# GATE | GATE-IT-2004 | Question 12

• Difficulty Level : Medium
• Last Updated : 19 Nov, 2018

Consider a system with 2 level caches. Access times of Level 1 cache, Level 2 cache and main memory are 1 ns, 10ns, and 500 ns, respectively. The hit rates of Level 1 and Level 2 caches are 0.8 and 0.9, respectively. What is the average access time of the system ignoring the search time within the cache?
(A) 13.0 ns
(B) 12.8 ns
(C) 12.6 ns
(D) 12.4 ns

Explanation: First, the system will look in cache 1. If it is not found in cache 1, then cache 2 and then further in main memory (if not in cache 2 also).

The average access time would take into consideration success in cache 1, failure in cache 1 but success in cache 2, failure in both the caches and success in main memory.

`Average access time = [H1*T1]+[(1-H1)*H2*T2]+[(1-H1)(1-H2)*Hm*Tm]`

where,

H1 = Hit rate of level 1 cache = 0.8
T1 = Access time for level 1 cache = 1 ns
H2 = Hit rate of level 2 cache = 0.9
T2 = Access time for level 2 cache = 10 ns
Hm = Hit rate of Main Memory = 1
Tm = Access time for Main Memory = 500 ns

So, Average Access Time   = ( 0.8 * 1 ) + ( 0.2 * 0.9 * 10 ) + ( 0.2 * 0.1 * 1 * 500)

= 0.8 + 1.8 + 10

= 12.6 ns

Thus, C is the correct choice.

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