Average Memory Access Time Problem

Question

A cache is having 60% hit ratio for read operation. Cache access time is 30 ns and main memory access time is 100 ns, 50% operations are read operation.
What will be the average access time for read operation?
(a) 50 ns      (b) 58 ns (c) 100 ns      (d) 70 ns

Comments

I think answer can be 70 ns only since they have mention avg access time for read instructions... 50*70/50 = 70... if they would have asked average read access time for each instruction then it can be ( (50*70) +(50*0) ) /100  = 35ns.. is it correct ??...

Answer 1

Given Hit Ratio for Reading = 60/100 = 3/5

 

∴ Miss Ratio m1 for Reading = 1-(3/5) = 2/5 = 0.4

 

 Given access time of cache t1 = 30ns & main memory = 100ns

 

Average Memory Access Time for Reading = t1 + m1*t2 = 30 ns + (0.4 * 100 ns) = 70ns

 

Here, there nothing to do with 50% and access time for writing. Hence, all things are given in terms of reading only.

Answer 2

read operations*( cache hit*cache access+cache miss*(cache access + main memory access))

0.5 [ 0.6*30+0.4*(30+100)]=0.5*(18+52)=35ns

Comments

@bikram sir

is it right answer?

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@  hem chandra joshi 

Yes correct..

Avg access time = Hit ratio * cache access time + miss ratio * ( Cache access time + MM access time )

Given, Cache access time is 30 ns and main memory access time is 100 ns.

Hit ratio = 60% so Miss ratio = 40% 

so Avg access time = (0.6 * 30 ) + 0.4 * ( 30 + 100 ) 

=  18 + 52

= 70 

so average access time is 70 ns.

Now it says 50% operations are read operation so average access time for read operation is 

= 0.5 * 70 

= 35 

average access time for read operation is 35 ns .

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But 35 is not in option

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in question it is not mentioned it is strict hierarchy

Answer 3

Answer: d
Explanation:
Avg access time (for read operation) = Hit ratio * cache access time + miss ratio *
( Cache access time + MM access time )
Given, Cache access time is 30 ns and main memory access time is 100 ns.
Hit ratio = 60% so Miss ratio = 40%
so Avg access time = (0.6 * 30 ) + 0.4 * ( 30 + 100 )
= 18 + 52
= 70
so average access time for read operation is 70 ns.