Speedup

Speedup in latency is defined by the following formula:[2]

${\displaystyle S_{\text{latency}}={\frac {L_{1}}{L_{2}}}={\frac {T_{1}W_{2}}{T_{2}W_{1}}},}$

where

• S_latency is the speedup in latency of the architecture 2 with respect to the architecture 1;
• L₁ is the latency of the architecture 1;
• L₂ is the latency of the architecture 2.

Speedup in latency can be predicted from Amdahl's law or Gustafson's law.

Speedup in throughput is defined by the formula:[3]

${\displaystyle S_{\text{throughput}}={\frac {Q_{2}}{Q_{1}}}={\frac {\rho _{2}A_{2}T_{1}W_{2}}{\rho _{1}A_{1}T_{2}W_{1}}}={\frac {\rho _{2}A_{2}}{\rho _{1}A_{1}}}S_{\text{latency}},}$

where

• S_throughput is the speedup in throughput of the architecture 2 with respect to the architecture 1;
• Q₁ is the throughput of the architecture 1;
• Q₂ is the throughput of the architecture 2.

We are testing the effectiveness of a branch predictor on the execution of a program. First, we execute the program with the standard branch predictor on the processor, which yields an execution time of 2.25 seconds. Next, we execute the program with our modified (and hopefully improved) branch predictor on the same processor, which produces an execution time of 1.50 seconds. In both cases the execution workload is the same. Using our speedup formula, we know

${\displaystyle S_{\text{latency}}={\frac {L_{\text{old}}}{L_{\text{new}}}}={\frac {2.25~\mathrm {s} }{1.50~\mathrm {s} }}=1.5.}$

Our new branch predictor has provided a 1.5x speedup over the original.

We can also measure speedup in cycles per instruction (CPI) which is a latency. First, we execute the program with the standard branch predictor, which yields a CPI of 3. Next, we execute the program with our modified branch predictor, which yields a CPI of 2. In both cases the execution workload is the same and both architectures are not pipelined nor parallel. Using the speedup formula gives

${\displaystyle S_{\text{latency}}={\frac {L_{\text{old}}}{L_{\text{new}}}}={\frac {3~{\text{CPI}}}{2~{\text{CPI}}}}=1.5.}$

We can also measure speedup in instructions per cycle (IPC), which is a throughput and the inverse of CPI. Using the speedup formula gives

${\displaystyle S_{\text{throughput}}={\frac {Q_{\text{new}}}{Q_{\text{old}}}}={\frac {0.5~{\text{IPC}}}{0.33~{\text{IPC}}}}=1.5.}$

We achieve the same 1.5x speedup, though we measured different quantities.