Latihan Soal File Pile

Soal!

Parameter harddisk

• Putaran disk = 6000 rpm
• Seek time (s) = 5 ms
• Transfer rate (t) = 2048 byte/s
• T_RW = 2 ms

Parameter penyimpanan

• Metode blocking = variable length spanned blocking
• Ukuran block (B) = 1024 byte
• Ukuran pointer block (P) = 8 byte
• Ukuran interblock gap (G) = 512 byte

Parameter file

• Jumlah record difile (n) = 10600 record
• Jumlah rata-rata atribut (a') = 5 byte
• Jumlah rata-rata field (A) = 7 byte
• Jumlah rata-rata nilai (V) = 15 byte

Parameter reorganisasi

• Jumlah penambahan record (o) = 1000 record
• Jumlah record ditandai sebagai dihapus (d) = 200 byte

Carilah R, T_F, T_N, T_I, T_U, T_X, T_Y!

Jawaban:

1. R = a' (A + V + 2)
   R = 5 (7 + 15 + 2)
   R = 120
   
   
2. T_F, menggunakan metode Spanned Blocking
   Bfr = (B - P) / (R + P)
   Bfr = (1024 - 8) / (120 + 8)
   Bfr = 7,93
   
   W = P + (P + G) / Bfr
   W = 8 + (8 + 512) / 7,93
   W = 73,57
   
   t' = (^t/₂) * {R / (R + W)}
   t' = (²⁰⁴⁸/₂) * {120 / (120 + 73,57)}
   t' = 634,8
   
   T_F = ¹/₂n (R / t')
   T_F = ¹/₂10600 (120 / 634,8)
   T_F = 1001,89
   
   
3. T_N = T_F
   T_N = 1001,89
   
   
4. T_I
   r = ¹/₂ * (^{(60 * 1000)}/_RPM)
   r = ¹/₂ * (^{(60 * 1000)}/₆₀₀₀)
   r = 5
   
   Btt = ^B/_t
   Btt = ¹⁰²⁴/2048
   Btt = 0,5
   
   T_I = s + r + btt + T_RW
   T_I = 5 + 5 + 0,6 + 2
   T_I = 12,5
   
   
5. T_U = T_F + T_RW + T_I
   T_U = 1001,89 + 2 + 12,5
   T_U = 1016,39
   
   T_U = T_F + T_RW
   T_U = 1001,89 + 2
   T_U = 1003,89
   
   
6. T_X = 2T_F
   T_X = 2*1001,89
   T_X = 2003,78
   
   
7. T_Y = (n + o)(R + t') + (n + o - d)(R / t')
   T_Y = (10600 + 1000)(120 + 634,8) + (10600 + 1000 - 200)(120 / 634,8)
   T_Y = 4140

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