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A Stackelberg equilibrium for a missile procurement problem
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European Journal of Operational Research 193 (2009) 238–249 www.elsevier.com/locate/ejor
O.R. Applications
A Stackelberg equilibrium for a missile procurement problem Ryusuke Hohzaki b
a,*
, Shinichi Nagashima
b
a Department of Computer Science, National Defense Academy, 1-10-20 Hashirimizu, Yokosuka 239-8686, Japan Defense Planning and Policy Department, Air Staff Office, Ministry of Defense, 5-1 Ichigaya-honmura, Sinjuku-ku, Tokyo 162-8804, Japan
Received 28 August 2006; accepted 12 October 2007 Available online 26 October 2007
Abstract This paper deals with a procurement problem of missiles involving the efficient assignment of the missiles to some targets. Within a fixed amount of budget, a leader purchases several types of missiles, by which he aims to damage as much value as possible a follower hides into some facilities later. The effectiveness of the missile depends on the type of missile and facility. A payoff of the game is the expected amount of destroyed value. The problem is generalized as a two-person zero-sum game of distributing discrete resources with a leader and a follower. Our problem is to derive a Stackelberg equilibrium for the game. This type of game has an abundance of applications. The problem is first formulated into an integer programming problem with a non-separable objective function of variables and it is further equivalently transformed into a maximin integer knapsack problem. We propose three exacts methods and an approximation method for an optimal solution. Ó 2007 Elsevier B.V. All rights reserved. Keywords: Missile allocation; Stackelberg equilibrium; Game theory; Two-person zero-sum game; Integer programming
1. Introduction In this paper, we deal with a procurement problem of missiles involving the efficient assignment of the missiles to some targets. Within a fixed amount of budget, a leader purchases several types of missiles, which he assigns to facilities of a follower to damage as much value as possible the follower hides there later. The effectiveness of the missile depends on the type of missile and facility. A payoff of the game is the expected amount of destroyed value. Table 1 shows the budgeted military expenditure of some countries for recent several years, published in SIPRI (Stockholm International Peace Research Institute) Year Book 2005. The numbers of expenditure are written in local currency in the upper and the increment ratios of the expenditure to the preceding year are in the lower. Individual country has his own military and diplomatic policy, the international circumstance around him or economic situation, which affects his military expenditure. Since September 11 in 2001, the USA increases its military expenditure rapidly as a result of massive budgetary allocations for the global war on terrorism. However this rapid growth is a rare case among many countries although it is the case for Russia and China. The expenditures of many developed countries, the UK, Germany, France and Japan, do not change so drastically. For Japan, the expenditure has been staying at the same level and their expense for the procurement of military equipments is also staying around 18–19% of the gross expenditure. Without special incidents or reasons, the developed countries confront these circumstances in greater or lesser degree and the defense ministry is requested to make an efficient procurement plan for military equipments within the limited amount of budget.
*
Corresponding author. Tel.: +81 46 841 3810; fax: +81 46 844 5911. E-mail address: [email protected] (R. Hohzaki).
0377-2217/$ - see front matter Ó 2007 Elsevier B.V. All rights reserved. doi:10.1016/j.ejor.2007.10.033
R. Hohzaki, S. Nagashima / European Journal of Operational Research 193 (2009) 238–249
239
Table 1 Military expenditure Country
Currency
Financial year 1999
2000
2001
2002
2003
2004
Japan
Billion yen
USA
Million dollars
UK
Million pounds
France
Million euro
Germany
Million euro
Italy
Million euro
Russia
Million rubles
China
Billion yuan
4934 0.002 280969 0.024 22548 0.003 36510 0.014 30603 0.026 22240 0.056 167822 0.961 165 0.107
4935 0.000 301697 0.074 23532 0.044 36702 0.005 30554 0.002 24325 0.094 273106 0.627 182 0.103
4950 0.003 312743 0.037 24464 0.040 37187 0.013 30648 0.003 24592 0.011 365374 0.338 215 0.181
4956 0.001 356720 0.141 26227 0.072 38681 0.040 31168 0.017 25887 0.053 457640 0.253 251 0.167
4954 0.000 414400 0.162 31286 0.193 40212 0.040 30800 0.012 24421 0.057 567692 0.240 274 0.092
4916 0.008 466600 0.126 29868 0.045 41822 0.040 30515 0.009 25160 0.030 655787 0.155 305 0.113
The problem of missile procurement is described as follows. Player A is going to hide a total volume V of valuable material in T shelters. On the other hand, Player B is procuring m types of missiles using budget C. The price of a type j of missile is cj and its single shot kill probability (SSKP) against Shelter i is estimated to be bij . Some of the missiles are for special-purpose use so that they perform very effective against specific shelters but perform bad against other types of shelters. Some of the missiles are for general-purpose use. They work so against almost all types of shelters. The characteristics of their performance affect their prices. The special-purpose missile is comparatively expensive. On the other hand, the general-purpose missile is cheaper because it could be made through the standardized mass-production line in a factory. Player B would be concerned about the combination or the trade-off between these missiles in his procurement plan. After the purchase of missiles by Player B, Player A can observe a Player B’s assignment plan of missiles to the shelters of Player A from some intelligence information. Based on the observation, Player A can make the best plan to hide his material into T shelters so as to maximize the amount of material surviving from the possible attack of missiles by Player B in the future. Considering the best response of Player A, Player B has to make an efficient procurement plan of missiles to damage Player A’s valuable material as much as possible, prior to Player A’s decision making. The missile procurement problem is generalized to a two-person zero-sum game of distributing discrete resources with a leader and a follower. For the so-called resource allocation problem (RAP) with discrete resources, many researchers have studied one-sided optimization problems on Player B’s side so far [8]. Fruitful results have been accumulated mainly for the problems with constraints on the total amount of resources or separable objective functions of decision variables. As an application in the military field, they study so-called target assignment problem or missile allocation problem, and search problem. As a target assignment problem, Manne [5] deals with how to assign a fixed number of single type of missiles to multiple targets. Lemus and David [4] or Nickel and Mangel [6] extend Manne’s problem to a multi-type missile problem but their constraints are still placed on the number of missiles. The constraint on the number of discrete resources is easy to deal with mathematically compared with cost constraints. As an example of the search problem, we can take Kadane’s research [7], where a searcher decides which cell he searches knowing a distribution probability of a hidden target in cells. Every time during a fixed number of time points, he repeats making the decision with constraints on search cost to maximize the total detection probability. Hohzaki and Iida [3] generalize the target assignment problem and the search problem to a general RAP with cost constraints and propose methods for an optimal solution. We have reviewed one-sided optimization problems until now. In this paper, we are going to discuss a two-person zerosum game of distributing discrete resources with a leader and a follower, and to derive a Stackelberg equilibrium for it. One of applications of the resource allocation game is the missile procurement problem, described above. We can think of other military applications as follows: 1. (Example 1) An inspection game with uncertain placement of illegal material. An inspector (Player B) is going to dispatch an inspection team to an inspectee country (Player A). The inspection team consists of m types of specialists, e.g. chemists, physicists, engineers and so on. The dispatching cost depends on the specialties of members. The inspector wants to organize the team within an assigned inspection budget C in a cost-efficient way. Each inspection member is supposed to go to one of T facilities in the inspectee country and is expected to find illegal material if exist. The detection probability of the illegal material depends on the specialty of the dispatched member and the facility type. After observing the mem-
