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August 30, 2022

DECISION TREES for Risk Assessment One of the great advantages of decision trees is their interpretability. The rules learnt for classification are easy for a person to follow, unli

computer science

  

DECISION TREES for Risk Assessment

One of the great advantages of decision trees is their interpretability. The rules learnt for classification are easy for a person to follow, unlike the opaque "black box" of many other methods, such as neural networks. We demonstrate the utility of this using a German credit data set. You can read a description of this dataset at the UCI site. The task is to predict whether a loan approval is good or bad credit risk based on 20 attributes. We've simplified the data set somewhat, particularly making attribute names and values more meaningful.

1. Download the credit_Dataset.arff dataset and load it to Weka.

2. (5 Points) When presented with a dataset, it is usually a good idea to visualise it first. Go to the Visualise tab. Click on any of the scatter plots to open a new window which shows the scatter plot for two selected attributes. Try visualising a scatter plot of age and duration. Do you notice anything unusual? You can click on any data point to display all it's values.

3. (5 Points) In the previous point you should have found a data point, which seems to be corrupted, as some of its values are nonsensical. Even a single point like this can significantly affect the performance of a classifier. How do you think it would affect Decision trees? A good way to check this is to test the performance of each classifier before and after removing this datapoint.

4. (10 Points) To remove this instance from the dataset we will use a filter. We want to remove all instances, where the age of an applicant is lower than 0 years, as this suggests that the instance is corrupted. In the Preprocess tab click on Choose in the Filter pane. Select filters > unsupervised > instance > RemoveWithValues. Click on the text of this filter to change the parameters. Set the attribute index to 13 (Age) and set the split point at 0. Click Ok to set the parameters and Apply to apply the filter to the data. Visualise the data again to verify that the invalid data point was removed.

5. (20 Points) On the Classify tab, select the Percentage split test option and change its value to 90%. This way, we will train the classifiers using 90% of the training data and evaluate their performance on the remaining 10%. First, train a decision tree classifier with default options. Select classifiers > trees > J48 and click Start. J48 is the Weka implementation of the C4.5 algorithm, which uses the normalized information gain criterion to build a decision tree for classification.

6. (20 Points) After training the classifier, the full decision tree is output for your perusal; you may need to scroll up for this. The tree may also be viewed in graphical form by right-clicking in the Result list and selecting Visualize tree; unfortunately this format is very cluttered for large trees. Such a tree accentuates one of the strengths of decision tree algorithms: they produce classifiers which are understandable to humans. This can be an important asset in real life applications (people are seldom prepared to do what a computer program tells them if there is no clear explanation). Observe the output of the classifier and try to answer the following questions:

o How would you assess the performance of the classifier? Is the Percentage of Correctly Classified Instances a sufficient measure in this case? Why? Hint: check the number of good and bad cases in the test sample, using the confusion matrix. Each column of the matrix represents the instances in a predicted class, while each row represents the instances in an actual class. For example let us define an experiment from P positive instances and N negative instances. The four outcomes can be formulated in a 2 by 2 contingency table or confusion matrix. One benefit of a confusion matrix is that it is easy to see if the system is confusing two classes (i.e. commonly mislabeling one as another).

o Looking at the decision tree itself, are the rules it applies sensible? Are there any branches which appear absurd? At what depth of the tree? What does this suggest?
Hint: Check the rules applied after following the paths: (a) CheckingAccount = <0, Foreign = yes, Duration >11, Job = skilled, OtherDebtors = none, Duration <= 30 and (b) CheckingAccount = <0, Foreign = yes, Duration >11, Job = unskilled.

o How does the decision tree deal with classification in the case where there are zero instances in the training set corresponding to that particular path in the tree (e.g. those leaf nodes that have (0:0))?

7. (20 Points) Now, explore the effect of the confidenceFactor option. You can find this by clicking on the Classifer name (to the right of the Choose button on the Classify tab). On the Classifier options window, click on the More button to find out what the confidence factor controls. Try the values 0.1, 0.2, 0.3 and 0.5. What is the performance of the classifier at each case? Did you expect this given your observations in the previous questions? Why do you think this happens?

