Solution

The correct answer is 22.

Key Points

• According to the question, the total size of the main memory is 16 GB. The size of the virtual address space is 32 bits. The page size is 4KB. It is also given that the frame size and page size are the same for the given machine.
• Converting all the units into bits. So, 16 GB= (2⁴ x 2¹⁰ x 210 x 2¹⁰ x 2³) bits = 2³⁷ bits.
• 4KB= (2² x 210 x 23) bits = 2¹⁵ bits.
• So, the number of bits reserved for the frame offset is = (2³⁷/ 2¹⁵) = 2²² which is 22 units.

Thus the correct answer is 22.

Additional Information

• In an operating system, the page is referred to as a continuous memory block with a fixed memory size. But it is virtual memory. The page is the smallest data set unit in the virtual memory.
• The frame in the operating system is referred to as a fixed length of the physical block. The frame is often found on the RAM block.