Risky traveling salesman problem

European Journal of Operational Research 212 (2011) 69–73

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European Journal of Operational Research journal homepage: www.elsevier.com/locate/ejor

Production, Manufacturing and Logistics

Risky traveling salesman problem Nikolaos Papadakos a,⇑, George Tzallas-Regas a, Berç Rustem a, Joanne Thoms b a b

Imperial College London, Department of Computing, 180 Queen’s Gate, London SW7 2AZ, UK BAE Systems Integrated System Technologies Limited, Lyon Way, Frimley, Camberley, Surrey GU16 7EX, UK

a r t i c l e

i n f o

Article history: Received 15 December 2009 Accepted 15 January 2011 Available online 21 January 2011

a b s t r a c t In this paper we introduce a methodology for optimizing the expected cost of routing a single vehicle which has a probability of breaking down or failing to complete some of its tasks. More specifically, a calculus is devised for finding the optimal order in which each site should be visited. Ó 2011 Elsevier B.V. All rights reserved.

Keywords: Uncertainty Travelling salesman problem Stochastic vehicle routing

1. Introduction The traditional traveling salesman problem (TSP) is a combinatorial optimization problem where a single vehicle is used to satisfy the demand of a number of customers, while minimizing the operating cost. Stochastic modeling has been introduced in traditional TSP by several authors over the past years [6,12,13], where uncertainties are attributed in the presence of customers [1,2,8,9], the demand level [3], the travel time [10,11], and the service time at customers’ site [7]. From the cases considered so far it is not possible to model military missions where the vehicle used has the risk of being struck by the opponent, and may therefore be unable to continue with the rest of their tasks. Furthermore one should also consider that the vehicle routed has also a probability to fail to strike the opponent. In such cases one needs to consider if it is worth spending extra time to re-engage the opponent or not. In this paper we develop a calculus that finds the optimal order in which we need to engage each opponent, thereby simplifying the solution methodology. The usability of the model described above is obviously not restricted to military purposes. One may use it for routing a vehicle while considering the risk of it breaking-down or failing to deliver. Undoubtedly such a risk is higher in military operations and we will therefore assume this specific application throughout the paper. We also change the nomenclature in order to stress the focus on the risk of a routing. Therefore, instead of vehicle, customers, routings, and serving a customer we will talk respectively about asset, opponents, mission, and engaging an opponent. ⇑ Corresponding author. E-mail addresses: [email protected] (N. Papadakos), gt299@ doc.ic.ac.uk (G. Tzallas-Regas), [email protected] (B. Rustem), jo.thoms@baesystems. com (J. Thoms). 0377-2217/$ - see front matter Ó 2011 Elsevier B.V. All rights reserved. doi:10.1016/j.ejor.2011.01.025

This paper introduces a calculus that can be used to compute the order by which the opponents should be engaged to obtain the optimal expected cost. We focus on the objective function as this is the only aspect influenced by the introduced uncertainties. Our methodology is therefore left open for any specific application, any constraints, and any additional uncertainties one may wish to model. In Section 2 we define our problem and in Section 3 we develop the mission optimization calculus. Then in Section 4 we explain how that calculus is used to find optimal solutions, and give one example of a model where the calculus can be used. Finally, conclusions are drawn in Section 5. 2. Problem definition 2.1. Engagements and missions An engagement is the confrontation of an asset against an opponent. If the opponent is not struck, then re-engagements are possible. Opponents will be denoted by D, which symbol will also represent engagements with that opponent, and one will write Dk when an asset is scheduled to engage D at most k-times consecutively, in case D was not incapacitated after the first k 1 attempts. The opponents’ objective is usually to strike a center the asset is defending. Instead of using the terms routing and tour the term mission is used here to emphasize its risky nature. A mission is a list of ordered engagements that a vehicle has to carry. As an example for mission M we write M :¼ D1 D23 D1 D4 , declaring that the asset is scheduled to engage first opponent D1, then opponent D3 twice (if it was not incapacitated the first time), after that opponent D1 once more (if it was not incapacitated the first time), and finally opponent D4. The set of engagements contained in a mission can