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bers of the inspection team, the inspectee is going to keep illegal material in secret as much as possible after surviving the inspection if he takes an illegal action. The payoff of the game is the expected volume of material found by the inspection team. 2. (Example 2) A P hide-and-search game in cells. A target hides in T cells. He takes a mixed strategy of hiding in cell i with T probability pi ; i¼1 p i ¼ 1. At each time j of m time points, a searcher can look into any cell as many times as he wants, spending cost cj for one looking. The detection probability of the target by one looking depends on the time point and the cell if the target is there. The problem is to find an optimal looking plan for the searcher on a time axis within a fixed search budget C and to find an optimal mixed strategy of the target. For the sake of comprehensibility, we return back to the missile procurement problem although it has some equivalent problems, stated above as Examples 1 and 2. From now on, let us discuss a Stackelberg game of the procurement problem and propose some methods to derive its equilibrium. In the next section, we begin with an exact description of our mathematical model and formulate it into an integer programming problem with constraints on cost and an exponential form of objective function. We will show that the problem can be reduced to a maximin integer knapsack problem. In Section 3, we propose three exact methods and an approximation method for an optimal strategy. By numerical examples, we analyze a missile procurement problem from a cost-efficient point of view in Section 4 and conclude the paper with some remarks in Section 5. 2. Modeling and formulation Here we model a missile procurement game, which a leader, say Player B, and a follower, say Player A, play. (A1) Player B purchases m types of missiles at an expense of C to destroy valuable material that Player A will hide in his facilities later. A type j of missile has price cj . Using a j-type missile, Player B can demolish Facility i with probability bij > 0. (A2) Player A has a total V of continuously-divisible valuable material. After observing the missile allocation of Player B, he is going to divide the material into some parts and hide them in T facilities, e.g. shelters, warehouses or factories, numbered from i ¼ 1 to T, to keep the material safe from the attack of Player B’s missiles. (A3) The players take actions as follows. First, Player B procures some missiles and then Player A makes a decision about how he divides and hides his valuable material in his facilities. Player B desires to lose Player A’s material as much as possible and Player A as little as possible. Player B is a leader and Player A is a follower as decision makers in the game. We assume that C and ci are natural numbers and that V and bij are real numbers. We denote a strategy of Player A by vi P 0, which indicates the amount of the material hidden in Facility i and satisfies PT v i¼1 i ¼ V from Assumption (2). For a Player B’s strategy of assigning missiles, we use notation xij 2 Z as the number of type assigned to Facility i of Player A. We can represent a constraint on the expense of purchasing missiles by Pm j missiles PT c x 6 C. If we denote a set of facilities by I f1; . . . ; T g and a set of missile types by J f1; . . . ; mg, feasible j¼1 j i¼1 ij regions for the strategies of Player A and B, or D and A, are given by ( ) T X D ¼ ðv1 ; . . . ; vT Þ 2 RT vi P 0; i ¼ 1; . . . ; T ; vi ¼ V ; ð1Þ i¼1 ( ) m T X X IJ cj xij 6 C ; ð2Þ A ¼ fxij g 2 Z xij P 0; i ¼ 1; . . . ; T ; j ¼ 1; . . . ; m; j¼1 i¼1 respectively. A j-type missile can demolish Facility i with probability 0 < bij 6 1, which is referred to as the single shot kill probability (SSKP). bij mayP have another representation of bij 1 expðaij Þ, where aij > 0. Therefore, Qm x m 1 j¼1 ð1 bij Þ ij ¼ 1 expð j¼1 aij xij Þ is the kill probability of Facility i by an assignment plan of missiles fxij ; j 2 Jg. Now we can evaluate the expected amount of the material to be lost for all facilities by ( !) ! T m T m X X X X vi 1 exp aij xij vi exp aij xij : ¼V i¼1
j¼1
i¼1
j¼1
Because the first term of the above expression is a constant, we regard the second term Eðv; xÞ, which is the expected untouched amount, as a payoff function of the game from now on. The payoff Eðv; xÞ is given by
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Eðv; xÞ ¼
T X
vi exp
i¼1
m X
241
! ð3Þ
aij xij ;
j¼1
where v ¼ fvi ; i 2 Ig and x ¼ fxij ; i 2 I; j 2 Jg. Our problem is to find a minimax value and a Stackelberg equilibrium for the game with the payoff Eðv; xÞ. As a minimizer or a leader, Player B wants to minimize the payoff by a desirable assignment plan of missiles x while Player A decides an allocation of the material v as a maximizer or a follower so as to maximize the payoff. The minimax value of the game is given by ! T m X X G ¼ min max vi exp aij xij : fxij g2A fvi g2D
i¼1
j¼1
Considering the feasible region D, we can transform the maximization involved in the above problem as follows: ! ! T m m X X X vi exp aij xij ¼ V max exp aij xij : max fvi g2D
i¼1
j¼1
i
j¼1
P P m m We derive the right-hand side by letting vk ¼ 0 for k 2 I satisfying exp j¼1 akj xkj < maxi exp j¼1 aij xij . If we proceed to a minimization problem of the above result with respect to x, we have the following problem: ( ) ! m X ðP 0 ÞG ¼ min V kexp aij xij 6 k; i ¼ 1; . . . ; T ; fxij g 2 A : ð4Þ j¼1 P Pm m An inequality exp j¼1 aij xij 6 k is equivalent to log k 6 j¼1 aij xij . A replacement l log k leads us to ( ) m X V V G ¼ min aij xij ; i ¼ 1; . . . ; T ; fxij g 2 A ¼ ; ð5Þ l 6 exp l exp nC j¼1 where the number nC is given by ( ) m X ðP 1 ÞnC ¼ max ll 6 aij xij ; i ¼ 1; . . . ; T ; fxij g 2 A j¼1 ( ) X X m X ¼ max min aij xij c x 6 C; xij 2 Z þ ; i 2 I; j 2 J ; i2I i2I j2J j ij j¼1
ð6Þ
and Z þ is a set of all nonnegative integers. In the case that I consists of one element, the problem ðP 1 Þ becomes so-called integer knapsack problem (IKP). That is why we call the problem ðP 1 Þ the maximin integer knapsack problem or the maximin IKP. We can interpret our problem in the context of the classical IKP as follows. We pack m types of items into T knapsacks as long as the expense of purchasing the items does not go beyond budget C. A j-type item has a price cj and value aij when it is put into the ith knapsack. We want to increase the minimum packed value among each knapsack as much as possible so as to equalize the packed value among all knapsacks. From the above discussion, we can easily see the NP-hardness of our problem and we have the following theorem. Theorem 1. Problem ðP 0 Þ is NP-hard. An optimal strategy of Player B, xij , is given by anP optimal solution of the maximin IKP m or problem ðP 1 Þ. For Player A, he should not hide any material in Facility i for n < C j¼1 aij xij and should hide the entire Pm amount V of the material among facilities satisfying nC ¼ j¼1 aij xij in an arbitrary way of dividing V. From the theorem, we only need to focus on a solution for the problem ðP 1 Þ from now on. First of all, we explain a lemma, by which we can preclude self-evident solutions. We use the following notation. c0 is the minimum among unit prices of all kinds of items and J 0 is a set of types of items having the minimum price. For each i 2 I, a set j0 ðiÞ is defined as types of item j 2 J 0 having the maximum of coefficients aij . These are defined by c0 ¼ min cj ; j¼1;...;m
J 0 ¼ fj 2 Jjcj ¼ c0 g;
j0 ðiÞ ¼ arg max aij : j2J 0
ð7Þ
Lemma 1. (i) If C < Tc0 , an optimal value is nC ¼ 0 and one of optimal solutions is xij ¼ 0. (ii) If Tc0 6 C < 2Tc0 , an optimal value is nC ¼ min aij0 ðiÞ . One of optimal solutions is given by xij0 ðiÞ ¼ 1; xij ¼ 0; j 6¼ j0 ðiÞ i2I for each i 2 I.
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Proof. (i) is the case that budget C is too little that there is a knapsack at least that Player B cannot purchase even an item for, and an optimal value is zero. (ii) is the case that Player B can afford to buy an item with minimum price c0 for all knapsacks but cannot buy any more. An optimal value is determined by the minimum coefficient aij0 ðiÞ among all knapsacks i 2 I. h From now, we are going to discuss an optimal solution assuming C P 2Tc0 implicitly to avoid trivial cases of Lemma 1. 3. Optimal solution for the maximin integer knapsack problem 3.1. A dynamic programming method As mentioned in the preceding section, the maximin IKP contains the classical IKP as a subproblem. We can utilize some methods developed in the past for the solution of the IKP. We represent a maximal value of the IKP with budget D for knapsack n by ( ) X m m X þ ð8Þ ðIKPn ðDÞÞln ðDÞ ¼ max anj xnj c x 6 D; xnj 2 Z : j¼1 j nj fxnj ;j2J g j¼1 Similarly, we denote a feasible region for solutions and an optimal value for the maximin IKP with budget D and n knapsacks by S n ðDÞ and F n ðDÞ, respectively, which are defined as follows: ( ) X n X þ c x 6 D; xij 2 Z ; i ¼ 1; . . . ; n; j 2 J S n ðDÞ fxij ; i ¼ 1; . . . ; n; j 2 J g i¼1 j2J j ij F n ðDÞ max
m X
min
fxij g2S n ðDÞ i¼1;...;n
aij xij :
j¼1
P o m þ c x c x 6 C; x 2 Z , we have the following recursive formula with respect to the num j j j j j j¼1 j¼1
nP m
Using notation W ber of knapsacks.
