
Índice Secundário - Data_Locacao 
Fs = ⎣ (2 x 1024B)/(12B+16B) ⎦ = ⎣73,14⎦ = 73 
Bs = ⎡ n/Fs ⎤ = ⎡ (50.000.000/73) ⎤ = ⎡684.931,50⎤ = 684.932 
Us = t – (Fs x Rs) = 2048B – 73 x (12B+16B) = 2048B – 2.044B = 4B 
Ss = t x Bs = 2KB x 684.932 = 1,30GB 
As = ⎡ log Bs ⎤ = ⎡ log 684.932 ⎤ = ⎡19.38⎤ = 20 + 1 = 21 
 
 
 
(Atores) 
 
n = 10.000 registros 
r = 16B + 160B = 176B 
 
Chave Primária: Codigo -> 16B 
 
Arquivo Indexado 
F = ⎣ t/r ⎦ = ⎣ (2 x 1024B)/176B ⎦ ≈ ⎣ 11,63 ⎦ = 11 
B = ⎡ n/F ⎤ = ⎡ 10.000 / 11 ⎤ ≈ ⎡ 909,09 ⎤ = 910 
U = t - ( F x r ) = 2 x 1024B - ( 11 x 176B) = 112 
Consumo de espaço -> B x t = 910 x 2KB = 1.820 KB ≈ 1,77 MB 
 
Índice Primário 
Fp = ⎣ (2 x 1024B)/(16B+16B) ⎦ = ⎣64⎦ = 64 
Bp = ⎡ B/Fp ⎤ = ⎡ (910/64) ⎤ = ⎡14,21⎤ = 15 
Up = t – (Fp x Rp) = 2048B – 64 x (16B+16B) = 0B 
Sp = t x Bp = 2KB x 15 = 30KB 
Ap = ⎡ log Bp ⎤ = ⎡ log 15 ⎤ = ⎡3,90⎤ = 4 + 1 = 5 
 
 
 
(AtoresEmFilmes) 
 
n = 1.000.000 registros 
r = 16B + 16B = 32B 
 
Chave Primária: CodAtor -> 16B, CodFilme -> 16B 
 
Arquivo Indexado 
F = ⎣ t/r ⎦ = ⎣ (2 x 1024B)/32B ⎦ = ⎣ 64 ⎦ = 64 
B = ⎡ n/F ⎤ = ⎡ 1.000.000 / 64 ⎤ = ⎡ 15.625 ⎤ = 15.625 
U = t - ( F x r ) = 2 x 1024B - ( 64 x 32B) = 0B 
Consumo de espaço -> B x t = 15.625 x 2KB = 31.250 KB ≈ 30,51MB 
 
Índice Primário 
Fp = ⎣ (2 x 1024B)/(16B + 16B + 16B) ⎦ = ⎣42,66⎦ = 42 
Bp = ⎡ B/Fp ⎤ = ⎡ (15.625/51) ⎤ = ⎡372,02⎤ = 373 
Up = t – (Fp x Rp) = 2048B – 42 x (16B + 16B + 16B) = 32B 
Sp = t x Bp = 2KB x 373 = 746KB 
Ap = ⎡ log Bp ⎤ = ⎡ log 373 ⎤ = ⎡8,54⎤ = 9 + 1 = 10 
 
 
 
 
 
 
 
 
(Filmes) 
 
n = 2.000.000 registros 
r = 16B + 160B + 80B = 256B 
 
Chave Primária: Codigo -> 16B 
 
Arquivo Indexado 
F = ⎣ t/r ⎦ = ⎣ (2 x 1024B)/256B ⎦ = ⎣ 8 ⎦ = 8 
B = ⎡ n/F ⎤ = ⎡ 2.000.000 / 8 ⎤ = ⎡ 250.000 ⎤ = 250.000 
U = t - ( F x r ) = 2 x 1024B - ( 8 x 256B) = 0B 
Consumo de espaço -> B x t = 250.000 x 2KB = 500.000 KB ≈ 0,47 GB 
 
Índice Primário 
Fp = ⎣ (2 x 1024B)/(16B + 16B) ⎦ = ⎣64⎦ = 64 
Bp = ⎡ B/Fp ⎤ = ⎡ (250.000/64) ⎤ = ⎡3906,25⎤ = 3907 
Up = t – (Fp x Rp) = 2048B– 64 x (16B + 16B) = 0B 
Sp = t x Bp = 2KB x 3907 = 7,63MB 
Ap = ⎡ log Bp ⎤ = ⎡ log 3907 ⎤ = ⎡11.93⎤ = 12 + 1 = 13