One Multiply Scattering Junction

By factoring out $k_i$, we obtain

$$\begin{eqnarray*} f^{{+}}_i(t) &=& f^{{+}}_{i-1}(t-T) + f_{{\Delta}}(t) \nonumber \\ f^{{-}}_{i-1}(t+T) &=& f^{{-}}_i(t) + f_{{\Delta}}(t) \end{eqnarray*}$$

where

$$f_{{\Delta}}(t) \mathrel{\stackrel{\mathrm{\Delta}}{=}}k_i\left[f^{{+}}_{i-1}(t-T) - f^{{-}}_i(t) \right]$$

Only one multiplication and three additions necessary

Another one-multiply form:

$$\begin{eqnarray*} f^{{+}}_i(t) &=&f^{{-}}_i(t) + \alpha_i(t)\tilde{f_d}(t) \nonumber \\ f^{{-}}_{i-1}(t+T) &=& f^{{+}}_i(t) - \tilde{f_d}(t) \end{eqnarray*}$$

where

$$\begin{eqnarray*} \alpha_i(t) &\mathrel{\stackrel{\mathrm{\Delta}}{=}}& 1+k_i\no... ...stackrel{\mathrm{\Delta}}{=}}& f^{{+}}_{i-1}(t-T) - f^{{-}}_i(t) \end{eqnarray*}$$

Again one multiplication and three additions necessary