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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 _____.

+1 vote
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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But answer is given as : 76
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this question is from made easy test series? leave it answer is 50ns only.
https://gateoverflow.in/76502/me-co-write-through

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Yes.

Thank you .
0

https://gateoverflow.in/143595/cache#a143671

Can you please,check out this question on same type?

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@Nils I am also getting same answer as 80ns for that question
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what will be answer in case of write back policy?

pls explain.

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