Speedup = T_old / T_new

Performance Engineering

Meaning

Speedup quantifies how much faster a new implementation runs compared to an original one by taking the ratio of the old execution time to the new execution time. It helps developers assess the effectiveness of optimizations or hardware upgrades. The metric is meaningful only when both measurements are performed under comparable conditions.

Primary Function

Performance analysis

Communicative Purpose

Enables quantitative comparison of execution times to assess optimization impact.

Pattern

measure T_old → apply change → measure T_new → compute Speedup

Core Structure

Speedup = T_old / T_new

Função primária

Performance analysis

Propósito comunicativo

Enables quantitative comparison of execution times to assess optimization impact.

Situações de gatilho

Benchmarking: evaluating two algorithm versions; Optimization: measuring effect of code refactoring; System tuning: comparing hardware configurations

Contextos

Performance engineering, systems programming, data‑science pipelines, high‑performance computing

Padrão

measure T_old → apply change → measure T_new → compute Speedup

Estrutura central

Speedup = T_old / T_new

Colocados típicos

throughput
latency
efficiency
benchmark
profiling
scaling

Substituições comuns

inverse speedup (T_new / T_old) – yields slowdown
percentage improvement ((T_old - T_new) / T_old * 100) – expresses gain as a percent

Erros comuns

Swapping numerator and denominator: assuming speedup = T_new / T_old → yields values <1 for improvements, misleading conclusions; Using CPU time for T_old and wall‑clock time for T_new: mismatched metrics → inaccurate speedup; Ignoring warm‑up runs: measuring only first iteration → overestimates speedup; Forgetting to ensure T_new > 0: division by zero error; Comparing workloads with different input sizes: speedup reflects workload change rather than optimization

Similar / contraste

Throughput vs Speedup: throughput measures work per unit time, speedup measures time reduction; Latency vs Speedup: latency is absolute time, speedup is relative improvement

Interferências

Coming from JavaScript: assuming single‑threaded timing is sufficient – in Java multithreaded environments JIT warm‑up can skew T_new measurements

Família do chunk

Throughput
Latency
Efficiency
Amdahl's Law
Parallel speedup

Nuance

Do not use when workloads differ in size or nature, as the ratio becomes meaningless; Speedup >1 indicates improvement but may hide scaling bottlenecks, so interpret alongside resource usage; T_new must be greater than zero and measurements should be stable to avoid division errors and noise

Efeito pragmático

Allows teams to quantify optimization benefits, justify resource allocation, detect regressions early, and set realistic performance targets.

Dica de memória

Speedup is like a race’s finish‑line gap: it tells you how many times quicker the winner crossed compared to the starter.

Nota

Speedup assumes identical input data and environment; any change in these factors invalidates the ratio.

Frequência: Very highFormulaicidade: FixedTipo de construção: formulaPrioridade de aquisição: Automatic productionPrioridade de output: BothTag de espaçamento: Immediate

Examples

import time def old_algo(data): total = 0 for x in data: total += x ** 2 return total def new_algo(data): return sum(x ** 2 for x in data) data = list(range(100000)) t0 = time.perf_counter() old_algo(data) T_old = time.perf_counter() - t0 t0 = time.perf_counter() new_algo(data) T_new = time.perf_counter() - t0 speedup = T_old / T_new print(f"Speedup: {speedup:.2f}x")

import time, numpy as np size = 1000000 arr = np.random.rand(size) def loop_sum(a): s = 0.0 for v in a: s += np.sqrt(v) return s def vectorized_sum(a): return np.sqrt(a).sum() t0 = time.perf_counter() loop_sum(arr) T_old = time.perf_counter() - t0 t0 = time.perf_counter() vectorized_sum(arr) T_new = time.perf_counter() - t0 print("Speedup:", T_old / T_new)

import time, multiprocessing as mp def compute_chunk(start): total = 0 for i in range(start, start + 1000000): total += i * i return total def run_single(): return compute_chunk(0) def run_multi(): with mp.Pool(4) as pool: results = pool.map(compute_chunk, [0, 1000000, 2000000, 3000000]) return sum(results) t0 = time.perf_counter() run_single() T_old = time.perf_counter() - t0 t0 = time.perf_counter() run_multi() T_new = time.perf_counter() - t0 print("Speedup:", T_old / T_new)