Cache organization is often influenced by the desire to reduce the cache\'s ener
ID: 3776691 • Letter: C
Question
Cache organization is often influenced by the desire to reduce the cache's energy consumption. For that purpose we assume that the cache is physically distributed into a data array (holding the data), tag array (holding the tags), and replacement array (holding information needed by replacement policy). In addition, each one of these arrays is physically distributed into multiple sub-arrays (one per way) that can be individually accessed. For example, a four-way set associative LRU cache has four data sub-arrays, four tag sub-arrays, and four replacement sub-arrays. The replacement sub-arrays are accessed once every access when the LRU replacement is used. However, it is not needed when Random replacement is used. For a specific cache, it was determined that the accesses to the different arrays have the following energy consumption: Estimate the energy consumption for the following configurations. The cache is 4-way set associative. Main memory access and cache refill (although important) are not considered here. Provide answers for the LRU and Random replacement policies. A cache read hit. All arrays are read simultaneously. The cache access is now split across two cycles. In the first cycle, all tag sub-arrays are accessed. In the second cycle, only the sub-array whose tag matched will be accessed. Repeat part (a) for a cache read hit. Repeat part (b) for a cache read miss, assuming that no data array access in the second cycle for a cache miss.Explanation / Answer
Estimate the reserve control use in power units for the accompanying arrangements.
We expect the reserve is four-way set associative.Main memory get to poweralbeit critical is not considered here.
Give answers to the LRU,FIFO,and irregular substitution approaches.
a.A cache read hit.All arrays are read simultaneously..
we see the cache power control use as P .
For LRU: P =( 20 + 5 + 1) *4 = 104unit.
It is read hit.
So the FIFO and Random substitution approaches don't have to get to the Miscellaneous cluster.
So for FIFO and Random:
P = ( 20 + 5)* 4 = 100unit
b.Repeat part(a) assuming that the cache access is split across two cycles.In the ±rst cycle,all the tag sub-arrays are accessed.In the second cycle,only the sub-array whose tag matched will be accessed..
For LRU. In second cycle,we don't have to get to every one of the information sub-cluster so that the power use is
P = (5 + 1)* 4 + 20 = 44units
For FIFO and Radom,it's the same with LRU.
In second cycle,we don't have to get to every one of the information sub-exhibit so that the power use is
P = 5* 4 + 20 = 40units
c.Repeat part(c) for a cache read miss(no data array accesses in the second cycle)
for cache read miss
for reserve read miss
For LRU the power use is: P = ( 20 + 5 + 1)* 4 = 104units
For FIFO,it need to get to Miscellaneous cluster once.
P = ( 20 + 5)* 4 + 1 = 101units
For Random P = ( 20 + 5)* 4 = 100units
no dat utilized as a part of the in exhibit is
p=5*4+20+1=41units.