7 In a system with main memory size is 2048 bytes byte addre

7) In a system with main memory size is 2048 bytes (byte addressable) Page/Frame size is 32 bytes, process size is 128 bytes.

a). How many bits are there in the logical address?

b). How many bits are there for the offset?

c). How many bits are there for page number?

d). How many frames are there?

e). How many pages are there?

8) In the system described in question 7, given the following page table, translate the logical addresses to the physical addresses

8 a) 00000001010

4 b)00000110101

1 c)00001011011

2 d)00001111111

7 . Answer:

a). How many bits are there in the logical address?

It is equal to size of process = 2^7 = 7 bits

b). How many bits are there for the offset?

It is equal to page size = 2^5 = 5 bits

c). How many bits are there for page number?

It is equal to 2^7 / 2^5 = 2^2 = 2 bits

d).How many frames are there? It is 2^11 / 2^5 = 2^6 = 64 e). How many pages are there? It is equal to 2^7 / 2^5 = 2^2 = 4