課程名稱︰作業系統 課程性質︰必修 課程教師︰薛智文 開課學院：電機資訊學院 開課系所︰資訊工程學系 考試日期（年月日）︰100/11/10 考試時限（分鐘）：180 是否需發放獎勵金：是 （如未明確表示，則不予發放） 試題 : This is a open-book and open-own-note exam. Please do answer with your own SUCCINCT words in the order of question number in EXACTLY one given answer sheet. You can write in Chinese and keep this paper. If you would like to design your own project 3, send me a one-page proposal as soon as possible. Register your team members of project 3 in facebook by 11/24. Good luck. 1. When describing message passing, what is the difference between synchronous or asynchronous and blocking or non-blocking? [5%] 2. For a system of multiprogramming or multitasking, what kind of applications might be more suitable, respectively, in athe environment? Why? [10%] 3. What is the purpose of upcall? [5%] If your OS does not support upcall such as Linux, how do you implement the same function of upcall? [5%] 4. Vitualization enables multiple OSes running simultaneously on a physical machine. What is "system consolidation" in a virtualized environment? [5%] In what situation, we still need one OS running on a virtualized environment? [5%] 5. Exclude caching; describe 2 examples where Operating System and Computer Architecture help each other to improve performance? [10%] 6. What are the differences between kernel thread and user thread? [5%] In what situation, applications using kernel threads would always have a better performance than using user threads? [5%] 7. In what situations, thread pool is not beneficial? [5%] Give example application for each situation? [5%] 8. The shortest job first scheduling algorithm has the optimal average waiting time. Why it is not usally applicable? [5%] What can we do if we would like to achieve the shortest average waiting time? [5%] 9. Why timestamp-based protocol ensures conflict serializability? [5%] Why it also ensures freedom from deadlock? [5%] Hint: what cause conflicting operations and which in the four deadlock condictions can the protocol break? 10.The following codes with semaphores can be used to implement a monitor for synchronization, where F is an external procedure, x is a condition variable. Explain the purposes of using semaphores next [5%] and x-sem[5%]? semaphore mutex; // (initially = 1) semaphore next; // (initially = 0) int next-count = 0; semaphore x-sem; // (initially = 0) int x-count = 0; wait(mutex); ... body of F; ... if (next-count > 0) signal(next); else signal(mutex); ---------------------------------------- x.wait() ┌───────────────┐ │x-count++; │ │if (next-count > 0) │ │ signal(next); │ │else │ │ signal(mutex); │ │wait(x-sem); │ │x-count--; │ └───────────────┘ x.signal() ┌───────────────┐ │if (x-count > 0) { │ │ next-count++; │ │ signal(x-sem); │ │ wait(next); │ │ next-count--; │ │} │ └───────────────┘ 11.The following program uses a monitor to solve the dining-philosopher problem. Explain why it might have starvation problem. [5%] How to solve it? [10%] Hint: refer to question 10, no coding is required. Pseudo code might be. monitor DP { enum { THINKING; HUNGRY, EATING) state[5]; condition self[5]; void pickup(int i) { state[i] = HUNGRY; test(i); if (state[i] != EATING) self[i].wait; } void putdown(int i) { state[i] = THINKING; // test left and right neighbors test((i + 4) % 5); test((i + 1) % 5); } void test(int i) { if ((state[(i + 4) % 5] != EATING) && (state[i] == HUNGRY) && (state[(i + 1) % 5] != EATING)) { state[i] = EATING; self[i].signal(); } } initialization_code() { for (int i = 0; i < 5; i++) state[i] = THINKING; } } --

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