Compute the average load time

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A computer has a cache, main memory and a hard disk. If a referenced word is in the cache, it takes 15 ns to access it. If it is in main memory but not in the cache, it takes 85 ns to load (the block containing) it into the cache (this includes the time to originally check the cache), and then the reference lookup is started again. If the word is not in main memory, it takes 10 ms to load (the block containing) it from the disk into main memory, and then the reference lookup is started again. The cache hit ratio is 0.4. In the case of a cache miss, the probability that the word is in the main memory is 0.7. Compute the average load time.

My Answer

    Given:

    Cache access time = 15 ns
    Cache hit rate = 0.4
    Cache miss rate = 1 – 0.4 = 0.6
    RAM access time = 85 ns
    RAM hit rate = 0.7
    RAM miss rate = 1 – 0.7 = 0.3
    Disk access time = 10ms = 10000000 ns



    >  Average access time = (cache access time x cache hit rate) + (cache
    > miss rate) x (RAM access time + RAM hit rate) + (cache miss rate x ram
    > miss rate x disk access time)
    >     = (15*0.4) + (0.6)(85*0.7) + (0.6)(0.3)(10000000)
    >     = 1 800 041,7 ns

I hope you are enjoying the computer systems class at Birkbeck... ;P

I think you missed something though:

(1) You are assuming the 10ms include the initial checking of the cache (he specified it for the 85ns but not for the 10ms so would add that to be on the safe side)

(2) It says the reference lookup is started again after loading into cache and main memory respectively... So from the question I understand that words can only be accessed from the cache (otherwise why bother with the 85ns?). Hence, I think you need to add the time it takes to load it into the cache from main memory when retrieved from the disk initially. Also, although I am not entirely sure on this one as it is a bit ambiguous, I think you need to add another 15ns for the word to be accessed in the cache after its loaded from main memory...

Interested to hear some thoughts

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Kindly share the answers of first three questions in assignment here if you have done em.. :P

For this question answer is :

Cache access time = 15ns
Memory access time = 85ns +15ns = 100ns
Disk access time = 10x106 + 100ns = 10000100ns
Average load time = 0.4 x 15ns + 0.6[0.7 x 100ns + 0.3(10000100ns)]
Average load time =     6 +0.6(70 + 3000030) = 6 + 1800060
Average load time = 1800066ns = 1.8ms

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If it is in main memory but not in the cache, it takes 85 ns to load (the block containing) it into the cache (this includes the time to originally check the cache).

You dont't need to add 85 (Memory) and 15 (Cache)

For this question answer is :

Cache access time = 15ns
Memory access time = 85ns
Disk access time = 10x106 + 85ns = 1000085ns
Average load time = 0.4 x 15ns + 0.6[0.7 x 85ns + 0.3(1000085ns)]
Average load time =     6 +0.6(59.5 + 3000025.5) = 6 + 1800051
Average load time = 1800057ns = 1.8ms

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Comments
  • I'll appreciate any comment.
  • This question is purely a math problem. It's not plausible that any real system would have exactly 10 ms hard page faults, with enough precision that you could usefully compute an average that takes into account cache hit/miss as well. (At least not with such a high page miss rate dominating the average!). i.e. The CPU cache is making a negligible difference to the average access time with such a high miss rate. (I'm assuming a coherent CPU cache, so all accesses that hit in cache could have hit in RAM without causing a page miss. But the CPU cache hit rate is very low.)
  • Anyway yeah, looks right. Pretty terrible CPU cache, though. Such a low hit rate, and so such a slow hit time. Compare a modern 4GHz Skylake desktop where L1d has 4 to 5 cycle latency, so ~1 ns hit time, but still about ~85 ns for main memory is still about right. (DRAM access times in ns have stayed near constant even as bandwidth and CPU clock speeds have increased.)
  • you can mail me chxanu123@gmail.com