For the following problems assume 1 kilobyte (KB)1024 kilobytes1024 bytes and 1 megabyte (MB)For this problem, assume you have address translation hardware with the followingproperties:1Virtual addresses, physical addresses, and page table entries are 32 bits wide.The page size in the system is 4 KBA virtual address is a page number followed by a byte offset within the page.(a)How many bits of the virtual address must be used for the offset, so that everybyte in the page can have a unique address?(b)How many bits are left over in the virtual address to store the page number?(c)How many different page numbers does an address space in this systemsupport? (You can express this as a power of two)(d)If a page table consists of a page table entry for each page number in an addressspace, how much space in MB would the page table take up if it were stored inphysical memory?2Given the assumptions above and the page table below, translate the following virtualaddresses (expressed as base ten integers) to physical addresses (also base tenintegers). If the corresponding page is not resident in memory (NR), indicate a pagefaultHint: rather than translating these values to binary or hexadecimal, you can use thefollowing formulaspageNum = |virtualAddress/pageSize] where [x] is x rounded down to thenearest integeroffset virtualAddress mod pageSize where x mody is the remainder afterdividing x by yphysicalAddress frameNum x pageSize + offset.Show your work (a) VA 0(b) VA 4095PageFrame #0(c) VA 4096NR1(d) VA 10000202(e) VA 20000NR(f) VA 30000NR617NRLO

Question
Asked Nov 12, 2019
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For the following problems assume 1 kilobyte (KB)
1024 kilobytes
1024 bytes and 1 megabyte (MB)
For this problem, assume you have address translation hardware with the following
properties:
1
Virtual addresses, physical addresses, and page table entries are 32 bits wide.
The page size in the system is 4 KB
A virtual address is a page number followed by a byte offset within the page.
(a)
How many bits of the virtual address must be used for the offset, so that every
byte in the page can have a unique address?
(b)
How many bits are left over in the virtual address to store the page number?
(c)
How many different page numbers does an address space in this system
support? (You can express this as a power of two)
(d)
If a page table consists of a page table entry for each page number in an address
space, how much space in MB would the page table take up if it were stored in
physical memory?
2
Given the assumptions above and the page table below, translate the following virtual
addresses (expressed as base ten integers) to physical addresses (also base ten
integers). If the corresponding page is not resident in memory (NR), indicate a page
fault
Hint: rather than translating these values to binary or hexadecimal, you can use the
following formulas
pageNum = |virtualAddress/pageSize] where [x] is x rounded down to the
nearest integer
offset virtualAddress mod pageSize where x mody is the remainder after
dividing x by y
physicalAddress frameNum x pageSize + offset.
Show your work
help_outline

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For the following problems assume 1 kilobyte (KB) 1024 kilobytes 1024 bytes and 1 megabyte (MB) For this problem, assume you have address translation hardware with the following properties: 1 Virtual addresses, physical addresses, and page table entries are 32 bits wide. The page size in the system is 4 KB A virtual address is a page number followed by a byte offset within the page. (a) How many bits of the virtual address must be used for the offset, so that every byte in the page can have a unique address? (b) How many bits are left over in the virtual address to store the page number? (c) How many different page numbers does an address space in this system support? (You can express this as a power of two) (d) If a page table consists of a page table entry for each page number in an address space, how much space in MB would the page table take up if it were stored in physical memory? 2 Given the assumptions above and the page table below, translate the following virtual addresses (expressed as base ten integers) to physical addresses (also base ten integers). If the corresponding page is not resident in memory (NR), indicate a page fault Hint: rather than translating these values to binary or hexadecimal, you can use the following formulas pageNum = |virtualAddress/pageSize] where [x] is x rounded down to the nearest integer offset virtualAddress mod pageSize where x mody is the remainder after dividing x by y physicalAddress frameNum x pageSize + offset. Show your work

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(a) VA 0
(b) VA 4095
Page
Frame #
0
(c) VA 4096
NR
1
(d) VA 10000
2
0
2
(e) VA 20000
NR
(f) VA 30000
NR
6
1
7
NR
LO
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(a) VA 0 (b) VA 4095 Page Frame # 0 (c) VA 4096 NR 1 (d) VA 10000 2 0 2 (e) VA 20000 NR (f) VA 30000 NR 6 1 7 NR LO

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Expert Answer

Step 1

For solving this question, a user must know the meaning of the virtual address and the use of a virtual address in the page frame.  

Virtual Address: A digital address is a computer memory binary number that requires a program to use a primary storage location. In a machine that implements memory control, the digital address varies from the physical address, which is the position of information on an address bus which refers to a specific primary processing cell or to a particular register in an I / O (input/output) system that is memory-mapped.

Step 2

Formula Used:

  1. So, calculating of offset of the page size :

         offset bits = log2(page size)

 2. So, calculate the Number of bits for page number :

    No. of bits for page no. = virtual address - offset

   Given Data:

   1. Page Size:4kb

  2. Virtual address, Physical address and the table...

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