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Topic Wise Question Bank
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    • ☆
      1
      Operating SystemPaging
      Differentiate between Logical Address and Physical Address in context of Operating System.
      Combined Bank, AE(IT)/AME, 26 | AME/ANE/AE

      Difference Between Logical Address and Physical Address

      • Definition: Logical Address is generated by CPU; Physical Address is the actual location in memory (RAM).
      • Visibility: Logical Address is visible to the user/program; Physical Address is not directly visible to the user.
      • Generation: Logical Address is created during program execution; Physical Address is obtained after address translation.
      • Translation: Logical Address is converted to Physical Address using MMU (Memory Management Unit).
      • Address Space: Logical Address Space is virtual; Physical Address Space is real memory.
      • Access: CPU uses Logical Address; Memory hardware uses Physical Address.

    • ☆
      2
      Operating SystemPaging
      Consider a machine with 64 MB physical memory and a 32 bit virtual address space. If the page size is 4 KB, what is the approximate size of the page table.
      CB, AE(IT)/AHME/SO, 21 | AME/ANE/AE

      Physical Address Space = 64MB = \(2^{26}\text{B}\)

      Virtual Address = 32-bits, \(\therefore\) Virtual Address Space = \(2^{32}\text{B}\)

      Page Size = 4KB = \(2^{12}\text{B}\)

      Number of pages = \(\frac{2^{32}}{2^{12}} = 2^{20}\) pages.

      Number of frames = \(\frac{2^{26}}{2^{12}} = 2^{14}\) frames.

      \(\therefore\) Page Table Size = \(2^{20} \times 14\text{-bits} \approx 2^{20} \times 16\text{-bits} \approx 2^{20} \times 2\text{B} = 2\text{MB}\).



    • ☆
      3
      Operating SystemPaging
      A system utilizes a 32-bit logical address space and a page size of 220 bytes. If each entry (block) in the Translation Lookaside Buffer (TLB) requires 6 bytes, calculate the total memory required for the TLB to map the entire logical address space.
      Combined Bank, AP-23, 26 | Assistant Programmer

      Given:

      • Logical address space = 32-bit
      • Page size = 220 bytes
      • TLB entry size = 6 bytes per entry

      Step 1: Calculate Total Logical Address Space

      A 32-bit address can reference 232 unique bytes.
      Total logical address space = 232 bytes

      Step 2: Calculate Number of Pages

      Number of pages = Total address space ÷ Page size
      Number of pages = 232 ÷ 220 = 2(32 − 20) = 212 = 4096 pages

      Step 3: Calculate Total TLB Memory

      Each page requires one TLB entry to map the entire logical address space.

      Total TLB memory = Number of pages × Entry size
      Total TLB memory = 4096 × 6 = 24576 bytes

      In kilobytes: 24576 ÷ 1024 = 24 KB


    • ☆
      4
      Operating SystemPaging
      Consider a logical address space of 512 pages, each of 2-KB page size, mapped onto a physical memory containing 128 frames.
      a. How many bits are required in the logical address?
      b. How many bits are required in the physical address? (10 Marks)
      Combined Bank, SO-IT, 25 | Senior Officer (IT)

      Given
      Logical address space = 512 pages
      Page size = 2 KB = 2048 bytes
      Physical memory = 128 frames

      a) Bits required in the Logical Address
      Number of pages = 512 = 29 → Page number bits = 9 bits
      Page size = 2048 bytes = 211 → Offset bits = 11 bits
      So, Logical address bits = Page number bits + Offset bits = 9 + 11 = 20 bits

      b) Bits required in the Physical Address
      Number of frames = 128 = 27 → Frame number bits = 7 bits
      Offset remains same = 11 bits
      So, Physical address bits = Frame number bits + Offset bits = 7 + 11 = 18 bits


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