Given:
Loading time per dumper (\( L \)) = 3 minutes
Number of dumpers (\( N \)) = 8
Cycle time of a dumper (\( C \)) = 22 minutes (includes loading time)
Objective:
Find the waiting time (\( W \)) of a dumper before it is loaded again.
Approach:
Calculate the total time the shovel takes to load all dumpers once.
Determine when each dumper returns after completing its cycle.
Compute the time difference between when a dumper returns and when the shovel is available to load it again.
Step 1: Total Loading Time for All Dumpers
The shovel loads dumpers sequentially, taking \( L = 3 \) minutes per dumper. For \( N = 8 \) dumpers:
\[
\text{Total loading time} = N \times L = 8 \times 3 = 24\ \text{minutes}
\]
Step 2: Dumper Cycle Time
The cycle time (\( C = 22 \) minutes) includes:
\[
C = \text{Loading time} + \text{Travel, dumping, and return time}
\]
\[
22 = 3 + \text{Other activities time}
\]
\[
\text{Other activities time} = 22 - 3 = 19\ \text{minutes}
\]
Step 3: Dumper Return and Waiting Time
Each dumper returns after completing its cycle. The shovel takes 24 minutes to cycle through all dumpers.
For any dumper:
\[
\text{Time when loading starts} = t
\]
\[
\text{Time when dumper returns} = t + 22
\]
\[
\text{Time when shovel is free to load it again} = t + 24
\]
\[
\text{Waiting time} = (t + 24) - (t + 22) = 2\ \text{minutes}
\]
Verification:
Dumper 1:
Starts loading at \( t = 0 \), finishes at \( t = 3 \).
Returns at \( t = 22 \).
Shovel finishes loading all dumpers at \( t = 24 \).
Waits from \( t = 22 \) to \( t = 24 \): \( W = 2 \) minutes.
Dumper 2:
Starts loading at \( t = 3 \), finishes at \( t = 6 \).
Returns at \( t = 25 \).
Shovel loads Dumper 1 at \( t = 24 \), finishes at \( t = 27 \).
Waits from \( t = 25 \) to \( t = 27 \): \( W = 2 \) minutes.
This pattern holds for all dumpers.
General Formula:
The waiting time can also be derived as:
\[
W = \text{Total loading time} - \text{Cycle time} = 24 - 22 = 2\ \text{minutes}
\]
Final Answer:
The waiting time of a dumper is \(\boxed{2}\) minutes.