To find the number of overlapping cycles between the two iterations, we need to determine two specific points in time:
Start of Iteration 2: The cycle when the first instruction of the second iteration ($I_1$) enters the first stage ($S_1$).
End of Iteration 1: The cycle when the last instruction of the first iteration ($I_4$) finishes the last stage ($S_4$).
The overlap is the duration where the pipeline is processing instructions from both iterations simultaneously.
Step 1: Analyze the Flow of Iteration 1
We need to map the execution of instructions $I_1, I_2, I_3,$ and $I_4$ through stages $S_1$ to $S_4$.
Rules: An instruction can only enter stage $N$ if:
a) It has finished stage $N-1$.
b) Stage $N$ is free (previous instruction has finished using it).
Execution Trace:
Instruction $I_1$:
$S_1$: Cycle 1 (Duration 1)
$S_2$: Cycle 2 (Duration 1)
$S_3$: Cycles 3-4 (Duration 2)
$S_4$: Cycle 5 (Duration 1)
Instruction $I_2$:
$S_1$: Starts after $I_1$ finishes $S_1$ (cycle 1). Runs cycles 2-3 (Duration 2).
$S_2$: Starts after $I_2$ finishes $S_1$ (cycle 3) AND $S_2$ is free (cycle 2). Runs cycle 4 (Duration 1).
$S_3$: Starts after $I_2$ finishes $S_2$ (cycle 4) AND $S_3$ is free (cycle 4). Runs cycle 5 (Duration 1).
$S_4$: Starts after $I_2$ finishes $S_3$ (cycle 5) AND $S_4$ is free (cycle 5). Runs cycle 6 (Duration 1).
Instruction $I_3$:
$S_1$: Starts after $I_2$ finishes $S_1$ (cycle 3). Runs cycles 4-5 (Duration 2).
$S_2$: Starts after $I_3$ finishes $S_1$ (cycle 5) AND $S_2$ is free (cycle 4). Runs cycles 6-7 (Duration 2).
$S_3$: Starts after $I_3$ finishes $S_2$ (cycle 7) AND $S_3$ is free (cycle 5). Runs cycles 8-10 (Duration 3).
$S_4$: Starts after $I_3$ finishes $S_3$ (cycle 10) AND $S_4$ is free (cycle 6). Runs cycles 11-13 (Duration 3).
Instruction $I_4$:
$S_1$: Starts after $I_3$ finishes $S_1$ (cycle 5). Runs cycle 6 (Duration 1).
$S_2$: Starts after $I_4$ finishes $S_1$ (cycle 6) AND $S_2$ is free ($I_3$ finishes at cycle 7). Runs cycles 8-10 (Duration 3).
$S_3$: Starts after $I_4$ finishes $S_2$ (cycle 10) AND $S_3$ is free ($I_3$ finishes at cycle 10). Runs cycles 11-12 (Duration 2).
$S_4$: Starts after $I_4$ finishes $S_3$ (cycle 12) AND $S_4$ is free ($I_3$ finishes at cycle 13). Runs cycle 14 (Duration 1).
Iteration 1 End Time: Cycle 14 (when $I_4$ leaves $S_4$).
Step 2: Determine Start of Iteration 2
The second iteration begins immediately after the first. The first instruction of the second iteration is $I_1$ (let's call it $I_1'$).
$I_1'$ needs to enter stage $S_1$.
$S_1$ was last occupied by $I_4$ of the first iteration.
$I_4$ occupied $S_1$ during Cycle 6.
Therefore, $S_1$ is free starting Cycle 7.
Iteration 2 Start Time: Cycle 7.
Step 3: Calculate Overlap
The overlapped cycles are the cycles where Iteration 2 has started but Iteration 1 has not yet finished.
Visual Timeline:
Cycle 6: Iteration 1 is running ($I_4$ in $S_1$, $I_3$ in $S_2$, $I_2$ in $S_4$).
Cycle 7: Iteration 2 Starts ($I_1'$ enters $S_1$). Iteration 1 is still running ($I_3$ in $S_2$). (Overlap Cycle #1)
Cycle 8: Iteration 2 running. Iteration 1 running ($I_4$ enters $S_2$, $I_3$ enters $S_3$). (Overlap Cycle #2)
...
Cycle 14: Iteration 1 finishes ($I_4$ completes $S_4$). Iteration 2 is still running in earlier stages. (Overlap Cycle #8)
Calculation:
$$\text{Overlap} = (\text{End of Iteration 1}) - (\text{Start of Iteration 2}) + 1$$
$$\text{Overlap} = 14 - 7 + 1$$
$$\text{Overlap} = 8 \text{ cycles}$$
Answer:
There are 8 overlapped cycles between the iterations.
