17. A computer system implements a paged virtual memory system. Assume a 16-bit
ID: 3719996 • Letter: 1
Question
17. A computer system implements a paged virtual memory system. Assume a 16-bit virtual address space and a 24-bit physical address space. Assume that the first 6 bits of a virtual address index the page table and the rest of the bits are the page offset. A process has the following indexed page table. Index | Page Table Entry (PTE) 0 | 0x3800 1 | 0x3600 2 | 0x3200 3 | 0x1000 Each page table entry gives a hexadecimal page frame addresses. Translate the following two hexadecimal VM addresses into their corresponding hexadecimal physical address. Hint: Translate the VM address to binary first. Translate the binary VM address to a binary PM address. Translate the binary PM address into hexadecimal. VM Address 0x084B is _____________________ in Physical Memory VM Address 0x0C78 is _____________________ in Physical Memory
Explanation / Answer
Hexadecimal to binary conversion is simple for every hex character/digit there is corresponding 4 bits binary, we can replace hex character/digit from corresponding 4 bits binary number. (binary to hex does opposite binary string should be multiple of 4 digits long of not we add 0 in start of binary number to make total bits in binary number a multiple of 4)
0 = 0000
1 = 0001
2 = 0010
3 = 0011
4 = 0100
5 = 0101
6 = 0110
7 = 0111
8 = 1000
9 = 1001
A = 1010
B = 1011
C = 1100
D = 1101
E = 1110
F = 1111
Solution A:
0x084B =
Hex
0
8
4
B
Binary
0000
1000
0100
1011
Page Table index = 0000 10 = 2 in decimal
Now page table index 2 will maps to Page Table Entry => 0x3200 (frame starting at physical memory address)
Page Offset = 00 0100 1011 in binary = 0x04B in hex
Final Physical memory address = base physical address + offset of page
= 0x3200 + 0x04B
= 0x324B Ans
Solution B:-
0x0C78
Hex
0
C
7
8
Binary
0000
1100
0111
1000
Page Table Index = 0000 11 = 3 in decimal
Now page table index 3 will maps to Page Table Entry => 0x1000 (frame starting at physical memory address)
Page Offset = 00 0111 1000 in binary = 0x078 in hex
Final Physical memory address = base physical address + offset of page
= 0x1000 + 0x078
= 0x1078 Ans
0 = 0000
1 = 0001
2 = 0010
3 = 0011
4 = 0100
5 = 0101
6 = 0110
7 = 0111
8 = 1000
9 = 1001
A = 1010
B = 1011
C = 1100
D = 1101
E = 1110
F = 1111
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.