Solved Example Problems on Virtual Memory Concept

Example : 1

The logical address space in a computer system consists of 128 segments. Each segment can have up to 32 pages of 4 K words each. Physical memory consists of 4 K blocks of 4 K words in each. Formulate the logical and physical address formats.

Solution :

Number of bits for segment address = log2 128 = log2 27 = 7 bits

Number of bits for page address = log2 32 = log2 25 = 5 bits

Number of bits for word address = log2 4096 = log2 212 = 12 bits

Number of bits for block address = log2 4096 = log2 212 = 12 bits

Example : 2

An address space is specified by 32 bits and corresponding memory space by 24 bits.

i) How many words are there in the address space?

ii) How many words are there in the memory space?

iii) If a page consists of 4 K words, how many pages and blocks are there in the systems.

Solution :

i) Words in the address space = 232 = 4 G words

ii) Words in the memory space = 224 = 16 M words

iii) Number of pages = Words in address space / Words per page = 4 G words / 4 K words

= 1 M pages

iv) Number of blocks = Words in address space / (Words per page/block)

 = 16 M words / 4 K blocks = 4 K words

Example : 3

An address space is specified by 24 bits and the corresponding memory space by 16 bits. How many words are there in the virtual memory and in the main memory?

Solution :

Words in the address space, i.e., in the virtual memory

= 224= 16 M words

Words in the memory space, i.e., in the main memory

=224 = 64 K words