 Solution

Put $D=t_1+t_2$, $a=t_1{\rho }_{\mathrm{FM}}^L+t_2{\rho }_{\mathrm{NM}}$ and $b=t_1{\rho }_{\mathrm{FM}}^H+t_2{\rho }_{\mathrm{NM}}$. In the CIP thickness-averaged two-current model, assume spin is conserved and electron trajectories sample the layers, so each channel's effective resistivity is the thickness average: ${\bar{\rho }}_L=a/D$, ${\bar{\rho }}_H=b/D$. This uses averaged relaxation rates with comparable transport weights, neglecting additional interface resistance. Here ${\rho }_{\mathrm{NM}}$ is the per-spin-channel resistivity, twice the measured nonmagnetic bulk resistivity. The stated long mean free path motivates sampling more than one layer.

A four-layer period has thickness $2D$. For an in-plane square, $R_{\sigma }={\bar{\rho }}_{\sigma }L/(2DL)={\bar{\rho }}_{\sigma }/(2D)$, hence the two parallel spin conductances add:The square's side $L$ cancels. The layers are not being treated as four macroscopic resistors in series along the in-plane current; the averaging is over the traversing carriers' scattering histories.