( F n ðDÞ ¼ max min fxnj ;j2J g
m X
anj xnj ; F n1 D
j¼1
¼ max n d2W
!) cj xnj
j¼1
(
Pmax m
fxnj ;j2J g
m X
j¼1
cj xnj ¼d;xnj 2Z þ
o min
m X
) anj xnj ; F n1 ðD dÞ
j¼1
) # X m þ ¼ max min max anj xnj c x ¼ d; xnj 2 Z ; F n1 ðD dÞ d2W j¼1 j nj fxnj ;j2J g j¼1 ( ) " # X m m X þ ¼ max min max anj xnj c x 6 d; xnj 2 Z ; F n1 ðD dÞ ¼ max min fln ðdÞ; F n1 ðD dÞg d2W j¼1 j nj d¼1;...;D fxnj ;j2J g j¼1 "
¼
max c0 6d6Dðn1Þc0
(
m X
minfln ðdÞ; F n1 ðD dÞg
ð9Þ
The last expression is derived from Lemma 1 telling that knapsack n and other n 1 knapsacks need budget c0 and ðn 1Þc0 , respectively, to avoid trivial cases. To compute the optimal value F T ðCÞ, we only need to repeat applying the above dynamic programming (DP) (9) to D ¼ nc0 ; nc0 þ 1; . . . ; C for each n while increasing n by one from 1 to T. Initial conditions of the computation are F 1 ðDÞ ¼ 0;
D ¼ 1; . . . ; c0 1;
F 1 ðDÞ ¼ l1 ðDÞ;
D ¼ c0 ; . . . ; C:
From some past studies on the knapsack problem, we know that all optimal values ln ðDÞ; D ¼ 1; . . . ; C; are obtained in the process of solving a knapsack problem IKPn ðCÞ once. The mapping from budget D to optimal value ln ðDÞ is referred to as the integer knapsack function. If we solve the problem IKPn ðCÞ and save ln ðDÞ in advance, we can simply load ln ðdÞ for computation when we apply the dynamic programming method. Noting that the price system of items cj are same for all knapsacks, the computation time required to solve IKPn ðCÞ seems not to depend on the knapsack number n so much. If we denote the computation time by knapðCÞ, we may estimate the total time to solve the IKP for all knapsacks by T knapðCÞ. In the DP algorithm (9), the comparison operators ‘min’ and ‘max’ are transacted once for each d while the number d varies D ðn 1Þc0 c0 þ 1 ¼ D nc0 þ 1 times. Furthermore, index D changes within nc0 6 D 6 C. In the most outer
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243
repetition layer of the DP method, we change n from 2 to T. The total number of computational repetitions in the algorithm is T C X X
2ðD nc0 þ 1Þ ¼
n¼2 D¼nc0
c20 c0 ð2T 3 þ 3T 2 þ T 6Þ ð2C þ 3ÞðT 2 þ T 2Þ þ ðC þ 1ÞðC þ 2ÞðT 1Þ: 6 2
Now the order of computational complexity is estimated by OðTC 2 þ T 2 Cc0 þ T 3 c20 Þ. As the result, the complexity of the proposed dynamic programming method is the order of OðT knapðCÞ þ TC 2 þ T 2 Cc0 þ T 3 c20 Þ for the maximin IKP ðP 1 Þ. 3.2. Advance evaluation method Here we review some useful information about the classical IKP [1,2], which we are going to use for our problem. A target problem for our review is the following: m m X X ðIKPðdÞÞlðdÞ ¼ max aj x j ; s:t: cj xj 6 d; xj 2 Z þ ; j 2 J; ð10Þ x
j¼1
j¼1
where m items are sorted and numbered such that a1 =c1 P a2 =c2 P P am =cm . The relation between d and optimal value lðdÞ is called the integer knapsack function, which is usually a non-decreasing right-continuous step function for variable d. There are some algorithms developed to give discontinuity points, where the function jumps discontinuously. We sort these points of d in the increasing order like f ð1Þ < f ð2Þ < in a list and call it a step list. Let xj ðj ¼ 1; . . . ; mÞ be an optimal solution for IKPðdÞ and gðdÞ be a minimum index j for xj > 0 defined by minfjjxj > 0g; x 6¼ 0; gðdÞ ¼ ð11Þ m; x ¼ 0: Given the above situation, a solution ðx1 ; . . . ; xj1 ; xj 1; xjþ1 ; . . . ; xm Þ is optimal for IKPðd cj Þ if xj > 0 and then gðdÞ 6 gðd cj Þ holds for every j of xj > 0. Therefore, among j satisfying j 6 gðd cj Þ and d cj P 0, there is an index ~ for IKPðd cj Þ becomes optij such that a new solution generated by adding one to the jth element of optimal solution x mal for IKPðdÞ. It follows that lðdÞ ¼ maxfaj þ lðd cj Þjj 6 gðd cj Þ; d cj P 0g: j
ð12Þ
An index j to maximize the right-hand side gives gðdÞ. We also know that there is a certain budget D such that lðdÞ ¼ a1 þ lðd c1 Þ for all d of D 6 d. Namely, for IKPðdÞ with budget d larger than D , an optimal solution is always given by adding one to the first element x1 of an optimal solution for IKPðd c1 Þ. This property is referred to as the periodicity of the IKP. To confirm the periodicity, we only need to check gðdÞ ¼ 1; d ¼ D ; D þ 1; . . . ; D þ max cj 1 ð13Þ j
for a budget D . We have outlined the known properties related to the IKP. From now on, we are going to devise a method for our problem ðP 1 Þ based on these properties. Assume that we have optimal value nd and an optimal solution for budget d. To pull up the optimal value more, it is required to increase budget d i assigned to each knapsack i of li ðd i Þ ¼ nd by some adequate volume. For knapsack i 2 I, the necessary budget d i to increase li ðd i Þ can be evaluated by a step list fi ðÞ, which we obtain in the process of solving IKPi ðCÞ. The following algorithm incorporates the evaluation of the necessary increase of knapsack’s budget in advance of changing the total budget d. That is why we call the algorithm the advance evaluation method. Step 1. Solve IKPi ðCÞ for each knapsack i 2 I and obtain P an integer knapsack function li ðDÞ; D ¼ 1; . . . ; C and a step list fi ðÞ. Set initial conditions li ¼ 1; i 2 I; d i2I fi ðli Þ. In this initial step, d must be Tc0 . Step 2. Set l ¼ mini2IP li ðfi ðli ÞÞ; K ¼ fnjln ðfn ðln ÞÞ ¼ lg. Step 3. Calculate s ¼ n2K ðfn ðln þ 1Þ fn ðln ÞÞ. If d þ s > C, terminate. An optimal value is l. d i ¼ fi ðli Þ is an optimal budget to be assigned to knapsack i. Using d i , we solve IKPi ðd i Þ to obtain an optimal solution or optimal assignment of type-j items xij ; j 2 J for knapsack i. If d þ s 6 C, increase d by d ¼ d þ s; li ¼ li þ 1; i 2 K and return to Step 2. In Step 3, the algorithm increases a budget assignment for knapsack i in accordance with a step list fi ðli Þ. As the total budget d increases by s in Step 3, a corresponding optimal value l always moves up. We can verify that this method does work properly as follows. If we re-assign any part of fn ðln þ 1Þ fn ðln Þ of knapsack n 2 K to other knapsacks in Step 3, the value of the objective function stays at a tentative value l. If we decrease the assign-
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ment budget of knapsack n 62 K by some volume and add the volume to the assignment budget of other knapsacks, a current value ln ðfn ðln ÞÞ goes down to the previous value ln ðfn ðln 1ÞÞ. The assignment fn ðln 1Þ must have been increased at some time in Step 3, which means that ln ðfn ðln 1ÞÞ < l for the current tentative value l. It follows that the re-assignment of fn ðln þ 1Þ fn ðln Þ causes a smaller value of the objective function less than l. According to the above discussion, we can see that this algorithm adjusts the assignment of budget for each knapsack so as to maximize the value of the objective function. Every time budget d is renewed in Step 3, we are informed a discontinuity point, where the value of the objective function jumps up. 3.3. Sequential increment method The advance evaluation method uses the knapsack function, which is obtained after solving IKPi ðCÞ with budget C for each knapsack i 2 I. However, we can anticipate that optimal assignment budget to each knapsack can be rather smaller than C when we finally solve problem ðP 1 Þ. Solving IKPi ðCÞ in advance might be wasting. In a method proposed here, we increase a tentative assignment budget for each knapsack by using not the knapsack function but the information (11)–(13) sequentially. To do that, we incorporate the method proposed for a solution of the ordinary IKP in an algorithm here. Assume that for knapsack i, type numbers J are sorted in the decreasing order in terms of aij =cj ; j 2 J such that aili ð1Þ =cli ð1Þ P aili ð2Þ =cli ð2Þ P P aili ðmÞ =cli ðmÞ . Let us define cmax ¼ maxj2J cj . Step 1. Set initial values of parameters for every knapsack i 2 I, as follows: li ð0Þ ¼ li ð1Þ ¼ ¼ li ðc0 1Þ ¼ 0; li ðc0 Þ ¼ aij0 ðiÞ , gi ð0Þ ¼ gP i ð1Þ ¼ ¼ g i ðc0 1Þ ¼ m; gi ðc0 Þ ¼ minfkjcli ðkÞ ¼ c0 g, rðiÞ ¼ 0; d i ¼ c0 and Di ¼ c0 . ~ Step 2. Calculate d ¼ i2I d i . If d > C, go to Step 4 and terminate. Otherwise, save d i in d i , that is, set d~i ¼ d i ; i 2 I. Calculate l ¼ mini2I li ðd i Þ and K ¼ fnjln ðd n Þ ¼ lg. If d þ jKj > C, go to Step 4 and terminate. Step 3. After running the following procedure for each n 2 K, return to Step 2. Increase d n P by one, d n ¼ d n þ 1, and repeat calculating ln ðd n Þ by the following procedure until ln ðd n 1Þ < ln ðd n Þ. But if i2I d i > C in the middle of the procedure, go to Step 4 and terminate. (i) If rðnÞ ¼ 1, set ln ðd n Þ ¼ ln ðd n cln ð1Þ Þ þ anln ð1Þ . (ii) If rðnÞ ¼ 0, set j ¼ arg maxfln ðd n cln ðjÞ Þ þ anln ðjÞ jd n cln ðjÞ P 0; j 6 gn ðd n cln ðjÞ Þg; j
ð14Þ
ln ðd n Þ ¼ ln ðd n cln ðj Þ Þ þ anln ðj Þ ;
ð15Þ
gn ðd n Þ ¼ j :
ð16Þ
If j ¼ 1 and d n ¼ Dn þ cmax 1, set rðnÞ ¼ 1. If j 6¼ 1, renew Dn by Dn ¼ d n . Step 4. An optimal value is l. For each knapsack i 2 I, solve IKPi ðd~i Þ to obtain an optimal assignment of items fxij ; j 2 Jg as an optimal solution and terminate. gn ðd n Þ and Dn , which are initialized in Step 1 and renewed in Step 3, correspond to gðdÞ and D in condition (13). When the condition is verified, we mark a sign rðnÞ ¼ 1, which indicates the periodicity for optimality. The setting of (14) and (15) is corresponding to the calculation of (12) for knapsack n. In Step 2, K is a set of knapsacks with the tentative minimum objective value l. In Step 3, we increase the assignment of budget d n to each knapsack n in K until its value ln ðd n Þ is drawn up while increasing the total budget d sequentially. This main process of increasing the assignment budget to the knapsacks is as same as the advance evaluation method in essence. That is why the validity of this method is verified for its execution. 3.4. An approximation method We have discussed so far some exact methods for an optimal solution. Because our problem is NP-hard, even an approximation method also deserves the proposition. Here we are going to devise an approximation solution in the polynomial order of computational time. First of all, we relax integer valuables of problem ðP 1 Þ to real numbers. A resultant problem is ðRLP1 Þ^ l ¼ max min i2I
s:t:
m X
T X m X i¼1
aij xij ;
ð17Þ
cj xij 6 C;
ð18Þ
j¼1
j¼1
xij P 0;
xij 2 R;
i 2 I;
j 2 J:
For the relaxed problem, an optimal solution is given by the following theorem.
