SQL
October 13, 2023Computer-Organization
October 13, 2023Computer-Organization
| Question 1 |
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?
| 13.0 ns | |
| 12.8 ns | |
| 12.6 ns | |
| 12.4 ns |
Question 1 Explanation:
Average access time = [H1 * T1] + [(1 – H1) * Hm * Tm]
H1 = 0.8, (1 – H1) = 0.2
H2 = 0.9, (1 – H2) = 0.1
T1 = Access time for level 1 cache = 1ns
T2 = Access time for level 2 cache = 10ns
Hm = Hit rate of main memory = 1
Tm = Access time for main memory = 500ns
Average access time = [(0.8 * 1) + (0.2 * 0.9 * 10) + (0.2)(0.1) * 1 * 500]
= 0.8 + 1.8 + 10
= 12.6ns
H1 = 0.8, (1 – H1) = 0.2
H2 = 0.9, (1 – H2) = 0.1
T1 = Access time for level 1 cache = 1ns
T2 = Access time for level 2 cache = 10ns
Hm = Hit rate of main memory = 1
Tm = Access time for main memory = 500ns
Average access time = [(0.8 * 1) + (0.2 * 0.9 * 10) + (0.2)(0.1) * 1 * 500]
= 0.8 + 1.8 + 10
= 12.6ns
Correct Answer: C
Question 1 Explanation:
Average access time = [H1 * T1] + [(1 – H1) * Hm * Tm]
H1 = 0.8, (1 – H1) = 0.2
H2 = 0.9, (1 – H2) = 0.1
T1 = Access time for level 1 cache = 1ns
T2 = Access time for level 2 cache = 10ns
Hm = Hit rate of main memory = 1
Tm = Access time for main memory = 500ns
Average access time = [(0.8 * 1) + (0.2 * 0.9 * 10) + (0.2)(0.1) * 1 * 500]
= 0.8 + 1.8 + 10
= 12.6ns
H1 = 0.8, (1 – H1) = 0.2
H2 = 0.9, (1 – H2) = 0.1
T1 = Access time for level 1 cache = 1ns
T2 = Access time for level 2 cache = 10ns
Hm = Hit rate of main memory = 1
Tm = Access time for main memory = 500ns
Average access time = [(0.8 * 1) + (0.2 * 0.9 * 10) + (0.2)(0.1) * 1 * 500]
= 0.8 + 1.8 + 10
= 12.6ns
Subscribe
Login
0 Comments