Solving the probabilistic profitable tour problem on a line

Abstract

Among the most important variants of the traveling salesman problem (TSP) are those relaxing the constraint that every locus should necessarily get visited, rather taking into account the revenues (prizes) from the visits into the objective function. In the Profitable Tour Problem (PTP) we seek for a tour visiting a suitable subset of customers with the target to maximize the net gain (profit) defined as the difference between the total revenue collected from the visited customers and the incurred traveling costs. The metric TSP can be modeled as a PTP with large revenues. As such, PTP is well-known to be NP-hard and also APX-hardness follows. Nevertheless, PTP is solvable in polynomial time on particular graph structures like lines, trees and circles. In line with the recent emphasis on robust optimization, and motivated by the current flourishing of retailed delivery services, in this paper we initiate the study of the Probabilistic Profitable Tour Problem (PPTP), the probabilistic generalization of PTP in which the customers will show up with a known probability, in their respective loci, only after the tour has been designed and committed to. Here, the selection of the customers has to be made a priori, before knowing if a customer will actually submit his request or not. While the tour has to be designed and committed to without this knowledge, the revenues will be collected only from customers who will require the service. The objective is to maximize the expected net gain obtained visiting only the customers that show up. We offer a polynomial time algorithm computing and characterizing the space of optimal solutions for the special case of the PPTP where customers are distributed on a line.

Appendix

Proof of Proposition 1

• Proof of formula (6)

  $$\begin{aligned} R(S)=\sum _{v_{i}\in S}\pi _{i}p_{i} =\left( \sum _{v_{i}\in S^{\prime }}\pi _{i}p_{i}\right) + \pi ^{S}p^{S} = R(S^{\prime })+\pi ^{S}p^{S}; \end{aligned}$$

• Proof of formula (7)

  $$\begin{aligned} L(S)&= \sum _{v_{i}\in S}\left[ l_{i}\pi _{i}\cdot {\displaystyle \prod _{{v_j}\in S,\ j>i}}(1-\pi _{j})\right] \\&= \sum _{v_{i}\in S^{\prime }}\left[ l_{i}\pi _{i}\cdot {\displaystyle \prod _{{v_j}\in S,\ j>i}}(1-\pi _{j})\right] + l^{S}\pi ^{S}\cdot 1\\&= (1-\pi ^{S})\cdot \sum _{v_{i}\in S^{\prime }}\left[ l_{i}\pi _{i}\cdot {\displaystyle \prod _{{v_j}\in S^{\prime },\ j>i}}(1-\pi _{j})\right] + l^{S}\pi ^{S}\cdot 1\\&= (1-\pi ^{S})\cdot L(S^{\prime })+l^{S}\pi ^{S}; \end{aligned}$$

• Proof of formula (8)

  $$\begin{aligned} C(S,c)&=cL(S)\\&=c(1-\pi ^{S})L(S^{\prime })+\pi ^{S}l^{S}c\\&=C(S^{\prime },c(1-\pi ^{S}))+\pi ^{S}l^{S}c; \end{aligned}$$

• Proof of formula (9)

  $$\begin{aligned} G(S,c)&=R(S)-C(S,c)\\&=R(S^{\prime })+\pi ^{S}p^{S}-(C(S^{\prime },c(1-\pi ^{S}))+\pi ^{S}l^{S}c)\\&=\left[ R(S^{\prime })-(1-\pi ^{S})cL(S^{\prime })\right] +\pi ^{S} (p^{S}-l^{S}c)\\&=G(S^{\prime },(1-\pi ^{S})c)+\pi ^{S}(p^{S}-l^{S}c). \end{aligned}$$

$\square$

Proof of Proposition 2

Let $S_1$ and $S_2$ be two set of customers such that $S_1 \subseteq S_2$. we show that $L(S_1)\le L(S_2)$ by induction on cardinality of $S_2$ using equality (7).

Base cases. If $|S_2|=0$, both $S_1$ and $S_2$ are empty and we get $L(S_1)=0\le L(S_2)=0$. If $|S_2|=1$, then either $S_1=\emptyset$ or $S_1=S_2$; and we get $L(S_1)=0\le L(S_2);$ and $L(S_1)=L(S_2)$, respectively.

Induction hypothesis. Let us assume that the property holds for all $|S_2|\le n$ and show that it hold also for $|S_2|=n+1$.

Induction step. Let $n>1$ be the cardinality of $S_2$. Then either $v^{S_2}\in S_1$ or $v^{S_2} \notin S_1$.

