On a system using demand-paged memory, it takes 120 ns to satisfy a memory reque
ID: 3866322 • Letter: O
Question
On a system using demand-paged memory, it takes 120 ns to satisfy a memory request if the page is in memory. If the page is not in memory, the request takes (on average) 5 ms.
a. What would the page fault rate need to be to achieve an effective access time of 1 ms? Assume the system is only running a single process and the CPU is idle during page swaps.
b. What would be the factor of slowdown if one in every 1000 accesses causes a page fault?
c. If a performance degradation of 15% is allowed, what would the maximum bound on the page fault rate be?
Explanation / Answer
a.Average Memory Access Time =
Hit time (no page fault) +
Miss rate (page fault rate) x Miss penalty
1 * 10-3 = 120 * 10-9 + page fault rate * 5 * 10-3
page fault rate = 19.9%
b.
avg time = 120ns * 999 + 1*5*106 ns = 5119880ns for the given case
noramlly it would be 120000 ns --- it would be 42.66 times slower than usual
c.performance degradation of 15% means it would be 1.15 times slower so it will be 1.15 times more
consider for 1000 access so it will be 120*1.15 = 138 ns
so 138ns = 120 ns * (1-p) + p * 5 *106ns
p = 0.0000036
page fault rate = 0.00036 %