Re: [理工] [OS] I/O operation

[dendrobium (石斛蘭)]

※ 引述《flyguava (紅芭樂)》之銘言：
: Consider a file currently consisting of 200 blocks. Assume that the file
: control block (and the index block, in the case of indexed allocation) is
: already in memory. Calculate how many disk I/O operations are required for
: contiguous, linked, and indexed (single-level) allocation strategies. Assume
: that each disk I/O operation accesses only one block. In the contiguous
: allocation case, there is no room to grow in the beginning, but there is room
: to grow in the end. Moreover, the block information to be added is already
: stored in memory.
: (a). The block is added at the beginning.
contiguous : 200(讀) + 201(寫) = 401
linked     : 1(寫)             = 1
indexed    : 1(寫)             = 1

: (b). The block is added in the middle.
contiguous : 100(讀) + 101(寫) = 201
linked     : 100(讀) + 2(寫)   = 102
indexed    : 1(寫)             = 1

: (c). The block is added at the end.
contiguous : 1(寫)             = 1
linked     : 1(讀) + 2(寫)     = 3
indexed    : 1(寫)             = 1


: (d).The block is removed from the beginning.
contiguous : 199(讀) + 199(寫) = 398
linked     : 1(讀)             = 1
indexed    : 0                 = 0

: (e).The block is removed from the middle.
contiguous : 100(讀) + 100(寫) = 200
linked     : 101(讀) + 1(寫)   = 102
indexed    : 0                 = 0

參考看看^^a

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推 flyguava:先謝謝^^  但我想問e 小題 contiguoos 為何不是99+99?
→ flyguava:linked 不是100+1 呢?  謝謝
→ dendrobium:因為我想的middle 是第101個block (從1開始算XD
推 flyguava:恩恩  懂了  感恩^^
推 fef92:請問(c)linked 為什麼只要讀1次就知道最後一個block在哪
→ dendrobium:因為physical directory有紀錄最後一個block的位置