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A hierarchical memory system that uses cache memory has cache access time of 80 nanoseconds, main memory access time of 200 nanoseconds, 85% of memory requests are for read, hit ratio of 0.9 for read access and the write-through scheme is used. What will be the average access time of the system both for read and write requests ?
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given,

memory access time (Tm)=200 ns

cache access time (Tc) = 80 ns

hit ratio for read access =0.9

for write through we generally use no write allocate technique .

so on write average memory access time =  Tm =200 ns

on read average memory access time= Tc+(1-0.9) Tm

                                                              =80+0.1*200 =100 ns

now 85% operation are memory read .

so average memory access time=0.85*100 +0.15*200 = 115 ns

https://gateoverflow.in/14480/Formula-write-back-write-through-access-time-parallel-serial?show=14502#a14502

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To calculate the average access time (AAT) of the system, we need to consider the access time of the cache, main memory, and the time taken to write the data back to memory.

Given:

  • Cache access time (Tc) = 80 nanoseconds
  • Main memory access time (Tm) = 200 nanoseconds
  • Percentage of read requests (Pr) = 85%
  • Hit ratio for read access (Hr) = 0.9
  • Write-through scheme is used

To calculate the average access time for read requests, we can use the following formula:

AAT_read = Tc + (1 - Hr) * Tm

Here, (1 - Hr) represents the miss ratio for read requests.

Substituting the given values, we get:

AAT_read = 80 + (1 - 0.9) * 200 = 100 nanoseconds

To calculate the average access time for write requests, we need to take into account the time taken to write the data back to memory. Since write-through scheme is used, the write request must update both the cache and main memory.

AAT_write = Tc + Tm

Therefore, the average access time for write requests is:

AAT_write = 80 + 200 = 280 nanoseconds

To calculate the overall average access time (AAT), we need to take into account the percentage of read and write requests. Since 85% of requests are for read and 15% are for write, we can use the following formula:

AAT = Pr * AAT_read + (1 - Pr) * AAT_write

Substituting the values, we get:

AAT = 0.85 * 100 + 0.15 * 280 = 121 nanoseconds

Therefore, the average access time of the system for both read and write requests is 121 nanoseconds.

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0.85[ 0.9(80) + 0.1(80+200) ] + 0.15 [ 0.9(80) + 0.1(80+2*200) ]

= 100.3 ns

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