In a system with main memory size is 4096 bytes bytes addres

In a system with main memory size is 4096 bytes (bytes addressabe) Page/Frame size is 32 bytes, a process can have up to 16 segments.

a. How many bits are there in a physical address?

b. How many bits are there in a logical address?

c. How many bits for the segment number, page number, and offset?

Solution

Answer:

Main memory is 2^(12) bytes of size, therefore we need 12 bits to address main memory. There are 16 segments which can be represented by 4 bits , & each segment is of size 32 bytes ( it can be represented by 5 bits ) . Bits for Logical address = bits for segments + bits for segment size . => Bits of logical address = 4 bits for segment + 5 bit for size = 9 bits to represent logical address.