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?
(A) 5, 3
(B) 10, 10
(C) 15, 13
(D) 15, 15
Explanation: Logical address space has 8 pages, size of each page, offset = 1024 words
Number if bits in page# field of the logical address = log28 bits = 3 bits
Offset bits = log21024 = 10 bits
So, Logical address = 3 + 10 = 13 bits
Physical memory has 32 frames, offset = 1024 words
Number of bits in frame# field of the Physical address = log232 bits = 5 bits
Physical address = 5 + 10 = 15 bits
Option (C) is correct.
Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.