2. Problem formulation
• This is classification/Regression problem, this presentation will only
cover classification of electrical grid stability into two classes;
1. Stable
2. Unstable
3. Introduction to dataset
• It is collected from UCI machine learning repository named as Electrical
Grid Stability Simulated Data.
• Electrical Grid data set in which we have different attributes for examine
the stability of the system.
• The analysis is performed for different sets of input values using the
methodology Naïve Bayes, Random forest and decision tree for classify
the system stability.
• Further, if you are interested about the dataset refer to; Schäfer,
Benjamin, et al. 'Taming instabilities in power grid networks by
decentralized control.
• We will examine the response of the system stability depends on 10,000
observations and 13 attributes, 1 classes attribute (stab).
4. Attributes Information
• Total - 14 predictive attributes, 1target attribute(2classes)
• tau[x] : Reaction time of participant (real from the range [0.5,10]s).
• p[x] : Nominal power consumed(negative)/produced(positive)(real).
• g[x] : Coefficient (gamma) proportional to price elasticity (real from
the range [0.05,1]s^-1).
• stab: The maximal real part of the characteristic equation root (if
positive - the system is linearly unstable)(real)
• stabf: The stability label of the system (stable/unstable)
7. Preprocessing
• Zero N/A values
• Problem: Overfitting of the
same attributes
• Sensitivity of P(x) cannot be
ignored
• Method of choosing attributes
is analyzing summary
characteristics, mean and
quartile values are considered.
8. Preprocessing Continue
• Null the following attributes by analyzing the summary characteristics:
• Tau4
• P3
• G2
• Considered attributes; summary (Talk about p4-p2);
13. Naïve Bayes Results
• Accuracy = 97.03%
• Precision [How many selected items are
relevant OR TP/(TP+FP)]= = 0.9558
• Recall [How many items are selected
relevant OR TP/TP+FN]= 0.9589
• F1 (Harmonic mean of precision and recall)=
(2 * 0.9558 * 0.9589) / (0.9558 + 0.9589) =
0.9573475
14. Random Forest Results
• Accuracy = 100%
• Precision [How many selected
items are relevant OR
TP/(TP+FP)]= 1
• Recall [How many items are
selected relevant OR TP/TP+FN]=
1
• F1 = (2 * 1* 1) / (1 + 1) = 1
15. Decision Tree Results
• Accuracy = 100%
• Precision [How many selected
items are relevant OR
TP/(TP+FP)]= 1
• Recall [How many items are
selected relevant OR TP/TP+FN]=
1
• F1 = (2 * 1 * 1) / (1 + 1) = 1
16. Conclusion
• Data is being divided in 95:5 for training and testing dataset.
• Naïve Bayes gave accuracy of 97.03%, while predicting versus actual
results to determine the class.
• Random forest gave 100% accuracy.
• Decision tree as well gave 100% accuracy.
• So for this dataset, prediction of class is done accurately by Random
forest and decision tree.