Question

An Example CPU Burst 4 5 6 10 4 12 The Guess 10 7 6 6 Ti All quantities in ms with a = 1/2 To = 10 Tn+1 = Qtn +(1 - «) Tn TiSlide 9 is given aboveUsing the exponential average formula and the example on slide 9 of Lecture 7, compute the next estimated burst time, using t

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Answer #1

tn = The length of the nth CPU Burst

\tau(n+1) = Predicted value for the CPU Burst

\tau(0 )= 10

\tau(n+1) = \alpha tn + (1-\alpha)\tau(n)


1. \alpha = 0

   \tau (n+1 ) = (0)tn + (1-0)\tau(n)

                = 0+ (1) \tau (n)

                 \tau (n+1 ) = \tau (n)

(n=0)     \tau 1 =\tau0 = 10                                                         (given \tau 0 = 10)

(n=1)     \tau 2 =\tau1 = 10

(n=2,3,4,5)            Similarly, \tau 6=\tau5=\tau4=\tau3=\tau2=10

ti t0 t1 t2 t3 t4 t5
4 5 6 10 4 12
\taui \tau0 \tau1 \tau2 \tau3 \tau4 \tau5 \tau6
10(given) 10 10 10 10 10 10

2. \alpha = 1

    \tau (n+1 )= (1)tn + (1-1) \tau (n)

                = tn+ (0) \tau (n)

               \tau (n+1 )= tn


(n=0)     \tau 1 = t0 = 4

(n=1)     \tau 2 = t1 = 5

(n=2)     \tau 3 = t2= 6

(n=3)    \tau4 = t3= 10

(n=4)     \tau 5 = t4 = 4

(n=5)     \tau 6 = t5 = 12
   

ti t0 t1 t2 t3 t4 t5
4 5 6 10 4 12
\taui \tau0 \tau1 \tau2 \tau3 \tau4 \tau5 \tau6
10(given) 4 5 6 10 4 12

3. \alpha = 0.75

   \tau (n+1 ) = (0.75)tn + (1-0.75) \tau (n)

                = (0.75)tn+ (0.25) \tau (n)

       \tau (n+1 ) = (0.75)tn+ (0.25) \tau (n)


(n=0)    \tau1 = (0.75)t0+ (0.25)\tau0

                   = (0.75)(4) + (0.25)(10)                                                                (given \tau 0 = 10)

                   = 3+2.5

                   = 5.5

(n=1)     \tau 2 = (0.75)t1+ (0.25)\tau1

                   = (0.75)(5)+ (0.25)(5.5)

                   = 3.75 + 1.375

                   = 5.125

(n=2)     \tau 3 = (0.75)t2+ (0.25)\tau2

                   = (0.75)(6)+ (0.25)(5.125)

                   = 4.5 + 1.28125

                   = 5.78125

(n=3)    \tau4 = (0.75)t3+ (0.25)\tau3

                   = (0.75)(10)+ (0.25)(5.78125)

                   = 7.5 + 1.4453125

                   = 8.9453125

(n=4)     \tau 5 = (0.75)t4+ (0.25)\tau4

                   = (0.75)(4)+ (0.25)(8.9453125)

                   = 3 + 2.236328125

                   = 5.236328125

(n=5)     \tau 6 = (0.75)t5+ (0.25)\tau5

                   = (0.75)(12)+ (0.25)(5.236328125)

                   = 9 + 1.309082031

                   = 10.309082031

ti t0 t1 t2 t3 t4 t5
4 5 6 10 4 12
\taui \tau0 \tau1 \tau2 \tau3 \tau4 \tau5 \tau6
10(given) 5.5 5.125 5.78125 8.9453125 5.236328125 10.309082031


4. \alpha = 0.25

   \tau (n+1 ) = (0.25)tn + (1-0.25) \tau (n)

                = (0.25)tn+ (0.75) \tau (n)

    \tau (n+1 ) = (0.25)tn+ (0.75) \tau (n)


(n=0)     \tau 1 = (0.25)t0+ (0.75)\tau0

                   = (0.25)(4) + (0.75)(10)                                                                (given \tau 0 = 10)

                   = 1+7.5

                   = 8.5

(n=1)    \tau2 = (0.25)t1+ (0.75)\tau1

                   = (0.25)(5)+ (0.75)(8.5)

                   = 1.75 + 6.375

                   = 8.125

(n=2)     \tau 3 = (0.25)t2+ (0.75)\tau2

                   = (0.25)(6)+ (0.75)(8.125)

                   = 1.5 + 6.09375

                   = 7.59375

(n=3)     \tau 4 = (0.25)t3+ (0.75)\tau3

                   = (0.25)(10)+ (0.75)(7.59375)

                   = 2.5 + 5.6953125

                   = 8.1953125

(n=4)     \tau 5 = (0.25)t4+ (0.75)\tau4

                   = (0.25)(4)+ (0.75)(8.1953125)

                   = 1 + 6.146484375

                   = 7.146484375

(n=5)     \tau 6 = (0.25)t5+ (0.75)\tau5

                   = (0.25)(12)+ (0.75)(7.146484375)

                   = 3 + 5.3598632813

                   = 8.3598632813

ti t0 t1 t2 t3 t4 t5
4 5 6 10 4 12
\taui \tau0 \tau1 \tau2 \tau3 \tau4 \tau5 \tau6
10(given) 8.5 8.125 7.59375 8.1953125 7.146484375 5.3598632813
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