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+1 vote

here when we calculate time if the write operation has to be performed then

according to me we have to calculate it in following way-

EMAT =0.8*(0.9*(100)+0.1*(100+1000)) + 0.2(100 +1000)

 =380 ns

i have included time to write in main memory as well as cache, because if we have to perform write operation then the cache has to update too.

here is the question and they have not included cache time,

is there anything wrong in my approach if it is the please correct me

asked in Operating System by Active (1.6k points)
recategorized by | 64 views
As write through policy is used:
average time = read + write
                        = 0.8(0.9(100)+0.1(1100))+0.2(0.9(1000)+0.1(1000))
                        = 160 + 200
                        = 360
thank you for the link it was very useful to understand the concept.

but i think here at the time of writing we do not need to use probability of accessing chache because we must write  in memory and cache these writes should be simultaneously(write through policy),

so answer should be evaluate in this way-

EMAT=0.8*(0.9*(100)+0.1*(1100)) + 0.2*(1000)

        =160 +200


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