Map virtual address to physical address Virtual address=32 bits segment number=9
ID: 3565554 • Letter: M
Question
Map virtual address to physical address
Virtual address=32 bits
segment number=9 bits
page number=10 bits
offset within page=9 bits
leading 4 bits of VA are unused
size of segement table =512 words, size of page table=1024 words, size of each program =512 words
Physical memory=array of 524,288 integers
1024 frames of size 512 words
physical address=19 bits
Physical memory setup
PT of segment 2 starts at address 2048
page 0 of segment 2 starts at address 512
page 1 of segment 2 starts at address -1
Virtual address to be translated
1)1048576
2)1048586
provide me step by step explanations. copying of wiki/not answering to my question in your post leads to immediate report to chegg team
Explanation / Answer
Virtual address format is 4 bits unused 9 bits segment num 10 bits page num 9 bits offset
1048576 in hex is 0010 0000
in binary 0000 000000010 0000000000 000000000 separated as per VA format which is
Hence for 1048576 segment number = 2 page num = 0 offset =0
form Physical memory setup we know that Page Table of segment 2 starts at address 2048 and page 0 of segment 2 starts at address 512.
Hence for 104876 physical address is 512 (page nym =0 and offset =0)
1048586 in hex is 0010 000A
in binary 0000 000000010 0000000000 000001010 separated as per VA format which is
Hence for 1048576 segment number = 2 page num = 0 offset =10
form Physical memory setup we know that Page Table of segment 2 starts at address 2048 and page 0 of segment 2 starts at address 512.
Hence for 104876 physical address is 512+10 =522 (page nym =0 and offset =10)