※ 引述《aweila75 (David)》之铭言： : 9.下列何种排程演算法不可能导致饥饿现象： : a.FIFO b.FILO c.SJF d.Longest Job First : 答案应该是d吧？ : 11.网路卡位於OSI哪一层？ : a. 2 b. 3 c.4 d.5 : 答案是a吗？ : 13.下列何种装置不是OSI第一层装置？ : a.Hub b.Transceiver c.bridge d.repeater : (不知道) : 16.下列何者是使用动态连结的优点？ : a.较小的执行档案 b.较短的执行时间 c.较容易维修 d.较依赖外部档案 : 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) : 好像知道他问什麽又好像不知道，我觉得答案是d？ : 烦请解释题目一下。 : 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不知道对吗？ : 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)语汇分析阶段→语法分析阶段→解释阶段→与机器无关的最佳化阶段→储存位置的 : 分配阶段→数码产生阶段→组合及输出阶段 : (答案是这个吗？) : (b)???请解答ꄊ: (c)linker是将编译过的程式连结其他子程式与程式库，产生可执行模组。 : loader是在linker完成可执行模组後将其载入主记忆体执行的角色。(这题没写错吧?) 补充： 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 资料量: 2sector = 2*512 byte = 1KB 1/2401sec = 0.4164 ms 1) + 2) + 3) = 7ms + 4.16ms + 0.4164ms = 11.5764ms 这样算是否正确？ --

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