SME
Operating Systems · Problem type
Effective access time with page faults
Use this for “Memory takes 100 ns, a page fault takes 8 ms to service, and 1 access in 100 000 faults. What is the effective access time?”
This is an example. Type your own values to start a new problem.
EAT (ns)
=
(1
−
p
)
×
m (ns)
+
p
×
F (ms)
Formulas
EAT
=
(1
−
p
)
×
m
+
p
×
F
179.999
=
(1
−
(1
×
10
-5
))
×
100
+
(1
×
10
-5
)
×
8,000,000
Formulas are worked in ns. The step-by-step shows the conversions.
Assumptions
•
The fault rate is the share of memory accesses that find their page on disk.
•
The service time covers the whole fault: the trap, the disk read and restarting the access.
•
A fault is thousands of times slower than a memory read, so a tiny p still dominates EAT.
Step-by-step
WE KNOW
p = 1 × 10⁻⁵, m = 100 ns, F = 8 ms
FIND
effective access time (EAT)
Step 1 · Convert to the formula’s units
The formulas don’t work directly in the units you chose, so convert first (to ns).
F = 8 ms = 8,000,000 ns (1 ms = 10⁶ ns)
Step 2 · Find effective access time (EAT)
Use EAT = (1 − p) × m + p × F
Weigh each case’s time by how often it happens: a memory read, or a fault’s service time.
EAT
=
(1
−
10
-5
)
×
100
+
10
-5
×
8,000,000
EAT = 1 × 100 + 80
EAT = 99.999 + 80
EAT = 179.999 ns
Check
Put the numbers back in. Both sides should match.
179.999
=
(1
−
10
-5
)
×
100
+
10
-5
×
8,000,000
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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