\\
    0, \,\,  &\text{if}\,\,  j \leq i 
    \end{cases}\label{eqn:power_gamma_relation}$$ where $$\begin{aligned}
    \gamma_{1}&=&\frac{P_{T}×}{1+g_{0,1}\sum_{k=2}^{N+1} \frac{(g_{0,k-1}-g_{0,k})}{g_{0,k}g_{0,k-1}\sum_{l=0}^{k-1}\frac{1}{g_{0,l}×}×}  ×} \label{eqn:gamma_one}\\
     \gamma_{j}&=&\frac{g_{0,1}\frac{(g_{0,j-1}-g_{0,j})}{g_{0,j}g_{0,j-1}\sum_{l=0}^{j-1}\frac{1}{g_{0,l}×}×}×}{1+g_{0,1}\sum_{k=2}^{N+1}
      \frac{(g_{0,k-1}-g_{0,k})}{g_{0,k}g_{0,k-1}\sum_{l=0}^{k-1}\frac{1}{g_{0,l}×}×}  ×} P_{T} \,\,\,\, \forall \, j \geq 2 \label{eqn:gamma_k}\nonumber\end{aligned}$$

2.  The achievable rate optimized over the power allocation for a given placement of nodes is given by: $$R^{opt}_{P_T}(y_1,y_2,\cdots,y_N)=C \bigg( \frac{\frac{P_{T}}{\sigma^{2}×}}
    {\frac{1}{g_{0,1}×}+\sum_{k=2}^{N+1}
     \frac{(g_{0,k-1}-g_{0,k})}{g_{0,k}g_{0,k-1}\sum_{l=0}^{k-1}\frac{1}{g_{0,l}×}×}  ×} \bigg)\label{eqn:capacity_multirelay}$$

The basic idea is to choose the power levels (i.e., $P_{i,j},\, 0 \leq i<j \leq N+1$) in (\[eqn:achievable\_rate\_multirelay\]) so that all the terms in the $\min\{\cdot\}$ in (\[eqn:achievable\_rate\_multirelay\]) become equal. We provide explicit expressions for $P_{i,j},\,0 \leq i<j \leq N+1$ and the achievable rate (optimized over power allocation) in terms of the power gains. See Appendix \[appendix:proof\_of\_multirelay\_channel\_capacity\_theorem\_after\_power\_allocation\] for the detailed proof. A result on the equality of certain terms under optimal power allocation has also been proved in [@reznik-etal04degraded-gaussian-multirelay-channel] for the coding scheme used in [@reznik-etal04degraded-gaussian-multirelay-channel]. But it was proved in the context of a degraded Gaussian multi-relay channel, and the proof depends on an inductive argument, whereas our proof utilizes LP (linear programming) duality.

Recalling the exponential path-loss parameter $\rho$, and the source-sink distance $L$, let us define $\lambda := \rho L$, which can be treated as a measure of attenuation in the line.

Let us now comment on 