→ aweila75:有誰可以跟我說一下，我算的是否正確嗎？ 05/24 23:25
※ 引述《aweila75 (David)》之銘言：
: 9.下列何種排程演算法不可能導致飢餓現象：
: a.FIFO b.FILO c.SJF d.Longest Job First
: 答案應該是d吧？
: 11.網路卡位於OSI哪一層？
: a. 2 b. 3 c.4 d.5
: 答案是a嗎？
: 13.下列何種裝置不是OSI第一層裝置？
: a.Hub b.Transceiver c.bridge d.repeater
: (不知道)
: 16.下列何者是使用動態連結的優點？
: a.較小的執行檔案 b.較短的執行時間 c.較容易維修 d.較依賴外部檔案
: 答案是a吧？
: 18.what is the lower bound of comparison sorts under the von neumann
: architecture?
: a. O(n^2) b.O(nlogn) c.O(n) d.O(logn)
: 好像知道他問什麼又好像不知道，我覺得答案是d？
: 煩請解釋題目一下。
: 3.Suppose the access times for main memory and cache are 50 ns and 5 ns,
: What is the overall average memory access time if the cache hit rate is 80%?
: 我算是11不知道對嗎？
: 5.(a)What are the phases (in correct order) in the compilation processo f
: converting a C program into an executable binary program?
: (b) In which phase of part(a), may an assembler be used?
: (c) What are the roles of loader and linker in staring up an executable
: program?
: (a)語彙分析階段→語法分析階段→解釋階段→與機器無關的最佳化階段→儲存位置的
: 分配階段→數碼產生階段→組合及輸出階段
: (答案是這個嗎？)
: (b)???請解答ꄊ: (c)linker是將編譯過的程式連結其他子程式與程式庫，產生可執行模組。
: loader是在linker完成可執行模組後將其載入主記憶體執行的角色。(這題沒寫錯吧?)
補充：
4. A hard disk drive has the following characteristics:
rotation speed = 7200 rev/min
access arm movement time = 1ms fixed start-up time + 0.02 ms for each track
crossed ( the 1 ms time is a constant no matter how far the arm moves.)
Number of surfaces = 10
Number of tracks per surface = 1024
Number of sectors per track = 40
Number of bytes per sector = 512
(a) What is the capacity of this disk?
(b) Assume that the read/write head moves for about 300 tracks on average for
every seek. What is the average access time for transferring 2 consecutive
sectors of data?
我算的如下：
(a) 10*1024*40*512bytes = 200MB
(b) 1)seek time: 1ms + 0.02ms*300 = 7ms
2)rotation time: (60/7200)*1/2 = 4.16ms
3)data transfer time: 40*512bytes = 20480 bytes
20480bytes/0.00833(sec) = 2458583 bytes/sec
= 2401KB/sec
資料量: 2sector = 2*512 byte = 1KB
1/2401sec = 0.4164 ms
1) + 2) + 3) = 7ms + 4.16ms + 0.4164ms = 11.5764ms
這樣算是否正確？
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