demand paging

Question

In a demand paging memory system, page table is held in registers. The time taken to service a page fault is 8 m.sec. if an empty frame is available or if the replaced page is not modified, and it takes 20 m.secs., if the replaced page is modified. What is the average access time to service a page fault assuming that the page to be replaced is modified 70% of the time ?

What is the maximum acceptable page-fault rate for an effective access time of no more than 200
nanoseconds?

Answer 1

8ms when page is clean.

20ms when page is dirty.

page table access time negligible as available in registers. Normally not available in registers as page table size is much bigger than the available registers. We use register only for storing base address of page table.

Tavg = 0.7*20 +0.3*8

          =  14+2.4

           =16.4ms

Let H is page fault rate.

200 ns >=  [H(20) + 1-H(8)] ms

solve this for answer of second question.