GATE Overflow Test Series | Computer Organization and Architecture | Test 1 | Question: 26

Question

Assume a main memory access time of $64\;ns$ and a memory system capable of a sustained transfer rate of $16$ GBps. If the block size is $64$ bytes, what is the maximum number of outstanding misses we need to support assuming that we can maintain the peak bandwidth given the request stream and that accesses never conflict. For simplicity, ignore the time between misses.

Answer 1

Assuming that we can maintain the peak bandwidth, the memory system can support $(16 \times 10^9)/64 = 250$ million block references per second.

Since, a cache miss takes $64\;ns,$ in worst case (assume all accesses to be misses) we need to support $250 \times 10^6 \times 64 \times 10^{-9} = 16$ outstanding misses.

Comments

@gatecse 

 memory system capable of a sustained transfer rate of 16 GBps

This transfer rate is between the processor and first-level cache?

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If you draw a block with all cache levels and main memory included in it, then 16GBps will be the speed at which data can go out from here.

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I understood that ,250 million block references per sec. are possible  and  64 ns is the miss penalty for cache .

outstanding misses  = missed blocks which are not yet retrieved and such misses are stored.

as assumed in the answer that . let all the block references are misses , then miss penalty for all those block ref. will be 250 million per sec * 64 ns

but this is independent of time, whats wrong here , not getting it? how are we actually calculating the outstanding misses.

@hetgate23

Answer 2

In addition to the best answer

Outstanding means, it hasn't been resolved. In the case of cache misses, that means the data request has not been serviced yet. When a cache miss happens, it has to fetch the data from the next level cache/memory, until this data arrives, you need to hold this request somewhere. You've got two options in this situation. Either block the entire cache, until data from the next level arrives. This is the easiest to implement, but quite bad in terms of performance. On the other hand, you could save the information about this data request somewhere, and carry on servicing others. This is is much better in terms of performance. Now, coming to this "place" where you are going to store information about the pending cache miss, this is a hardware structure. So you can only have finite number of "outstanding" misses. Number of entries in this buffer, is what is being provided as the number of outstanding cache misses.

Extra read: https://courses.cs.washington.edu/courses/csep548/06au/lectures/cacheAdv.pdf