Advanced Interconnect Optimizations

Timing Driven Buffering Problem Formulation

• Given– A Steiner tree– RAT at each sink– A buffer type– RC parameters– Candidate buffer locations

• Find buffer insertion solution such that the slack at the driver is maximized

Candidate Buffering Solutions

Candidate Solution Characteristics

• Each candidate solution is associated with– vi: a node

– ci: downstream capacitance

– qi: RAT

vi is a sinkci is sink capacitance

v is an internal node

Van Ginneken’s Algorithm

Candidate solutions are propagated toward the source

Dynamic Programming

Solution Propagation: Add Wire

• c2 = c1 + cx

• q2 = q1 – rcx2/2 – rxc1

• r: wire resistance per unit length

• c: wire capacitance per unit length

(v1, c1, q1)(v2, c2, q2)x

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Solution Propagation: Insert Buffer

• c1b = Cb

• q1b = q1 – Rbc1

• Cb: buffer input capacitance

• Rb: buffer output resistance

(v1, c1, q1)(v1, c1b, q1b)

Solution Propagation: Merge

• cmerge = cl + cr

• qmerge = min(ql , qr)

(v, cl , ql) (v, cr , qr)

Solution Propagation: Add Driver

• q0d = q0 – Rdc0 = slackmin

• Rd: driver resistance

• Pick solution with max slackmin

(v0, c0, q0)(v0, c0d, q0d)

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Example of Merging

Left candidates

Right candidates

Merged candidates

Solution Pruning

• Two candidate solutions– (v, c1, q1)

– (v, c2, q2)

• Solution 1 is inferior if – c1 > c2 : larger load

– and q1 < q2 : tighter timing

Pruning When Insert Buffer

They have the same load cap Cb, only the one with max q is kept

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Generating Candidates

(1)

(2)

(3)

From Dr. Charles Alpert

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Pruning Candidates

(3)

(a) (b)

Both (a) and (b) “look” the same to the source.Throw out the one with the worst slack

(4)

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Candidate Example Continued

(4)

(5)

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Candidate Example ContinuedAfter pruning

(5)

At driver, compute which candidate maximizesslack. Result is optimal.

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Merging Branches

Right Candidates

Left Candidates

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Pruning Merged Branches

Critical

With pruning

Basic Data Structure

(c1, q1) (c2, q2) (c3, q3)

Sorted list such that

• c1 < c2 < c3

• If there is no inferior candidates q1 < q2 < q3

Worse load cap

Better timing

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Prune Solution List

(c1, q1) (c2, q2) (c3, q3)

Increasing c

q1 < q2 ?

(c4, q4)

q3 < q4 ?

Y

NPrune 2 q1 < q3 ?

q2 < q3 ?

Y

q3 < q4 ?

Y

Prune 3 q1 < q4 ?

N Prune 3

N

N Prune 4N Prune 4

q2 < q4 ?

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Pruning In Merging

(cl1, ql1)

(cl2, ql2)

(cl3, ql3)

(cr1, qr1)

(cr2, qr2)

ql1 < ql2 < qr1 < ql3 < qr2

Merged candidate

s

(cl1+cr1, ql1)

(cl2+cr1, ql2)

(cl3+cr1, qr1)

(cl3+cr2, ql3)

(cl1, ql1)

(cl2, ql2)

(cl3, ql3)

(cr1, qr1)

(cr2, qr2)

(cl1, ql1)

(cl2, ql2)

(cl3, ql3)

(cr1, qr1)

(cr2, qr2)

(cl1, ql1)

(cl2, ql2)

(cl3, ql3)

(cr1, qr1)

(cr2, qr2)

Left candidate

s

Right candidate

s

Van Ginneken Complexity

• Generate candidates from sinks to source

• Quadratic runtime

– Adding a wire does not change #candidates

– Adding a buffer adds only one new candidate

– Merging branches additive, not multiplicative

– Linear time solution list pruning