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Faculty Assignment2 Q3 · 28×28 CNN

Faculty Assignment 2 · worked practice

Question

Question. Input28×28×1; conv5,6valid; pool2stride2; conv5,16valid; pool2stride2; dense120,dense84,dense10. Find layer dimensions and parameters. (Faculty assignment: no marks printed)

Answer — step by step

Given

Input=[28, 28, 1]

Layer	Kernel	Stride	Filters / units
Conv	5	1	6
Pool	2	2	0
Conv	5	1	16
Pool	2	2	0
Dense	0	0	120
Dense	0	0	84
Dense	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	28×28×1	⌊(28+2×0−5)/1⌋+1=24; parameters=(5×5×1+1)×6=156	24×24×6	156
Pool	24×24×6	⌊(24−2)/2⌋+1=12; channels stay 6; no trainable weights	12×12×6	0
Conv	12×12×6	⌊(12+2×0−5)/1⌋+1=8; parameters=(5×5×6+1)×16=2416	8×8×16	2416
Pool	8×8×16	⌊(8−2)/2⌋+1=4; channels stay 16; no trainable weights	4×4×16	0
Dense	4×4×16	Flatten/input length=4×4×16=256; parameters=(256+1)×120=30840	120	30840
Dense	1×1×120	Flatten/input length=1×1×120=120; parameters=(120+1)×84=10164	84	10164
Dense	1×1×84	Flatten/input length=1×1×84=84; parameters=(84+1)×10=850	10	850

Step 2. Sum only trainable parameters

156 + 2416 + 30840 + 10164 + 850 = 44426

Final answer

Total parameters=44426. Shape and parameter count are separate results.