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,66=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,66=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,66=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