Deep Learning · TEE · section

R25 Q4a · 64×64 CNN shape and parameter table

February 2025 re-exam · complete solved paper

Question

Question. For input64×64×3: conv3×3,32filters,P1,S1; maxpool2×2,S2; conv3×3,64,P1,S1; pool2,S2; conv3×3,128,P1,S1; pool2,S2; FC256; softmax43. Find each shape and all weights/biases. (10 marks)

Answer — step by step

Given

Input=[64, 64, 3]

Layer	Kernel	Stride	Padding	Filters / units
Conv	3	1	1	32
Pool	2	2	0	0
Conv	3	1	1	64
Pool	2	2	0	0
Conv	3	1	1	128
Pool	2	2	0	0
Dense	0	0	0	256
Dense	0	0	0	43

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+2×1−3)/1⌋+1=64; parameters=(3×3×3+1)×32=896	64×64×32	896
Pool	64×64×32	⌊(64−2)/2⌋+1=32; channels stay 32; no trainable weights	32×32×32	0
Conv	32×32×32	⌊(32+2×1−3)/1⌋+1=32; parameters=(3×3×32+1)×64=18496	32×32×64	18496
Pool	32×32×64	⌊(32−2)/2⌋+1=16; channels stay 64; no trainable weights	16×16×64	0
Conv	16×16×64	⌊(16+2×1−3)/1⌋+1=16; parameters=(3×3×64+1)×128=73856	16×16×128	73856
Pool	16×16×128	⌊(16−2)/2⌋+1=8; channels stay 128; no trainable weights	8×8×128	0
Dense	8×8×128	Flatten/input length=8×8×128=8192; parameters=(8192+1)×256=2097408	256	2097408
Dense	1×1×256	Flatten/input length=1×1×256=256; parameters=(256+1)×43=11051	43	11051

Step 2. Sum only trainable parameters

896 + 18496 + 73856 + 2097408 + 11051 = 2201707

Final answer

Total parameters=2201707. Shape and parameter count are separate results.