# Cyber Firefly Algorithm Based on Adaptive Memory Programming for Global Optimization
## 4 Experimental Results and Discussions
### Performance Evaluation
#### 4.3.1 Comparison with CFA Counterparts
A classic statistical test based on the central limit theorem for comparing two proportions is to approximate the mean \(X_{a}-X_{b}\) as a normal distribution if the collected number of samples  
\begin{table}
\begin{tabular}{c c c c c c c c} \hline \hline
**Functions** & **(a) GSO** & **(b) FA** & **(c) Min (a, b)** & **(d) CFA** & **Merit (d, a)** & **Merit (d, b)** & **Merit (d, c)** \\ \hline [PERSON] (2) & \(-0.999999\) & \(-0.999999\) & \(-0.999999\) & \(-1.0\) & \(1.0\) & \(1.0\) & \(1.0\) \\ [PERSON] (2) & \(-184.059055\) & \(-91.464488\) & \(-184.059055\) & \(-186.723\) & \(9.86\times 10^{-1}\) & \(4.90\times 10^{-1}\) & \(9.86\times 10^{-1}\) \\ [PERSON] (2) & \(0.000007\) & \(0.0\) & \(0.0\) & \(0.0\) & \(6.50\times 10^{-2}\) & \(1.0\) & \(1.0\) \\ [PERSON] (2) & \(0.0\) & \(0.0\) & \(0.0\) & \(0.0\) & \(1.0\) & \(1.0\) & \(1.0\) \\ [PERSON] (3) & \(0.0\) & \(0.0\) & \(0.0\) & \(0.0\) & \(1.0\) & \(1.0\) & \(1.0\) \\ [PERSON] (4, 5) & \(-6.883340\) & \(-3.339854\) & \(-6.883340\) & \(-8.09613\) & \(8.50\times 10^{-1}\) & \(4.13\times 10^{-1}\) & \(8.50\times 10^{-1}\) \\ [PERSON] (4, 7) & \(-6.624701\) & \(-2.593236\) & \(-6.624701\) & \(-10.2611\) & \(6.46\times 10^{-1}\) & \(2.53\times 10^{-1}\) & \(6.46\times 10^{-1}\) \\ [PERSON] (4, 10) & \(-7.770920\) & \(-2.211454\) & \(-7.770920\) & \(-10.3554\) & \(7.50\times 10^{-1}\) & \(2.14\times 10^{-1}\) & \(7.50\times 10^{-1}\) \\ [PERSON] (10) & \(0.02167\) & \(0.000045\) & \(0.000045\) & \(0.0\) & \(2.31\times 10^{-5}\) & \(1.11\times 10^{-2}\) & \(1.11\times 10^{-2}\) \\ [PERSON] (10) & \(1.658934\) & \(1.070588\) & \(1.070858\) & \(0.002511\) & \(1.51\times 10^{-3}\) & \(2.35\times 10^{-3}\) & \(2.35\times 10^{-3}\) \\ [PERSON] (10) & \(2.499270\) & \(11.099298\) & \(2.499270\) & \(0.646338\) & \(2.59\times 10^{-1}\) & \(5.82\times 10^{-2}\) & \(2.59\times 10^{-1}\) \\ [PERSON] (10) & \(0.154192\) & \(17.666979\) & \(0.154192\) & \(0.01645\) & \(1.07\times 10^{-1}\) & \(9.31\times 10^{-4}\) & \(1.07\times 10^{-1}\) \\ [PERSON] (10) & \(0.003303\) & \(0.000115\) & \(0.000115\) & \(0.0\) & \(1.51\times 10^{-3}\) & \(4.31\times 10^{-3}\) & \(4.31\times 10^{-3}\) \\ [PERSON] (20) & \(0.085764\) & \(0.00264\) & \(0.000264\) & \(0.0\) & \(5.83\times 10^{-6}\) & \(1.89\times 10^{-3}\) & \(1.89\times 10^{-3}\) \\ [PERSON] (20) & \(8.473378\) & \(21.006502\) & \(8.473378\) & \(0.26705\) & \(3.15\times 10^{-2}\) & \(1.27\times 10^{-2}\) & \(3.15\times 10^{-2}\) \\ [PERSON] (20) & \(5.550878\) & \(32.150683\) & \(5.550878\) & \(2.466102\) & \(4.44\times 10^{-1}\) & \(7.67\times 10^{-2}\) & \(4.44\times 10^{-1}\) \\ [PERSON] (20) & \(0.5535463\) & \(6.000160\) & \(0.553463\) & \(0.007213\) & \(1.30\times 10^{-2}\) & \(1.20\times 10^{-2}\) & \(1.30\times 10^{-2}\) \\ [PERSON] (20) & \(0.002574\) & \(0.001387\) & \(0.001387\) & \(0.0\) & \(1.94\times 10^{-4}\) & \(3.60\times 10^{-4}\) & \(3.60\times 10^{-4}\) \\ [PERSON] (30) & \(0.151057\) & \(0.00761\) & \(0.000761\) & \(0.0\) & \(3.31\times 10^{-6}\) & \(6.57\times 10^{-4}\) & \(6.57\times 10^{-4}\) \\ [PERSON] (30) & \(23.373028\) & \(31.808803\) & \(23.373028\) & \(0.801094\) & \(3.43\times 10^{-2}\) & \(2.52\times 10^{-2}\) & \(3.43\times 10^{-2}\) \\ [PERSON] (30) & \(9.135285\) & \(59.807848\) & \(9.135285\) & \(4.68069\) & \(5.12\times 10^{-1}\) & \(7.83\times 10^{-2}\) & \(5.12\times 10^{-1}\) \\ [PERSON] (30) &is sufficiently large. Therefore, 30 independent runs of each competing algorithm were conducted. For each run, the online best function value obtained at a particular FE, say \(e\), is a sample for \(X_{a}(e)\), the random variable for the result obtained by method A at FE \(e\). We tally the samples at every instance of FEs during the whole duration of executing the algorithm. By examining the mean curve of \(X_{a}(e)\) and \(X_{b}(e)\) as \(e\) increases during the evolution, we can testify if method A well outperforms method B. For clear illustration, the boundary and mean curves over the 30 runs are plotted.