Jayy6y1yfredh Jayy6y1yfredh

12-02-2017
Computers and Technology

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Consider a physical address with a page frame size of 2kb. how many bits must be used to represent the page-frame offset of the physical address?

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rsmith6559 rsmith6559

13-02-2017

11 bits would be enough for 0 - 2047 decimal, 2048 (2k) possible values. 2048 itself would need a 12 bit, which would allow from 0 - 4095 possible values. Useful tidbit to remember: 1024, 1k is ten bits.

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