ð19Þ
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245
^ of the relaxed problem ðRLP 1 Þ are given by ^ and an optimal solution x Theorem 2. An optimal value l C
^ ¼ PT l
k¼1 csðkÞ =aksðkÞ
^xisðiÞ ¼ PT
ð20Þ
;
C=aisðiÞ
k¼1 csðkÞ =aksðkÞ
;
i 2 I; ^xij ¼ 0; i 2 I; j 6¼ sðiÞ; j 2 J;
ð21Þ
where sðiÞ arg maxj2J faij =cj g. Proof. The replacement of qij ¼ cj xij changes the original problem ðRLP1 Þ to ( ) m T X m X aij X ^ ¼ max min l q q 6 C; qij P 0; i 2 I; j 2 J : fqij g i2I cj ij i¼1 j¼1 ij j¼1 P By additional substitutions of cij aij =cj and d^i mj¼1 qij , we can transfer the problem as follows: ( ) X m T m X X ^ ^ ¼ max min cij qij qij ; qij P 0; i 2 I; j 2 J ; d 6 C; d i ¼ i¼1 i fd^i ;qij g i2I j¼1 j¼1 ( ) X T ¼ max min cisðiÞ qisðiÞ d^ 6 C; d^i ¼ qisðiÞ P 0; i 2 I ; i¼1 i fd^i ;qisðiÞ g i2I ( ) X T ^ ^ ^ ¼ max min cisðiÞ d i d 6 C; d i P 0; i 2 I ; i2I i¼1 i fd^ i g ( ) T X ¼ max ll 6 cisðiÞ d^i ; i 2 I; d^i ¼ C; d^i P 0; i 2 I : fd^ i ;lg i¼1
ð22Þ
An optimal value for the last reduced problem is given by T X ^ ¼ cisðiÞ d^i ; i 2 I; l d^i ¼ C: i¼1
This simultaneous system of equations has a solution of C C ¼ PT ; 1=c c ksðkÞ k¼1 k¼1 sðkÞ =aksðkÞ
^ ¼ PT l
csðiÞ =aisðiÞ d^i ¼ C PT : k¼1 csðkÞ =aksðkÞ
Transformation (22) is possible by the setting of d^i ¼ qisðiÞ ; i 2 I; qij ¼ 0; j 6¼ sðiÞ; j 2 J. That confirms us that the theorem gives us an optimal solution f^xij g. h Using the theorem, we can propose a simple greedy algorithm as an approximation method. The first step is to derive an integer solution fxij g ¼ fb^xij cg by rounding off a solution f^xij g of the relaxed problem ðRLP1 Þ. Residual budget PP C i j cj xij is used for the purchase of the cheapest items with price c0 for each knapsack in K, which is defined in Step 2 of the advance evaluation method and the sequential increment method. An entire set of the greedy algorithm is as follows: Step 1. By Theorem 2, solve the relaxed problem P P ðRLP1 Þ to obtain P a real solution f^xij g. Round off the solution by fxij g ¼ fb^xij cg. Calculate C R ¼ C i2I j2J cj xij and li ¼ j2J aij xij ; i 2 I. Step 2. l ¼ mini2I li ; K ¼ fnjln ¼ lg. Step 3. If jKjc0 > C R , terminate with an approximation value l and an approximation solution fxij g. Otherwise, modify a tentative solution fxnj g for each n 2 K as follows: xnj0 ðnÞ ¼ xnj0 ðnÞ þ 1;
ln ¼ ln þ anj0 ðnÞ ;
where j0 ðnÞ is given by definition (7). Return to Step 2 after renewing C R ¼ C R jKjc0 .
4. Numerical examples Here we consider a concrete budgeting problem of procuring missiles from the cost-efficient point of view. Player A is going to hide a total amount V of valuable material in T ¼ 4 shelters. On the other hand, Player B is going to purchase
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m ¼ 5 types of missiles using budget C. The price cj of a type j of missile and an estimated SSKP of a j-type missile against Shelter i, bij , are illustrated in Table 2. A type j ¼ 1; . . . ; 4 of missile is specifically effective for the same number of shelter j. But a type-5 of missile is for general-purpose use and effective to every shelter to some extent. The general-purpose missile costs cheaper than special-purpose one. The prices of special-purpose missiles have an order of c1 6 c2 6 c3 6 c4 . Player A can observe a Player B’s assignment plan of missiles aiming at the shelters of Player A from some intelligence information and he can make a best hiding plan of his material into m shelters. Considering this situation comprehensively, Player B is requested to make a cost-efficient procurement plan of missiles, especially taking account of the efficient combination of special-purpose and general-purpose missiles. The value V = exp nC given by Eq. (5) is the minimax expected amount of Player A’s material surviving the Player B’s attack and 1= exp nC is the minimax ratio of the survived amount to the total, i.e. the minimax survivorship ratio, where nC is an optimal value of the minimax IKP ðP 1 Þ. Fig. 1 illustrates the survivorship ratios in percentage while changing an integer total budget C from 0 through 50. The curve has a step-function form as the knapsack function does. Because of the form, some ratios can be the same for different budgets, e.g. we can see the same ratios from C ¼ 26 through 40. Considering the sensitivity of the budget, Player B asks some contractors or makers of missiles to compare the following eight alternates besides of the basic plan of Table 2. 1. 1SU: Upgrade the basic specification of the special-purpose types of missile j ¼ 1; . . . ; 4 and increase their prices, i.e. improve ajj of the j-type missile and price up cj by one for j ¼ 1; . . . ; 4. 2. 1GU: Upgrade the basic specification of the general-purpose missile j ¼ 5 and increase its price, i.e. improve ai5 ; i ¼ 1; . . . ; 4 and price up c5 by one. 3. 2SU: Price up cj of the special-purpose missiles j ¼ 1; . . . ; 4 by two and improve their basic performances ajj more over Alternate 1SU. 4. 2GU: Price up c5 of the general-purpose missile by two and improve its performance ai5 ; i ¼ 1; . . . ; 4 more over 1GU. 5. 1SD: Price down cj of the special-purpose missiles j ¼ 1; . . . ; 4 by one and downgrade their performances ajj . 6. 1GD: Price down c5 of the general-purpose missile by one and downgrade its performance ai5 ; i ¼ 1; . . . ; 4. 7. 2SD: Price down cj of the special-purpose missiles j ¼ 1; . . . ; 4 by two and downgrade their performances ajj more over 1SD. 8. 2GD: Price down c5 of the general-purpose missile by two and downgrade its performance ai5 ; i ¼ 1; . . . ; 4 more over 1GD. The naming of the alternates has a rule as follows. Alphabet ‘S’ or ‘G’ stands for ‘‘special-purpose’’ or ‘‘general-purpose’’, respectively. ‘U’ or ‘D’ indicates ‘‘upgrade’’ or ‘‘downgrade’’. Number ‘1’ or ‘2’ preceding those alphabets represents the extent to which the missile is upgraded or downgraded, or the missile is priced up or down. Performance parameter a0ij corresponding to the basic SSKP bij in Table 2 is calculated by bij ¼ 1 expða0ij Þ. Other parameters ajj ; j ¼ 1; . . . ; 4, for Alternate 1SU and 2SU, and ai5 ; i 2 I for 1GU and 2GU are calculated by the following evaluation formulas. 1 1 1U 0 2U 1U aij ¼ aij 1 þ ; aij ¼ aij 1 þ : ð23Þ cj þ 1 cj þ 2 Similarly the performance parameters corresponding to downgraded alternates are evaluated by 1 1 0 2D 1D ¼ a 1 ¼ a 1 a1D ; a : ij ij ij ij cj 1 cj 2
ð24Þ
The SSKPs reduced from formulas (23) and (24) are shown in Table 3. For the upgraded alternates, more expensive missile of 2SU or 2GU is improved less than less expensive one of 1SU or 1GU, compared with the increment of price. For the Table 2 Single shot kill probabilities Shelters
1 2 3 4 ci
Types of missile 1
2
3
4
5
0.60 0.01 0.01 0.01 5
0.01 0.60 0.01 0.01 6
0.01 0.01 0.60 0.01 7
0.01 0.01 0.01 0.60 8
0.20 0.20 0.20 0.20 4
Ratios(%)
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247
100 90 80 70 60 50 40 30 20 10 0 0
5
10
15
20
25
30
35
40
45
50
C Fig. 1. The minimax expected survivorship ratio by basic-performance missiles.