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N. Papadakos et al. / European Journal of Operational Research 212 (2011) 69–73

be corresponded to that mission, for instance for the previously mentioned example M ! fD1 ; D3 ; D4 g. When set theoretic notation is used for a mission it is implied that this acts on the corresponding set of engagements. Of course given a mission with re-engagements, the asset will only re-engage the opponents if they have not been previously incapacitated. A mission might have several constraints, either geographic restrictions or resource consumption. Resource consumption could be for example the usage of fuel or the number of available missiles. The set of all feasible missions will be denoted as M. Additionally throughout this paper missions will be denoted by K; L; M; N . Given two missions M1 and M2 , a new mission is defined as M :¼ M1 M2 , if and only if M 2 M, and where for this combined mission M the asset will first go through the engagements of M1 and then through those of M2 . 2.2. Risks and expected cost All engagements are considered, in the generic case, to be of a stochastic nature, as there is a risk of being incapacitated or failing to incapacitate the opponent. Therefore the mission expected cost or simply the expected cost is defined as the expected cost of all possible events in the given mission. The objective is to find an optimal mission for which the expected cost of damages is optimal. If it is a defensive mission one maximizes the expected damages avoided, and if it is an offensive mission one maximizes the damages incurred on the opponent. In order to develop a mission calculus the expected cost of several useful missions is computed hereafter. In all these computations p will be the probability of the asset surviving the engagement, and q the probability of incapacitating the opponent. Additionally, we assume that the opponent, upon succeeding his hostile mission, is able to incur a cost cD on us by hitting the defence center. Likewise the opponent is able to destroy one of the assets with cost c (cost of being incapacitated). Therefore cD and c will need to have opposite signs. Furthermore, it is quite convenient to use a different symbol for the probability of surviving while failing to incapacitate the opponent, which is defined as,

^ :¼ pð1 qÞ: p

ð1Þ

Finally, for all the previously mentioned probabilities and costs, subscripts will be used when one needs to indicate to which engagement or mission they refer. All the above probabilities and costs are considered known and one needs to find the optimal sequence of engagements as well as the number of engagements per opponent, that minimizes the expected cost. For a single opponent D the expected cost of a single engagement is:

EðDÞ ¼ ð1 pÞð1 qÞc þ ð1 pÞqðc þ cD Þ þ pð1 qÞ0 þ pqcD ¼ ð1 pÞc þ qcD :

ð2Þ ð3Þ

In equality (2) all four possible outcomes were considered, respectively the asset being incapacitated while the opponent escapes, the asset being incapacitated while incapacitating the opponent, the asset surviving while the opponent escapes, and finally the asset surviving while incapacitating the opponent. If the expected cost of a mission is not positive, it is not worth undergoing such a mission. For this reason it is assumed hereafter that M 2 M, if and only if EðMÞ P 0. If the opponent D is engaged k P 0 times, then the expected cost is: k

EðD Þ ¼

k1 X i¼0

^k 1p p EðDÞ ¼ EðDÞ; ^ 1p î

ð4Þ

where the first equality stems from the fact that the asset is reengaged if and only if it failed to incapacitate the opponent in all the previous engagements and has also managed to survive. The second equality is due to the geometric progression. Now it is straightforward to see that for a mission MN 2 M, with M \ N ¼ ; and probability pM of surviving through out mission M, the expected cost is:

EðMN Þ ¼ EðMÞ þ pM EðN Þ:

ð5Þ

It would be instructive here to give an example of how one may compute pM . Take a mission Dk M 2 M, with D R M, based on Eq. (6) the expected cost is: k

k2 X

k

EðD MÞ ¼ EðD Þ þ pD qD

! îD þ pD p ^Dk1 EðMÞ p

i¼0

^k1 1p D ^Dk1 EðMÞ: þp ¼ EðDk Þ þ pD qD ^D 1p

ð6Þ

The first equality stems from the fact that if the asset failed to incapacitate opponent D in the: îD ) then the asset i attempt (0 6 i 6 k 2 and with probability p will proceed with mission M if and only if in the i + 1 attempt it survives the engagement and also incapacitates D (with probability pDqD), ^Dk1 ) then the asset will proceed k 1 attempt (with probability p with mission M if and only if it survives the k engagement (with probability pD) irrespective of whether it incapacitated the opponent or not. With the help of Eqs. (3), (4), (6), one may apply Eq. (5) recursively in order to compute the expected cost of a mission Dk11 Dknn 2 M, with Di – Dj for i – j:

