A computer has 64-bit virtual addresses, byte-addressed memory, 1 Kbyte page siz

Question

A computer has 64-bit virtual addresses, byte-addressed memory, 1 Kbyte page size, 16 M page frames in main memory, and has 8 bits of metadata about each page.

1 .A program for this computer uses a total of 8 consecutive pages of virtual memory to hold all of its instructions and data. How many bytes of main memory will be consumed for the page table of this program if the page table is two level and the question is what is the lower bound for table space. The second level table holds the page metadata, not the first-level table. However, all page table entries are aligned in memory as 32-bit integers.

2. how about three level table?

Explanation / Answer

ANSWER FOR 1:

>>   For 8 consecutive pages there will be 8 entries in the page table.

>>   In each entry frame number and meta data will be stored.

>> we know frame number is 24 bits and given meta data is 8 bits andpage table entry size is = 24 + 8 = 32 bits and

>> If the page table is two level then in 1st level we will have 2 entries

>> in 2nd level we will have 4 entries for a page table.Hence Page table size is = 4 * 32 bits = 16 bytes.

>> If the page table is three level then in 1st level we will have 2 entriesand in 2nd level we will have 2 entries and in 3rd level

>> we will have 2 entries for a page table.Therefore Page table size is = 2 * 32 bits = 8 bytes.

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