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Suppose you have a set $S = \{r_1, ..., r_n :\, r_k \in (1, \infty)\, \forall \,k \in \{1,...,n\}\}$. Find a bijective mapping $f: \{0,...,n-1\}\rightarrow \{1,...,n\}$ that minimizes \begin{align*} \sum_{k=0}^{n-1}r_{f(k)}^{k} \end{align*} Solution: Let $f$ be a mapping that satisfies $k< j \, \rightarrow \, r_{f(k)} \ge r_{f(j)}$. In other words, $f$ sorts $S$ in non-increasing order. To prove optimality, consider a bijective mapping $g$ defined on the same domain and range, and suppose $g$ minimizes the sum. Further, suppose $g$ differs from $f$. Therefore, there must be some inversion: an index $i$ such that $r_{g(i)} < r_{g(i+1)}$. We will show that swapping these two images decreases the total sum, contradicting the optimality of $g$. Notice that if we only swap these two, then the remaining terms of the sum are unchanged. To show that the swap decreases the total sum, we must verify \begin{align*} r_{g(i+1)}^{i}+r_{g(i)}^{i+1} &< r_{g(i)}^{i}+r_{g(i+1)}^{i+1}\\ r_{g(i)}^{i+1} -r_{g(i)}^{i}&< r_{g(i+1)}^{i+1}-r_{g(i+1)}^{i}&\iff \\ r_{g(i)}^i(r_{g(i)} - 1)&<r_{g(i+1)}^i(r_{g(i+1)}-1)&\iff \end{align*} But this final inequality holds because each rate exceeds one and by assumption $r_{g(i)} < r_{g(i+1)}$. This contradicts the optimality of $g$, hence any mapping that deviates from sorting the rates in descending order is not optimal.

Now suppose you have a set $S = \{r_1, ..., r_n :\, r_k \in (0, 1)\, \forall \,k \in \{1,...,n\}\}$. Find a bijective mapping $f: \{0,...,n-1\}\rightarrow \{1,...,n\}$ that maxmizes \begin{align*} \sum_{k=0}^{n-1}r_{f(k)}^{k} \end{align*} Note: It is interesting to point out that the naïve attempt of sorting $S$ in ascending order fails to maximize the sum in certain cases. Take for example the set \begin{align*} \biggr\{\frac{1}{10}, \frac{1}{2}, \frac{4}{7}, \frac{3}{5}, \frac{2}{3}\biggr\}&\\ \biggr(\frac{1}{10}\biggr)^0 + \biggr(\frac{1}{2}\biggr)^1 + \biggr(\frac{4}{7}\biggr)^2 + \biggr(\frac{3}{5}\biggr)^3 + \biggr(\frac{2}{3}\biggr)^4&\approx 2.240061476\\ \biggr(\frac{1}{10}\biggr)^0 + \biggr(\frac{4}{7}\biggr)^1 + \biggr(\frac{2}{3}\biggr)^2 + \biggr(\frac{3}{5}\biggr)^3 + \biggr(\frac{1}{2}\biggr)^4&\approx 2.294373016 \end{align*}

You can verify that this second sum is optimal in Mathematica, using the following code

S = {1/10, 1/2, 4/7, 3/5, 2/3};

p = Range[0, 4];

perm = Permutations[S];

totals = Total[#^p] & /@ perm;

maxPos = First[Flatten[Position[totals, Max[totals]]]]

perm[[maxPos]]

So my question is: can you think of an efficient rule to maximize the second sum? Currently my best solution is to literally enumerate all $n!$ possible ways to order the terms. Also, why is this problem so much harder? I feel it has something to do with the unboundness of the first problem. A colleague of mine suggested that the solution space of first problem is convex while the second is not, though I do not have a good grasp on why that is. (I know the definition of convexity, but what part of the problem formulation makes the solution space second problem not convex?)

Any help / input would be greatly appreciated.