Final answer:
The speedup of the pipeline is calculated by comparing the total non-pipelined time per instruction (35 ns) to the pipelined cycle time (10 ns). Under ideal conditions, the pipeline achieves an approximate 3.5 times speedup compared to the non-pipelined implementation.
Step-by-step explanation:
To calculate the speedup of a pipeline, we compare the time it takes to process a sequence of instructions in both pipelined and non-pipelined architectures. In a non-pipelined implementation, each instruction would be processed sequentially, thus taking the sum of all stage times plus the inter-stage register delays.
The total time for a non-pipelined implementation would be:
- 7 ns for stage 1
- 4 ns for stage 2
- 3 ns for stage 3
- 8 ns for stage 4
- 5 ns for stage 5
- Inter-stage register delay: 4 * 2 ns (since we have four inter-stage registers)
Therefore, the total non-pipelined time = 7 + 4 + 3 + 8 + 5 + (4 * 2) = 27 + 8 = 35 ns.
In the pipelined implementation, in steady state, a new instruction can be initiated after the longest stage time. Since stage 4 is the longest stage at 8 ns, and the stage is followed by an inter-stage register of 2 ns, we get a clock cycle of 10 ns for the pipeline.
To determine the speedup, we compare the non-pipelined time per instruction to the pipelined cycle time:
Speedup = Non-pipelined time / Pipelined cycle time = 35 ns / 10 ns = 3.5
Therefore, under ideal conditions, the pipeline would achieve approximately a 3.5 times speedup compared to the non-pipelined implementation.