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Organisasi File Sequential

Organisasi File Sequential. Sequential File. Sequential Access. File Sequential. Adanya keberurutan record – record di dalam file menurut kriteria  Ordered File Karakteristik : Record berisi semua nilai atribut dengan posisi yang sama .

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Organisasi File Sequential

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  1. Organisasi File Sequential

  2. Sequential File File Sequential

  3. Sequential Access File Sequential

  4. File Sequential • Adanyakeberurutan record – record didalam file menurutkriteria Ordered File • Karakteristik : • Record berisisemuanilaiatributdenganposisi yang sama. • Adanyaaturan / kriteriatertentu yang menjadikuncipengurutan data. Kuncibersifatunik. • Umumnyaterdapatpada media yang lama (cards, tapes). • Secarafisik, record diurutkanberdasarkankunci primary. • Tidakdapat direct access terhadap record yang diinginkan. • Pencariansekuensialsampai record ditemukan. • Pencarianbinerdigunakanuntukmempercepatakses (harusdiketahuiukuran file danposisitengah file). File Sequential

  5. File Sequential • Namaatributtidakperluditulispadasetiap record, tapimunculpada file header. • Denganadanyakonstrainsekuensdan record tetapmakaterjadipeningkatanefisiensi, tetapiadapenurunanfleksibilitas. • Record – record harusdijagaberdasarkanatributkunci. • Penyisipandilakukandiakhir file ataudi slot kosongakibatpenghapusan record. File Sequential

  6. File Sequential • Penyisipandilakukandenganmenggunakan file transaction log. Jikaukuran file log sudahcukupbesar, makadilakukanreorganisasi. • Secaraperiodikdilakukan merge antara file log dan file utama / master file. • Komponen : • File utama • File transaction log  berupastruktur Pile File Sequential

  7. Penyisipan Record • Penyisipan • Lambat • Pencarian sequential untukmencariposisi yang akanditempati record. • Jikaadatempat yang cukuppadahalaman yang dicari, makatulis record. • Jikatidakcukuptempat, makaakandipindahkansejumlah record kehalamanberikutnya. • Jikatidakadatempat yang kosong, makaakandilakukanpenyusunan yang berulang-ulangsampaiditemukantempat yang cukup. • Dapatmenggunakan “overflow” untukmempersingkatwaktu. File Sequential

  8. ModifikasidanPenghapusan Record • Modifikasi • Lambat • Pencarian sequential • Melakukanmodifikasi • Penulisanulang record • Penghapusan • Lambat • Pencarian sequential • Memberitandapada record ataumengosongkantempatdari record yang dihapus • Penulisanulang record File Sequential

  9. Kinerja File Sequential • R = a. V a : Jumlahatributpada 1 record V : Panjang rata – rata nilaiatribut (byte) • Fetch Record (TF) • Pencarianmenggunakanatribut non-kunci • Belumada file log  rata – rata ½ file akanditelusuri TF = ½ waktupencarianseluruhblok TF = ½ b. B / t’ = ½ n R / t’ • Sudahada file log TFo = ½ o R / t’ TF = ½ (n + o) R / t’ File Sequential

  10. Kinerja File Sequential • Fetch Record (TF) • Pencarianmenggunakanatributkunci (pencarianbiner) • Belumterbentuk log TF = 2log (b) (s + r + btt + c) TF = 2log (n / Bfr) (s + r + btt + c) • Sudahterbentuk log TF = 2log (n / Bfr) (s + r + btt + c) + TFO TF = 2log (n / Bfr) (s + r + btt + c) + ½ o (R / t’) • Waktuuntukmendapatkan 1 record berikutnya (TN) TN = waktu transfer 1 blok x peluangditemukannya record dalamblok yang sama TN = btt / Bfr = R / t File Sequential

  11. Kinerja File Sequential • Waktupenyisipan record baru (TI) • Cari, geser, sisip TI = TF + ½ (n / Bfr) (btt + TRW) • Memakai log file TI = s + r + TRW + (TY / o) • Waktu Update (TU) • Bukankunci TU = TF + TRW • Terhadapkunci : cari record, hapus record, sisipkan record TU = TF (main) + TI (file log) File Sequential

