A machine has a 32bit address space and an 8KB page The page

A machine has a 32-bit address space and an 8-KB page. The page table is entirely in hardware, with one 32-bit word per entry. When a process starts, the page table is copied to the hardware from memory, at the rate of one word every 100 nsec. If each process runs for 100 msec (including the time to load the page table), what fraction of the CPU time is devoted to loading the page tables? (Assume that each process uses its entire virtual address space during execution.)

Solution

If there are 2^32 addresses with 8K page sizes, then there are 2^32 - 2^13 = 2^19 pages.
Each 32b-word entry in the page table takes 100ns to load: 2^19 * 100ns = 52428800ns = 52.4288ms.
Total CPU time = 100ms.
Percent of CPU time taken to load page table: 52.4288ms/100ms = 52% (more than half).