合肥生活安徽新聞合肥交通合肥房產生活服務合肥教育合肥招聘合肥旅游文化藝術合肥美食合肥地圖合肥社保合肥醫院企業服務合肥法律

        代做comp3511、代寫Python/Java編程

        時間:2023-11-13  來源:合肥網hfw.cc  作者:hfw.cc 我要糾錯



         
        When accessing the logical address at <segment # = 2, offset = 400>, the results after 
        address translation is: 
        A) No such segment 
        B) Return address 4400 
        C) Invalid permission 
        D) Address out of range 
        9) In a system with **-bit address, given the logical address 0x0000F1AE (in 
        hexadecimal) with a page size of 256 bytes, what is the page offset? 
        A) 0xAE 
        B) 0xF1 
        C) 0xA 
        D) 0xF100 
         
        10) A computer based on dynamic storage memory allocation has a main memory with 
        the capacity of 55MB (initially empty). A sequence of operations including main 
        memory allocation and release is as follows: 1. allocate 15MB; 2. allocate 30MB; 3. 
        release 15MB; 4. allocate 8MB; 5. allocate 6MB. Using the best-fit algorithm, what is the 
        size of the largest leftover hole in the main memory after the above five operations? 
        A) 7MB 
        B) 9MB 
        C) 10MB 
        D) 15MB 
         
        2. [20 points] Synchronization 

        You are asked to implement the second reader-writer solution, in which once a writer 
        is ready, it needs to perform update as soon as possible. There are two classes of 
        processes accessing shared data, readers and writers. Readers never modify data, thus 
        multiple readers can access the shared data simultaneously. Writers modify shared 
        data, so at most one writer can access data (no other writers or readers). This solution 
        gives priority to writers in the following manner: when a reader tries to access shared 
        data, if there is a writer accessing the data or if there are any writer(s) waiting to access 
        shared data, the reader must wait. In another word, readers must wait for all writer(s), 
        if any, to update shared data, or a reader can access shared data, only when there is no 
        writer either accessing or waiting. 
        The following variables will be used: 
        semaphore mutex =1; /* lock for accessing shared variables */ 
        semaphore readerlock=0; /* readers waiting queue */ 
        semaphore writerlock=0; /* writers waiting queue */ 
        int R_count = 0; /* number of readers accessing data */ 
        int W_count = 0; /* number of writer accessing data */ 
        int WR_count = 0; /* number of readers waiting */ 
        int WW_count = 0; /* number of writers waiting*/ 
        TA Yixin YANG ( yyangfe@cse.ust.hk ) is responsible for questions before the deadline 
        TA Peng YE ( pyeac@cse.ust.hk ) is responsible for grading and appeal handling after the deadline 

        Please fill in the blanks to design Writer  s and Reader  s code. 

        Writer() { 
        // Writer tries to enter 
        wait(mutex); 
        while (BLANK1) { // Is it safe to write? 
        WW_count++; 

        BLANK2 

        WW_count--; 

        W_count++; // Writer inside 
        signal(mutex); 
         
        // Perform actual read/write access 
         
        // Writer finishes update 

        BLANK3 

        W_count--; // No longer active 
        if (BLANK4){ // Give priority to writers 
        signal(writerlock); // Wake up one writer 
        } else if (WR_count > 0) { // Otherwise, wake reader 

        BLANK5 


        signal(mutex); 


        Reader() { 
        // Reader tries to enter 
        wait(mutex); 
        while (BLANK6) { // writer inside or waiting? 

        BLANK7 
        wait(readerlock); // reader waits on readerlock 

        BLANK8 


        Rcount++; // Reader inside! 
        signal(mutex); 
         
        // Perform actual read-only access 
         
        // Reader finishes accessing 
        wait(mutex); 
        R_count--; // No longer active 
        if (BLANK9) // No other active readers 

        BLANK10 

        signal(mutex); 


         

        3. [30 points] Deadlocks 

        Consider the following snapshot of a system: 
        TA Yixin YANG ( yyangfe@cse.ust.hk ) is responsible for questions before the deadline 
        TA Peng YE ( pyeac@cse.ust.hk ) is responsible for grading and appeal handling after the deadline 

         Allocation Max Available 
         A B C D A B C D A B C D 
        P0 2 0 0 1 4 2 3 4 3 3 2 1 

        P1 3 1 2 1 5 2 3 2 

        P2 2 1 0 3 2 3 1 6 

        P3 1 3 1 2 1 4 2 4 

        P4 1 4 3 2 3 6 6 5 
         
        Please answer the following questions using the banker  s algorithm: 
        1) (5 points) What is the content of the matrix Need denoting the number of resources 
        needed by each process? 

         
        2) (10 points) Is the system in a safe state? Why? 


