0 votes
31 views
A process takes 20 ns on a cache hit and 400 μs on a cache miss to read an instruction. Approximately 20% of the time read request is found in the cache. Then the average access time is ________ μs.
(Upto 3 decimal placces)
| 31 views
0
320.004?
0
Yes how

1 Answer

+2 votes
Best answer
Note:
We will not use conventional Hierarchical access formula here, as they have clearly given time required when cache miss occurs and time when hit occurs.

AMAT =cacheHitRate*(time when cache hit) + cacheMissRate*(TimewhenCacheMiss)

AMAT =0.2(20ns) +0.8(400$\mu sec$)= 320.004$\mu sec$
by (269 points)
selected
+1

Good question - good answer.

VERY important to note

A process takes 20 ns on a cache hit and 400 μs on a cache miss to read an instruction

Does this mean that we should assume hierarchical or simultaneous organization ? NEITHER. Because it doesn't matter. Irrespective of which scheme, the question gives a specific number to use.

0

Thanks @shashin,@tp21

if nothing is mentioned about hierarchical or simultaneous access we should used direct method what given above answer (simultaneous)?

0

If nothing is given, going with hierarchical is the convention.

if nothing is mentioned about hierarchical or simultaneous access we should used direct method what given above answer

Please be careful while thinking about this. This 'answer' does not give anything. Please don't pick up new concepts from solutions. Hierarchical or simultaneous are the only standard ways of accessing cache. Anything else is a problem-specific thing. Don't learn something from one numerical and try to apply it everywhere.