Amdahl's law

If 30% of the execution time may be the subject of a speedup, p will be 0.3; if the improvement makes the affected part twice as fast, s will be 2. Amdahl's law states that the overall speedup of applying the improvement will be:

${\displaystyle S_{\text{latency}}={\frac {1}{1-p+{\frac {p}{s}}}}={\frac {1}{1-0.3+{\frac {0.3}{2}}}}=1.18.}$

For example, assume that we are given a serial task which is split into four consecutive parts, whose percentages of execution time are p1 = 0.11, p2 = 0.18, p3 = 0.23, and p4 = 0.48 respectively. Then we are told that the 1st part is not sped up, so s1 = 1, while the 2nd part is sped up 5 times, so s2 = 5, the 3rd part is sped up 20 times, so s3 = 20, and the 4th part is sped up 1.6 times, so s4 = 1.6. By using Amdahl's law, the overall speedup is

${\displaystyle S_{\text{latency}}={\frac {1}{{\frac {p1}{s1}}+{\frac {p2}{s2}}+{\frac {p3}{s3}}+{\frac {p4}{s4}}}}={\frac {1}{{\frac {0.11}{1}}+{\frac {0.18}{5}}+{\frac {0.23}{20}}+{\frac {0.48}{1.6}}}}=2.19.}$

Notice how the 5 times and 20 times speedup on the 2nd and 3rd parts respectively don't have much effect on the overall speedup when the 4th part (48% of the execution time) is accelerated by only 1.6 times.

Assume that a task has two independent parts, A and B. Part B takes roughly 25% of the time of the whole computation. By working very hard, one may be able to make this part 5 times faster, but this reduces the time of the whole computation only slightly. In contrast, one may need to perform less work to make part A perform twice as fast. This will make the computation much faster than by optimizing part B, even though part B's speedup is greater in terms of the ratio, (5 times versus 2 times).

For example, with a serial program in two parts A and B for which T_A = 3 s and T_B = 1 s,

• if part B is made to run 5 times faster, that is s = 5 and p = T_B/(T_A + T_B) = 0.25, then
  $${\displaystyle S_{\text{latency}}={\frac {1}{1-0.25+{\frac {0.25}{5}}}}=1.25;}$$

  
• if part A is made to run 2 times faster, that is s = 2 and p = T_A/(T_A + T_B) = 0.75, then
  $${\displaystyle S_{\text{latency}}={\frac {1}{1-0.75+{\frac {0.75}{2}}}}=1.60.}$$

  

Therefore, making part A to run 2 times faster is better than making part B to run 5 times faster. The percentage improvement in speed can be calculated as

${\displaystyle {\text{percentage improvement}}=100\left(1-{\frac {1}{S_{\text{latency}}}}\right).}$

• Improving part A by a factor of 2 will increase overall program speed by a factor of 1.60, which makes it 37.5% faster than the original computation.
• However, improving part B by a factor of 5, which presumably requires more effort, will achieve an overall speedup factor of 1.25 only, which makes it 20% faster.

If the non-parallelizable part is optimized by a factor of ${\displaystyle O}$, then

${\displaystyle T(O,s)=(1-p){\frac {T}{O}}+{\frac {p}{s}}T.}$

It follows from Amdahl's law that the speedup due to parallelism is given by

${\displaystyle S_{\text{latency}}(O,s)={\frac {T(O)}{T(O,s)}}={\frac {(1-p){\frac {1}{O}}+{p}}{{\frac {1-p}{O}}+{\frac {p}{s}}}}.}$

When ${\displaystyle s=1}$, we have ${\displaystyle S_{\text{latency}}(O,s)=1}$, meaning that the speedup is measured with respect to the execution time after the non-parallelizable part is optimized.

When ${\displaystyle s=\infty }$,

${\displaystyle S_{\text{latency}}(O,\infty )={\frac {T(O)}{T(O,s)}}={\frac {(1-p){\frac {1}{O}}+{p}}{{\frac {1-p}{O}}+{\frac {p}{s}}}}=1+{\frac {p}{1-p}}O.}$

If ${\displaystyle 1-p=0.4}$, ${\displaystyle O=2}$ and ${\displaystyle s=5}$, then:

${\displaystyle S_{\text{latency}}(O,s)={\frac {T(O)}{T(O,s)}}={\frac {{0.4}{\frac {1}{2}}+0.6}{{\frac {0.4}{2}}+{\frac {0.6}{5}}}}=2.5.}$

Next, we consider the case wherein the non-parallelizable part is reduced by a factor of ${\displaystyle O'}$, and the parallelizable part is correspondingly increased. Then

${\displaystyle T'(O',s)={\frac {1-p}{O'}}T+\left(1-{\frac {1-p}{O'}}\right){\frac {T}{s}}.}$

It follows from Amdahl's law that the speedup due to parallelism is given by

${\displaystyle S'_{\text{latency}}(O',s)={\frac {T'(O')}{T'(O',s)}}={\frac {1}{{\frac {1-p}{O'}}+\left(1-{\frac {1-p}{O'}}\right){\frac {1}{s}}}}.}$