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Consider a two level memory hierarchy. L1(cache) has an accessing time of 10 ns  and main memory has an accessing time of 20ns. Writing or updating contents into their memory takes 20 ns and 30 ns for L1 and main memory respectively. Assume L1 gives misses 80% of the time. The average writing time for system (in ns) if it uses write-through technique is _____________

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It is write through, both cache and MMU update at same time.

L1 gives misses 80% of the time. That means L1 gives hit 20% of the time . Memory is accessed 80% of time and cache is accessed 20 % of time.

H = 0.2 , (1- H) = 0.8 

mem update time  =  30 ns

TAvg writing Time  = H * T mem update time + ( 1 - H ) * ( T mem update time )  

= 0.2 *  30 + 0.8 * 30 

= 6 + 24

= 30 

In write through both cache and MMU update same time that's why only MMU update time is consider here .

So, average writing time is 30 ns .

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