Deep Learning · TEE · section

F25 Q1b

Final examination December 2025 fully solved

Question

Question (5 marks). For automated leaf-disease image classification, discuss factors affecting loss selection. For classes Normal, Mild, Severe with target [0,1,0] and predicted probabilities [0.1,0.8,0.1], compute and interpret categorical cross-entropy.

Answer — step by step

Given

Classes = Normal,Mild,Severe; target = [0,1,0].

Predicted probabilities = [0.1,0.8,0.1]; use natural logarithms.

To find

Calculate categorical cross-entropy and interpret the result.

Formula — which one and why

Categorical cross-entropy. For mutually exclusive classes with a one-hot target, all zero-target terms disappear. The loss is the negative log of the probability assigned to the true class.

yⱼ = target entry; pⱼ = predicted probability; c = index of the true class. Use natural logarithms unless another base is specified.

$L=-\sum_j y_j\ln p_j$ LaTeX: L=-\sum_j y_j\ln p_j

$L=-\ln p_c$ LaTeX: L=-\ln p_c

Yaad rakhne ka cue: One-hot mein sirf correct class ki probability ka minus ln.

Solution — step by step

Select a loss according to the task and output meaning: real-valued regression often uses MSE/MAE, binary classification uses binary cross-entropy, and one-of-K classification uses categorical cross-entropy. Consider class imbalance, noisy labels/outliers, whether outputs are calibrated probabilities, and differentiability for optimization.

Step 1: expand the categorical cross-entropy sum.

$L=-[0\ln(0.1)+1\ln(0.8)+0\ln(0.1)]$ LaTeX: L=-[0\ln(0.1)+1\ln(0.8)+0\ln(0.1)]

Step 2: keep the term for the true class.

$L=-\ln(0.8)$ LaTeX: L=-\ln(0.8)

Step 3: calculate its natural logarithm.

$\ln(0.8)\approx-0.223143551$ LaTeX: \ln(0.8)\approx-0.223143551

Step 4: negate the logarithm.

$L\approx0.223143551$ LaTeX: L\approx0.223143551

Final answer: loss ≈ 0.223144 nats. The true class received probability 0.8, so the loss is relatively small; probability 1 would give loss zero.