Recalling that by Foster’s first formula
$$\sum_{ e_{i} \in E} p_{i} \cdot x_{i} =(n-1), $$
and that
$$\frac{2k}{n \sqrt{K}} \le x_{i} \le\frac{K}{\sqrt{k}}, \quad\mbox{for all } 1 \le i \le m, $$
let us now consider the set
$$S= \Biggl\{ \mathbf{x} \in\mathbb{R}^{m}: \sum _{i=1}^{m} x_{i} p_{i} = (n-1), \frac{K}{\sqrt{k}} \ge x_{1} \ge x_{2} \ge\cdots\ge x_{m} \ge\frac{2k}{n \sqrt{K}} \Biggr\} . $$
The function
\(CW(G)=f(x_{i},p_{i})\) is
p-Schur-convex and thus its lower and upper bounds on
S are attained at the minimum and maximum element of
S with respect to the
p-majorization order, respectively.
From Remark
5 we know that the minimal element of
S is
\(\mathbf{x}_{\mathbf{\ast p}}(S)= [(\frac{n-1}{\sum_{i=1}^{m} p_{i}} )]^{m}\). Thus the lower bound is
$$f(\mathbf{x}_{\mathbf{\ast p}}, \mathbf{p})= \sum_{i=1}^{m} p_{i} \cdot \frac {\sum_{i=1}^{m} p_{i}}{n-1} - C = \frac{(C')^{2}}{n-1} -C. $$
The maximal element
\(\mathbf{x}^{\mathbf{\ast p}}\) of the set
S can be computed with Corollary
4, yielding
$$\mathbf{x}^{\mathbf{\ast p}} = \biggl[\underbrace{\frac{K}{\sqrt{k}}, \ldots, \frac{K}{\sqrt{k}}}_{l\mbox{-}\mathrm{times}}, \theta, \underbrace{\frac{2k}{n \sqrt{K}}, \ldots,\frac{2k}{n \sqrt{K}} }_{(m-l-1)\mbox{-}\mathrm{times}} \biggr], $$
where
l is the first integer such that
$$ \frac{K}{\sqrt{k}} \sum_{i=1}^{l} p_{i} + \frac{2k}{n \sqrt{K}} \sum_{i=l+1}^{m} p_{i} \le(n-1) < \frac{K}{\sqrt{k}} \sum_{i=1}^{l+1} p_{i} + \frac{2k}{n \sqrt{K}} \sum_{i=l+2}^{m} p_{i} $$
(14)
and
\(\theta= p_{l+1} \cdot (n-1 -\frac{K}{\sqrt{k}} \sum_{i=1}^{l} p_{i} - \frac{2k}{n \sqrt{K}} \sum_{i=l+2}^{m} p_{i} )\). Let
\(D=\sum_{i=1}^{l} p_{i} \) and
$$H=\frac{n-1- \frac{2k}{n \sqrt{K}} C'}{\frac{K}{\sqrt{k}} - \frac {2k}{n \sqrt{K}}}. $$
From (
14) easy computations show that
$$0 \le(H-D) < p_{l+1}. $$
Moreover,
$$f\bigl(\mathbf{x}^{\mathbf{\ast p}},\mathbf{p}\bigr)= \biggl( \frac{n \sqrt{K}}{{2k}} - \frac{\sqrt{k}}{K} \biggr) (H-D) + \frac{1}{ ( \frac{K}{\sqrt {k}} - \frac{{2k}}{n \sqrt{K}} )(H-D) + \frac{2k}{n \sqrt{K}} \cdot p_{l+1}} + T, $$
where
$$T= \biggl( \frac{n \sqrt{K}}{2k}+ \frac{\sqrt{k}}{K} \biggr) C' - \frac{n \sqrt{K}}{2k} \cdot p_{l+1} - \frac{n^{2}-n}{2 \sqrt{k} \sqrt {K}} -C. $$
Let
\(y=H-D\) and consider the function
$$h(y)= \biggl( \frac{n \sqrt{K}}{{2k}} - \frac{\sqrt{k}}{K} \biggr) y + \frac{1}{ ( \frac{K}{\sqrt{k}} - \frac{{2k}}{n \sqrt{K}} )y + \frac{2k}{n \sqrt{K}} \cdot p_{l+1}} + T. $$
The first derivative is
$$h^{\prime}(y)= - \frac{\frac{K}{\sqrt{k}} - \frac{{2k}}{n \sqrt {K}}}{ [ ( \frac{K}{\sqrt{k}} - \frac{{2k}}{n \sqrt{K}} )y + \frac{2k}{n \sqrt{K}} \cdot p_{l+1} ]^{2}} + \frac{n \sqrt{K}}{{2k}} - \frac{\sqrt{k}}{K} $$
and the only nonnegative stationary point is
$$\widehat{y} = \frac{ \sqrt{ \frac{2 \sqrt{kK}}{n} } - \frac{2k}{n \sqrt{K}} \cdot p_{l+1}}{\frac{K}{\sqrt{k}} - \frac{{2k}}{n \sqrt{K}}}. $$
Assuming, without loss of generality that
\(k \le1 \le K\), we can also be assured that
\(\widehat{y} < p_{l+1}\). The stationary point
\(\widehat {y}\) turns out to be a minimum. Thus the maximum value of the function
h is attained at the extremum of the interval
\([0, p_{l+1}]\). We have
$$\begin{aligned}& h(0)= \frac{1}{\frac{2k}{n \sqrt{K}} \cdot p_{l+1}} +T,\\& h ( p_{l+1} )= \frac{1}{\frac{K}{\sqrt{k}} \cdot p_{l+1} } + \biggl(\frac{n \sqrt{K}}{2k} - \frac{\sqrt{k}}{K} \biggr) \cdot p_{l+1} + T. \end{aligned}$$
We can get rid of
\(p_{l+1}\) by using the bounds on the resistances. We obtain
$$\begin{aligned}& h(0) \le\frac{nK}{2k} + T,\\& h ( p_{l+1} ) \le\frac{\sqrt{k}}{\sqrt{K}} + \frac{n \sqrt{K}}{2k\sqrt{k}} - \frac{1}{K} + T. \end{aligned}$$
The assertion easily follows from the bound
$$T \le \biggl( \frac{n \sqrt{K}}{2k} + \frac{\sqrt{k}}{K} \biggr) C' - \frac{n}{2k} - \frac{n^{2}-n}{2 \sqrt{k} \sqrt{K}} - C. $$
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