  12. Kinerja File Sequential • Waktupembacaanseluruh record (TX) TX = Tsort (o) + (n + o) R / t’ • Waktureorganisasi File (TY) TY = Tsort (o) + nold (R / t’) + o (R / t’) + nnew (R / t’) TY = Tsort (o) + 2 (n + o) (R / t’) • Tsort (o) = 2log (o / btt) File Sequential

  13. Soal Latihan • Diketahui file sequential : • Putaran disk = 8000 rpm • Seek time (s) = 5 ms = 0,005 s • Transfer rate (t) = 2048 byte/s • TRW = 2r • Ukuran blok (B) = 4096 byte • Ukuran pointer blok (P) = 8 byte • IBG (G) = 1024 byte • Jumlah record pada file (n) = 100000 record • Jumlah field (a) = 8 field • Panjang nilai (V) = 25 byte • Jumlah record file log (o) = 5000 record • Waktu pemrosesan (c) = 2 ms = 0,002 s File Sequential

  14. Soal Latihan • Hitung : R, TF, TN, TI, TU, TX, TY jika metode blocking : • Fixed • Variable-length spanned • Variable-length unspanned File Sequential

  15. Pembahasan Soal Latihan Metode Fixed Blocking • R = a. V = 8. 25 = 200 byte • TF (Non Kunci ) = ½ (n + o) R / t’ Bfr = B / R = 4096 / 200 = 20,48 = 20 record W = G / Bfr = 1024 / 20 = 51 byte t’ = (t / 2) (R / (R + W)) = (2048 / 2) (200 / (200 + 51)) = 1024 (0,797) = 815,94 s TF = ½ (n + o) R / t’ = ½ (100000 + 5000) (200 / 815,94) = 52500 (0,245) = 12868,59 s File Sequential

  16. Pembahasan Soal Latihan • TF (Kunci) = 2log (n / Bfr) (s + r + btt + c) + TFO r = (60 . 1000) / (2 rpm) = 60000 / (2. 8000) = 3,75 ms = 0,00375 s btt = B / t = 4096 / 2048 = 2 s TFo = ½ o R / t’ = ½ (5000) (200 / 815,94) = 612,79 s TF = 2log (n / Bfr) (s + r + btt + c) + TFO = 2log (100000 / 20) (0,005 + 0,00375 + 2 + 0,002) + 612,79 = 2log (5000) (2,01075) + 612,79 = 12,29 (2,01075) + 612,79 = 637,5 s • TN = btt / Bfr = 2 / 20 = 0,1 s File Sequential

  17. Pembahasan Soal Latihan • TI = s + r + TRW + (TY / o) Tsort (o) = 2log (o / btt) = 2log (5000 / 2) = 11,29 s TY = Tsort (o) + 2 (n + o) (R / t’) = 11,29 + 2 (100000 + 5000) (200 / 815,94) = 11,29 + 210000 (0,245) = 51461,29 s TI = s + r + TRW + (TY / o) = 0,005 + r + 2r + (51461,29 / 5000) = 0,005 + 3r + 10,29 = 0,005 + 3 (0,00375) + 10,29 = 10,31 s • TU (Non Kunci) = TF + TRW = 12868,59 + 2 (0,00375) = 12868,6 s File Sequential

  18. Pembahasan Soal Latihan • TU (Kunci) = TF (main) + TI (file log) TF (main) = 2log (n / Bfr) (s + r + btt + c) = 2log (100000 / 20) (0,005 + 0,00375 + 2 + 0,002) = 2log (5000) (2,01075) = 24,71 s TU = TF (main) + TI (file log) = 24,71 + 10,31 = 35,02 s • TX = Tsort (o) + (n + o) R / t’ = 11,29 + (100000 + 5000) (200 / 815,94) = 11,29 + 105000 (0,245) = 25736,29 s • TY = 51461,29 s File Sequential

  19. Pembahasan Soal Latihan • Lanjutkan penyelesaian latihan dengan metode Variable-Length Spanned dan Variable-Length Unspanned File Sequential

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