 Solution

An $A\to B$ hop can occur only while the charge occupies $A$. Hence its stationary number per unit time isThe reverse count is $R_{B\to A}=f_3{p}_B^{\mathrm{st}}$, which equals this result. There is therefore no signed stationary current along that edge. The individual hopping traffic need not equal the contact-arrival traffic; their ratio is $R_{A\to B}/R_{\mathrm{in}}=f_2/f_4$. A charge can make several internal visits before leaving, so neither conservation nor the single-charge constraint forces those two one-way rates to coincide.