EAT – Page Faults

EAT = Effective Access Time

MA = Memory Access

PFT = Page Fault Time

P = Probability of page fault (0 <= P <= 1)

EAT = (1 – P) * MA + P * PFT

Example 1:

Typical hard disk has average latency of 8 milliseconds, seek time of 15 milliseconds, and transfer time of 1 millisecond.

So assume PFT = 8 + 15 + 1 = 25 milliseconds.

Also assume MA = 100 nanoseconds

EAT = (1 – P) * 100 + P * 25,000,000

100 – 100*P + 25,000,000*P

100 – 24,999,900*P

Example 2:

Assume: PFT = 25 milliseconds

MA = 100 nanoseconds

What would the frequency of page faults be in order to achieve a 10 percent or less degradation in Memory Access (MA) time?

MA = 100 ns

10% of 100 ns is 10 ns

We desire an EAT less than or equal to 100 ns + 10 ns = 110 ns.

110 > (1 – P)*MA + P * PFT

110 > (1 – P)*100 + P * 25,000,000

110 > 100 – 100*P + 25,000,000*P

110 – 100 > 25,000,000*P – 100*P

10 < 24,999,900*P

P < 10/24,999,900

P < 1/2,499,990

Therefore, to achieve an EAT of 110, you can have no more than 1 page fault in 2,499,990 memory accesses.