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When a cache is 10 times faster than main memory , and cache can be used 90% of the time , how much speed we gain by using cache ?
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+1 vote
USING AMDHALS LAW,

SpeedupMAX = 1/((1-F)+(F/S)) (here F=0.9,S=10)

=5.263
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Can you explain why we used Ambhals law here and not the regular method
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Thank you, it really helped
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are you preparing for gate.
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Yes
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Are you a repeater or final year student
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Repeater

Let the access time of main memory be m units.

Then access time of cache = 0.1m units (Since cache is 10 times faster)

Old access time = m [only main memory used]

New access time (with cache) = 90% * (0.1m) + 10% * (0.1m + m)

= 0.9 * (0.1m) + 0.1 * (0.1m + m)

= 0.09m + 0.11m = 0.2m

Therefore speedup = $\frac{Old access time}{New access time} = \frac{m}{0.2m} = 5$ (ans)

by (29 points)