Table 3 Single shot kill probabilities for alternates
b11
b22
b33
b44
bi5 ; i 2 I
0.705 – 0.657 – 0.600 0.497 – 0.368 –
0.692 – 0.649 – 0.600 0.520 – 0.423 –
0.682 – 0.643 – 0.600 0.534 – 0.457 –
0.674 – 0.639 – 0.600 0.544 – 0.480 –
– 0.268 – 0.235 0.200 – 0.138 – 0.072
Ratio(%)
2SU 2GU 1SU 1GU Basic plan 1SD 1GD 2SD 2GD
SSKPs
100 90 80 70 60 50 40 30 20 10 0
1SU 1GU 2SU 2GU
0
5
10
15
20
25
30
35
40
45
50
C Fig. 2. The minimax expected survivorship ratio by upgraded plans.
Ratio(%)
Alternates
100 90 80 70 60 50 40 30 20 10 0
1SD 1GD 2SD 2GD
0
5
10
15
20
25
30
35
40
45
50
C Fig. 3. The minimax expected survivorship ratio by degraded plans.
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Ratio(%)
downgraded alternates, a cheaper missile of 2SD or 2GD is more downgraded than a missile of 1SD or 1GD, compared with the decrement of price. Fig. 2 illustrates the minimax expected survivorship ratios for the upgraded alternates and Fig. 3 for the downgraded alternates. The ratios for all alternates, including the basic plan, are shown in Fig. 4. We can improve the sensitivity of the survivorship ratio to budget C by comparing several alternates. The improvement of the sensitivity enables us to budget for the construction of weapons with elaborate consideration from the cost-efficient point of view. In Table 4, each small table has some information about an optimal solution for a given budget C: the best alternate, the minimax expected survivorship ratio (%) and an optimal assignment of missiles to facilities fxij ; i 2 I; j 2 Jg, in detail. In
100 90 80 70 60 50 40 30 20 10 0
0
5
10
15
20
25
30
35
40
45
50
C Fig. 4. The minimax expected survivorship ratio by all plans.
Table 4 Optimal assignment of missiles to shelters Shelters
Types of missile
C = 8–11
2GD
1 2 3 4
0 0 0 0
C = 18–20
2SD
1 2 3 4
1 0 0 0
C = 26–29
Basic
1 2 3 4
1 0 0 0
C = 38–41
2SU
1 2 3 4
1 0 0 0
C = 44–47
1SD
1 2 3 4
2 0 0 0
1
2
3
4
0 0 0 0
0 0 0 0
1 1 1 1
63.3 0 1 0 0
0 0 1 0
0 0 0 1
0 0 0 0
40.0 0 1 0 0
0 0 1 0
0 0 0 1
0 0 0 0
31.8 0 1 0 0
0 0 1 0
0 2 0 0
0 0 2 0
0 0 0 1
0 0 0 1
25.3 0 0 0 2
Types of missile
C = 12–14
1GD
1 2 3 4
0 0 0 0
C = 21
2SD
1 2 3 4
2 0 0 0
C = 30–33
1SU
1 2 3 4
1 0 0 0
C = 42
2SU
1 2 3 4
1 0 0 0
C = 48–50
1SD
1 2 3 4
3 0 0 0
5
92.8 0 0 0 0
Shelters
0 0 0 0
1
2
3
4
0 0 0 0
0 0 0 0
1 1 1 1
57.7 0 1 0 0
0 0 1 0
0 0 0 1
0 0 0 0
36.1 0 1 0 0
0 0 1 0
0 0 0 1
0 0 0 0
30.8 0 1 0 0
0 0 1 0
0 2 0 0
0 0 2 0
0 0 0 1
0 0 1 1
23.1 0 0 0 2
Types of missile
C = 15–17
2SD
1 2 3 4
1 0 0 0
C = 22–25
1SD
1 2 3 4
1 0 0 0
C = 34–37
2SU
1 2 3 4
1 0 0 0
C = 43
2SD
1 2 3 4
3 0 0 0
5
86.2 0 0 0 0
Shelters
0 0 0 0
1
2
3
4
5
80.0 0 1 0 0
0 0 0 0
0 0 0 0
0 0 1 1
50.3 0 1 0 0
0 0 1 0
0 0 0 1
0 0 0 0
32.6 0 1 0 0
0 0 1 0
0 0 0 1
0 0 0 0
29.5 0 3 0 0
0 0 2 0
0 0 0 2
0 0 0 0
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249
each small table, the range of budget C, the best alternate and the minimax ratio are written in the first row. In below rows, an optimal assignment of missiles are tabled. We can analyze the results as follows. (1) 0 6 C 6 7: For such low budgets as C < Tc0 ¼ 8, Player B cannot construct enough missile to be assigned to all shelters of Player A. (2) 8 6 C 6 11: Player B cannot afford to purchase any expensive special-purpose missile and comes to buy some general-purpose missiles by the most degraded plan 2GD because the missiles are comparatively cheaper. (3) 12 6 C 6 14: As similar to the preceding case, Player B buys only general-purpose missiles but he use a moderately degraded plan 1GD. (4) 15 6 C 6 17: By Plan 2SD, Player B constructs special-purpose missiles j ¼ 1; 2, which are assigned to Shelter i ¼ 1; 2, because their prices are comparatively cheaper. To each of Shelter i ¼ 3; 4, he assigns a general-purpose missile with basic performance. (5) 18 6 C 6 21: Player B has enough budget to assign special-purpose missiles to all shelters although he uses the most degraded plan 2SD. (6) 22 6 C 6 25: Player B uses a similar assignment to the preceding case but he adopts the moderately degraded plan 1SD. (7) 26 6 C 6 29: As similar to the preceding case, Player B assigns a special-purpose missile with basic performance to each shelter. He can reach the same effectiveness by Alternate 1GU, 2GU, 1GD or 2GD. (8) 30 6 C 6 33: Player B uses the upgraded plan 1SU of assigning only special-purpose missiles. (9) 34 6 C 6 42: Player B uses the most upgraded plan 2SU and assigns the highest-quality missiles to every shelter. He complements an excess budget by purchasing some general-purpose missiles. (10) C ¼ 43: Player B purchases several special-purpose missiles for every shelter by the degraded plan 2SD. The assignment of multiple special-purpose missiles is effective for shelters, even though the missiles are downgraded. (11) 43 6 C: Player B still focuses on the multiple assignment of special-purpose missiles but he uses the moderately degraded plan 1SD.
5. Conclusions In this paper, we deal with a Stackelberg equilibrium for a missile procurement problem. The problem can be regarded as a distribution game of general discrete resources with a leader and a follower. This type of problem has a wide variety of applications in the fields of the military or international affairs, as mentioned in Introduction. As we did in this study, there are many needs for the extension of classical resource allocation problems (RAPs) to games. Especially, it is more difficult to propose practical solution methods for the discrete optimization problem of the RAP than other types of the RAP. Therefore, we cannot avoid the development of practical algorithms for the discrete RAP formulated in the game. This paper is a trial for the development. Although for the integer programming problem formulated in this paper, its objective function is not separable for variables, we can transform the problem into a maximin integer knapsack problem (a maximin IKP) equivalently. This paper utilizes mathematical theory known about the IKP to propose some solution methods by incorporating the theory. One of main purposes of this paper is to propose methods for an equilibrium of a resource allocation game with discrete resources. To improve algorithms for the maximin-type of IKP further or develop more efficient algorithms from a computational point of view is another. This would be a future study for us. We deal with a Stackelberg-type of game in this paper but we leave the extension of it to a standard game with simultaneous decisions of players as our future work. References [1] P.C. Gilmore, R.E. Gomory, The theory and computation of Knapsack function, Operations Research 14 (1966) 1045–1074. [2] R.S. Garfinkel, G.L. Nemhauser, Integer Programming, John Wiley & Sons, 1972. [3] R. Hohzaki, K. Iida, An integer resource allocation problem with cost constraint, Journal of the Operations Research Society of Japan 41 (1998) 470– 482. [4] F. Lemus, K.H. David, An optimal allocation of different weapons to a target complex, Operations Research 11 (1963) 787–794. [5] A.S. Manne, A target-assignment problem, Operations Research 7 (1959) 258–260. [6] R.H. Nickel, M. Mangel, Weapon acquisition with target uncertainty, Naval Research Logistics 32 (1985) 567–588. [7] J.B. Kadane, Discrete search and the Neyman–Pearson lemma, Journal of Mathematical Analysis and Applications 22 (1968) 156–171. [8] T. Ibaraki, N. Katoh, Resource Allocation Problem, The MIT Press, London, 1988.
