| 1 |
What is the primary objective of landslide susceptibility mapping as described in the article?
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To mitigate the economic and environmental damage by predicting areas at risk. |
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Landslide susceptibility mapping identifies areas that are more likely to experience landslides. This supports planning and risk reduction before damage occurs. |
The answer is based on hazard-risk mapping, which combines environmental factors to classify areas by their likelihood of landslide occurrence. |
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| 2 |
Which machine learning algorithm was noted for having the highest success rate according to the article?
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Random Forest |
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According to the article, Random Forest achieved the highest success rate among the machine-learning models evaluated for landslide susceptibility mapping. |
Random Forest is an ensemble-learning method that combines many decision trees to improve prediction accuracy and reduce overfitting. |
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| 3 |
If the area of Chattogram district is 75% susceptible to landslides, and the highly susceptible zone covers approximately 12% of the district, what is the area (in percentage) that is not highly susceptible?
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63% |
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The highly susceptible zone covers 12% of the district. Therefore, the area that is not highly susceptible is 100% − 12% = 88%. However, because the question states that 75% of the district is susceptible, the intended answer is likely 63% if it asks for the susceptible area that is not highly susceptible: 75% − 12% = 63%. |
Percentage of susceptible but not highly susceptible area = total susceptible area − highly susceptible area = 75% − 12% = 63% |
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| 4 |
Considering that the total number of analyzed landslides is 255, and 80% were used for training the models, how many landslide instances were used for testing?
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51 |
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If 80% of 255 landslide instances were used for training, the remaining 20% were used for testing = 255 x 0.20 = 51. |
Testing data = total data x (1 − training proportion) = 255 x (1 − 0.80) = 51. |
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| 5 |
If the total area of Chattogram district is 7,000 km² and the very high susceptible zone covers 9% of the district, what is the area of the very high susceptible zone in km²?
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630 km² |
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The very high susceptible zone is 9% of 7,000 km^3. =7,000 x 0.09 = 630 km^3. |
Area = total area x percentage = 7,000 x 9/100 = 630 km^3. |
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| 6 |
Assuming the false positive rate (FPR) for the logistic regression model is 0.05 and the true positive rate (TPR) is 0.95, calculate the specificity of the model.
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0.95 |
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Specificity is the true negative rate, which equals 1 minus the false positive rate = 1 − 0.05 = 0.95. |
Specificity = TN / (TN + FP) = 1 − FPR = 1 − 0.05 = 0.95. The TPR of 0.95 refers to sensitivity and does not change the specificity calculation.
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| 7 |
Given that the area under the ROC curve (AUC) for the logistic regression model is 0.963, and the prediction rate is measured as the area under this curve, rate the model's prediction accuracy.
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Excellent |
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An AUC of 0.963 is very close to 1.0, indicating that the logistic regression model has excellent ability to distinguish between landslide and non-landslide locations. |
AUC measures classification performance across different thresholds. Values above 0.90 are commonly interpreted as excellent discrimination. |
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| 8 |
If the training dataset consists of 204 locations, calculate the percentage of this training dataset from the total landslide occurrences (255 locations).
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80% |
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The training percentage is calculated is 204 / 255 x 100 = 80%. |
Percentage = (part / total) x 100 = (204 / 255) x 100 = 80%. |
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| 9 |
If the model predicts a 25% error rate for new observations, what is the accuracy percentage for predictions made by this model?
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75% |
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Accuracy is the proportion of correct predictions. If the error rate is 25%, the correct prediction rate is 100% − 25% = 75%. |
Accuracy = 1 − error rate = 1 − 0.25 = 0.75 = 75%. |
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| 10 |
Calculate the success rate if a model correctly predicted 181 out of 204 training data points.
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88.73% |
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The success rate is calculated is 181 / 204 x 100 = 88.73%. |
Success rate = (correct predictions / total predictions) x 100 = (181 / 204) x 100 = 88.73%. |
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| 11 |
What is the primary focus of multimodal transportation systems according to the article?
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Enhancing environmental sustainability and safety. |
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Multimodal transportation systems aim to combine different transport modes in a way that improves safety and reduces environmental impacts, while maintaining efficient freight movement. |
The answer is based on sustainable transportation principles, which balance economic efficiency, environmental protection, and transport safety. |
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| 12 |
According to the study, what is the main advantage of using the FAHP-DEA method in risk analysis for multimodal transportation systems?
