Which is the correct answer?....

[kumar_eee]

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

 

[eecs4ever]

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

 

[neils_arm_strong]

I agree with eecs4ever

 

[Davood Amerion]

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