If $v^{S_2}\in S_1$, we have $v^{S_2}=v^{S_1}=v^{S}$, and $L(S_1)=(1-\pi ^{S}\cdot L(S_1^{\prime })+ l^{S}\pi ^{S}\le L(S_2^{\prime })+ l^{S}\pi ^{S}=L(S_2)$; the last inequality comes from induction hypothesis because $|S_2^{\prime }|=|S_2|-1=n$.

If $v^{S_2} \notin S_1$, we have $S_1\subseteq S_2$ and $L(S_1)\le L(S_2^{\prime })\le L(S_2^{\prime })+l^{S_2}\pi ^{S_2}=L(S_2)$; the first inequality comes from induction hypothesis because $|S_2^{\prime }|=|S_2|-1=n$. $\square$

Proof of Proposition 3

• Proof of formula (10) Equality follows straightforwardly from the following decompositions:

  $$\begin{aligned} R(S_1)&=\sum _{v_{i}\in S_1\backslash S_2}\pi _{i}p_{i}+\sum _{v_{i}\in S_1\cap S_2} \pi _{i}p_{i},\\ R(S_2)&=\sum _{v_{i}\in S_2\backslash S_1}\pi _{i}p_{i}+\sum _{v_{i}\in S_1\cap S_2} \pi _{i}p_{i},\\ R(S_1\cup S_2)&=\sum _{v_{i}\in S_1\backslash S_2}\pi _{i}p_{i}+\sum _{v_{i}\in S_2\backslash S_1}\pi _{i}p_{i}+\sum _{v_{i}\in S_1\cap S_2}\pi _{i}p_{i},\\ R(S_1\cap S_2)&=\sum _{v_{i}\in S_1\cap S_2}\pi _{i}p_{i}; \end{aligned}$$

• Proof of formula (11) We first discuss two special cases.

  • (a) When $S_2\subseteq S_1$ (or $S_1\subseteq S_2$) the property is trivially true since $S_1\cup S_2=S_1$ and $S_1\cap S_2=S_2$ (or $S_1\cup S_2=S_2$ and $S_1\cap S_2=S_1$) and equality boils down to an identity.

  • (b) When $S_1\cap S_2=\emptyset$ the property boils down to the inequality $L(S_1\cup S_2)\le L(S_1)+L(S_2)$ which comes from the comparison of the following expressions:

    $$\begin{aligned} L(S_1)&=\sum _{v_{i}\in S_1}\pi _{i}l_{i} {\displaystyle \prod _{{j}\in S_1,\ j>i}} (1-\pi _{j}),\\ L(S_2)&=\sum _{v_{i}\in S_2}\pi _{i}l_{i} {\displaystyle \prod _{{j}\in S_2,\ j>i}} (1-\pi _{j})\\ L(S_1\cup S_2)&=\sum _{v_{i}\in S_1}\pi _{i}l_{i} {\displaystyle \prod _{{j}\in S_1\cup S_2,\ j>i}} (1-\pi _{j})+\sum _{v_{i}\in S_2}\pi _{i}l_{i} {\displaystyle \prod _{{j}\in S_1\cup S_2,\ j>i}} (1-\pi _{j}). \end{aligned}$$

    where inequalities

    $$\begin{aligned} \sum _{v_{i}\in S_1}\pi _{i}l_{i}{\displaystyle \prod _{{j}\in S_1\cup S_2,\ j>i}}(1-\pi _{j})&\le \sum _{v_{i}\in S_1}\pi _{i}l_{i}{\displaystyle \prod _{{j}\in S_1,\ j>i}}(1-\pi _{j}),\\ \sum _{v_{i}\in S_2}\pi _{i}l_{i}{\displaystyle \prod _{{j}\in S_1\cup S_2,\ j>i}}(1-\pi _{j})&\le \sum _{v_{i}\in S_2}\pi _{i}l_{i}{\displaystyle \prod _{{j}\in S_2,\ j>i}}(1-\pi _{j}) \end{aligned}$$

    hold because each term in the left summations has more factor less than 1.

We conduct the proof by induction on the cardinality of $S=S_1\cup S_2$ and make use of notation of equation (7) which we recall here for reader’s convenience

$$\begin{aligned} L(S)=\left( 1-\pi ^{S}\right) L\left( S^{\prime }\right) +\pi ^{S} l^{S}; \end{aligned}$$

where we indicate as $v^{S}$ the farthest customer of a given set S (with location $l^{S},$ profit $p^{S}$ and probability $\pi ^{S}$), and define $S^{\prime }=S\backslash \{v^{S}\}$.

Base cases. If $|S|\le 2$ we have (w.l.o.g.) $S_2\subseteq S_1$ and/or $S_1\cap S_2=\emptyset$. Thus, we fall in one of cases a,b discussed above.

