1989
DOI: 10.1145/59287.59291
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Scheduling expressions on a pipelined processor with a maximal delay of one cycle

Abstract: Consider a pipelined machine that can issue instructions every machine cycle. Sometimes, an instruction that uses the result of the instruction preceding it in a pipe must he delayed to ensure that a program computes a right value. We assume that issuing of such instructions is delayed by at most one machine cycle. For such a machine model, given an unbounded number of machine registers and memory locations, an algorithm to find a shortest schedule of the given expression is presented and analyzed. The propose… Show more

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Cited by 50 publications
(27 citation statements)
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“…[3,19]) but differs from those used in more recent articles targeting also multiple-issue processors, in particular [9,16,7]. For a precedence (i, j) ∈ R with associated latency ℓ(i, j), if t i is the cycle where instruction i is scheduled, then j can be scheduled earliest in cycle t i + ℓ(i, j) + 1 (and not…”
Section: Formal Problem Statementmentioning
confidence: 99%
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“…[3,19]) but differs from those used in more recent articles targeting also multiple-issue processors, in particular [9,16,7]. For a precedence (i, j) ∈ R with associated latency ℓ(i, j), if t i is the cycle where instruction i is scheduled, then j can be scheduled earliest in cycle t i + ℓ(i, j) + 1 (and not…”
Section: Formal Problem Statementmentioning
confidence: 99%
“…These precedence and latency constraints make instruction scheduling an N P-hard combinatorial optimization problem [2], even for single-issue processors that allow only at most one instruction to be inserted into the pipeline (issued ) in every clock cycle. Polynomial-time solvability is known only for the very restrictive case that the maximum occurring latency is one clock cycle [3].…”
Section: Introductionmentioning
confidence: 99%
“…By using induction and the properties of forward scheduling and backward scheduling, we can prove the fol- v [5,12] v [6,15] v [6,15] v [8,15] 6 v [6, 15] v [8,10] v [8,10] v [4,8] v [5,13] v [2,6] v [3,7] Proof Suppose that there exists a feasible schedule σ ′ , but a schedule σ computed by our algorithm is not feasible. Let v k be the first late instruction and t the earliest integer time point satisfying 1) there are m k (σ(v k ) − t) instructions scheduled in the time interval [t, σ(v k )) on m k functional unit of type R(v k ) in σ, where m k is the number of functional units of type R(v k ), and 2) for each instruction …”
Section: Compute Dmentioning
confidence: 99%
“…As a result, no feasible exists for the original problem instance P (v i ). v [3,14] v [3,12] v [3,15] v [5,15] v [5,15] v [5,12] v [6,15] v [6,15] v [5,14] v [8,15] v [3,10] 6 v [6, 15] v [8,10] v [8,10] v [4,9] v [2,6] Figure 2. The edge-consistent release times and deadlines in P Example 1 Consider a problem instance P with 14 instructions and an ILP processor with two heterogeneous functional units F 1 and F 2 .…”
mentioning
confidence: 99%
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