A computed load power is not accepted as regulated when headroom fails. Exact arithmetic here means exact results for the stated model inputs; measured inputs still carry uncertainty and significant-figure limits.

highlighted = computed this step

Power arithmetic is still gated by headroom

The load arithmetic asks for 10 W, but the input minimum must also clear the required input.

Vin,minVout+VhV_{\text{in,min}}\ge V_{\text{out}}+V_h
Failed headroom setupThe power claim is present, but the gate is still checked.Vin 5 VVout 5 VVh 1 VVreq 6 Vgate fail

The same load claim fails until input reaches the boundary

The load power row stays the same; only the input minimum decides whether the regulated claim may proceed.

Vin,minVrequiredPloadgate5 V6 V10 Wfail6 V6 V10 Wpass7 V6 V10 Wpass\begin{array}{c|c|c|c}V_{\text{in,min}}&V_{\text{required}}&P_{\text{load}}&\text{gate}\\5\ \text{V}&6\ \text{V}&10\ \text{W}&\text{fail}\\6\ \text{V}&6\ \text{V}&10\ \text{W}&\text{pass}\\7\ \text{V}&6\ \text{V}&10\ \text{W}&\text{pass}\\\end{array}

Computed load power is blocked by failed headroom

The arithmetic load power still exists, but the regulated supply claim is rejected because headroom fails.

5 V2 A=10 Wwhile5 V<6 V5\ \text{V}\cdot2\ \text{A}=10\ \text{W}\quad\text{while}\quad5\ \text{V}<6\ \text{V}
Failed headroom blocks acceptancePower arithmetic does not override the headroom gate.Vin 5 VVout 5 VVh 1 VVreq 6 Vgate fail