Running Head Computer Analogy Please Discuss In Detail ✓ Solved

Running Head Computer Analogy21 Please Discuss In Detail The Diffe

Consider a logical address space of 64 pages each 1024 bytes in size, mapped onto a physical memory consisting of 32 frames, each 1024 bytes.

a. Calculate the number of bits in the logical address.

b. Calculate the number of bits in the physical address.

Given a logical memory and page table, demonstrate the mapping of pages in the physical memory.

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Running Head Computer Analogy21 Please Discuss In Detail The Diffe

Running Head Computer Analogy21 Please Discuss In Detail The Diffe

Understanding the structure of logical and physical addresses in computer memory systems is essential in operating systems design, particularly in paging mechanisms. This essay explores the calculation of address bits based on given memory parameters and illustrates how logical pages map onto physical memory frames through page tables.

Address Space and Address Bits Calculation

Logical Address Space and Bits

Given that the logical address space contains 64 pages, each 1024 bytes in size, we first determine the total logical address space size. This can be computed as:

  • Total logical memory: 64 pages * 1024 bytes per page = 65,536 bytes.

Since addressing is byte-oriented, the total number of addresses is 65,536, which equates to 2^16. This directly implies that the logical address requires 16 bits to uniquely identify each byte within the logical address space.

Moreover, logical addresses are typically divided into a page number and an offset within that page. The number of bits for the page number is determined by the total number of pages:

  • Number of bits for page number: log2(64) = 6 bits.
  • Number of bits for offset within a page: log2(1024) = 10 bits.

Thus, the logical address is composed of 6 bits for the page number and 10 bits for the offset, totaling 16 bits.

Physical Address Bits

Given the physical memory has 32 frames of 1024 bytes each, the total physical address space size is:

  • Total physical memory: 32 frames * 1024 bytes = 32,768 bytes.

This equates to 2^15 addresses, requiring 15 bits for the full physical address.

Similarly, physical addresses are divided into a frame number and an offset within the frame. The number of bits for the frame number is:

  • Number of bits for frame number: log2(32) = 5 bits.
  • Number of bits for offset: 10 bits (same as logical offset, since frame size matches page size).

Therefore, the physical address comprises 5 bits for the frame number and 10 bits for the offset, totaling 15 bits.

Mapping Logical Pages to Physical Memory

Given Page Table and Memory Layout

Assuming a hypothetical page table mapping logical pages to physical frames, for example:

  • Logical page 0 → Frame 3
  • Logical page 1 → Frame 0
  • Logical page 2 → Frame 7
  • Logical page 3 → Frame 12

Using this page table, logical pages are mapped onto specific physical frames in memory. For example, logical page 0 maps into frame 3, which occupies the physical address range from byte 31024 to 31024 + 1023.

Similarly, logical page 1 maps into frame 0, occupying bytes 0 to 1023, and so forth. This mapping informs the hardware how to translate logical addresses to physical addresses during memory access.

Conclusion

This process illustrates the role of address translation in virtual memory systems. By calculating the number of bits in logical and physical addresses and understanding page-to-frame mappings, we grasp how operating systems manage memory efficiently, enabling processes to operate in an abstracted address space while interacting with physical hardware.

References

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  • Chen, L., & Liu, J. (2018). Memory Management Techniques and Performance Analysis. International Journal of Computer Applications, 180(12), 17-22.

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