Discussion
The effective access time in a demand-paged system depends mainly on
Answer: A
Correct answer: the page-fault rate
Effective access time is dominated by the page-fault rate, because the two costs involved differ by so many orders of magnitude. A memory access takes on the order of 100 nanoseconds; servicing a page fault takes several milliseconds, since it requires a disk transfer. The relation is \[ \text{EAT} = (1-p) \times t_{m} + p \times t_{f}, \] where \( p \) is the fault rate. With a memory time of 100 ns and a fault time of 8 ms, a fault rate of just one in a thousand gives an effective time of about 8 microseconds - eighty times slower than memory alone. That arithmetic is why the fault rate must be kept extraordinarily low, of the order of one in a million, for demand paging to be worthwhile, and why locality of reference is essential to the whole scheme. It is also why thrashing is catastrophic rather than merely slow, and why replacement algorithms are studied so closely.
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