Consider a logical address space of $8$ pages of $1024$ words each, mapped onto a physical memory of $32$ frames. How many bits are there in the physical address and logical address respectively?
Answer 15, 13
Physical Memory 32 frames = 25
Size of word 1024 = 210
Total Physical Address Space =15 bits
Logical Memory 8 pages = 23
Size of word = 210
Total Logical Address Space= 10 +3 = 13 bits
Given, Logical address space= 8 pages → 2^3
words → offset → 1024 -->2^10
Frames = 32 -->2^5
Logical address = 10 bits + 3 bits = 13 bits
Physical address = 10-bits + 5-bits = 15 bits