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Consider a system with byte-addressable memory, 31 bit logical addresses, 4 kilobyte page size and page table entries of 4 bytes each. The size of the page table in the system in megabytes is _____?
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Consider a system with byte-addressable memory, 31 bit logical address...
Calculating the Size of the Page Table


To calculate the size of the page table, we need to consider the number of pages and the size of each page table entry.

Calculating the Number of Pages


The logical address space is given as 31 bits, which means we have a total of 2^31 possible logical addresses.

Since the page size is 4 kilobytes, we can calculate the number of pages as follows:
Number of pages = (Total logical address space) / (Page size)
= (2^31) / (4 * 1024)
= (2^31) / (2^12)
= 2^31-12
= 2^19

Therefore, we have a total of 2^19 pages in the system.

Calculating the Size of Each Page Table Entry


The size of each page table entry is given as 4 bytes.

Calculating the Size of the Page Table


To calculate the size of the page table, we multiply the number of pages by the size of each page table entry:
Size of page table = (Number of pages) * (Size of each page table entry)
= (2^19) * (4 bytes)
= (2^19) * (4 * 8 bits)
= (2^19) * (32 bits)
= (2^19) * (2^5)
= 2^24 bytes

Since we are asked to express the size of the page table in megabytes, we can convert the size to megabytes as follows:
Size of page table in megabytes = (Size of page table) / (2^20)
= (2^24 bytes) / (2^20 bytes)
= 2^4
= 16 megabytes

Therefore, the size of the page table in the system is 16 megabytes.
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Consider a system with byte-addressable memory, 31 bit logical addresses, 4 kilobyte page size and page table entries of 4 bytes each. The size of the page table in the system in megabytes is _____?
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