Overlapping work without rushing it

Computing · Computer Systems · ages 18-20

Name ______________________ Date ____________

One instruction still passes how many steps?
- One
- Five
- Nine
What does pipelining raise?
- Finished instructions per round
- Volts per transistor
- Letters per opcode
Forwarding wires a fresh result straight back to cut waiting.
Circle one: True False
An instruction adds two numbers the prior load has not delivered. What happens?
- The line retires early
- The chip orders lunch
- The line stalls until the value lands
Five instructions, five steps each. Type the pipelined total in rounds.
Answer: ______________
A branch guess proves wrong. What happens to the half-done steps?
- They are thrown away
- They are framed as art
- They run twice as fast
A mispredicted branch keeps its half-done steps and counts them as finished.
Circle one: True False
Five instructions save sixteen rounds over running alone. Throughput rises, yet one lag stays five. Why?
- Transistors tire with age
- Each instruction still crosses five steps
- Nine is larger than five

Answer key

For grown-ups. Fold this page away before handing over the rest.

Five · Each instruction still crosses all five steps.
Finished instructions per round · Overlap raises throughput, finished work per round.
True · That shortcut feeds neighbours early.
The line stalls until the value lands · A missing input forces a stall.
9 · Five for the first plus one per follower makes nine.
They are thrown away · Wrong-path work is dumped and the line refills.
False · Wrong-path steps are dumped, not counted.
Each instruction still crosses five steps · Latency per instruction never shrank.