coa

Question

Hit ratio of the cache memory read request is 85% and the cache memory is 5 times faster than main memory. Block size in memory organization is 4 words. The access time of the main memory is 72 ns per word. Write through protocol (simultaneous memory organization) is used in the system. CPU generates 60% of the read requests to read the data and the remaining for write operation. What is the average access time (in ns) of the memory when considering both read and write operations?

Comments

my ans is 132.105 nsec

plzz check and tell correct or wrong ..

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@ Arjun Sir,

I got a slightly different answer(69.12 ns) by using the concepts given by one of the professor and now I am in great dilemma when it comes to write access problems.

There he told that write miss penalty is always going to be same ( i.e memory time if considering only one cache) and when he calculated the average access time he even included the write access time of cache also .

For example= 

one of the formula given is 

AMAT= 1 + %read*(1-h)*(M) + % write*(M)  ( 1 is the access time of cache in clock cycles )

actually it is  AMAT = 1 + avg. stalls per access.

And this was given under the concept Write Through , No write Allocate , No Write Buffer... Please refer it and tell me whether any thing wrong in this ... I am not able to upload the photo . I tried many times..Waiting for your response Sir. Thank You.. 

Answer 1

Average Memory Access Time = 0.6 * Avg. time for read + 0.4 * Average Time for write

Main memory access time = 72 ns per word

Cache access time = Main memory access time/5 (Given in question) = 14.4 ns

Avg. time for read = 0.85 * 14.4 + 0.15(14.4+ * 72 * 4) (On a cache miss entire cache block is taken from main memory)

= 57.6 ns

 (simultaneous memory organization  is given in question for writing as all writes are going to memory irrespective of cache hit/miss. So, we can assume read is hierarchical access though it won't change the answer much)

Avg. time for write = 72 ns (No need to update the whole cache block- we are given main memory access time per word meaning a word can be directly accessed and there is no point updating the whole block in write through cache)

So, Average memory access time = 0.6 * 57.6 + 0.4 * 72 = 63.36 ns

Comments

@Arjun Sir

Here https://gateoverflow.in/14480/formula-write-back-write-through-access-time-parallel-serial

Tread=H×Tcache+(1−H)×(Tmemory_block)

But above question,you are usig

Tread=H×Tcache+(1−H)×(Tcache+Tmemory_block),Pleae tell why Tcache is there for simultaneous organization?

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@ Arjun Sir,

I got a slightly different answer(69.12 ns) by using the concepts given by one of the professor and now I am in great dilemma when it comes to write access problems.

There he told that write miss penalty is always going to be same ( i.e memory time if considering only one cache) and when he calculated the average access time he even included the write access time of cache also .

For example= 

one of the formula given is 

AMAT= 1 + %read*(1-h)*(M) + % write*(M)  ( 1 is the access time of cache in clock cycles )

actually it is  AMAT = 1 + avg. stalls per access.

And this was given under the concept Write Through , No write Allocate , No Write Buffer... Please refer it and tell me whether any thing wrong in this ... I am not able to upload the photo . I tried many times..Waiting for your response Sir. Thank You.. 

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It is written simultaneous organization is used for system.. Then why you have taken simultaneous for write only.. Why not for read also?? Plz clear this doubt

Answer 2

read hit (h_r) =0.85

write hit (h_w)= 1 because of simultaneous memory orgⁿ

^{read frequency (f_r)= 0.6     write frequency (f_w)= 1-0.6}=0.4

mm access time (t_m)= 72 ns per word

so time to access a block = 4 * 72= 288 ns per block

s= t_m/t_c

5=288/t_c

t_c=288/5=57.6  so cm access time= 57.6 ns per block

so t_{avg read} =h_r*t_c+(1- h_r)*tm   = 0.6*57.6 +0.4*288 =92.16 ns  (block wise access)

   t_{avg write} =h_w*t_m = 1* 72 =72 ns   (because in simultaneous orgⁿ  in case of write opⁿ  hit ratio is always one.because data block need not required write in cm . you go directly at mm and wriite your required word not block be careful )

total t_avg= f_r* t_{avg read} + f_w*t_{avg write} =0.6*92.16 + 0 .4*72 =84.09 ns

Comments

cache memory is 5 times faster than main memory

So, per word access time of cache must be 5 times faster 

Answer 3

http://www.techtud.com/doubt/average-access-time

Comments

somebody solve here and tell .? that explanation does not suit me well.??