Timing Boundary Depth Scans

Delay Repartition Three-Row Scan

Different stage splits close identically when their checked sum is the same.

Different stage splits close identically when their checked sum is the same. Exact arithmetic here means exact results for the stated model inputs; measured inputs still carry uncertainty and significant-figure limits.

highlighted = computed this step

One-two-three stages exactly spend the budget

Stage delays 1, 2, and 3 s total 6 s. The budget is 6 s, so slack is 0 s.

1+2+3=6 s,6−6=0 s1+2+3=6\ \text{s},\quad 6\mathord{-}6=0\ \text{s}
Delay repartition scanThe checked closure uses the sum, not stage position.period 8 sedges 3edge 1 at 8 slast 16 sd1 1 sd2 2 sd3 3 stotal 6 srequired at 14 sbudget 6 sslack 0 saccepted

Evenly split stages spend the same budget

Stage delays 2, 2, and 2 s total 6 s. The budget is 6 s, so slack is 0 s.

2+2+2=6 s,6−6=0 s2+2+2=6\ \text{s},\quad 6\mathord{-}6=0\ \text{s}
Delay repartition scanThe checked closure uses the sum, not stage position.period 8 sedges 3edge 1 at 8 slast 16 sd1 2 sd2 2 sd3 2 stotal 6 srequired at 14 sbudget 6 sslack 0 saccepted

A late long stage still closes when the sum is six

Stage delays 1, 1, and 4 s total 6 s. The budget is 6 s, so slack is 0 s.

1+1+4=6 s,6−6=0 s1+1+4=6\ \text{s},\quad 6\mathord{-}6=0\ \text{s}
Delay repartition scanThe checked closure uses the sum, not stage position.period 8 sedges 3edge 1 at 8 slast 16 sd1 1 sd2 1 sd3 4 stotal 6 srequired at 14 sbudget 6 sslack 0 saccepted