Midsem 2025 solved

This page solves the IIIT Kottayam midsem paper of Oct 2025. It covers output size, learnable parameters, threshold choice, and optimizer fixes. After this page you can solve each midsem numerical without help.

IIIT Kottayam held the mid semester exam in Oct 2025. The paper is IOE414 Deep Learning. Maximum marks are 50. This page solves Q1, Q2, and Q3 with full steps.

Q1 Convolution output and learnable parameters (10M)Search on YouTube

A convolution layer takes a 64 by 64 by 16 input. It applies 32 kernels (also called filters) of size 3 by 3. Stride is 2 and padding is 1. The task has two parts. Find the feature map size. Count the learnable parameters. See convolution output size and parameter counts.

(a) Output dimensions Search on YouTube

Output size measures the spatial size of each feature map. It depends on input size, kernel size, stride, and padding.

O = \left\lfloor \frac{W - K + 2P}{S} \right\rfloor + 1

$W$ is input width. $K$ is kernel size. $P$ is padding. $S$ is stride. $O$ is output width. The same rule gives output height.

Step 1. Insert the numbers.

\begin{aligned} O &= (64 - 3 + 2 \times 1)/2 + 1 \\ &= (64 - 3 + 2)/2 + 1 \\ &= 63/2 + 1 \\ &= 31.5 + 1 \\ &= 32.5 \rightarrow \mathbf{32} \end{aligned}

Step 2. Build the full shape.

\begin{aligned} \text{width} &= \mathbf{32} \\ \text{height} &= \mathbf{32} \\ \text{depth} &= \mathbf{32} \\ \text{output} &= \mathbf{32 \times 32 \times 32} \end{aligned}

Input has 16 channels. Each kernel spans all 16 channels. The layer uses 32 kernels. Depth is 32.

Do not keep 32.5 as the output size. Feature map size must be an integer. Floor the division before adding 1. Slides use the integer formula.

Write the substitution line in the exam. Examiners give marks for the formula even when arithmetic slips. State the floor step in words.

(b) Learnable parameter count Search on YouTube

Learnable parameters mean weights and biases only. Each kernel has weights for every input channel. Each kernel has one bias.

P = (K \times K \times C_{in} + 1) \times C_{out}

$K$ is kernel size. $C_{in}$ is input channel count. $C_{out}$ is kernel count. The 1 means one bias per kernel.

Step 1. Count one kernel.

\begin{aligned} \text{weights per kernel} &= 3 \times 3 \times 16 = 144 \\ \text{bias per kernel} &= 1 \\ \text{values per kernel} &= 144 + 1 = 145 \end{aligned}

Step 2. Count all kernels.

\begin{aligned} P &= 145 \times 32 = \mathbf{4640} \\ \text{weights} &= 144 \times 32 = \mathbf{4608} \\ \text{biases} &= 32 \end{aligned}

The layer has 4640 learnable values. It has 4608 weights. It has 32 biases.

Q2 Recruitment neuron and threshold (10M)Search on YouTube

A single neuron scores job candidates. Five test scores act as inputs. Weights combine them into one number. A threshold decides recruit or reject. See perceptron threshold.

(a) Neuron model Search on YouTube

The neuron forms a weighted sum. A step activation function compares the sum with a threshold. Bias equals negative threshold in the bias form. This page uses threshold form for clarity.

z = \sum_{i=1}^{5} w_{i} x_{i}

a = 1 \text{ if } z \ge t \text{ else } 0

$z$ is the weighted sum. $w_{i}$ is the weight for input $i$ . $x_{i}$ is input $i$ . $t$ is the threshold. $a$ is the output label.

(b) Threshold from labelled samples Search on YouTube

Eight labelled candidates fix the threshold. Recruited means label 1. Rejected means label 0.

ID	Score	Label
ID1	0.8000	1
ID2	0.5425	0
ID3	0.7900	1
ID4	0.3400	0
ID5	0.7055	1
ID6	0.5120	0
ID7	0.9265	1
ID8	0.4050	0

0.5425 \lt t \le 0.7055

$t$ is the threshold. Any $t$ in this range separates the samples.

