Weapon target assignment problem: Difference between revisions

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{{More citations needed|date=April 2024}}

The '''weapon target assignment problem''' ('''WTA)''') is a class of [[combinatorial optimization]] problems present in the fields of [[Optimization (mathematics)|optimization]] and [[operations research]]. It consists of finding an optimal assignment of a set of [[weapon]]s of various types to a set of targets in order to maximize the total expected damage done to the opponent.

The basic problem is as follows:

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Notice that as opposed to the classic [[assignment problem]] or the [[generalized assignment problem]], more than one agent (i.e., weapon) can be assigned to each ''task'' (i.e., target) and not all targets are required to have weapons assigned. Thus, we see that the WTA allows one to formulate optimal assignment problems wherein tasks require cooperation among agents. Additionally, it provides the ability to model probabilistic completion of tasks in addition to costs.

Both static and dynamic versions of WTA can be considered. In the static case, the weapons are assigned to targets once. The dynamic case involves many rounds of assignment where the state of the system after each exchange of fire (round) inis considered in the next round. While the majority of work has been done on the static WTA problem, recently the dynamic WTA problem has received more attention.

In spite of the name, there are nonmilitary applicationapplications of the the WTA. The main one is to search for a lost object or person by heterogeneous searchers. The agents mightassets besuch dogas teamsdogs, aircraft, walkers, etc. The goalproblem is to minimizeassign the nondetection probability by assigning the agentsassets to a partition of the space in which the lost object might beis located to minimize the probability of not finding the object. SearchThe software"value" <ref>WinCasie</ref>,of foreach example,element includesof thisthe partition is the probability that the object is located capabilitythere.

==Formal mathematical definition==

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== Algorithms and generalizations ==

It has long been known that assignment problems are [[NP-hard]]{{citation needed|date=September 2013}}. Nonetheless, anAn exact solution can be found using [[branch and bound]] techniques which utilize [[relaxation (approximation)]].<ref>{{cite journal |last1=Andersen |first1=A.C. |last2=Pavlikov |first2=K. |last3=Toffolo |first3=T.A.M. |year=2022 |title=Weapon-Target Assignment Problem: Exact and Approximate Solution Algorithms |journal=Annals of Operations Research |volume=312 |issue=2 |pages=581–606 |doi=10.1007/s10479-022-04525-6|url=https://findresearcher.sdu.dk/ws/files/204132463/WTA.pdf }}</ref> Many [[heuristic algorithm]]s have been proposed which provide near-optimal solutions in [[polynomial time]].<ref>{{cite journal |last1=Ahuja, R|first1=Ravindra K. et|last2=Kumar |first2=Arvind |last3=Jha |first3=Krishna alC. |last4=Orlin |first4=James B. |year=2007 |title=Exact and Heuristic Algorithms for the Weapon-Target Assignment Problem. |journal=Operations Research |volume=55( |issue=6), pp. |pages=1136–1146, 2007|doi=10.1287/opre.1070.0440}}</ref>

==Example==

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! Weapon Type !! <math> V_{1} = 5 </math> !! <math> V_{2} = 10 </math> !! <math> V_{3} = 20 </math>

|-

| Tank || 0.3 || 0.2 || 0.055

|-

| Aircraft || 0.1 || 0.6 || 0.5

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| Sea Vessel || 0.4 || 0.5 || 0.4

|}

One feasible solution is to assign the sea vessel and one aircraft to the highest valued target (3). This results in an expected survival value of <math> 20(0.6)(0.5)= 6 </math>. One could then assign the remaining aircraft and 2 tanks to target #2, resulting in expected survival value of <math> 10 (0.4)(0.8)^2 = 2.56 </math>. Finally, the remaining 3 tanks are assigned to target #1 which has an expected survival value of <math> 5 (0.7)^3 = 1.715 </math>. Thus, we have a total expected survival value of <math> 6 + 2.56 + 1.715 = 10.275 </math>. Note that a better solution can be achieved by assigning 3 tanks to target #1, 2 tanks and sea vessel to target #2 and 2 aircraft to target #3, giving an expected survival value of <math> 5(0.7)^3 +10(0.5)(0.8)^2 + 20(0.5)^2 = 9.915 </math>.

==See also==

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[[Category:Combinatorial optimization]]

[[Category:Matching (graph theory)]]

[[Category:Combat modeling]]