Review

Why can average request time be used to determine the work-load required for saturation?

Time (ms) Type A requests : 2ms CPU time (occur 90% of the time) Type B requests : 200ms CPU time (occur 10% of the time) Average Request Time = (0.9 * 2) + (0.1 * 200) = 21.8ms

Reduce inter-request arrival period to 2oms from 100ms

Average Request Time = (0.9 * 2) + (0.1 * 200) = 21.8ms Time (ms) Type A requests : 2ms CPU time (occur 90% of the time) Type B requests : 200ms CPU time (occur 10% of the time) requests in queue at 200 ms At 200ms we have to process 10 requests within 20ms. 10 requests will take milliseconds. We cannot process all requests in the queue before the next 10ms request comes in. This means from now on queue size will keep increasing at a constant rate, meaning response time will increase at a constant rate for upcoming incoming requests. Note that throughput stays the same and does not increase to cope with increased load (increased load => 1 request every 20ms). That is what we mean by saturation.

Average request must complete before another request arrives in order to prevent system saturation i.e.: Average request time < Inter-request arrival period