EðDk11 Dknn Þ ¼

n X i¼1

k

EðDi i Þ

i1 Y

0

i1 Y j¼1

pj @qj

j¼1

p

kj Dj

¼

n X ^ki i 1p ½ð1 pi Þc þ qi ci î 1p i¼1 1

^kj j 1 1p k 1 ^j j A; þp ^j 1p

ð7Þ

î :¼ p ^Di . Eq. (7) is basically the objective funcwhere pi :¼ pDi and p tion of the given optimization problem. Based on Eq. (5) and for the sake of completeness the empty mission O is introduced, with the intuitive values pO ¼ 1 and qO ¼ 0, and therefore EðOÞ ¼ 0. Furthermore M0 O is defined, and the empty set ; is corresponded to O. 3. Mission optimization calculus In this section we develop the mission optimization calculus for a single vehicle. We start by noticing that Eq. (7) implies that the expected cost of a mission depends on the order of the engagements. Therefore finding the optimal mission partly depends on ordering the engagements. There should first be a distinction between engagements that have zero and non-zero survival probability. If the asset is engaged with zero survival probability it will not be able to carry out any further engagements. Let D0 be the optimal engagement – i.e. the one with the highest E(D0) value – among those with zero survival probability. Let also M be the optimal mission among those containing only engagements with non-zero survival probability. Now if:

EðMÞ þ pM EðD0 Þ > EðD0 Þ;

ð8Þ

then mission MD0 is the optimal one. Otherwise, D0 is the optimal one. Based on this one now only needs a method for sorting every

71


engagement D with non-zero survival probability (pD > 0). Such a method is developed next. Notice that Eq. (8) can be re-written as fM > EðD0 Þ, where the mission priority is defined as:

fM :¼

EðMÞ qM ¼cþ cM 2 R :¼ ½1; þ1: 1 pM 1 pM

ð9Þ

As shown by the next theorem, mission priority fM indicates which mission should precede in order to achieve an optimal result.

Otherwise, if L – O, take L ¼ Mi K for some i 2 {2, . . . , k + 1}. Notice that from Lemma 4, fMi K < fMi < fM1 , and from Theorem 1 it follows that EðM1 LÞ > EðLM1 Þ. From this, and because pM1 L ¼ pLM1 – being merely engagement permutations – one obtains with the help of Lemma 3:

EðM1 LN Þ > EðLM1 N Þ ¼ EðM0 Þ:

ð10Þ

Additionally, from the inductive assumption EðM2 Mkþ1 Þ P EðLN Þ, and therefore from Lemma 2 it follows that EðM1 Mkþ1 Þ P EðM1 LN Þ. Then with the assistance of Eq. (10) one obtains the desired result. h

Theorem 1. Let M \ N ¼ ;. EðMN Þ > EðNMÞ if and only if fM > fN .

With the above theorem one partitions all missions in classes according to their priority values. In case they have equal priorities there is no reason to have one or the other preceding.

Proof. Using Eq. (5) one can see that, EðMN Þ > EðNMÞ () EðMÞ þ pM EðN Þ > EðN Þ þ pN EðMÞ () EðMÞð1 pN Þ > EðN Þð1 pM Þ () fM > fN . h

3.2. Re-engagements

3.1. Positive survival probability Next it is useful to see what happens if one prefixes or suffixes two different missions M; N with the same mission K. Lemma 2. Let pK > 0, K \ M ¼ ;, and K \ N ¼ ;. EðKMÞ > EðKN Þ if and only if EðMÞ > EðN Þ. pK >0

Proof. Using Eq. (5) it is possible to see that, EðMÞ > EðN () EðKÞ þ pK EðMÞ > EðKÞ þ pK EðN Þ () EðKMÞ > EðKN Þ. h Lemma 3. Let K \ M ¼ ; and K \ N ¼ ;. EðN Þ and pM P pN , then EðMKÞ > EðNKÞ.

If

EðMÞ >

Proof. Using Eq. (5) one can see that, EðMÞ > EðN Þ and pM P pN ) EðMÞ þ pM EðKÞ > EðN Þ þ pN EðKÞ ) EðMKÞ > EðNKÞ. h The following lemma will also prove useful later. Lemma 4. Let M \ N ¼ ;. If fM > fN , then fM > fMN > fN . Proof. Since M \ N ¼ ;, the survival probability is pMN ¼ pM pN . From the definitions of fM and fN one can deduce

fMN :¼

EðMÞ þ pM EðN Þ fM ð1 pM Þ þ pM fN ð1 pN Þ ¼ ; 1 pM pN 1 pM pN

from which it is straightforward to show that fM > fMN > fN .

h

As proved by the next theorem, given several missions one can sort the mission priorities in order to find their order in the optimal case. Theorem 5. Let M1 ; . . . ; Mn missions with \ni¼1 Mi ¼ ;, pM1 ; . . . ; pMn > 0, fM1 > > fMn , and M :¼ M1 Mn 2 M. Then for every M0 :¼ Mi1 Min 2 M with M0 – M, and all i1, . . . , in different, EðMÞ > EðM0 Þ. Proof. Induction will be used. The case n = 2 holds due to Theorem 1. Now assume that it holds for n = k, and take M0 ¼ LM1 N . If L ¼ O, then from the inductive assumption EðM2 Mkþ1 Þ > EðN Þ, and from Lemma 2 it follows that EðM1 Mkþ1 Þ > EðM1 N Þ ¼ EðM0 Þ.

