Finding the cheapest buy order with fixed inflation for each product

Question

Let's say we have a set of products $M$, a total of $|M|=n$ that we want to buy. However, we can only buy one product at a time, so that we need a total of $n$ time-units to buy all items.

Each product $p\in M$ has a base price $b_p$, as well as an inflation rate $r_p$.
At time unit $t$, product $p$ therefore has the price $b_p \cdot {r_p}^t $.
We can assume $b_p\in \mathbb{R}$, $r_p>1$.

I'm looking for an efficient (i.e. polytime) algorithm that returns the optimal buying order which minimizes the total price.

There's an easier variant of the problem (all $b_p=1$) which can be solved using the greedy algorithm
"Always buy the one product with the highest inflation".

Given that the base prices can be vastly different, this algorithm can't be directly transferred, but knowing that an easier version of the problem had a linear-time solution, this one probably still isn't $\mathsf{NP}$-complete.

If we view every product as a function of time $p_i(t) = b_p \cdot {r_p}^t$, the greedy algorithm above tells us, that if for $t_0$ holds $p_i(t_0) = p_j(t_0)$, then from the point once $t_0$ has passed we should always pick of $p_i, p_j$ the one with the higher inflation.

I'd be open for both hints and solutions.

Answer 1

What is the problem in the question?

The price of product $p$ at time $t$ is $b_p{r_p}^t $, where $p$ and $t$ are integers between 1 and $n$.. We want to find a permutation $f$ of $1,2,\cdots,n$ such that $\sum_{p=1}^nb_p{r_p}^{f(p)}$ is minimum.

A more general problem and polynomial-time algorithms

The price of product $p$ at time $t$ is $c(p,t)$, where $p$ and $t$ are integers between 1 and $n$. We want to find a permutation $f$ of $1,2,\cdots,n$ such that $\sum_{p=1}^nc(p,f(p))$ is minimum.

The problem above is none other than the famous assignment problem. There are various polynomial-time algorithms to solve it. For example, this version of Hungarian algorithm runs in $O(n^4)$ time.

For the problem in the question, it can also be solved in polynomial time since we can compute all $b_p{r_p}^t$ for $1\le p, t\le n$ in $O(n^2)$ time.

Answer 2

some thoughts but haven't been mathematically proven. to solve the problems we need to decide the priority to buy each product

1. if $ p_i(t) = p_j(t) $ has crossing point at $ t_{ij} > 0 $, we have two cases:
2. when $ t < t_{ij} $ and $ p_i(t) > p_j(t) $, then we want to buy $ p_j $ sooner than $ p_i $
3. the opposite case of 1.

this can partially define the purchase order we want for those have crossing points at $ t > 0 $

2. to decide the purchase order of those don't have crossing point at $ t > 0 $, we can rely on their initial price $ p_i(0) $, e.g. if $ p_i(0) > p_j(0) $, we want to buy $ p_i $ before $ p_j $.

3. the above two steps define a partial order of all the products, to finalize an optimal purchase order, we can enumerate all the tsort order of the graph(nodes are products, partial order as edges) to find the optimal purchase order.

time complexity:

solving 1. and 2. takes $ O(n\log(n)) $ (solving each exponential equations takes $ O(\log(n)) $ using binary search)

solving 3. takes $ O(V+E) $ where $ V=n, E\le n^2 $