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Chapter 4 Retiming

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**Definitions Retiming Purposes**

Retiming is a mapping from a given DFG, G to a retimed DFT, Gr such that the corresponding transfer function of G and Gr differ by a pure delay z-L. Purposes To facilitate pipelining to reduce clock cycle time To reduce number of registers needed. (C) by Yu Hen Hu

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**Cut-set Retiming Delay transfer theorem Feed-forward cut-set:**

Feed-back cut-set Delay transfer theorem Adding arbitrary non-negative number of delays to each edge of a feed-forward cut-set of a DFG will not alter its output, except the output timing will be delayed. Transfer the same amount of delays from edges of the same direction across a feed-back cut set of a DFG to all edges of opposing edges across the same cut set will not alter the output, but its timing. (C) by Yu Hen Hu

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**Feed-forward Cut-Set Retiming**

Consider the FIR digital filter and its DFG: y(n) = b0x(n) + b1x(n-1) Critical path length = TM+TA Select a cut set Insert a delay each to each edge in the cut set. Retiming: ynew(n) = b0x(n-1) + b1x(n-2) ynew(n) = y(n-1) Critical path = Max(TM, TA) D x(n) x(n-1) X b0 X b1 D x(n) x(n-1) + y(n) X b0 X b1 D D + y(n) (C) by Yu Hen Hu

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**Feed-back Cut Set Retiming**

Consider an IIR digital filter y(n) = a·y(n-2) + x(n) loop bound = (TM+TA)/2 clock cycle = TM+TA Shift 1 delay to the other edge across a feed-back cut set Filter remains unchanged. loop bound = (TM+TA)/2 clock cycle = Max(TM ,TA) x(n) y(n) x(n) y(n) + + 2D D D a a (C) by Yu Hen Hu

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Timing Diagram Assume tM = tA = 1 t.u. Before retiming After retiming x(1) x(2) x(3) x(4) MAC 1 2 3 4 y(1) y(2) y(3) y(4) x(1) x(2) x(3) x(4) x(5) x(6) x(7) x(7) Add 1 2 3 4 5 6 7 8 y(1) y(5) y(6) y(7) y(7) y(2) y(3) y(4) a y(1) Mul 1 2 3 4 5 6 7 8 (C) by Yu Hen Hu

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**Feed-back Cut Set Retiming**

Consider an IIR digital filter y(n) = ay(n-1) + x(n) loop bound = (TM+TA) throughput = 1/(TM+TA) x(2k-1)=x(k) x(2k) = 0 Clock period = (TM+TA) Throughput = 1/[2(TM+TA)] x(n) y(n) + x(m) y(m) + D 2D a a (C) by Yu Hen Hu

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**Slowdown + Retiming + + Start with y(n) = a y(n-1) + x(n)**

clock cycle = Max(TM ,TA) Throughput = 1/[2max(TM,TA)] Start with y(n) = a y(n-2) + x(n) loop bound = (TM+TA)/2 clock cycle = Max(TM ,TA) throughput = 1/ Max(TM ,TA) x(n) y(n) x(m) y(m) + + D D D D a a (C) by Yu Hen Hu

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**Example 3.2.1 Node delay = 1 t.u. Before retiming:**

Critical path: a3 a4 a5 a6 Clock cycle time = 4 2 delay units After cut-set retiming Critical path: a3 a5, a4 a6 Clock cycle time = 2 6 delay units After additional retiming Critical path: none Clock cycle time = 1 11 delay units a5 a3 D a1 a2 a3 a4 a5 a6 2D a4 a2 D D a6 2D a1 D D D 2D a3 a5 (C) by Yu Hen Hu

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**Slow Down for Cut-Set Retiming**

(C) by Yu Hen Hu

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**Node Retiming v v … Retiming equation: e v u**

Transfer delay through a node in DFG: r(v) = # of delays transferred from out-going edges to incoming edges of node v w(e) = # of delays on edge e wr(e) = # of delays on edge e after retiming Retiming equation: subject to wr(e) 0. Let p be a path from v0 to vk then e u v D 3D 2D r(v) = 2 v v 2D D 3D v0 e0 v1 e1 … vk ek p (C) by Yu Hen Hu

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Invariant Properties Retiming does NOT change the total number of delays for each cycle. Retiming does not change loop bound or iteration bound of the DFG If the retiming values of every node v in a DFG G are added to a constant integer j, the retimed graph Gr will not be affected. That is, the weights (# of delays) of the retimed graph will remain the same. (C) by Yu Hen Hu

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**Node Retiming Examples**

(C) by Yu Hen Hu

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**DFG Illustration of the Example**

T = max. {(1+2+1)/2, (1+2+1)/3} = 2 Cr. Path delay = 2+1 = 3 t.u T = max. {(1+2+1)/2, (1+2+1)/3} = 2 Cr. Path Delay = max{2,2,1+1} = 2 t.u (C) by Yu Hen Hu