Step 1. Find the border values.

\begin{aligned} \text{recruited scores} &= 0.8000, 0.7900, 0.7055, 0.9265 \\ \min \text{ recruited} &= \mathbf{0.7055} \\ \text{rejected scores} &= 0.5425, 0.3400, 0.5120, 0.4050 \\ \max \text{ rejected} &= \mathbf{0.5425} \end{aligned}

Step 2. Pick the threshold.

\begin{aligned} \text{midpoint} &= (0.5425 + 0.7055)/2 \\ &= 1.2480/2 \\ &= \mathbf{0.6240} \\ t &= \mathbf{0.6} \end{aligned}

Value 0.6 lies inside the clean gap. It labels all eight samples correctly. It leaves margin on both sides.

Do not pick the edge value 0.7055 without reason. Noise can push a future score across the edge. Pick a value inside the gap. Value 0.6 keeps distance from both borders.

List the two border scores in the exam. Write the interval form. Any value inside the interval earns full marks. Justify the choice in one line.

(c) Score two new candidates Search on YouTube

Two new candidates need scores. Candidate A has inputs 0.82, 0.50, 0.65, 0.58, and 0.60. Candidate B has inputs 0.70, 0.78, 0.80, 0.55, and 0.58. Compare each weighted sum with 0.6. Weights are 0.20, 0.15, 0.10, 0.25, and 0.30.

Candidate A.

\begin{aligned} 0.82 \times 0.20 &= 0.1640 \\ 0.50 \times 0.15 &= 0.0750 \\ 0.65 \times 0.10 &= 0.0650 \\ 0.58 \times 0.25 &= 0.1450 \\ 0.60 \times 0.30 &= 0.1800 \\ z_{A} &= 0.1640 + 0.0750 + 0.0650 + 0.1450 + 0.1800 \\ &= 0.4490 + 0.1800 \\ &= \mathbf{0.6290} \end{aligned}

Value 0.6290 exceeds 0.6. Candidate A gets label 1. The neuron recruits A.

Candidate B.

\begin{aligned} 0.70 \times 0.20 &= 0.1400 \\ 0.78 \times 0.15 &= 0.1170 \\ 0.80 \times 0.10 &= 0.0800 \\ 0.55 \times 0.25 &= 0.1375 \\ 0.58 \times 0.30 &= 0.1740 \\ z_{B} &= 0.1400 + 0.1170 + 0.0800 + 0.1375 + 0.1740 \\ &= 0.4745 + 0.1740 \\ &= \mathbf{0.6485} \end{aligned}

Value 0.6485 exceeds 0.6. Candidate B gets label 1. The neuron recruits B.

Value 0.6485 exceeds 0.6290. B ranks higher than A. Both cross 0.6. Both get recruit labels.

Q3 Slow training and poor minima (5M)Search on YouTube

Training stalls in a poor local minimum. Loss stays high across epochs. Gradients stay small in flat regions. Saddle points slow plain gradient descent. Minibatch noise helps escape saddles. See optimizers.

\begin{aligned} v &\leftarrow \beta v - \eta \nabla_{\theta} L \\ \theta &\leftarrow \theta + v \end{aligned}

$v$ is velocity. $\beta$ is the momentum factor. $\eta$ is the learning rate. $\theta$ is the parameter vector. $L$ is the loss.

Use momentum or NAG for ravines. Use RMSprop or Adam for adaptive rates. Adam combines momentum with per-weight scaling. Prefer minibatch updates over full batch updates. The paper marks PTO on this question. Continuation after PTO is unknown.

Name two optimizers in the exam. State one line for each choice. Mention minibatch noise for saddle points.

Recap Search on YouTube

Compute conv output with floor((W-K+2P)/S)+1 and set depth to kernel count.
Count learnable values with (K by K by Cin+1) by Cout, weights plus one bias each.
Form neuron score with weighted sum and step rule at threshold t.
Pick threshold inside the clean gap and keep margin from both borders.
Escape poor minima with momentum, NAG, Adam, adaptive rates, and minibatch noise.

Q1 Convolution output and learnable parameters (10M)Search on YouTubeBookmarkSaved

(a) Output dimensionsSearch on YouTubeBookmarkSaved

(b) Learnable parameter countSearch on YouTubeBookmarkSaved

Q2 Recruitment neuron and threshold (10M)Search on YouTubeBookmarkSaved

(a) Neuron modelSearch on YouTubeBookmarkSaved

(b) Threshold from labelled samplesSearch on YouTubeBookmarkSaved

(c) Score two new candidatesSearch on YouTubeBookmarkSaved

Q3 Slow training and poor minima (5M)Search on YouTubeBookmarkSaved

RecapSearch on YouTubeBookmarkSaved

On this page