        3) (5 points) If a request from process P1 arrives for (1, 1, 0, 0), can the request be 

         4) (10 points) If a request from process P4 arrives for (0, 0, 2, 0), can the request be 
        granted immediately? Why? 

        TA Yixin YANG ( yyangfe@cse.ust.hk ) is responsible for questions before the deadline 
        TA Peng YE ( pyeac@cse.ust.hk ) is responsible for grading and appeal handling after the deadline 

        4. [30 points] Memory Management 

        1) (15 points) Consider the segment table shown in Table A. Translate each of the 
        virtual addresses in Table B into physical addresses. Indicate errors (out of range, no 
        such segment) if an address cannot be translated. 

        Table A 

        Segment number Starting address Segment length 
        0 260 ** 
        1 1466 160 
        2 2656 130 
        3 146 50 
        4 2064 370 
         

        Table B 

        Segment number Offset 
        0 420 
        1 144 
        2 198 
        3 296 
        4 50 
        5 ** 

        2) (15 points) Consider a virtual memory system providing 128 pages for each user 
        program; the size of each page is 8KB. The size of main memory is 256KB. Consider 
        one user program occupied 4 pages, and the page table of this program is shown as 
        below: 
         
        Logical page number Physical block number 
        TA Yixin YANG ( yyangfe@cse.ust.hk ) is responsible for questions before the deadline 
        TA Peng YE ( pyeac@cse.ust.hk ) is responsible for grading and appeal handling after the deadline 

        Assume there are three requests on logical address 040AFH, 140AFH, 240AFH. First 
        please describe the format of the logical address and the physical address. Then please 
        illustrate how the virtual memory system will deal with these requests. 
         
        請加QQ:99515681 或郵箱:99515681@qq.com   WX:codehelp

        掃一掃在手機打開當前頁
      1. 上一篇:COMP3173代做、代寫C/C++程序設計
      2. 下一篇:代做CEG3136、代寫C/C++程序語言
      3. 無相關信息
        合肥生活資訊

        合肥圖文信息
        急尋熱仿真分析?代做熱仿真服務+熱設計優化
        急尋熱仿真分析?代做熱仿真服務+熱設計優化
        出評 開團工具
        出評 開團工具
        挖掘機濾芯提升發動機性能
        挖掘機濾芯提升發動機性能
        海信羅馬假日洗衣機亮相AWE  復古美學與現代科技完美結合
        海信羅馬假日洗衣機亮相AWE 復古美學與現代
        合肥機場巴士4號線
        合肥機場巴士4號線
        合肥機場巴士3號線
        合肥機場巴士3號線
        合肥機場巴士2號線
        合肥機場巴士2號線
        合肥機場巴士1號線
        合肥機場巴士1號線
      4. 短信驗證碼 酒店vi設計 deepseek 幣安下載 AI生圖 AI寫作 aippt AI生成圖片

        關于我們 | 打賞支持 | 廣告服務 | 聯系我們 | 網站地圖 | 免責聲明 | 幫助中心 | 友情鏈接 |

        Copyright © 2025 hfw.cc Inc. All Rights Reserved. 合肥網 版權所有
        ICP備06013414號-3 公安備 42010502001045

        主站蜘蛛池模板: 春暖花开亚洲性无区一区二区| 无码人妻精一区二区三区| 无码一区二区三区免费| 久久一区二区三区精品| 香蕉免费一区二区三区| 中文字幕一区二区三区5566| 精品人妻少妇一区二区三区不卡| 国模无码一区二区三区| 国产乱码精品一区二区三| 天堂资源中文最新版在线一区| 国产一区二区三区韩国女主播 | 亚洲国产高清在线一区二区三区| 久久精品中文字幕一区| 精品视频一区二区三三区四区| 国产精品第一区揄拍| 亚州国产AV一区二区三区伊在| 丰满爆乳一区二区三区| 国产一区二区三区韩国女主播| 无码人妻一区二区三区免费视频 | 亚洲中文字幕无码一区二区三区| 无码aⅴ精品一区二区三区浪潮 | 国产一区二区三区在线视頻| 午夜福利一区二区三区在线观看 | 美女视频一区三区网站在线观看| 国产一区二区三区在线免费| 美女视频在线一区二区三区| 日本一区二区三区爆乳| 久久无码一区二区三区少妇| 亚洲大尺度无码无码专线一区| 成人区人妻精品一区二区不卡视频| 亚洲综合在线成人一区| 国产精品毛片a∨一区二区三区| 韩国福利影视一区二区三区| 肥臀熟女一区二区三区| 性色AV一区二区三区天美传媒| 国精品无码A区一区二区| 国产精品视频一区国模私拍| 久久精品国产一区二区三区肥胖| 深夜福利一区二区| 人妻在线无码一区二区三区| 亚洲AV无码一区二区乱子伦|