 Solution

Divide the primordial tensor power spectrum by the slow-roll curvature power spectrum:Differentiating the scalar-field Friedmann equation and using the inflaton equation gives $\dot{H}=-{\dot{\varphi }}^2/(2M_{\mathrm{pl}}^2)$. Hence $r=16{ϵ}_H\simeq 16ϵ$. For the quadratic model at 50 remaining e-folds,This is above the supplied upper bound, so the model fails the stated tensor constraint, despite its satisfactory scalar tilt and adjustable scalar amplitude. Calibrating $m$ does not reduce $r$, since the coupling cancels from the ratio.