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Consider a two level memory hierarchy L1 (cache) has an accessing time of 20 nsec and
main memory has accessing time 40 nsec. Writing or updating contents into their
memory takes 40 nsec and 50 nsec for L1 and main memory respectively. Assume that
L1 gives misses 65% of the total time. The average writing time for system (in nsec) if it
uses write-through technique _____.

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76 ns.

Simultaneous write (Tw )= max (40,50) 

Avg write time = H​​​​​wT​​​​w + (1-Hw)(T​​​​​​m + T​​​​​​w) = 0.35*50 + 0.65*(40+50) = 76ns

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L1 A.T = 20ns
M/M A.T = 40ns
L1 Write Time = 40ns
M/M Write Time = 50ns

Write Through Technique: on each write request by CPU, memory will be updated everytime.
In Write Through we use "No Write Allocate" techniqiue, it means if there is a  write miss in cache we directly write memory instead of bringing a block first into cache and then write it.

Avg Write Time = .65*max(40,50)ns + .35*50 = 50ns
edited

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