European Journal of Operational Research 193 (2009) 238–249 www.elsevier.com/locate/ejor
O.R. Applications
A Stackelberg equilibrium for a missile procurement problem Ryusuke Hohzaki b
a,*
, Shinichi Nagashima
b
a Department of Computer Science, National Defense Academy, 1-10-20 Hashirimizu, Yokosuka 239-8686, Japan Defense Planning and Policy Department, Air Staff Office, Ministry of Defense, 5-1 Ichigaya-honmura, Sinjuku-ku, Tokyo 162-8804, Japan
Received 28 August 2006; accepted 12 October 2007 Available online 26 October 2007
Abstract This paper deals with a procurement problem of missiles involving the efficient assignment of the missiles to some targets. Within a fixed amount of budget, a leader purchases several types of missiles, by which he aims to damage as much value as possible a follower hides into some facilities later. The effectiveness of the missile depends on the type of missile and facility. A payoff of the game is the expected amount of destroyed value. The problem is generalized as a two-person zero-sum game of distributing discrete resources with a leader and a follower. Our problem is to derive a Stackelberg equilibrium for the game. This type of game has an abundance of applications. The problem is first formulated into an integer programming problem with a non-separable objective function of variables and it is further equivalently transformed into a maximin integer knapsack problem. We propose three exacts methods and an approximation method for an optimal solution. Ó 2007 Elsevier B.V. All rights reserved. Keywords: Missile allocation; Stackelberg equilibrium; Game theory; Two-person zero-sum game; Integer programming
1. Introduction In this paper, we deal with a procurement problem of missiles involving the efficient assignment of the missiles to some targets. Within a fixed amount of budget, a leader purchases several types of missiles, which he assigns to facilities of a follower to damage as much value as possible the follower hides there later. The effectiveness of the missile depends on the type of missile and facility. A payoff of the game is the expected amount of destroyed value. Table 1 shows the budgeted military expenditure of some countries for recent several years, published in SIPRI (Stockholm International Peace Research Institute) Year Book 2005. The numbers of expenditure are written in local currency in the upper and the increment ratios of the expenditure to the preceding year are in the lower. Individual country has his own military and diplomatic policy, the international circumstance around him or economic situation, which affects his military expenditure. Since September 11 in 2001, the USA increases its military expenditure rapidly as a result of massive budgetary allocations for the global war on terrorism. However this rapid growth is a rare case among many countries although it is the case for Russia and China. The expenditures of many developed countries, the UK, Germany, France and Japan, do not change so drastically. For Japan, the expenditure has been staying at the same level and their expense for the procurement of military equipments is also staying around 18–19% of the gross expenditure. Without special incidents or reasons, the developed countries confront these circumstances in greater or lesser degree and the defense ministry is requested to make an efficient procurement plan for military equipments within the limited amount of budget.
*
Corresponding author. Tel.: +81 46 841 3810; fax: +81 46 844 5911. E-mail address: [email protected] (R. Hohzaki).
0377-2217/$ - see front matter Ó 2007 Elsevier B.V. All rights reserved. doi:10.1016/j.ejor.2007.10.033
R. Hohzaki, S. Nagashima / European Journal of Operational Research 193 (2009) 238–249
239
Table 1 Military expenditure Country
Currency
Financial year 1999
2000
2001
2002
2003
2004
Japan
Billion yen
USA
Million dollars
UK
Million pounds
France
Million euro
Germany
Million euro
Italy
Million euro
Russia
Million rubles
China
Billion yuan
4934 0.002 280969 0.024 22548 0.003 36510 0.014 30603 0.026 22240 0.056 167822 0.961 165 0.107
4935 0.000 301697 0.074 23532 0.044 36702 0.005 30554 0.002 24325 0.094 273106 0.627 182 0.103
4950 0.003 312743 0.037 24464 0.040 37187 0.013 30648 0.003 24592 0.011 365374 0.338 215 0.181
4956 0.001 356720 0.141 26227 0.072 38681 0.040 31168 0.017 25887 0.053 457640 0.253 251 0.167
4954 0.000 414400 0.162 31286 0.193 40212 0.040 30800 0.012 24421 0.057 567692 0.240 274 0.092
4916 0.008 466600 0.126 29868 0.045 41822 0.040 30515 0.009 25160 0.030 655787 0.155 305 0.113
The problem of missile procurement is described as follows. Player A is going to hide a total volume V of valuable material in T shelters. On the other hand, Player B is procuring m types of missiles using budget C. The price of a type j of missile is cj and its single shot kill probability (SSKP) against Shelter i is estimated to be bij . Some of the missiles are for special-purpose use so that they perform very effective against specific shelters but perform bad against other types of shelters. Some of the missiles are for general-purpose use. They work so against almost all types of shelters. The characteristics of their performance affect their prices. The special-purpose missile is comparatively expensive. On the other hand, the general-purpose missile is cheaper because it could be made through the standardized mass-production line in a factory. Player B would be concerned about the combination or the trade-off between these missiles in his procurement plan. After the purchase of missiles by Player B, Player A can observe a Player B’s assignment plan of missiles to the shelters of Player A from some intelligence information. Based on the observation, Player A can make the best plan to hide his material into T shelters so as to maximize the amount of material surviving from the possible attack of missiles by Player B in the future. Considering the best response of Player A, Player B has to make an efficient procurement plan of missiles to damage Player A’s valuable material as much as possible, prior to Player A’s decision making. The missile procurement problem is generalized to a two-person zero-sum game of distributing discrete resources with a leader and a follower. For the so-called resource allocation problem (RAP) with discrete resources, many researchers have studied one-sided optimization problems on Player B’s side so far [8]. Fruitful results have been accumulated mainly for the problems with constraints on the total amount of resources or separable objective functions of decision variables. As an application in the military field, they study so-called target assignment problem or missile allocation problem, and search problem. As a target assignment problem, Manne [5] deals with how to assign a fixed number of single type of missiles to multiple targets. Lemus and David [4] or Nickel and Mangel [6] extend Manne’s problem to a multi-type missile problem but their constraints are still placed on the number of missiles. The constraint on the number of discrete resources is easy to deal with mathematically compared with cost constraints. As an example of the search problem, we can take Kadane’s research [7], where a searcher decides which cell he searches knowing a distribution probability of a hidden target in cells. Every time during a fixed number of time points, he repeats making the decision with constraints on search cost to maximize the total detection probability. Hohzaki and Iida [3] generalize the target assignment problem and the search problem to a general RAP with cost constraints and propose methods for an optimal solution. We have reviewed one-sided optimization problems until now. In this paper, we are going to discuss a two-person zerosum game of distributing discrete resources with a leader and a follower, and to derive a Stackelberg equilibrium for it. One of applications of the resource allocation game is the missile procurement problem, described above. We can think of other military applications as follows: 1. (Example 1) An inspection game with uncertain placement of illegal material. An inspector (Player B) is going to dispatch an inspection team to an inspectee country (Player A). The inspection team consists of m types of specialists, e.g. chemists, physicists, engineers and so on. The dispatching cost depends on the specialties of members. The inspector wants to organize the team within an assigned inspection budget C in a cost-efficient way. Each inspection member is supposed to go to one of T facilities in the inspectee country and is expected to find illegal material if exist. The detection probability of the illegal material depends on the specialty of the dispatched member and the facility type. After observing the mem-
