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kumar_eee
Joined: 22 Sep 2004
Posts: 516
Helped: 16
Location: Bangalore,India
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 11 Feb 2006 23:27 Which is the correct answer?.... |
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Consider a two-level memory hierarchy system M1 & M2. M1 is accessed first and on miss M2 is accessed. The access of M1 is 2 nanoseconds and the miss penalty (the time to get the data from M2 in case of a miss) is 100 nanoseconds. The probability that a valid data is found in M1 is 0.97. The average memory access time is:
a) 4.94 nanoseconds
b) 3.06 nanoseconds
c) 5.00 nanoseconds
d) 5.06 nanoseconds
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11 Feb 2006 23:27 Ads |
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eecs4ever
Joined: 31 Jan 2006
Posts: 181
Helped: 26
Location: Analog Environment
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12 Feb 2006 0:19 Which is the correct answer?.... |
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c) 5.0 nanoseconds
E(AccessTime) = AccessTime given a hit * Pr(hit)
+ AccessTime given a miss * Pr(miss).
Pr (hit) = 0.97
Pr (miss) = 1- 0.97 = 0.03
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neils_arm_strong
Joined: 05 Jan 2006
Posts: 274
Helped: 20
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12 Feb 2006 12:55 Which is the correct answer?.... |
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| I agree with eecs4ever
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Davood Amerion
Joined: 01 Mar 2005
Posts: 589
Helped: 90
Location: Persia
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12 Feb 2006 16:16 Which is the correct answer?.... |
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average access time= T (M1) + P(miss)* T(M2)
= 2nS + 0.03*100= 5 nS
Added after 1 seconds:
average access time= T (M1) + P(miss)* T(M2)
= 2nS + 0.03*100= 5 nS
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