The condition under which re-engagements of an opponent occur is now studied. The following useful lemma shows that the mission priority remains the same no matter how many times an opponent is re-engaged. Lemma 6. Let k P 1, then fDk ¼ fD . Proof. With the assistance of Eqs. (4), (6), and definition (9) one can see that: ^k 1p

fDk

D EðDk Þ EðDÞ ^ EðDÞ 1p :¼ ¼ ¼: fD : D1p^k1 ¼ 1 pDk 1 p q 1 pD k1 D ^ þ p D D 1p ^D D

ð11Þ

The following theorem concerning re-engagements is now proved. Theorem 7. Let k1, . . . , kn be natural numbers, \ni¼1 Mi ¼ ;, and take missions M1 ; . . . ; Mnþ1 63 D, for which fD > fMi ; pMi > 0; 8i 2 f2; . . . ; n þ 1g, then EðM1 Dk1 M2 Dk2 Mn Dkn Mnþ1 ¼ EðM1 Dk1 þ...þkn M2 M3 Mnþ1 Þ. Proof. Induction will be used in order to prove the inequality. Based on Theorem 5 and Lemma 6 one can show that EðDk2 M2 M3 Þ > EðM2 Dk2 M3 Þ, from which it is easy to show that the present theorem holds for n = 2. Assume that it holds for n = m 1. Now for n = m:

EðM1 Dk1 M2 Dk2 M3 Dkm Mmþ1 Þ ¼ EðM1 Dk1 Þ þ pM1 Dk1 qDk1 EðM2 M3 Mmþ1 Þ i þ 1 qDk1 ÞEðM2 Dk2 M3 Dkm Mmþ1 Þ ;

ð12Þ

where qDk1 is the probability to incapacitate D in the k1th attempt. Furthermore, we have:

EðM1 Dk1 þþkm M2 M3 Mmþ1 Þ ¼ EðM1 Dk1 Þ þ pM1 Dk1 qDk1 EðM2 M3 Mmþ1 Þ i þ 1 qDk1 ÞEðDk2 þþkmþ1 M2 M3 Mmþ1 Þ :

ð13Þ

Now notice that from the assumption for n = m1:

EðM2 Dk2 þþkmþ1 M3 Mmþ1 Þ > EðM2 Dk2 M3 Dkm Mmþ1 Þ;

ð14Þ

and from Theorem 5, because fD > fM2 and based on Lemma 6:

EðDk2 þþkmþ1 M2 M3 Mmþ1 Þ > EðM2 Dk2 þþkmþ1 M3 Mmþ1 Þ: ð15Þ The result now follows from Eqs. (12)–(15). h


4. Practical applications In this section we consider a straightforward generalization to a homogeneous fleet of assets. 4.1. Mission and recourse decisions Problems like the one presented in this paper are typically solved by stochastic programming in two-stages. In the first stage one decides the mission that the vehicle will follow. Then after some random event occurs affecting the outcome of the first-stage decision, a recourse decision is made in the second stage that compensates for any bad effects that might have been experienced as a result of the first-stage decision. The optimal policy from such a model is a single first-stage policy and a collection of recourse decisions – a decision rule – defining which second-stage action should be taken in response to each random outcome. For the present model it is trivial to see that the following recourse decisions need to be taken for each possible random event: If an asset was assigned by the mission to re-engage an opponent, but that opponent was struck, no re-engagement is needed and the asset should continue with the next assigned opponent. If an asset was struck by opponent D, then another asset engaging the opponent with the lowest mission priority should now engage opponent D and basically take over the rest of the mission assigned to the asset that was struck. Once that mission is over the vehicle can continue with the mission that was previously assigned to it. 4.2. Solving an example In this paper we introduced a calculus for finding the order of engaging opponents that optimizes the expected cost. More specifically Theorems 5 and 7, and Remark 8 imply that if fD1 > > fDn , the optimal mission will be Dx11 Dxnn 2 M for some x1, . . . , xn, with expected cost given by Eq. (7), and one needs to find the values x1, . . . , xn for which the expected cost is optimal. So far we focused on the objective function and no specific constraints were studied. It may now be useful to give an example of an application of the calculus. We therefore consider the case where one asset has a maximum of m missiles and needs to engage n opponents, and one missile is used per engagement or re-engagement: x 1

min

n i1 ^j j X Y 1p ^xi i 1p ^xj j 1 ½ð1 pi Þc þ qi ci p j qj þp î ^j 1 p 1 p i¼1 j¼1

!

