Question

) 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

Answer 1

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.