Average access time

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?

Arul @innovwelt
27 Jan 2015 11:06 pm

I am not quite sure. can you tell me the answer?

I get approximately 49.104 ns

vishal @vishal92
28 Jan 2015 12:50 pm

innowelt date write is always in the main memory 

Shreyas Dawkhare @shreyasdawkhare
27 Jan 2015 11:42 pm

this is what they explained. not getting how Tavg calculated.

Mahesh Kumar @maheshkumars
27 Jan 2015 11:50 pm

I think we can directly read a word from MM as they said simultaneous,as there is no need of Block transfer.Only block transfer is needed in strict hierarchy(where only cache access is available).They are combining both.

vishal @vishal92
28 Jan 2015 12:57 pm

write through and write back are updation technique
smultaneous and hierachial are memory access technique
both are diffrent
write through and write back comes only in picture when memory write is there

Kunal Chalotra @kunalchalotr
19 Jul 2015 12:54 pm

remember from cache always word are read therefoer i think we have to do like this 72/5=14.4 nsec time for cahe and for main memory we have to consider block size which contain 4 words  i.e 72*4=288

Mahesh Kumar @maheshkumars
27 Jan 2015 11:51 pm






Rosni K V @rosnikv
28 Jan 2015 10:22 am

(0.85*57.6)+ (0.15*288)=taveR
since it is write through, T aveW= max(T main mem, T cache) ie, defenitly T main mem.
Now they are given 60% read req, using that frequency we can find average access time.

vishal @vishal92
28 Jan 2015 12:53 pm

as it is write through cache 
while reading 
1-if it gets the word in cache it read from cache
2-if there is read fault ,reads from the main memory,we havt to add the cache access time when miss it is write through not simultaneous memory access
while writing
1-if content in the cache it have to side by side updated in the main memory
2-so there is hit or miss in both cases main memory is accessed
so T(avg)=T(avg read)+T(avg write)

Arul @innovwelt
28 Jan 2015 01:44 pm

@vishal: since memory is organized into 4 word blocks. we have to consider main memory access for whole block instead of "per word".

Shreyas Dawkhare @shreyasdawkhare
28 Jan 2015 02:31 pm

Does anybody have good material regarding how data is accessed in the system having 2 levels  of cache , memory and disk. please share. I am totally confused now... Please share.....