Question

In a certain computation, 90% of the work is vectorizable. Of the remaining 10%, half is parallelizable for an MIMD machine. What are the speedups over sequential execution for a 10 PE SIMD machine and a 10 processor MIMD machine on this computation?

Answer 1

Answer:

The SIMD only gets the effect of 5.3 processors from 10 processors while the MIMD only gets the effect of 6.9 processors out of 10 processors.

Explanation:

According to Amdahl's law for an SIMD machine, the fraction parallel work is 90% while the fraction of serial work is 10% for a time $T_{10} =0.1$, $T_{1}+0.9$, $\frac{T _{1}}{10}$

The speedup = $\frac{T_{1}}{(0.1+\frac{0.9}{10})T_{1} }=\frac{1}{0.19} =5.26$

for an MIMD machine, the fraction parallel work is 95%, since vectorized work can also be paralleled by multiple processors, the time for 10 processors is

$T_{10} =0.05$, $T_{1}+0.95$, $\frac{T _{1}}{10}$

The speedup = $\frac{T_{1}}{(0.05+\frac{0.95}{10})T_{1} }=\frac{1}{0.145} =6.9$

The SIMD only gets the effect of 5.3 processors from 10 processors while the MIMD only gets the effect of 6.9 processors out of 10 processors.