Solution

The correct answer is 2 MB.

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. It is also given that each page table entry is of size 2 bytes.
• Converting all the units into bits. So, the physical memory size is = 2³⁴ Bytes.
• frame size = page size= 4KB= (2² x 2¹⁰) Bytes = 2¹² Bytes.
• So, the number of pages = (2³²/ 2¹²) Bytes = 2²⁰ Bytes.
• So, the size of the page table =  number of pages x page table entry = 2²⁰ x 2 Bytes = 2²⁰ x 16bits ≈ 2MB.

Thus the correct answer is 2 MB.

Additional Information

• The page is referred to as a continuous memory block with a fixed memory size in an operating system. 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.