8. (20 Points) Suppose that it is worse to classify a customer as good when they are bad, than it is to classify a customer as bad when they are good. Which value would you pick for the confidence factor? Which performance measure would you base your decision on?

9.  (20 Points)Finally we will create a random decision forest and compare the performance of this classifier to that of the decision tree and the decision stump. The random decision forest is an ensemble classifier that consists of many decision trees and outputs the class that is the mode of the class's output by individual trees. Again set the test option Percentage split to 90%. Select classifiers > trees > RandomForest and hit Start. Again, observe the output. How high can you get the performance of the classifier by changing the number of trees (numTrees) parameter? How does the random decision forest compare performance wise to the decision tree and decision stump?

  • attachment

    Module03-Assignment.docx

  • attachment

    credit_Dataset.arff

Assignment

(100 Points)

CYBR 7240 – Cyber Analytics and Intelligence Module 03: Decision Trees

DECISION TREES for Risk Assessment

One of the great advantages of decision trees is their interpretability. The rules learnt for classification are easy for a person to follow, unlike the opaque "black box" of many other methods, such as neural networks. We demonstrate the utility of this using a German credit data set. You can read a description of this dataset at the UCI site. The task is to predict whether a loan approval is good or bad credit risk based on 20 attributes. We've simplified the data set somewhat, particularly making attribute names and values more meaningful.

1. Download the credit_Dataset.arff dataset and load it to Weka.

2. (5 Points) When presented with a dataset, it is usually a good idea to visualise it first. Go to the Visualise tab. Click on any of the scatter plots to open a new window which shows the scatter plot for two selected attributes. Try visualising a scatter plot of age and duration. Do you notice anything unusual? You can click on any data point to display all it's values.

3. (5 Points) In the previous point you should have found a data point, which seems to be corrupted, as some of its values are nonsensical. Even a single point like this can significantly affect the performance of a classifier. How do you think it would affect Decision trees? A good way to check this is to test the performance of each classifier before and after removing this datapoint.

4. (10 Points) To remove this instance from the dataset we will use a filter. We want to remove all instances, where the age of an applicant is lower than 0 years, as this suggests that the instance is corrupted. In the Preprocess tab click on Choose in the Filter pane. Select filters > unsupervised > instance > RemoveWithValues. Click on the text of this filter to change the parameters. Set the attribute index to 13 (Age) and set the split point at 0. Click Ok to set the parameters and Apply to apply the filter to the data. Visualise the data again to verify that the invalid data point was removed.

5. (20 Points) On the Classify tab, select the Percentage split test option and change its value to 90%. This way, we will train the classifiers using 90% of the training data and evaluate their performance on the remaining 10%. First, train a decision tree classifier with default options. Select classifiers > trees > J48 and click Start. J48 is the Weka implementation of the C4.5 algorithm, which uses the normalized information gain criterion to build a decision tree for classification.

6. (20 Points) After training the classifier, the full decision tree is output for your perusal; you may need to scroll up for this. The tree may also be viewed in graphical form by right-clicking in the Result list and selecting Visualize tree; unfortunately this format is very cluttered for large trees. Such a tree accentuates one of the strengths of decision tree algorithms: they produce classifiers which are understandable to humans. This can be an important asset in real life applications (people are seldom prepared to do what a computer program tells them if there is no clear explanation). Observe the output of the classifier and try to answer the following questions:

· How would you assess the performance of the classifier? Is the Percentage of Correctly Classified Instances a sufficient measure in this case? Why? Hint: check the number of good and bad cases in the test sample, using the confusion matrix. Each column of the matrix represents the instances in a predicted class, while each row represents the instances in an actual class. For example let us define an experiment from P positive instances and N negative instances. The four outcomes can be formulated in a 2 by 2 contingency table or confusion matrix. One benefit of a confusion matrix is that it is easy to see if the system is confusing two classes (i.e. commonly mislabeling one as another).

· Looking at the decision tree itself, are the rules it applies sensible? Are there any branches which appear absurd? At what depth of the tree? What does this suggest? Hint: Check the rules applied after following the paths: (a) CheckingAccount = <0, Foreign = yes, Duration >11, Job = skilled, OtherDebtors = none, Duration <= 30 and (b) CheckingAccount = <0, Foreign = yes, Duration >11, Job = unskilled.

