Consider a system with 2 level cache. Access times of Level 1, Level 2 cache and main memory are 0.5 ns, 5 ns and 100 ns respectively. The hit rates of Level 1 and Level 2 caches are 0.7 and 0.8 respectively. What is the average access time of the system ignoring the search time within cache?
7.55 ns
Under the stated convention of ignoring failed-cache search time, the probabilities of service by L1, L2 and main memory are 0.7, \(0.3\times0.8=0.24\) and \(0.3\times0.2=0.06\), respectively. The L2 hit rate is conditional on an L1 miss.
\(T=0.7(0.5)+0.24(5)+0.06(100)=7.55\text{ ns}.\)
Here only the serving level’s access time is charged. Charging every sequential lookup instead would give a different model.
Important Questions from Memory Hierarchy
A computer has a memory hierarchy consisting of two-level cache (L1 and L2) and a main memory. If the processor needs to access data from memory, it first looks into L1 cache. If the data is not found in L1 cache, it goes to L2 cache. If it fails to get the data from L2 cache, it goes to main memory, where the data is definitely available. Hit rates and access times of various memory units are shown in the figure. The average memory access time in nanoseconds (ns) is ________. (rounded off to two decimal places)
- Match List I with List II
List I List II A. Address Space I. Associative Mapping B. Memory Space II. Logical address C. Cache Memory III. physical address D. Segmented Program IV. Virtual address
Choose the correct answer from the options given below: - What unit of memory is largest out of the following?
Related to computers, what is 'Wetware'?
____ number of lines are required to select __________ memory locations.
