Consider a computer system with a 64-bit logical address and 8-KB page size. The system supports...

Consider a computer system with a 64-bit logical address and 8-KB page size. The system supports up to 1024 MB of physical memory:

How many entries are there in each of the following in the page table?

Describe how a logical address may be used to obtain the frame number. You may use the virtual address 14533956 to illustrate your answer.

make it short and clear , please type in your keyboard.

Expert Solution

Logical address=64 bit.So address space is 2⁶⁴.

page size is given as 8KB.

Physical memory size=1024 MB.

Pages and pageframes are of equal size.So page frame size=8KB

So total number of pages in physical memory

=1024MB/8KB=(1024*2¹⁰*2¹⁰)/8*2¹⁰=1024*1024*1024/8*1024=1024*1024/8=128K number of pages in total.

Consider the virtual address 14533956, convert this into 64bit binary address.

0000000100000100000001010000001100000011000010010000010100000110. This Address is divided into page number and an offset.Page number refers to one entry in page table and the offset is copied as like this to get the physical address.

Here total number of pages are 128K.That is 2ⁿ=2⁷*2¹⁰,that is n=17.

17 bits are required out of 64 bits to refer to page table.This entry will provides the frame number.

00000001000001000 this will point to page table,which contains the frame number.Remaining bits are taken as offset.

venereology answered 6 months ago

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