What is average Acess Time of system considering only memory read cycles?

Question

The access time of cache memory is 100 ns and that of main memory 
1000 ns. it is estimated that 80% of the memory requests are for read 
and the remaining 20% for write. the hit ratio for read access only is 
0.9. A write -through procedure is used. 
a- what is the average access time of the system considring only 
memory read cycles? 
b-what is the average access time of the system for both read asn 
write requests? 
c-what is the hit ratio taking into consideration the write cycles? 
 

Answer 1

hit ration h=0.8 the miss ratio =0.1

since it is write through

so avg access time = h*cache access time + (1-h)*(cache access time+main memory access time)

for only read cycles :

0.8*((0.9*100)+0.1*(100+1000)) = 0.8 * 200 =160

for only write :

0.2*((0.9*100)+0.1*(100+1000)) = 0.2 * 200 =40

for both read and write :

avg access time = 160 + 40

                           =200

Comments

@cse23. I disagree with you.

When they have asked only for read, it means that the probablity of read operations is 1.

So, only for read operations, effective read time

= (hit ratio for cache)(cache access time) + (cache miss ratio)(Cache time + M.M time)

= 0.9 * 100   +    (1 - 0.9)(100 + 1000)

= 200ns

SImilarly, only for write, effective write time = 1000 ns( because its the write through cache and largest of the cache time and M.M time is considered).

Now, effective access time(considering a mix of read and write)

= (probablity of read operations)* (effective read operation time)

   + (probablity of write operations)* (effective write operation time)

= 0.8 * 200 + 0.2 * 1000

= 360 ns

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yes, u r right.

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@Sushant Gokhale , @Kapil why have u considered cache access time also on the cache miss......By default shouldn't we consider parallel accessing instead of Hierarchical access?

Correct me if I'm wrong.

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How did you know that it's a cache miss? You checked the cache ,right?

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What is the meaning of write through procedure ??

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But write thorough uses simultaneously access, right?

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yeah cse23 solution is totally wrong

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i guess answer should be
for read , t avg = h*tcm+(1-h)(tcm+tmm)=0.9(100)+0.(100+1000)=200ns
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for write , t avg = max(tcm,tmm) = 1000ns

(as it's write through policy , updates/writes parallelly in both CM and MM  )
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for read +write , t avg = 0.8(200)+0.2(1000)=360ns

 for c part , I’m not so sure that reqs of write will be counted as hits, but if so, then
suppose 100 reqs are there ,out of which 80 are for read . And out of this 80 , 90% are hits, so 90% of 80 = 72
write reqs if all hits then total hit should be 92% (or 0.92) , otherwise if considering  as miss then 72 %

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Write thorugh procedure uses simultaneous access memory organization so this solution is wrong

Answer 2

let suppose there are 100 instruction then

80 instruction will be read only instruction and

20 instruction are write instruction

hit ratio considering write cycles=0.9 x 80 + 1 x 20 (since there will be always hit on write )

                                                  = 92%

Comments

As it’s write  through policy but won’t it consume the time of updating in Main Memory ?