20:20 Tc=tpd+(tsetup+tpcq) tpd = d(c11) and suppose d(c12)>d(c11) so dc11 + (tsetup+ tpcq)= Tc 11 dc12+ ("+")=Tc12 we will take Tc12 as final Tc. It can accomodate the slowest path which means it can accomodate Tc11 as well as
Continuing with your brick analogy, hold time could be understood as the time the second brick should not be placed on the pedestal before the old man picks up the first brick which was already present on the pedestal.