Pipelining — علوم الحاسوب الجامعية

مفردات

English

The assembly line

Pipelining overlaps instructions in five stages — IF, ID, EX, MEM, WB — like an assembly line: instruction $i+1$ fetches while $i$ decodes. Once full, one instruction retires per cycle.
$n$ instructions, $k$ stages take $n + k - 1$ cycles: five instructions in five stages take $5 + 5 - 1 = 9$ cycles, not $25$; at $n = 100$: $104$ cycles.

{exercise:uni-cs0603-formula}

تدريب

Ten instructions on a five-stage pipeline take —

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English

The stage time is the slowest stage: with IF 250, ID 150, EX 200, MEM 250, WB 100 ps, every stage costs 250 ps — 250 ps per instruction against the single cycle's 900 ps.
Ideal CPI is 1; stalls raise it. A 1000-instruction program, 25% loads, half load-use (one stall each): $1000 + 125 + 4 = 1129$ cycles, CPI $1.129$.

{exercise:uni-cs0603-cpi}

تدريب

1000 instructions, 25% loads, half load-use: total cycles —

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English

A read-after-write hazard: add R1, R2, R3 then sub R4, R1, R1 — the sub reads R1 in ID one cycle before add writes it in WB.
Forwarding routes the ALU output straight back to the next ALU's inputs, hiding the wait — except when the producer is a load: lw R1, …; add R3, R1, R2 still stalls one cycle (data exists only after MEM). A taken branch, resolved in EX, costs one bubble too.

{exercise:uni-cs0603-hazard}

تدريب

Forwarding hides ALU-to-ALU waits, but load-use still costs —

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