Question

Evaluating the maximum number of pages needed, if a system supports 16 bit address line and 1K page size.

Answer 1

the answer is (2^32)/(2^12) = 2^20. Good job! Here 2^32 is the total number of bytes in memory. 2^12 is total number of bytes in a page.

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Answer 2

Conclusion

Bits

A bit is a binary digit, which is the smallest data increment on a computer. A bit can only contain one of two values: 0 or 1, which correspond to the electrical values off or on. Because bits are so little, working with information one bit at a time is uncommon.

Bytes

A byte is a unit of data that is eight binary digits long in most computer systems. Most computers use a byte to represent a character such as a letter, integer, or typographic sign. Each byte can store a string of bits that must be combined to form a bigger unit for application purposes.

Main Content

$(\frac{2^{32} }{2^{20}}) = 2^{20}$ The total amount of bytes in memory is $2^{32}$ in this case. The total amount of bytes in a page is $2^{12}$

To, learn more about Bits and Bytes

https://brainly.in/question/3005170

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