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Faculty Assignment2 Q6 · 64×64 two-convolution CNN

Faculty Assignment 2 · worked practice

Question

Question. Input64×64×3; conv5,16valid; pool2stride2; conv3,32same; pool2stride2; dense128; output10. Find shapes and parameters. (Faculty assignment: no marks printed)

Answer — step by step

Given

Input=[64, 64, 3]

Layer	Kernel	Stride	Padding	Filters / units
Conv	5	1	0	16
Pool	2	2	0	0
Conv	3	1	1	32
Pool	2	2	0	0
Dense	0	0	0	128
Dense	0	0	0	10

To find

Each output shape and all trainable weights plus biases.

Formula: which one and why

$O=\left\lfloor\frac{N+2P-F}{S}\right\rfloor+1$ LaTeX: O=\left\lfloor\frac{N+2P-F}{S}\right\rfloor+1

$P_{conv}=(F_hF_wC_{in}+1)C_{out},\quad P_{dense}=(n_{in}+1)n_{out}$ LaTeX: P_{conv}=(F_hF_wC_{in}+1)C_{out},\quad P_{dense}=(n_{in}+1)n_{out}

Each filter/unit has one bias. Pooling has zero learned parameters. Flatten preserves the number of values, then a dense layer connects every input value to every output unit.

Step 1. Work layer by layer

Layer	Input	Full substitution and reason	Output	Parameters
Conv	64×64×3	⌊(64+2×0−5)/1⌋+1=60; parameters=(5×5×3+1)×16=1216	60×60×16	1216
Pool	60×60×16	⌊(60−2)/2⌋+1=30; channels stay 16; no trainable weights	30×30×16	0
Conv	30×30×16	⌊(30+2×1−3)/1⌋+1=30; parameters=(3×3×16+1)×32=4640	30×30×32	4640
Pool	30×30×32	⌊(30−2)/2⌋+1=15; channels stay 32; no trainable weights	15×15×32	0
Dense	15×15×32	Flatten/input length=15×15×32=7200; parameters=(7200+1)×128=921728	128	921728
Dense	1×1×128	Flatten/input length=1×1×128=128; parameters=(128+1)×10=1290	10	1290

Step 2. Sum only trainable parameters

1216 + 4640 + 921728 + 1290 = 928874

Final answer

Total parameters=928874. Shape and parameter count are separate results.