Every wall you have stacked so far pretended the insulation covered all of it. It does not. Wood studs, plates and headers take a fifth to a quarter of a typical wall's area, and a stud is a far better conductor than the batt beside it. Heat, being opportunistic, takes the easy path — so the wall performs well below its label. That gap is the framing penalty, and the whole-wall number after it is counted is the effective R-value.

These two paths are parallel, not in series: a given watt goes through the cavity OR through the stud, never both. So they combine as conductances, area-weighted: $\dfrac{1}{R_{eff}} = \dfrac{f_{fr}}{R_{fr}} + \dfrac{1 - f_{fr}}{R_{cav}}$ — read aloud one over R-effective equals f over R-framing plus one minus f over R-cavity.

The subscripts carry the whole idea, so fix them now. $R_{cav}$ is the R-value of the cavity path — straight through the insulated bay. $R_{fr}$ is the R-value of the framing path — straight through the stud. $f_{fr}$ is the framing factor: the fraction of the wall's face that is wood, a bare number between 0 and 1 with no units, typically 0.20 to 0.25. Whatever is left, $1 - f_{fr}$, is the cavity's share. $R_{eff}$ is what comes out: the effective whole-wall R-value in m²·K/W.

Here is the mistake, and it is committed in real submittals: area-weighting the R-VALUES. Take RSI 3.5 cavity, RSI 1.0 studs, 25% framing. Weighted R-values give $0.25(1.0) + 0.75(3.5) = 2.88$. Weighted conductances give $\dfrac{1}{0.25/1.0 + 0.75/3.5} = 2.15$. The first number is a third too generous, and it is generous in the direction that fails a blower-door test. Average what FLOWS, never what resists — and note that the honest answer always comes out lower, because heat finds the easy path whether or not the spreadsheet does.