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In a computer system, the physical address space is 32 bits. The size of the pages is 4 KB. The maximum size of the page table of a process is 72 MB. Each table entry of the page contains 2 permission bit, 1 valid bit, 1 dirty bit along with the translation bits. What is the length of the virtual address supported by the given computer system (answer in bits)?
    Correct answer is '37'. Can you explain this answer?
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    In a computer system, the physical address space is 32 bits. The size ...
    Given Information:
    - Physical address space: 32 bits
    - Page size: 4 KB
    - Maximum size of page table: 72 MB
    - Each page table entry contains: 2 permission bits, 1 valid bit, 1 dirty bit, and translation bits

    Calculating the number of entries in the page table:
    Since the maximum size of the page table is given as 72 MB, which is the same as 72,000 KB, we can calculate the number of entries in the page table.

    1. Convert the page table size to bytes: 72,000 KB * 1024 bytes/KB = 73,728,000 bytes
    2. Calculate the number of entries in the page table: 73,728,000 bytes / (4 KB/page) = 18,432 entries

    Calculating the number of bits required for the page table:
    Since each page table entry contains 2 permission bits, 1 valid bit, 1 dirty bit, and translation bits, we need to calculate the total number of bits required for each entry.

    1. Calculate the number of translation bits required: 32 bits - 12 bits (4 KB) = 20 bits
    2. Calculate the number of bits required for each entry: 2 permission bits + 1 valid bit + 1 dirty bit + 20 translation bits = 24 bits

    Calculating the length of the virtual address:
    The virtual address is divided into two parts: the page number and the page offset.

    1. The page number is determined by the number of entries in the page table: log2(18,432) = 14 bits
    2. The page offset is determined by the page size: log2(4 KB) = 12 bits

    Total length of the virtual address:
    The total length of the virtual address is the sum of the page number bits and the page offset bits.

    14 bits (page number) + 12 bits (page offset) = 26 bits

    Explanation of the correct answer:
    The given computer system has a 32-bit physical address space. The page table size is limited to 72 MB, which results in a page table with 18,432 entries. Each entry requires 24 bits. The virtual address is divided into a page number and a page offset, with 14 bits for the page number and 12 bits for the page offset. Therefore, the total length of the virtual address is 26 bits. However, the correct answer is 37 bits, which suggests that additional information or calculations may be missing or incorrect in the given question.
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    Community Answer
    In a computer system, the physical address space is 32 bits. The size ...
    Given,
    Physical address space (PAS)  = 32 bits
    Page size = 4 KB
    Page table size = 72 MB
    As we know,
    Number of frames = 
    Now,
    Number of frames  =  = 220
    Translation =20 bits
    A byte is a unit of digital information that most commonly consists of eight bits. So, we have
    Entry of page table = 2 + 1 + 1 + 20 = 24 bits
    =3 byte
    Number of pages  =  
    =
    =24 MB
    Virtual address = number of pages × page size = 24 × 220 × 4 KB
    Virtual address =96×230  B
    Length of virtual address space =log2⁡ 96 × 230
    =37 bits
    Hence, the correct answer is 37.
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    A processor uses 2-level page tables for virtual to physical address translation. Page tables for both levels are stored in the main memory. Virtual and physical addresses are both 32 bits wide. The memory is byte addressable. For virtual to physical address translation, the 10 most significant bits of the virtual address are used as index into the first level page table while the next 10 bits are used as index into the second level page table. The 12 least significant bits of the virtual address are used as offset within the page. Assume that the page table entries in both levels of page tables are 4 bytes wide. Further, the processor has a translation look-aside buffer (TLB), with a hit rate of 96%. The TLB caches recently used virtual page numbers and the corresponding physical page numbers. The processor also has a physically addressed cache with a hit rate of 90%. Main memory access time is 10 ns, cache access time is 1 ns, and TLB access time is also 1 ns. Suppose a process has only the following pages in its virtual address space: two contiguous code pages starting at virtual address 0x00000000, two contiguous data pages starting at virtual address 000400000, and a stack page starting at virtual address 0FFFFF000. The amount of memory required for storing the page tables of this process is

    A processor uses 2-level page tables for virtual to physical address translation. Page tables for both levels are stored in the main memory. Virtual and physical addresses are both 32 bits wide. The memory is byte addressable. For virtual to physical address translation, the 10 most significant bits of the virtual address are used as index into the first level page table while the next 10 bits are used as index into the second level page table. The 12 least significant bits of the virtual address are used as offset within the page. Assume that the page table entries in both levels of page tables are 4 bytes wide. Further, the processor has a translation look-aside buffer (TLB), with a hit rate of 96%. The TLB caches recently used virtual page numbers and the corresponding physical page numbers. The processor also has a physically addressed cache with a hit rate of 90%. Main memory access time is 10 ns, cache access time is 1 ns, and TLB access time is also 1 ns. Assuming that no page faults occur, the average time taken to access a virtual address is approximately (to the nearest 0.5 ns)

    A processor uses 2-level page tables for virtual to physical address translation. Page tables for both levels are stored in the main memory. Virtual and physical addresses are both 32 bits wide. The memory is byte addressable. For virtual to physical address translation, the 10 most significant bits of the virtual address are used as index into the first level page table while the next 10 bits are used as index into the second level page table. The 12 least significant bits of the virtual address are used as offset within the page. Assume that the page table entries in both levels of page tables are 4 bytes wide. Further, the processor has a translation look-aside buffer (TLB), with a hit rate of 96%. The TLB caches recently used virtual page numbers and the corresponding physical page numbers. The processor also has a physically addressed cache with a hit rate of 90%. Main memory access time is 10 ns, cache access time is 1 ns, and TLB access time is also 1 ns. Assuming that no page faults occur, the average time taken to access a virtual address is approximately (to the nearest 0.5 ns)

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    In a computer system, the physical address space is 32 bits. The size of the pages is 4 KB. The maximum size of the page table of a process is 72 MB. Each table entry of the page contains 2 permission bit, 1 valid bit, 1 dirty bit along with the translation bits. What is the length of the virtual address supported by the given computer system (answer in bits)?Correct answer is '37'. Can you explain this answer?
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    In a computer system, the physical address space is 32 bits. The size of the pages is 4 KB. The maximum size of the page table of a process is 72 MB. Each table entry of the page contains 2 permission bit, 1 valid bit, 1 dirty bit along with the translation bits. What is the length of the virtual address supported by the given computer system (answer in bits)?Correct answer is '37'. Can you explain this answer? for Computer Science Engineering (CSE) 2024 is part of Computer Science Engineering (CSE) preparation. The Question and answers have been prepared according to the Computer Science Engineering (CSE) exam syllabus. Information about In a computer system, the physical address space is 32 bits. The size of the pages is 4 KB. The maximum size of the page table of a process is 72 MB. Each table entry of the page contains 2 permission bit, 1 valid bit, 1 dirty bit along with the translation bits. What is the length of the virtual address supported by the given computer system (answer in bits)?Correct answer is '37'. Can you explain this answer? covers all topics & solutions for Computer Science Engineering (CSE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for In a computer system, the physical address space is 32 bits. The size of the pages is 4 KB. The maximum size of the page table of a process is 72 MB. Each table entry of the page contains 2 permission bit, 1 valid bit, 1 dirty bit along with the translation bits. What is the length of the virtual address supported by the given computer system (answer in bits)?Correct answer is '37'. Can you explain this answer?.
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