Question

Question 27 7 pts Suppose the cache access time is 10ns, main memory access time is 200ns, and the cache hit rate is 90%. Assuming parallel (overlapped) access, what is the average access time for the processor to access an item? 29ns

Answer 1

Effective Access time (EAT)  is a weighted average that takes into account the hit ratio and relative access times of successive levels of memory

The EAT for a two-level memory is given by:

where H is the cache hit rate and AccessC and Access_MM are the access times for cache and main memory, respectively

Average Access Time for the processor to access an item :

This is because, when there is a miss, the processor will look into the main memory ( Access_MM ) and then add that information in the cache, and then retrieve that information from cache ( Access_c). Therefore the total access time when its a MISS= Access_{MM +} Access_c.

But in our case, access overlap take place, in this the processor will access the memory and cache parallely, so we will add only the access time of the memory here. Because the access time of cache is nullified in access overlap case as it doesn't have to access cache after memory. rather it will access parallely without any time delay

NOTE: IF THE ACCESS OVERLAP DOESNT OCCUR, THEN IN ESTIMATE ACCESS TIME, WE HAVE TO ADD BITH ACCESS TIME FOR CACHE AND MAIN MEMORY WHEN MULTIPLIED BY (1-H). THEREFORE , AT THAT TIME