Latihan Soal Organisasi File Sequential

Soal!

Parameter harddisk

• Putaran disk = 8000 rpm
• Seek time (s) = 5 ms
• Transfer rate (t) = 2048 byte/s
• T_RW = 2 ms

Parameter penyimpanan

• Metode blocking = fixed blocking
• Ukuran block (B) = 4096 byte
• Ukuran pointer block (P) = 8 byte
• Ukuran interblock gap (G) = 1024 byte

Parameter file

• Jumlah record file (n) = 100000 record
• Jumlah field (a) = 8 byte
• Jumlah nilai (V) = 25 byte

Parameter reorganisasi

• File log transaksi (o) = 0 record

Parameter pemrosesan

• Waktu untuk pemrosesan block (c) = 2 ms

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

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Jawaban:

1. R = a.V
   R = 8.25
   R = 200
   
   
2. T_F, menggunakan metode fixed blocking
   Bfr = ^B/_R
   Bfr = ⁴⁰⁹⁶/₂₀₀
   Bfr = 20,48
   
   W_G = ^G / _Bfr
   W_G = ¹⁰²⁴ / _20,48
   W_G = 50
   
   W_R = ^B / _Bfr
   W_R = ⁴⁰⁹⁶ / _20,48
   W_R = 200
   
   W = W_G + W_R
   W = 50 + 200
   W = 250
   
   t' = (^t/₂) * {R / (R + W)}
   t' = (²⁰⁴⁸/₂) * {200 / (200 + 250)}
   t' = 450,56
   
   T_F = ¹/₂n * (^R/_t')
   T_F = ¹/₂100000 * (²⁰⁰/_450,56)
   T_F = 22000
   
   
3. T_N
   Btt = ^B/_t
   Btt = ⁴⁰⁹⁶/₂₀₄₈
   Btt = 2
   
   T_N = ^btt / _Bfr
   T_N = ² / _20,48
   T_N = 0,09
   
   
4. T_I = T_F + ¹/₂(^N/_Bfr)(^btt/_TRW)
   T_I = 22000 + ¹/₂(¹⁰⁰⁰⁰⁰/_20,48)(²/₂)
   T_I = 24441,4
   
   
5. T_U = T_F + T_RW
   T_U = 22000 + 2
   T_U = 22002
   
   
6. T_X = T_sort(o) + (n + o)(^R/_t')
   T_X = 0 + (100000 + 0)(²⁰⁰/_450,56)
   T_X = 44000
   
   
7. T_Y = T_sort (o) + 2 (n + o)(^R/_t')
   T_Y = 0 + 2 (100000 + 0)(²⁰⁰/_450,56)
   T_Y = 88000
   
   

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