GATE CSE 2015 | Set 2 | Question: 24

Question

Assume that for a certain processor, a read request takes $50\:\text{nanoseconds}$ on a cache miss and $5\:\text{nanoseconds}$ on a cache hit. Suppose while running a program, it was observed that $80\%$ of the processor's read requests result in a cache hit. The average read access time in nanoseconds is ______.

Comments

I think we don't need to think whether it is simultaneous access or hierarchical, $50ns$ will take care.

Standard Formula is : $T_{avg} = H*T_h + (1-H)*T_{missPenalty}$ ,

$T_{missPenalty}$ includes Cache access in hierarchical case and exclude cache access in simultaneous. So whether it is simultaneous or not is not our concern. The given miss penalty is taking care of that.

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@Bikram sir

I think we must see access times from processor's point of view i.e after how much time the processor gets the requested data. Suppose that in a question we are explicitly given cache access time is 2ns and main memory access time is 30ns then here we will have to explicity calculate after how much time processor will get requested data. Lets say processor makes a request, now if the data is present in cache then processor will get the requested data in 2ns after cache is accessed else if it is not present then processor will get it in 30ns after main memory access (if simultaneous access) or 32ns after cache + main memory access (if hierarchichal access).

But here in the question we are directly given that processor gets requested data in 50ns in case of a cache miss, so we need not worry whether simultaneous or hierarchical access policy is used, here that information is abstracted away from us.

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NOTE: Simultaneous Access is a Misconception..

The given $50ns$ is Miss Time, Not Miss Penalty.

Detailed Video Explanation with Correct Concepts: www.youtube.com/watch?v=v49Qv5CrgdY&t=8735s

Average read time $= 0.80 \times 5 + 0.20 \times 50 = 14$ nsec.

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MISCONCEPTION of Simultaneous Access is discussed here: https://youtu.be/v49Qv5CrgdY?si=o42qpaOaqiySQHpk

Answer 1

ok first  Remember ,

Miss Time = Miss Penalty + Hit Time 

here three main formula all are same 

AMAT = hit time +miss rate * miss penalty

AMAT = Hit ratio * hit time + (1-hit time)* miss time 

AMAT = H (Tc) + (1-H)(Tc + Tm)  

where 

H =hit ratio 

Tc = cache access time 

Tm = miss penalty 

In the given question it  directly  mention the miss time, means Miss penalty+Hit time = 50 nsec
         

 

by putting this values in given formula 1 
Miss penalty = 50 - hit time = 50 - 5 = 45 

AMAT  = 5 + (1 - 0.8)*45 = 5+ 9 = 14 ns




by formula 2

AMAT  = 0.8 * 5 + ( 1- 0.8 ) * 50 = 4 + 10 = 14ns 




by formula 3 

Tc= 5 ns 

Tm = Miss Penalty = 45 

ATMT = 0.8 * 5 + (1 - 0.8 )(5 + 45)  = 14 ns