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Consider a memory-management system based on paging. The total size of the physical address space 64 MB, Pages of size 4 KB and the Logical address space of 4GB. a. No of pages b. No of Frames c. No of entries in Page Table d. Size of Page Table e. No of bits in Physical Address f. No of Bits in Logical Address

User Pr Shadoko
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Final answer:

The answers to the given questions are:

a. 262,144 pages, b. 16,384 frames,

c. 262,144 entries, d. 262,144 * sizeof(entry),

e. 26 bits, and) f. 30 bits

Step-by-step explanation:

a. Number of pages:

To calculate the number of pages in the logical address space, we divide the total logical address space by the page size:
Number of pages = Logical address space / Page size

= 4GB / 4 KB = 1024 MB / 4 KB

= 1024 * 1024 KB / 4 KB

= 262,144 pages

b. Number of frames:

Since the total size of the physical address space is 64 MB and the page size is 4 KB, we can calculate the number of frames as:

Number of frames = Physical address space / Page size

= 64 MB / 4 KB = 64 * 1024 KB / 4 KB

= 16,384 frames

c. Number of entries in Page Table:

Since each page has a corresponding entry in the page table, the number of entries in the page table is equal to the number of pages:

Number of entries in Page Table = Number of pages

= 262,144 entries

d. Size of Page Table:

To calculate the size of the page table, we multiply the number of entries in the page table by the size of each entry:

Size of Page Table = Number of entries in Page Table * Size of each entry

= 262,144 entries * sizeof(entry)

= 262,144 * sizeof(entry)

e. Number of bits in Physical Address:

Since there are 16,384 frames in the physical address space and each frame has a size of 4 KB, we can calculate the number of bits in the physical address as:

Number of bits in Physical Address = log2(Number of frames) + log2(Frame size)

= log2(16,384) + log2(4 KB)

= log2(2^14) + log2(2^12)

= 14 + 12

= 26 bits

f. Number of Bits in Logical Address:

Since there are 262,144 pages in the logical address space and each page has a size of 4 KB, we can calculate the number of bits in the logical address as:

Number of bits in Logical Address = log2(Number of pages) + log2(Page size)

= log2(262,144) + log2(4 KB)

= log2(2^18) + log2(2^12)

= 18 + 12

= 30 bits

User Stedman Blake
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