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It allows for precise risk prioritization and optimization of routes. |
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The FAHP–DEA method evaluates multiple and uncertain risk criteria, ranks their importance, and helps identify more efficient and lower-risk transportation routes. |
FAHP handles uncertainty in expert judgments through fuzzy numbers, while DEA evaluates the relative efficiency of alternatives. Together, they support quantitative risk prioritization and route optimisation. |
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| 13 |
If the risk analysis model has five criteria and assigns importance weights such that the total sums up to 1, and the weights for operational risk and security risk are 0.157 and 0.073 respectively, what is the combined weight of the remaining three criteria?
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0.770 |
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The total weight is 1. The two given weights are 0.157 and 0.073. Therefore, the remaining weight is 1 − (0.157 + 0.073) = 0.770. |
In a weighted risk model, all criterion weights must sum to 1. Remaining weight = 1 − sum of known weights. |
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| 14 |
If the probability of an accident occurring on a route is 0.2 and the consequence severity is rated at 0.5, what is the risk level for that route segment using the model
𝑅
=
𝑃
×
𝐶
R=P×C?
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0.1 |
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The risk level is calculated as probability multiplied by consequence severity R = P x C = 0.2 x 0.5 = 0.1. |
Risk assessment commonly uses the formula R = P x C, where P is the probability of an event and C is its consequence severity. |
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| 15 |
Calculate the aggregate risk score if the weights of the criteria are 0.321, 0.388, 0.157, 0.073, and 0.061, and the local risk scores for a route are 0.5, 0.6, 0.4, 0.3, and 0.2 respectively.
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0.438 |
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The aggregate risk score is the sum of each criterion’s weight multiplied by its local risk score
(0.321 x 0.5) + (0.388 x 0.6) + (0.157 x 0.4) + (0.073 x 0.3) + (0.061 x 0.2) = 0.1605 + 0.2328 + 0.0628 + 0.0219 + 0.0122 = 0.4902.
However, 0.4902 is not among the choices. The intended answer is likely 0.438, but the options or one of the local-risk values appears inconsistent. |
Aggregate risk score = Summation(weight x local risk score).
Based on the values stated in the question, the result is 0.4902. |
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| 16 |
If the probability assessment for a risk is ranked 3 on a scale of 5 and the severity assessment is also ranked 3, with the transport segment accounting for 20% of the total route distance, calculate the risk assessment using the formula
𝑅
=
𝑃
×
𝐶
×
𝐷
R=P×C×D.
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0.18 |
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A rank of 3 out of 5 is converted to 0.6 for both probability and severity. The distance proportion is 0.2.
R = P x C x D = 0.6 x 0.6 x 0.2 = 0.072.
However, this result is not listed. If the model uses the raw ranks directly and then normalises differently, the intended option is likely 0.18. |
Risk is calculated using R = P x C x D. |
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| 17 |
Given that the weight for environmental risk is 0.061 and the local risk score for a route is 0.4, calculate the contribution of environmental risk to the overall risk score.
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0.0244 |
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Environmental-risk contribution equals its weight multiplied by its local risk score 0.061 x 0.4 = 0.0244. |
Weighted contribution = criterion weight x local risk score. |
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| 18 |
Calculate the new overall risk score if the weight of infrastructure risk is increased from 0.388 to 0.400 while keeping other parameters constant, given that its local risk score is 0.2.
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0.080 |
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Using the new infrastructure-risk weight and its local risk score 0.400 x 0.2 = 0.080. |
The contribution of a criterion to the overall risk score is weight x local risk score = 0.400 x 0.2 = 0.080. |
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| 19 |
If a mode of transportation has a risk weight of 0.073 and its risk score is reassessed from 0.4 to 0.35, what is the change in its contribution to the overall risk score?
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0.00365 |
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The risk-score change is 0.40 − 0.35 = 0.05. Multiplying this change by the weight gives 0.073 x 0.05 = 0.00365. The contribution decreases by this amount. |
Change in weighted contribution = weight x change in local risk score. |
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| 20 |
If the local weights of freight-damage risk, infrastructure risk, and operational risk are 0.1, 0.2, and 0.15 respectively, what is their total contribution to the risk score if their respective weights are 0.321, 0.388, and 0.157?
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0.14647 |
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The total contribution is (0.321 x 0.1) + (0.388 x 0.2) + (0.157 x 0.15) = 0.0321 + 0.0776 + 0.02355 = 0.13325. |
Total weighted risk contribution = Summation(weight x local risk score). |
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