ð16Þ subject to

n X

xi 6 m;

ð17Þ

i¼1

xi 2 f0; 1; 2; . . . ; mg;

8i 2 f1; . . . ; ng:

ð18Þ

In the above model, assuming that fD1 > > fDn , the optimal expected cost is given with the help of the objective function (16) that

was derived by Eq. (7). Moreover constraint (17) ensures that no more than m missiles are used, and finally constraint (18) ensures that the number of engagements is a non-negative integer. No further constraints or costs are considered and the asset can access any opponent directly from any other opponent. Model (16)–(18) is a Integer Non-Linear Problem (INLP), and can be solved using the MINLP_BB software package, which employs branch-and-bound [5]. The solver guarantees to find global solutions, if the problem is convex. MINLP_BB is also effective to solve non-convex MINLP problems. Even though no guarantee can be given that a global solution is found in this case, the solver is more robust than outer approximation or Benders Decomposition which usually cut away large parts of the feasible region. The package implements a branch-and-bound scheme (e.g. [4]) using a depthfirst-search. The resulting NLP relaxations are solved using filterSQP. The user can influence the branching decision by supplying priorities for the integer variables. By default, the solver branches on the variable with the highest priority first. If there is a tie, then the variable with the largest fractional part is selected for branching. Computational experiments were conducted for model (16)– (18) in order to evaluate the efficiency of the non-linear algorithm discussed above, for several randomly generated instances. The experiments were performed on an IntelÒ CoreTM 2 CPU 6400 processor at 2.13 GHz. There were 20 random instances generated using the uniform distribution for a fixed number of opponents (n) and a fixed number of maximum engagements (m). The results are displayed in Figs. 1–3 which correspond to 60, 70, and 80 number of opponents. In these figures one can see candlesticks depicting the quartiles of the CPU time in seconds of the 20 randomly generated instances for a given maximum number of assignments. The computational results of the random instances show that most of the second and third quartiles are below 100 seconds for

CPU time (secs)

Remark 8. Notice that the implementation of mission L ¼ M1 Dk1 þþkn M2 M3 Mnþ1 requires the asset to move back and forth commuting between engagement D and missions Mi again and again. Therefore in the generic case where resources are modeled, L will consume more resources than N ¼ M1 Dk1 M2 Dk2 Mn Dkn Mnþ1 and it therefore it follows that if N 2 M then L 2 M. This obviously will depend on any specific application and may not always hold true.

900 800 700 600 500 400 300 200 100 0

0

20 40 80 60 m (maximum number of assignments)

100

Fig. 1. Number of opponents n = 60.

2400 2100 CPU time (secs)

72

1800 1500 1200 900 600 300 0

0

20 40 80 60 m (maximum number of assignments) Fig. 2. Number of opponents n = 70.

100

CPU time (secs)


2000 1800 1600 1400 1200 1000 800 600 400 200 0

73

breaking down and another probability of failing to complete the alotted task at a node. A calculus was developed that computes the optimal order in which a vehicle needs to be routed. An obvious application concerns the routing of military vehicles with the goal of engaging certain opponents. We mostly focused on the objective function so more specific models need to be studied in the future. References

0

20 40 80 60 m (maximum number of assignments)

100

Fig. 3. Number of opponents n = 80. Table 1 Average time with respect to number of opponents. Opponents n Average CPU time (seconds)

40 18.1

45 18.5

50 33.6

55 91.9

60 89.7

65 168

70 258

75 271

80 321

m = 60 opponents, below 300 seconds for m = 70 opponents, and below 400 seconds for m = 80 opponents. This means that for a given maximum number of assignments (m), in most cases the algorithm becomes slower for larger number of opponents (n). This is also supported by the data presented in Table 1, showing the average CPU times of 20 random instances for each m = 5, 10, 15, . . . , 95, 100 and n. A slow down with respect to n was of course expected as the number of variables of model (16)–(18) grows as a function of n. There does not seem be any pattern with respect to m for a fixed n. 5. Conclusions In this paper we presented for the first time a single-vehicle routing problem with a given probability of having the vehicle

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