# Operating System: Performance of Demand Paging

I am reading Operating systems from the book "Operating System Concepts" by Peter Baer Galvin, 7th edition.

Performance of Demand Paging:

Let p be the probability of a page fault (0 $\leqslant$ p $\leqslant$1). We would expect p to be close to zero—that is, we would expect to have only a few page faults. The effective access time is then effective access time = (1 - p) x ma + p x page fault time.

To compute the effective access time, we must know how much time is needed to service a page fault.

Example If we take an average page-fault service time of 8 milliseconds and a memory-access time of 200 nanoseconds, then the effective access time in nanoseconds is

effective access time = (1 - p) x (200) + p (8 milliseconds)

= (1 - p) x 200 + p x 8.00(1000

= 200 + 7,999,800 x p.

My doubt when there is a page table hit it we should multiply p by (ma + ma) because, we need one memory access to access page table from memory and one more memory access to access the actual page from memory. In my opinion formula should be effective access time = (1 - p)x(ma+ma) + p x page fault time.

Where ma is memory access time

Please let me where am I wrong?