) Given the following actual CPU burst for a task, {3, 4, 5, 3, 11, 12, 11, 11}, and an initial "best guess" at the burst as 5, apply the following “exponential averaging” formula to predict the length of the task's next CPU burst. Apply α’s of 0.2, 0.6, and 0.8. Which choice of the 3 α’s most closely tracks the actual burst sequence?
τ = α t + (1 – α) τ
n+1 n n
Given actual CPU bursts for a task {t0, t1, t2, t3, t4, t5, t6, t7} : {3, 4, 5, 3, 11, 12, 11, 11}.
And initial "best guess" at the burst, T0 = 5.
If α is 0.2:
T1 = α t0 + (1 – α) T0 = 0.2*3 + 0.8*5 = 4.6
T2 = α t1 + (1 – α) T1 = 0.2*4 + 0.8*4.6 = 4.48
T3 = α t2 + (1 – α) T2 = 0.2*5 + 0.8*4.48 = 4.584
T4 = α t3 + (1 – α) T3 = 0.2*3 + 0.8*4.584 = 4.2672
T5 = α t4 + (1 – α) T4 = 0.2*11 + 0.8*4.2672 = 5.61376
T6 = α t5 + (1 – α) T5 = 0.2*12 + 0.8*5.61376 = 6.8910
T7 = α t6 + (1 – α) T6 = 0.2*11 + 0.8*6.891 = 7.71280
T8 = α t7 + (1 – α) T7 = 0.2*11 + 0.8*7.7128 = 8.3702
If α is 0.6:
T1 = α t0 + (1 – α) T0 = 0.6*3 + 0.4*5 = 3.8
T2 = α t1 + (1 – α) T1 = 0.6*4 + 0.4*3.8 = 3.92
T3 = α t2 + (1 – α) T2 = 0.6*5 + 0.4*3.92 = 4.568
T4 = α t3 + (1 – α) T3 = 0.6*3 + 0.4*4.568 = 3.6272
T5 = α t4 + (1 – α) T4 = 0.6*11 + 0.4*3.6272 = 8.05088
T6 = α t5 + (1 – α) T5 = 0.6*12 + 0.4*8.05088 = 10.420352
T7 = α t6 + (1 – α) T6 = 0.6*11 + 0.4*10.420352 = 10.76814
T8 = α t7 + (1 – α) T7 = 0.6*11 + 0.4*10.76814 = 10.90725
If α is 0.8:
T1 = α t0 + (1 – α) T0 = 0.8*3 + 0.2*5 = 3.4
T2 = α t1 + (1 – α) T1 = 0.8*4 + 0.2*3.4 = 3.88
T3 = α t2 + (1 – α) T2 = 0.8*5 + 0.2*3.88 = 4.776
T4 = α t3 + (1 – α) T3 = 0.8*3 + 0.2*4.776 = 3.3552
T5 = α t4 + (1 – α) T4 = 0.8*11 + 0.2*3.3552 = 9.47104
T6 = α t5 + (1 – α) T5 = 0.8*12 + 0.2*9.47104 = 11.494208
T7 = α t6 + (1 – α) T6 = 0.8*11 + 0.2*11.494208 = 11.0988416
T8 = α t7 + (1 – α) T7 = 0.8*11 + 0.2*11.0988416 = 11.01976832
Looks like with α = 0.8 most closely tracks the actual burst sequence.
) Given the following actual CPU burst for a task, {3, 4, 5, 3, 11, 12,...
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