A file server uses caching, and achieves a hit rate of 80%. File operations in the server cost 5 ms of CPU time when the server finds the requested block in the cache, and take an additional 15 ms of disk I/O time otherwise. Explaining any assumptions you make, estimate the server’s throughput capacity (average requests/sec) if it is:
i) single-threaded;
ii) two-threaded, running on a single processor;
iii) two-threaded, running on a two-processor computer.
80% of accesses cost 5 ms; 20% of accesses cost 20 ms.
average request time is 0.8*5+.2*20 = 4+4=8ms.
i) single-threaded: rate is 1000/8 = 125 reqs/sec
ii) two-threaded: serving 4 cached and 1 uncached requests takes 25 ms. (overlap I/O with computation). Therefore throughput becomes 1 request in 5 ms. on average, = 200 reqs/sec
iii) two-threaded, 2 CPUs: Processors can serve 2 rqsts in 5 ms => 400 reqs/sec. But disk can serve the 20% of requests at only 1000/15 reqs/sec (assume disk rqsts serialised). This implies a total rate of 5*1000/15 = 333 requests/sec (which the two CPUs can service).
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