1Parallel Prefix Network AddersLecture 5Conditional-Sum AdderHybrid AddersCarry-Select Adders2One-level k-bit Carry-Select AdderOne-level k-bit Carry-Select AdderCost & Latency3Two-level k-bit Carry Select AdderConditional-Sum Adder4Top-level Block of Conditional-Sum AdderOperation of a 16-bit Conditional-Sum Adder5Conditional-Sum AdderHybrid Adders6A Hybrid Ripple-Carry/Carry-Lookahead AdderA Hybrid Carry-Lookahead/Carry-Select Adder7Parallel Prefix Network AddersParallel Prefix Network AddersBasic component - Carry operator (1)B” B’g”p”g’p’g pg = g” + g’p”p = p’p”(g, p) = (g’, p’) ¢ (g”, p”) = (g” + g’p”, p’p”)B8Parallel Prefix Network AddersBasic component - Carry operator (2)B” B’g”p”g’p’g pg = g” + g’p”p = p’p”(g, p) = (g’, p’) ¢ (g”, p”) = (g” + g’p”, p’p”)BProperties of the carry operator ¢AssociativeNot commutative[(g1, p1) ¢ (g2, p2)] ¢ (g3, p3) = (g1, p1) ¢ [(g2, p2) ¢ (g3, p3)] (g1, p1) ¢ (g2, p2) ≠≠≠≠ (g2, p2) ¢ (g1, p1)9Parallel Prefix Network AddersMajor conceptGiven:Find:(g0, p0) (g1, p1) (g2, p2) …. (gk-1, pk-1)(g[0,0], p[0,0]) (g[0,1], p[0,1]) (g[0,2], p[0,2]) … (g[0,k-1], p[0,k-1])ci= g[0,i-1]+ c0p[0,i-1]Parallel Prefix Network AddersSimilar problemParallel Prefix Sum ProblemGiven:Find:Parallel Prefix Adder ProblemGiven:Find:x0x0+x1x0+x1+x2… x0+x1+x2+ …+ xk-1x0x1x2… xk-1x0x0 ¢ x1x0 ¢ x1 ¢ x2… x0 ¢ x1 ¢ x2 ¢ … ¢ xk-1x0x1x2… xk-1where xi= (gi, pi)10Parallel Prefix Sums Network I Parallel Prefix Sums Network I Cost AnalysisCost = C(k) = 2 C(k/2) + k/2 == 2 [2C(k/4) + k/4] + k/2 = 4 C(k/4) + k/2 + k/2 == …. == 2 log k-1C(2) + k/2 (log2k-1) == k/2 log2k2Example:C(16) = 2 C(8) + 8 = 2[2 C(4) + 4] + 8 == 4 C(4) + 16 = 4 [2 C(2) + 2] + 16 == 8 C(2) + 24 = 8 + 24 = 32 = (16/2) log216C(2) = 111Parallel Prefix Sums Network I Delay AnalysisDelay = D(k) = D(k/2) + 1 == [D(k/4) + 1] + 1 = D(k/4) + 1 + 1 == …. == log2kExample:D(16) = D(8) + 1 = [D(4) + 1] + 1 == D(4) + 2 = [D(2) + 1] + 2 == 4 = log216D(2) = 1Parallel Prefix Sums Network II12Parallel Prefix Sums Network II Cost AnalysisCost = C(k) = C(k/2) + k-1 == [C(k/4) + k/2-1] + k-1 = C(k/4) + 3k/2 - 2 == …. == C(2) + (2k - 2k/2(log k-1)) - (log2k-1) == 2k - 2 - log2kExample:C(16) = C(8) + 16-1 = [C(4) + 8-1] + 16-1 == C(2) + 4-1 + 24-2 = 1 + 28 - 3 = 26= 2·16 - 2 - log216C(2) = 12Parallel Prefix Sums Network II Delay AnalysisDelay = D(k) = D(k/2) + 2 == [D(k/4) + 2] + 2 = D(k/4) + 2 + 2 == …. == 2 log2k - 1Example:D(16) = D(8) + 2 = [D(4) + 2] + 2 == D(4) + 4 = [D(2) + 2] + 4 == 7 = 2 log216 - 1D(2) = 113Brent-Kung parallel prefix graph for 16 inputsKogge-Stone parallel prefix graph for 16 inputs14Parallel Prefix Network AddersComparison of architecturesHybridNetwork 2Brent-KungKogge-StoneCost(k)Delay(k)Cost(16)Delay(16)Cost(32)Delay(32)2 log2k - 22k - 2 - log2klog2k+1log2kk/2 log2kk log2k - k + 16542632498655780129Latency-Cost Trade-off15A Hybrid Brent-Kung/Kogge-Stone parallel prefix graph for 16
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