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**Retiming for Minimizing Clock Period**

Note that retiming will NOT alter iteration bound T. Iteration bound is the theoretical minimum clock period to execute the algorithm. Let edge e connect node u to node v. If the node computing time t(u) + t(v) > T, then clock period T > T. For such an edge, we require that To generalize, for any path from v0 to vk, we have In other words, for any possible critical path in the DFG that is larger than T, we require wr(e) 1. (C) by Yu Hen Hu

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**Retiming Example Revisited**

wr(e21) 0, since t(2)+t(1) = 2 = T. wr(e13) 1, since t(1)+t(3) = 3 > T. wr(e14) 1, since t(1)+t(4) = 3 > T. wr(e32) 1, since t(3)+t(2) = 3 > T. wr(e42) 1, since t(4)+t(2) = 3 > T. Use eq. wr(euv) = w(e) + r(v) – r(u), w(e21) + r(1) – r(2) = 1 + r(1) – r(2) 0 w(e13) + r(3) – r(1) = 1 + r(3) – r(1) 1 w(e14) + r(4) – r(1) = 2 + r(4) – r(1) 1 w(e32) + r(2) – r(3) = 0 + r(2) – r(3) 1 w(e42) + r(2) – r(4) = 0 + r(2) – r(4) 1 (C) by Yu Hen Hu

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Solution continues Since the retimed graph Gr remain the same if all node retiming values are added by the same constant. We thus can set r(1) = 0. The inequalities become 1 – r(2) 0 or r(2) 1 1 + r(3) 1 or r(3) 0 2 + r(4) 1 or r(4) –1 r(2) – r(3) 1 or r(3) r(2) - 1 r(2) – r(4) 1 or r(2) r(4) + 1 Since one must have r(2) = +1. This implies r(3) 0. But we also have r(3) 0. Hence r(3)=0. These leave –1 r(4) 0. Hence the two sets of solutions are: r(0) = r(3) = 0, r(2) = +1, and r(4) = 0 or -1. (C) by Yu Hen Hu

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**Systematic Solutions Given a systems of inequalities:**

r(i) – r(j) k; 1 i,j N Construct a constraint graph: Map each r(i) to node i. Add a node N+1. For each inequality r(i) – r(j) k, draw an edge eji such that w(eji) = k. Draw N edges eN+1,i = 0. The system of inequalities has a solution if and only if the constraint graph contains no negative cycles If a solution exists, one solution is where ri is the minimum length path from the node N+1 to the node i. Shortest path algorithms: (Applendix A) Bellman-Ford algorithm Floyd-Warshall algorithm (C) by Yu Hen Hu

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**Bellman-Ford Algorithm**

Find shortest path from an arbitrarily chosen origin node U to each node in a directed graphif no negative cycle exists. Given a direct graph w(m,n): weight on edge from node m to node n, = if there is no edge from m to n r(i,j): the shortest path from node U to node i within j-1 steps. r(i,1) = w(U,i), r(i,j+1) = min {r(k,j) + w(k,i)}, j = 1, 2, …, N-1 if max(r(:,n-1)-r(:,n))>0, then there is a negative cycle. Else, r(i,n-1) gives shortest cycle length from i to U. -3 1 2 1 1 1 2 4 3 Note that 1 > 0, hence there is at least one negative cycle. spbf.m (C) by Yu Hen Hu

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**Floyd-Warshall Algorithm**

-3 Find shortest path between all possible pairs of nodes in the graph provided no negative cycle exists. Algorithm: Initialization: R(1) =W; For k=1 to N R(k+1)(u,v) = min{R(k)(u,:) + R(k)(:,v)} If R(k)(u,u) < 0 for any k, u, then a negative cycle exist. Else, R(N+1)(u,v) is SP from u to v 1 2 1 2 1 2 4 3 (C) by Yu Hen Hu

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**Retiming Example For retiming example:**

Bellman-Ford Algorithm for Shortest Path -1 3 1 1 2 1 -1 4 5 (C) by Yu Hen Hu

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Retiming Example Floyd-Warshall algorithm (C) by Yu Hen Hu

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**Retiming to Reduce Registers**

Delay reduction Register Sharing When a node has multiple fan-out with different number of delays, the registers can be shared so that only the branch with max. # of delays will be needed. Register reduction through node delay transfer from multiple input edges to output edges (e.g. r(v) > 0) Should be done only when clock cycle constraint (if any) is not violated. (C) by Yu Hen Hu

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**Time Scaling (Slow Down)**

… x(3) x(2) x(1) … y(3) y(2) y(1) Transform each delay element (register) D to ND and reduce the sample frequency by N fold will slow down the computation N times. During slow down, the processor clock cycle time remains unchanged. Only the sampling cycle time increased. Provides opportunity for retiming, and interleaving. + D … -- x(3) -- x(2) -- x(1) … y(3) -- y(2) -- y(1) + 2D (C) by Yu Hen Hu

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