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A computer system has an $L1$ cache, an $L2$ cache, and a main memory unit connected as shown below. The block size in $L1$ cache is $4$ words. The block size in $L2$ cache is $16$ words. The memory access times are $2$ nanoseconds, $20$ nanoseconds and $200$ nanoseconds for $L1$ cache, $L2$ cache and the main memory unit respectively.

 

When there is a miss in both $L1$ cache and $L2$ cache, first a block is transferred from main memory to $L2$ cache, and then a block is transferred from $L2$ cache to $L1$ cache. What is the total time taken for these transfers?

  1. $222$ nanoseconds
  2. $888$ nanoseconds
  3. $902$ nanoseconds
  4. $968$ nanoseconds
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Miss in L1 cache and L2 cache results in the:

(200+20) for one word to get from main memory to l2 cache
(20+2) for one word to get from L2 cache to L1 cache

So total Time Needed is:   (200+20)*4  + (20+2)*4   = 968 

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Access time from main memory is 200. So total time to access is is 200+20 nanoseconds Similarly for L1 cache to access from L2 cache is 20+2 nanoseconds So total time is 4*(220 + 22) = 968 nanoseconds.
 
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Answer is A

Time for transferring from main memory to L2 is 200ns . time for transferring from L2 to L1 is 20ns. Time needed to write into L1 is 2ns. The total time taken is 200 + 20 + 2 = 222ns.
Answer:

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