📐 Gustafson's Law (Parallel Scaling)
Calculate scaled speedup for parallel applications where workload size grows with processing power.
⚡ Fortran 90 Engine
Double Precision (IEEE 754)
✓ ISO / ASME Validated
Cfd
📊 Solver Telemetry
● ACTIVE
📥 Parallel Fraction & Processor Count
📖 Mathematical Formulation (Gustafson, 1988):
$$S_{\text{scaled}}(N) = 1 + f(N - 1) = (1 - f) + f N$$
$$E_{\text{scaled}}(N) = \frac{S(N)}{N} = f + \frac{1 - f}{N}$$
$$S_{\text{Amdahl}}(N) = \frac{1}{(1 - f) + \frac{f}{N}} \quad \text{(Strong Scaling)}$$
13.00×
Gustafson Scaled Speedup
81.3%
Weak Scaling Efficiency
4.00×
Amdahl Fixed Speedup
3.3×
Weak vs Strong Gain
💡
Weak Scaling Diagnostic
If the CFD mesh size is enlarged proportionally with 16 processors, Gustafson's Law predicts an effective speedup of 13.00× (Efficiency: 81.3%). In contrast, keeping the problem size fixed (Amdahl) would bottleneck at only 4.00×.
📈 Scaling Comparison: Gustafson (Weak) vs Amdahl (Strong)
🟢 Gustafson | 🟠 Amdahl🔍 View Raw GNU Fortran Double-Precision Solver Output
MODE=1 MODE_NAME=Single Point F= 0.800000 N= 16 S_GUSTAFSON= 13.0000 E_GUSTAFSON= 0.8125 S_AMDAHL= 4.0000 E_AMDAHL= 0.2500
📘 Calculation Methodology: Gustafson's Law Weak Scaling Speedup
Mathematical Model & Theory
Gustafson's law models parallel speedup when problem size scales with processor count $N$ (weak scaling), reflecting memory-constrained high-performance simulations:
$$S(N) = N - (1 - p)(N - 1) = (1 - p) + p N$$
$$E(N) = \frac{S(N)}{N} = p + \frac{1 - p}{N}$$
Assumptions
- Weak scaling paradigm with constant wall-clock execution time.
- Parallel portion scales linearly with compute resources.
Academic References
- Gustafson, J. L. (1988): Reevaluating Amdahl's Law, CACM.
- Hager, G., & Wellein, G.: Introduction to HPC, CRC Press.
Worked Engineering Example
Problem Statement:
Scale a grid on 128 nodes with parallel fraction $p = 0.98$. Calculate speedup and efficiency.
Step-by-step Solution:
1. $S(128) = 128 - (1 - 0.98)(127) = 128 - 2.54 = 125.46$.
2. $E(128) = 125.46 / 128 = 98.02\%$.
Final Result:
Scaled speedup is 125.46x (98.0% efficiency).
Scale a grid on 128 nodes with parallel fraction $p = 0.98$. Calculate speedup and efficiency.
Step-by-step Solution:
1. $S(128) = 128 - (1 - 0.98)(127) = 128 - 2.54 = 125.46$.
2. $E(128) = 125.46 / 128 = 98.02\%$.
Final Result:
Scaled speedup is 125.46x (98.0% efficiency).