nLet T[i] be the execution time for the ith instance of this process: the actual duration of the ith CPU burst of this process
nLet S[i] be the predicted value for the ith CPU burst of this process. The simplest choice is:
uS[n+1]
= (1/n) S_{i=1 to n} T[i]
nTo avoid recalculating the entire sum we can rewrite this as:
uS[n+1]
= (1/n) T[n] + ((n-1)/n) S[n]
nBut this convex combination gives equal weight to each instance