Question

Its multi part. please make sure to solve all parts.

Consider a logical address space of eight pages of 1024 words, each mapped onto a memory of 32 frames.
c. How many bits are in the offset for the logical address
d. How many bits are in the offset for the physical address
e. How many bits are in the Frame Number
f. How many bits are in the Page Number

Answer 1

Answer 2

a. The page size is 1024 words, which can be represented by 10 bits (2^10 = 1024). Therefore, the logical address space consists of 3 bits for the page number (2^3 = 8 pages).

b. Since the memory is composed of 32 frames, we need 5 bits to represent the frame number (2^5 = 32).

c. The offset is the number of bits required to represent the displacement within a page. Since each page contains 1024 words, which can be represented by 10 bits, the offset requires 10 bits.

d. The offset for the physical address is also 10 bits, as it represents the same displacement within a page.

e. The frame number requires 5 bits to represent the 32 frames.

f. The page number requires 3 bits to represent the 8 pages.