※ 引述《privatewind (傷神客)》之銘言：
: ※ 引述《a8909132 (shin)》之銘言：
: : Memory Problem.
: : Media applications that play audio or video files are part of a class of
: : workloads called "streaming" workloads; i.e , they bring in large amounts of
: : data but do not reuse much of it. Consider a video streaming workload that
: : access a 512KB working set sequentially with the following address stream:
: : 0,4,8,12,16,20,24,28,32......
: : (1)Assume a 64KB direct-mapped cache with a 32-byte line. What is the miss
: : rate for the address stream above. How is this miss rate sensitive to the
: : size of the cache or the working set? How would you categorize the misses this
: : workload is experiencing, based on the 3C model.
: Compulsory misses.
所以說上面題目給的Address stream就是要用來判斷3c model囉？
: The miss rate is proportion to the block size.
: In a 64KB direct-mapped cache, there exists 512K / 32 = 16K misses.
: : (2)Recompute the miss rate when the cache line size is 16 byte , 64 bytes, and
: : 128 bytes? What Kind of locality is this workload exploiting?
: : I/O Problem.
: In 16byte, 512K/16 = 32K.
: In 64byte, 512K/64 = 8K.
: In 128byte, 512K/128= 4K.
: Spatial locality.
這邊懂了，謝謝囉^ ^
: : What is the bottleneck in the following system setup, the CPU , memory
: : bus, or the disk set?
: : 1. The user program continuously performs reads of 64KB blocks, and requires
: : 2 million cycles to process each block.
: : 2. The operating system requires 1 million cycles of overhead for ezch I/O
: : oeration.
: : 3. The clock rate is 3GHz.
: ^CPU^ ??
這邊張凡講義上是clock rate，不過有可能是它講義上有打錯。
: : 4. The maximum sustained transfer rate of the memory bus is 640MB/sec.
: : 5. The read/write bandwidth of the disk controller and the disk drives is
: : 64MB/sec, disk average seek plus rotational latency is 9ms.
: : 6. There are 20 disks attached to the bus each with its own controller.
: : (Assume that each disk can be controlled independently and ignore disk
: : conflicts.)
: CPU = 3G/3M = 100 IO/second
: memory bus = 640MB/64KB=10000 IO/second
: 1 disk = 1/( 9ms+64KB/64MB) = 1/(9ms+1ms) = 100 IO/second
: disk set = 20 * 100 = 2000 IO/second
: CPU is the bottleneck.
這部份我再想一下，有不懂再麻煩了，先在此說一聲謝謝，太感謝這位高手的幫忙^^
: : 以上問題看張凡講義上的解答看不懂，麻煩請各位高手幫幫忙，在此先說聲謝謝^^
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