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

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.

Step-by-step 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`,
`(T _(1))/(10)`

The speedup =
`(T_(1))/((0.1+(0.9)/(10))T_(1) )=(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`,
`(T _(1))/(10)`

The speedup =
`(T_(1))/((0.05+(0.95)/(10))T_(1) )=(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.