· How does the decision tree deal with classification in the case where there are zero instances in the training set corresponding to that particular path in the tree (e.g. those leaf nodes that have (0:0))?

7. (20 Points) Now, explore the effect of the confidenceFactor option. You can find this by clicking on the Classifer name (to the right of the Choose button on the Classify tab). On the Classifier options window, click on the More button to find out what the confidence factor controls. Try the values 0.1, 0.2, 0.3 and 0.5. What is the performance of the classifier at each case? Did you expect this given your observations in the previous questions? Why do you think this happens?

8. (20 Points) Suppose that it is worse to classify a customer as good when they are bad, than it is to classify a customer as bad when they are good. Which value would you pick for the confidence factor? Which performance measure would you base your decision on?

9. (20 Points)Finally we will create a random decision forest and compare the performance of this classifier to that of the decision tree and the decision stump. The random decision forest is an ensemble classifier that consists of many decision trees and outputs the class that is the mode of the class's output by individual trees. Again set the test option Percentage split to 90%. Select classifiers > trees > RandomForest and hit Start. Again, observe the output. How high can you get the performance of the classifier by changing the number of trees (numTrees) parameter? How does the random decision forest compare performance wise to the decision tree and decision stump?

Deliverable:

· Your report including the screenshots of your implementation for each section and the results.

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@RELATION GermanCredit
@ATTRIBUTE CheckingAccount {<0 <200 >=200 none}
@ATTRIBUTE Duration NUMERIC
@ATTRIBUTE CreditHistory {ok ok_at_this_bank ok_til_now past_delays critical}
@ATTRIBUTE Purpose {car_new car_used furniture television appliances repairs education vacation retraining business others}
@ATTRIBUTE CreditAmount NUMERIC
@ATTRIBUTE SavingsAccount {<100 <500 <1000 >=1000 unknown}
@ATTRIBUTE YearsEmployed {unemployed <1 <4 <7 >=7}
@ATTRIBUTE InstallmentRate NUMERIC
@ATTRIBUTE PersonalStatus {male_divorced female male_single male_married female_single}
@ATTRIBUTE OtherDebtors {none co guarantor}
@ATTRIBUTE ResidentSince NUMERIC
@ATTRIBUTE Property {real_estate savings car unknown}
@ATTRIBUTE Age NUMERIC
@ATTRIBUTE OtherPlans {bank stores none}
@ATTRIBUTE Housing {rent own free}
@ATTRIBUTE NumCreditsAtBank NUMERIC
@ATTRIBUTE Job {unemployed unskilled skilled management}
@ATTRIBUTE Dependents NUMERIC
@ATTRIBUTE Telephone {no yes}
@ATTRIBUTE Foreign {yes no}
@ATTRIBUTE Approve {good bad}
@DATA
<0 0 ok furniture -1000000000 unknown unemployed -10000 female_single none 0 car -293 none own -1000 skilled -185 yes no good
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<200 36 past_delays car_new 2225 <100 >=7 4 male_single none 4 unknown 57 bank free 2 skilled 1 yes yes bad
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<0 36 critical furniture 6229 <100 <1 4 female co 4 unknown 23 none rent 2 unskilled 1 yes yes bad
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<200 15 critical television 1537 unknown >=7 4 male_single guarantor 4 real_estate 50 none own 2 skilled 1 yes yes good
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none 36 ok_til_now television 2394 unknown <4 4 female none 4 car 25 none own 1 skilled 1 no yes good
none 36 ok_til_now car_used 8133 <100 <4 1 female none 2 savings 30 bank own 1 skilled 1 no yes good
none 7 critical television 730 unknown >=7 4 male_single none 2 savings 46 none rent 2 unskilled 1 yes yes good
<0 8 critical others 1164 <100 >=7 3 male_single none 4 unknown 51 bank free 2 management 2 yes yes good
<200 42 critical business 5954 <100 <7 2 female none 1 real_estate 41 bank own 2 unskilled 1 no yes good
<0 36 ok_til_now education 1977