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bers of the inspection team, the inspectee is going to keep illegal material in secret as much as possible after surviving the inspection if he takes an illegal action. The payoff of the game is the expected volume of material found by the inspection team. 2. (Example 2) A P hide-and-search game in cells. A target hides in T cells. He takes a mixed strategy of hiding in cell i with T probability pi ; i¼1 p i ¼ 1. At each time j of m time points, a searcher can look into any cell as many times as he wants, spending cost cj for one looking. The detection probability of the target by one looking depends on the time point and the cell if the target is there. The problem is to find an optimal looking plan for the searcher on a time axis within a fixed search budget C and to find an optimal mixed strategy of the target. For the sake of comprehensibility, we return back to the missile procurement problem although it has some equivalent problems, stated above as Examples 1 and 2. From now on, let us discuss a Stackelberg game of the procurement problem and propose some methods to derive its equilibrium. In the next section, we begin with an exact description of our mathematical model and formulate it into an integer programming problem with constraints on cost and an exponential form of objective function. We will show that the problem can be reduced to a maximin integer knapsack problem. In Section 3, we propose three exact methods and an approximation method for an optimal strategy. By numerical examples, we analyze a missile procurement problem from a cost-efficient point of view in Section 4 and conclude the paper with some remarks in Section 5. 2. Modeling and formulation Here we model a missile procurement game, which a leader, say Player B, and a follower, say Player A, play. (A1) Player B purchases m types of missiles at an expense of C to destroy valuable material that Player A will hide in his facilities later. A type j of missile has price cj . Using a j-type missile, Player B can demolish Facility i with probability bij > 0. (A2) Player A has a total V of continuously-divisible valuable material. After observing the missile allocation of Player B, he is going to divide the material into some parts and hide them in T facilities, e.g. shelters, warehouses or factories, numbered from i ¼ 1 to T, to keep the material safe from the attack of Player B’s missiles. (A3) The players take actions as follows. First, Player B procures some missiles and then Player A makes a decision about how he divides and hides his valuable material in his facilities. Player B desires to lose Player A’s material as much as possible and Player A as little as possible. Player B is a leader and Player A is a follower as decision makers in the game. We assume that C and ci are natural numbers and that V and bij are real numbers. We denote a strategy of Player A by vi P 0, which indicates the amount of the material hidden in Facility i and satisfies PT v i¼1 i ¼ V from Assumption (2). For a Player B’s strategy of assigning missiles, we use notation xij 2 Z as the number of type assigned to Facility i of Player A. We can represent a constraint on the expense of purchasing missiles by Pm j missiles PT c x 6 C. If we denote a set of facilities by I f1; . . . ; T g and a set of missile types by J f1; . . . ; mg, feasible j¼1 j i¼1 ij regions for the strategies of Player A and B, or D and A, are given by ( ) T X D ¼ ðv1 ; . . . ; vT Þ 2 RT vi P 0; i ¼ 1; . . . ; T ; vi ¼ V ; ð1Þ i¼1 ( ) m T X X IJ cj xij 6 C ; ð2Þ A ¼ fxij g 2 Z xij P 0; i ¼ 1; . . . ; T ; j ¼ 1; . . . ; m; j¼1 i¼1 respectively. A j-type missile can demolish Facility i with probability 0 < bij 6 1, which is referred to as the single shot kill probability (SSKP). bij mayP have another representation of bij 1 expðaij Þ, where aij > 0. Therefore, Qm x m 1 j¼1 ð1 bij Þ ij ¼ 1 expð j¼1 aij xij Þ is the kill probability of Facility i by an assignment plan of missiles fxij ; j 2 Jg. Now we can evaluate the expected amount of the material to be lost for all facilities by ( !) ! T m T m X X X X vi 1 exp aij xij vi exp aij xij : ¼V i¼1
j¼1
i¼1
j¼1
Because the first term of the above expression is a constant, we regard the second term Eðv; xÞ, which is the expected untouched amount, as a payoff function of the game from now on. The payoff Eðv; xÞ is given by
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Eðv; xÞ ¼
T X
vi exp
i¼1
m X
241
! ð3Þ
aij xij ;
j¼1
where v ¼ fvi ; i 2 Ig and x ¼ fxij ; i 2 I; j 2 Jg. Our problem is to find a minimax value and a Stackelberg equilibrium for the game with the payoff Eðv; xÞ. As a minimizer or a leader, Player B wants to minimize the payoff by a desirable assignment plan of missiles x while Player A decides an allocation of the material v as a maximizer or a follower so as to maximize the payoff. The minimax value of the game is given by ! T m X X G ¼ min max vi exp aij xij : fxij g2A fvi g2D
i¼1
j¼1
Considering the feasible region D, we can transform the maximization involved in the above problem as follows: ! ! T m m X X X vi exp aij xij ¼ V max exp aij xij : max fvi g2D
i¼1
j¼1
i
j¼1
P P m m We derive the right-hand side by letting vk ¼ 0 for k 2 I satisfying exp j¼1 akj xkj < maxi exp j¼1 aij xij . If we proceed to a minimization problem of the above result with respect to x, we have the following problem: ( ) ! m X ðP 0 ÞG ¼ min V kexp aij xij 6 k; i ¼ 1; . . . ; T ; fxij g 2 A : ð4Þ j¼1 P Pm m An inequality exp j¼1 aij xij 6 k is equivalent to log k 6 j¼1 aij xij . A replacement l log k leads us to ( ) m X V V G ¼ min aij xij ; i ¼ 1; . . . ; T ; fxij g 2 A ¼ ; ð5Þ l 6 exp l exp nC j¼1 where the number nC is given by ( ) m X ðP 1 ÞnC ¼ max ll 6 aij xij ; i ¼ 1; . . . ; T ; fxij g 2 A j¼1 ( ) X X m X ¼ max min aij xij c x 6 C; xij 2 Z þ ; i 2 I; j 2 J ; i2I i2I j2J j ij j¼1
ð6Þ
and Z þ is a set of all nonnegative integers. In the case that I consists of one element, the problem ðP 1 Þ becomes so-called integer knapsack problem (IKP). That is why we call the problem ðP 1 Þ the maximin integer knapsack problem or the maximin IKP. We can interpret our problem in the context of the classical IKP as follows. We pack m types of items into T knapsacks as long as the expense of purchasing the items does not go beyond budget C. A j-type item has a price cj and value aij when it is put into the ith knapsack. We want to increase the minimum packed value among each knapsack as much as possible so as to equalize the packed value among all knapsacks. From the above discussion, we can easily see the NP-hardness of our problem and we have the following theorem. Theorem 1. Problem ðP 0 Þ is NP-hard. An optimal strategy of Player B, xij , is given by anP optimal solution of the maximin IKP m or problem ðP 1 Þ. For Player A, he should not hide any material in Facility i for n < C j¼1 aij xij and should hide the entire Pm amount V of the material among facilities satisfying nC ¼ j¼1 aij xij in an arbitrary way of dividing V. From the theorem, we only need to focus on a solution for the problem ðP 1 Þ from now on. First of all, we explain a lemma, by which we can preclude self-evident solutions. We use the following notation. c0 is the minimum among unit prices of all kinds of items and J 0 is a set of types of items having the minimum price. For each i 2 I, a set j0 ðiÞ is defined as types of item j 2 J 0 having the maximum of coefficients aij . These are defined by c0 ¼ min cj ; j¼1;...;m
J 0 ¼ fj 2 Jjcj ¼ c0 g;
j0 ðiÞ ¼ arg max aij : j2J 0
ð7Þ
Lemma 1. (i) If C < Tc0 , an optimal value is nC ¼ 0 and one of optimal solutions is xij ¼ 0. (ii) If Tc0 6 C < 2Tc0 , an optimal value is nC ¼ min aij0 ðiÞ . One of optimal solutions is given by xij0 ðiÞ ¼ 1; xij ¼ 0; j 6¼ j0 ðiÞ i2I for each i 2 I.
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Proof. (i) is the case that budget C is too little that there is a knapsack at least that Player B cannot purchase even an item for, and an optimal value is zero. (ii) is the case that Player B can afford to buy an item with minimum price c0 for all knapsacks but cannot buy any more. An optimal value is determined by the minimum coefficient aij0 ðiÞ among all knapsacks i 2 I. h From now, we are going to discuss an optimal solution assuming C P 2Tc0 implicitly to avoid trivial cases of Lemma 1. 3. Optimal solution for the maximin integer knapsack problem 3.1. A dynamic programming method As mentioned in the preceding section, the maximin IKP contains the classical IKP as a subproblem. We can utilize some methods developed in the past for the solution of the IKP. We represent a maximal value of the IKP with budget D for knapsack n by ( ) X m m X þ ð8Þ ðIKPn ðDÞÞln ðDÞ ¼ max anj xnj c x 6 D; xnj 2 Z : j¼1 j nj fxnj ;j2J g j¼1 Similarly, we denote a feasible region for solutions and an optimal value for the maximin IKP with budget D and n knapsacks by S n ðDÞ and F n ðDÞ, respectively, which are defined as follows: ( ) X n X þ c x 6 D; xij 2 Z ; i ¼ 1; . . . ; n; j 2 J S n ðDÞ fxij ; i ¼ 1; . . . ; n; j 2 J g i¼1 j2J j ij F n ðDÞ max
m X
min
fxij g2S n ðDÞ i¼1;...;n
aij xij :
j¼1
P o m þ c x c x 6 C; x 2 Z , we have the following recursive formula with respect to the num j j j j j j¼1 j¼1
nP m
Using notation W ber of knapsacks.
