Consider a system with page fault service time (S) = 100 ns, main memory…

Concept

Under demand paging, a memory reference either finds its page already resident in main memory or it misses and must wait for the page to be brought in from secondary storage. The page-fault rate P is the probability of a miss. The Effective Memory Access Time (EMAT) is the probability-weighted average of the fast in-memory access and the slow fault-handling path:

EMAT = (1 - P) x M + P x S

Here M is the time for a normal main-memory access and S is the full page-fault service time (the complete cost of fetching the page and restarting the reference). The weights (1 - P) and P sum to 1, so EMAT always lies between M and S, moving toward S as the fault rate rises.

Application

1. Identify the given values: page-fault rate P = 0.65, main-memory access M = 20 ns, page-fault service time S = 100 ns.

2. Compute the no-fault contribution: (1 - P) x M = (1 - 0.65) x 20 = 0.35 x 20 = 7 ns.

3. Compute the fault contribution: P x S = 0.65 x 100 = 65 ns.

4. Add the two weighted terms: EMAT = 7 + 65 = 72 ns.

Cross-check

Sanity bound: EMAT must fall between M = 20 ns and S = 100 ns, and 72 ns does. Because the fault rate is high (0.65), the result should sit well above the midpoint of M and S and lean toward S, which 72 ns does, confirming the weighting is applied in the right direction.