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        代做comp3511、代寫Python/Java編程

        時(shí)間:2023-11-13  來源:合肥網(wǎng)hfw.cc  作者:hfw.cc 我要糾錯(cuò)



         
        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. 
         
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