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A system uses shortest job first scheduling $(SJF)$ and exponential average of the measured lengths of previous $CPU$ bursts is  $a =0.25$.
 
The initial value of the predicted CPU burst time, i.e. $T_{1}=4$ $unit$. The predicted time for 4$^{th}$ CPU burst $T_{4}$ for a process with burst times of $4$ $unit$, $12$ $unit$ and $8$ $unit$ respectively is _______.
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The formula :   Pn+1 = a * Tn + (1 - a) * Pn ...... (i)

Given, exponential average of the measured lengths of previous CPU bursts is,   a =0.25 . 

a = 0.25, T1 = 4, T2 = 12,  T= 8 and P= 4 all these values are provided, now put them into the formula (i)

So, we have 

P2 = 0.25 * 4 + 0.75 * 4 = 4

P3 = 0.25 * 12 + 0.75 * 4 = 6

P4 = 0.25 * 8 + 0.75 * 6 = 6.5

Answer is 6.5 units.

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