1. Review

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

3. Time (ms)100300500700900 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

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

5. Average Request Time = (0.9 * 2) + (0.1 * 200) = 21.8ms Time (ms) 20 60 80 Type A requests : 2ms CPU time (occur 90% of the time) Type B requests : 200ms CPU time (occur 10% of the time) 100150200 10 requests in queue at 200 ms At 200ms we have to process 10 requests within 20ms. 10 requests will take 18 + 200 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.

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