Induction hypothesis. We assume that the property holds for $|S|=|S_1\cup S_2|\le n$ and show that it holds also for $|S|=n+1$.

Induction step. Given $|S_1\cup S_2|=n+1$, we assume w.l.o.g. that $v^{S}\in S_1$. The following two cases may occur:

• (a) $v^{S}\in S_2$ (thus $v^{S}\in S_1\cap S_2$). From (7), we get:

  $$\begin{aligned} L(S_1\cup S_2)&=\pi ^{S} l^{S}+\left( 1-\pi ^{S}\right) L\left( S_1^{\prime }\cup S_2^{\prime }\right) \\ L(S_1\cap S_2)&=\pi ^{S} l^{S}+\left( 1-\pi ^{S}\right) L\left( S_1^{\prime }\cap S_2^{\prime }\right) \\ L(S_1)&=\pi ^{S} l^{S}+\left( 1-\pi ^{S}\right) L\left( S_1^{\prime }\right) \\ L(S_2)&=\pi ^{S} l^{S}+\left( 1-\pi ^{S}\right) L\left( S_2^{\prime }\right) \end{aligned}$$

  and thus $L(S_1\cup S_2)+L(S_1\cap S_2)=2\pi ^{S} l^{S}+(1-\pi ^{S})\left[ L(S_1^{\prime }\cup S_2^{\prime })+L(S_1^{\prime }\cap S_2^{\prime })\right]$, where $|S_1^{\prime }\cup S_2^{\prime }|=n.$ Then, by induction hypothesis:

  $$\begin{aligned} L(S_1\cup S_2)+L(S_1\cap S_2)&\le 2\pi ^{S} l^{S}+\left( 1-\pi ^{S}\right) \left[ L(S_1^{\prime })+L(S_2^{\prime })\right] \\&\le \left( \pi ^{S} l^{S}+\left( 1-\pi ^{S}\right) L(S_1^{\prime })\right) +\left( \pi ^{S} l^{S}+\left( 1-\pi ^{S} \right) L(S_2^{\prime })\right) \\&= L(S_1)+L(S_2). \end{aligned}$$

• (b) $v^{S}\notin S_2$, thus

  $$\begin{aligned} v^{S}\notin S_1 \cap S_2 \ \text {with } S^{\prime }=S_1^{\prime }\cup S_2 \ \text {and } |S^{\prime }| = n. \end{aligned}$$

  From (7), we get $L(S_1) =\pi ^{S} l^{S}+(1-\pi ^{S})L(S_1^{\prime })$, and then

  $$\begin{aligned} L(S_1^{\prime })=L(S_1)+\pi ^{S}\left( L(S_1^{\prime })-l^{S}\right) . \end{aligned}$$

  By using the induction hypothesis ($|S_1^{\prime }\cup S_2|=n$) we get

  $$\begin{aligned} L(S_1^{\prime }\cup S_2)+L(S_1^{\prime }\cap S_2)&\le L(S_1^{\prime })+L(S_2) = L(S_1)+\pi ^{S}\left( L(S_1^{\prime })-l^{S}\right) +L(S_2) \end{aligned}$$

  Now, the final result is obtained by substituting back and recalling that $L(S_1\cap S_2)=L(S_1^{\prime }\cap S_2)$ (from $v^{S}\notin S_2$) and $L(S_1^{\prime })\le L(S_1^{\prime }\cup S_2)$ (from Proposition 2):

  $$\begin{aligned} L(S_1\cup S_2)+L(S_1\cap S_2)&=\pi ^{S} l^{S}+(1-\pi ^{S})L(S_1^{\prime }\cup S_2)+L(S_1^{\prime }\cap S_2)\\&=\pi ^{S} l^{S}-\pi ^{S} L(S_1^{\prime }\cup S_2)+L(S_1^{\prime } \cup S_2)+L(S_1^{\prime }\cap S_2)\\&\le \pi ^{S} l^{S}-\pi ^{S} L(S_1^{\prime }\cup S_2)+L(S_1^{\prime })+L(S_2)\\&=\pi ^{S} l^{S}-\pi ^{S} L(S_1^{\prime }\cup S_2)+L(S_1) +\pi ^{S}\left( L(S_1^{\prime })-l^{S}\right) +L(S_2)\\&=\pi ^{S}\left( l^{S}-L(S_1^{\prime }\cup S_2)+L(S_1^{\prime }) -l^{S}\right) +L(S_1)+L(S_2)\\&=\pi ^{S}\left( L(S_1^{\prime })-L(S_1^{\prime }\cup S_2)\right) +L(S_1)+L(S_2)\\&\le L(S_1)+L(S_2). \end{aligned}$$

The same considerations hold for the case $v\in S_2\backslash S_1$$\square$.