Parallel Speedup

Performance of a Parallel Algorithm n : problem size (e.g.: sort n numbers) p : number of processors Tp : parallel time Ts : sequential time (optimal sequ. alg.) S = Ts / Tp : speedup (1Sp) S S=p super-linear linear sub-linear p

Speedup linear speedup S = p optimal super linear speedup S > p : impossible Proof. Assume that parallel algorithm A has a speedup S > p for processors, i.e. S = Ts / Tp > p. Hence: Ts > T·p. Simulate A on a sequential, single processor machine. Then Tp(1) = Tp · p < Ts. Hence, Ts was not optimal. Contradiction.

S p

Scaled Speedup Ts may be unknown (in fact, for most real experiments this is the case) Scaled speedup S’ = Tp(1) / Tp(p) S’  S

Efficiency e = S / p efficiency (0e1) optimal linear speedup S = p  e = 1 e’ = S’ / p scaled efficiency