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+1 vote
Consider a system with the main memory access time as 200 ns and cache access time as 10 ns. Hit ratio for
read request is 0.8 and 80% of the memory requests are for read. If write through policy is used, then the
average time considering both read and write requests is _______.
a. 169.6 ns
b. 192.4 ns
c. 78.4 ns
d. None of these
asked in CO & Architecture by (307 points)   | 51 views
c) 78.4??
Yes. Correct.

sir please  send  solution..

1 Answer

+3 votes
Avg time= read request ( read hit*cache access time+ read miss*(main memory+cache access time)) + write request ( main memory)

Here write request* main memory is done because of write through cache so both cache and main memory Will be updated simultaneously and since main memory has large delay we have taken it

Avg time=0.8(.8*10+.2*210)+0.2*200=80 ns

Option D

unless not mention we have to take cache access time while calculating cache miss
answered by Boss (6.7k points)  
edited by


we are taking 

Avg time= read request ( read hit*cache time+ read miss*(main memory+cache time)) + write request ( main memory).  As write-through works in multilevel(heirarchical) design.

why you have not taken cache time in solution.

yes you are correct
but in answer key  (c) is given as correct answer.
I am getting the same answer as 80, same as explained by Tesla.
OK. Thank You. Sir.
There answer is incorrect, they have to mention if main memory time is including cache time or not, if not then we add cache acess time also, you can see any gate question whenever it is not mentioned they add cache acess time in main memory time

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