Consider a system with a logical address space of 16 MB. The page size is 1 KB.
ID: 3805838 • Letter: C
Question
Consider a system with a logical address space of 16 MB. The page size is 1 KB. The physical memory in the system is 2 MB. The page table for a process shows:
Page 0 maps to Frame 1
Page 1 maps to Frame 2
Page 2 maps to Frame 4
Page 3 maps to Frame 16
Page 4 maps to Frame 9
a. How many bits are required for each logical address? How many bits are required for each physical address?
b. To which physical address will the logical address 1524 translate? (The address is expressed in decimal.)
c. Which logical address will translate to physical address 1024? (The address is expressed in decimal.)
Explanation / Answer
1) Logical address = 16MB = 24 * 220 = 224 so 24 bits
Physical address = 2MB = 21 * 220 = 221 so 21 bits
2) 1524 = 000001 0111110100
Page size is 1KB = 10 Bits are offset and rest other bits are Page numbers i.e Page 1 maps to frame 2
So It maps to frame 2,
So Physical Address is 00000000010 0111110100 = 2548
3) If Physical address is 1024
00000000001 0000000000
So It means Frame 1, with Offset as 00 i.e It refers to Page 0 in Logical memory
So Logical address is , 000000000000000000000 = 0000
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.