How to use the Pipeline Speedup Calculator
- 1. Enter stage delaysType one delay per stage. The count sets k, the number of stages.
- 2. Add the restGive the latch delay, number of instructions and any stall cycles. The non-pipelined time is optional.
- 3. See the workingGet the cycle time, both total times, speedup, efficiency, throughput and a space-time diagram.
Pipeline speedup formulas
In a k-stage pipeline, every stage moves forward on the same clock, so the cycle time tp = largest stage delay + latch delay. The first instruction needs k cycles and every later one finishes one cycle after the one before it, so n instructions take (k + n − 1) × tp. Stall cycles from hazards are added to the cycle count.
- Non-pipelined time = n × tn. If tn is not given, use the sum of the stage delays.
- Speedup S = (n × tn) ÷ ((k + n − 1) × tp).
- Efficiency = S ÷ k. With equal stages this equals n ÷ (k + n − 1).
- Throughput = n ÷ pipelined time. Maximum speedup as n → ∞ is tn ÷ tp.
Worked example
A 4-stage pipeline has stage delays of 60, 50, 90 and 80 ns and a 10 ns latch. The cycle time is 90 + 10 = 100 ns. For 1,000 instructions the pipeline takes (4 + 1000 − 1) × 100 = 100,300 ns. Without pipelining, each instruction takes 60 + 50 + 90 + 80 = 280 ns, so 1,000 take 280,000 ns.
Speedup = 280,000 ÷ 100,300 = 2.79. The largest possible speedup is 280 ÷ 100 = 2.8, not 4, because the stages are unequal and the latch adds delay. Efficiency = 2.79 ÷ 4 ≈ 69.8%.
Common exam traps
Use the slowest stage for the clock, not the average. Add the latch (buffer) delay to every cycle when the question gives one. For an ideal pipeline with equal stages and no stalls, speedup tends to k for large n. A branch penalty of p cycles with a branch frequency f adds f × p stall cycles per instruction, so for n instructions enter n × f × p as total stalls. For more practice, try the computer organisation MCQ quiz.
Frequently asked questions
Why is speedup less than the number of stages?
Unequal stages make the clock wait for the slowest one, latches add delay, and filling the pipeline costs k − 1 extra cycles. Stalls lower it further.
Should I add the latch delay to the non-pipelined time?
No. A non-pipelined CPU has no stage registers, so tn is just the sum of stage delays unless the question gives another value.
How do I find throughput?
Divide the number of instructions by the total pipelined time. For large n it approaches one instruction per cycle.
What is CPI for an ideal pipeline?
Close to 1, because one instruction finishes every cycle once the pipeline is full. Stalls raise the average CPI above 1.