Question

Consider a logical address space of 64 pages of 1,024 words each, mapped onto a physical memory of 32 frames. Show your work. You can either type the solution or write

Answer 1

Logical address = 16 bits

Physical address = 15 bits

Step-by-step explanation:

Logical space, number of pages = 64 = 2^6

Word size per page = 1024 = 2^10

Size of logical address :

Number of pages * words per page

2^6 * 2^10 = 2^(6+10) = 2^16 = 16 bits

For the physical address :

Number of frames = 32 = 2^5 frames

Word size = 1024 = 2^10

Number of frames * word size

2^5 * 2^10 = 2^(10+5) = 15 bits