Step 1: Shortlist the routes that look short by inspection.
Instead of a formal labeling algorithm, list out the routes from A to G that avoid obviously long detours (like the $20$-minute A-B-H edge) and add up their distances directly.
Step 2: Route through C and E.
$A - C - E - G$: $8 + 10 + 15 = 33$ minutes.
Step 3: Route through B, D and H.
$A - B - D - H - G$: $10 + 5 + 7 + 12 = 34$ minutes.
Step 4: Route through C and D.
$A - C - D - F - G$: $8 + 9 + 6 + 8 = 31$ minutes.
Step 5: Route through B, D and F.
$A - B - D - F - G$: $10 + 5 + 6 + 8 = 29$ minutes.
This beats every other route checked so far.
Step 6: Check that no shorter combination is possible.
Any route reaching $G$ must pass through either $F$ ($8$ min from $G$) or $H$ ($12$ min from $G$) or $E$ ($15$ min from $G$), since these are the only three nodes directly linked to $G$. Getting to $F$ takes at least $10+5+6=21$ minutes (via $B-D-F$), so the cheapest approach to $G$ through $F$ is $21+8=29$. Getting to $H$ or $E$ costs at least $22$ or $18$ minutes respectively before adding their (larger) links to $G$, both of which give totals above $29$.
Step 7: Conclude.
The minimum time to commute from $A$ to $G$ is 29 minutes, along $A - B - D - F - G$.
\[ \boxed{29 \text{ minutes}} \]