AVERAGE REQUEST TIMEReview

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

Time (ms)100 300 500 700 900

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)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.

Average request must complete before another request arrives in order to prevent system saturation i.e.:

Average request time < Inter-request arrival period