( F n ðDÞ ¼ max min fxnj ;j2J g
m X
anj xnj ; F n1 D
j¼1
¼ max n d2W
!) cj xnj
j¼1
(
Pmax m
fxnj ;j2J g
m X
j¼1
cj xnj ¼d;xnj 2Z þ
o min
m X
) anj xnj ; F n1 ðD dÞ
j¼1
) # X m þ ¼ max min max anj xnj c x ¼ d; xnj 2 Z ; F n1 ðD dÞ d2W j¼1 j nj fxnj ;j2J g j¼1 ( ) " # X m m X þ ¼ max min max anj xnj c x 6 d; xnj 2 Z ; F n1 ðD dÞ ¼ max min fln ðdÞ; F n1 ðD dÞg d2W j¼1 j nj d¼1;...;D fxnj ;j2J g j¼1 "
¼
max c0 6d6Dðn1Þc0
(
m X
minfln ðdÞ; F n1 ðD dÞg
ð9Þ
The last expression is derived from Lemma 1 telling that knapsack n and other n 1 knapsacks need budget c0 and ðn 1Þc0 , respectively, to avoid trivial cases. To compute the optimal value F T ðCÞ, we only need to repeat applying the above dynamic programming (DP) (9) to D ¼ nc0 ; nc0 þ 1; . . . ; C for each n while increasing n by one from 1 to T. Initial conditions of the computation are F 1 ðDÞ ¼ 0;
D ¼ 1; . . . ; c0 1;
F 1 ðDÞ ¼ l1 ðDÞ;
D ¼ c0 ; . . . ; C:
From some past studies on the knapsack problem, we know that all optimal values ln ðDÞ; D ¼ 1; . . . ; C; are obtained in the process of solving a knapsack problem IKPn ðCÞ once. The mapping from budget D to optimal value ln ðDÞ is referred to as the integer knapsack function. If we solve the problem IKPn ðCÞ and save ln ðDÞ in advance, we can simply load ln ðdÞ for computation when we apply the dynamic programming method. Noting that the price system of items cj are same for all knapsacks, the computation time required to solve IKPn ðCÞ seems not to depend on the knapsack number n so much. If we denote the computation time by knapðCÞ, we may estimate the total time to solve the IKP for all knapsacks by T knapðCÞ. In the DP algorithm (9), the comparison operators ‘min’ and ‘max’ are transacted once for each d while the number d varies D ðn 1Þc0 c0 þ 1 ¼ D nc0 þ 1 times. Furthermore, index D changes within nc0 6 D 6 C. In the most outer
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243
repetition layer of the DP method, we change n from 2 to T. The total number of computational repetitions in the algorithm is T C X X
2ðD nc0 þ 1Þ ¼
n¼2 D¼nc0
c20 c0 ð2T 3 þ 3T 2 þ T 6Þ ð2C þ 3ÞðT 2 þ T 2Þ þ ðC þ 1ÞðC þ 2ÞðT 1Þ: 6 2
Now the order of computational complexity is estimated by OðTC 2 þ T 2 Cc0 þ T 3 c20 Þ. As the result, the complexity of the proposed dynamic programming method is the order of OðT knapðCÞ þ TC 2 þ T 2 Cc0 þ T 3 c20 Þ for the maximin IKP ðP 1 Þ. 3.2. Advance evaluation method Here we review some useful information about the classical IKP [1,2], which we are going to use for our problem. A target problem for our review is the following: m m X X ðIKPðdÞÞlðdÞ ¼ max aj x j ; s:t: cj xj 6 d; xj 2 Z þ ; j 2 J; ð10Þ x
j¼1
j¼1
where m items are sorted and numbered such that a1 =c1 P a2 =c2 P P am =cm . The relation between d and optimal value lðdÞ is called the integer knapsack function, which is usually a non-decreasing right-continuous step function for variable d. There are some algorithms developed to give discontinuity points, where the function jumps discontinuously. We sort these points of d in the increasing order like f ð1Þ < f ð2Þ < in a list and call it a step list. Let xj ðj ¼ 1; . . . ; mÞ be an optimal solution for IKPðdÞ and gðdÞ be a minimum index j for xj > 0 defined by minfjjxj > 0g; x 6¼ 0; gðdÞ ¼ ð11Þ m; x ¼ 0: Given the above situation, a solution ðx1 ; . . . ; xj1 ; xj 1; xjþ1 ; . . . ; xm Þ is optimal for IKPðd cj Þ if xj > 0 and then gðdÞ 6 gðd cj Þ holds for every j of xj > 0. Therefore, among j satisfying j 6 gðd cj Þ and d cj P 0, there is an index ~ for IKPðd cj Þ becomes optij such that a new solution generated by adding one to the jth element of optimal solution x mal for IKPðdÞ. It follows that lðdÞ ¼ maxfaj þ lðd cj Þjj 6 gðd cj Þ; d cj P 0g: j
ð12Þ
An index j to maximize the right-hand side gives gðdÞ. We also know that there is a certain budget D such that lðdÞ ¼ a1 þ lðd c1 Þ for all d of D 6 d. Namely, for IKPðdÞ with budget d larger than D , an optimal solution is always given by adding one to the first element x1 of an optimal solution for IKPðd c1 Þ. This property is referred to as the periodicity of the IKP. To confirm the periodicity, we only need to check gðdÞ ¼ 1; d ¼ D ; D þ 1; . . . ; D þ max cj 1 ð13Þ j
for a budget D . We have outlined the known properties related to the IKP. From now on, we are going to devise a method for our problem ðP 1 Þ based on these properties. Assume that we have optimal value nd and an optimal solution for budget d. To pull up the optimal value more, it is required to increase budget d i assigned to each knapsack i of li ðd i Þ ¼ nd by some adequate volume. For knapsack i 2 I, the necessary budget d i to increase li ðd i Þ can be evaluated by a step list fi ðÞ, which we obtain in the process of solving IKPi ðCÞ. The following algorithm incorporates the evaluation of the necessary increase of knapsack’s budget in advance of changing the total budget d. That is why we call the algorithm the advance evaluation method. Step 1. Solve IKPi ðCÞ for each knapsack i 2 I and obtain P an integer knapsack function li ðDÞ; D ¼ 1; . . . ; C and a step list fi ðÞ. Set initial conditions li ¼ 1; i 2 I; d i2I fi ðli Þ. In this initial step, d must be Tc0 . Step 2. Set l ¼ mini2IP li ðfi ðli ÞÞ; K ¼ fnjln ðfn ðln ÞÞ ¼ lg. Step 3. Calculate s ¼ n2K ðfn ðln þ 1Þ fn ðln ÞÞ. If d þ s > C, terminate. An optimal value is l. d i ¼ fi ðli Þ is an optimal budget to be assigned to knapsack i. Using d i , we solve IKPi ðd i Þ to obtain an optimal solution or optimal assignment of type-j items xij ; j 2 J for knapsack i. If d þ s 6 C, increase d by d ¼ d þ s; li ¼ li þ 1; i 2 K and return to Step 2. In Step 3, the algorithm increases a budget assignment for knapsack i in accordance with a step list fi ðli Þ. As the total budget d increases by s in Step 3, a corresponding optimal value l always moves up. We can verify that this method does work properly as follows. If we re-assign any part of fn ðln þ 1Þ fn ðln Þ of knapsack n 2 K to other knapsacks in Step 3, the value of the objective function stays at a tentative value l. If we decrease the assign-
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ment budget of knapsack n 62 K by some volume and add the volume to the assignment budget of other knapsacks, a current value ln ðfn ðln ÞÞ goes down to the previous value ln ðfn ðln 1ÞÞ. The assignment fn ðln 1Þ must have been increased at some time in Step 3, which means that ln ðfn ðln 1ÞÞ < l for the current tentative value l. It follows that the re-assignment of fn ðln þ 1Þ fn ðln Þ causes a smaller value of the objective function less than l. According to the above discussion, we can see that this algorithm adjusts the assignment of budget for each knapsack so as to maximize the value of the objective function. Every time budget d is renewed in Step 3, we are informed a discontinuity point, where the value of the objective function jumps up. 3.3. Sequential increment method The advance evaluation method uses the knapsack function, which is obtained after solving IKPi ðCÞ with budget C for each knapsack i 2 I. However, we can anticipate that optimal assignment budget to each knapsack can be rather smaller than C when we finally solve problem ðP 1 Þ. Solving IKPi ðCÞ in advance might be wasting. In a method proposed here, we increase a tentative assignment budget for each knapsack by using not the knapsack function but the information (11)–(13) sequentially. To do that, we incorporate the method proposed for a solution of the ordinary IKP in an algorithm here. Assume that for knapsack i, type numbers J are sorted in the decreasing order in terms of aij =cj ; j 2 J such that aili ð1Þ =cli ð1Þ P aili ð2Þ =cli ð2Þ P P aili ðmÞ =cli ðmÞ . Let us define cmax ¼ maxj2J cj . Step 1. Set initial values of parameters for every knapsack i 2 I, as follows: li ð0Þ ¼ li ð1Þ ¼ ¼ li ðc0 1Þ ¼ 0; li ðc0 Þ ¼ aij0 ðiÞ , gi ð0Þ ¼ gP i ð1Þ ¼ ¼ g i ðc0 1Þ ¼ m; gi ðc0 Þ ¼ minfkjcli ðkÞ ¼ c0 g, rðiÞ ¼ 0; d i ¼ c0 and Di ¼ c0 . ~ Step 2. Calculate d ¼ i2I d i . If d > C, go to Step 4 and terminate. Otherwise, save d i in d i , that is, set d~i ¼ d i ; i 2 I. Calculate l ¼ mini2I li ðd i Þ and K ¼ fnjln ðd n Þ ¼ lg. If d þ jKj > C, go to Step 4 and terminate. Step 3. After running the following procedure for each n 2 K, return to Step 2. Increase d n P by one, d n ¼ d n þ 1, and repeat calculating ln ðd n Þ by the following procedure until ln ðd n 1Þ < ln ðd n Þ. But if i2I d i > C in the middle of the procedure, go to Step 4 and terminate. (i) If rðnÞ ¼ 1, set ln ðd n Þ ¼ ln ðd n cln ð1Þ Þ þ anln ð1Þ . (ii) If rðnÞ ¼ 0, set j ¼ arg maxfln ðd n cln ðjÞ Þ þ anln ðjÞ jd n cln ðjÞ P 0; j 6 gn ðd n cln ðjÞ Þg; j
ð14Þ
ln ðd n Þ ¼ ln ðd n cln ðj Þ Þ þ anln ðj Þ ;
ð15Þ
gn ðd n Þ ¼ j :
ð16Þ
If j ¼ 1 and d n ¼ Dn þ cmax 1, set rðnÞ ¼ 1. If j 6¼ 1, renew Dn by Dn ¼ d n . Step 4. An optimal value is l. For each knapsack i 2 I, solve IKPi ðd~i Þ to obtain an optimal assignment of items fxij ; j 2 Jg as an optimal solution and terminate. gn ðd n Þ and Dn , which are initialized in Step 1 and renewed in Step 3, correspond to gðdÞ and D in condition (13). When the condition is verified, we mark a sign rðnÞ ¼ 1, which indicates the periodicity for optimality. The setting of (14) and (15) is corresponding to the calculation of (12) for knapsack n. In Step 2, K is a set of knapsacks with the tentative minimum objective value l. In Step 3, we increase the assignment of budget d n to each knapsack n in K until its value ln ðd n Þ is drawn up while increasing the total budget d sequentially. This main process of increasing the assignment budget to the knapsacks is as same as the advance evaluation method in essence. That is why the validity of this method is verified for its execution. 3.4. An approximation method We have discussed so far some exact methods for an optimal solution. Because our problem is NP-hard, even an approximation method also deserves the proposition. Here we are going to devise an approximation solution in the polynomial order of computational time. First of all, we relax integer valuables of problem ðP 1 Þ to real numbers. A resultant problem is ðRLP1 Þ^ l ¼ max min i2I
s:t:
m X
T X m X i¼1
aij xij ;
ð17Þ
cj xij 6 C;
ð18Þ
j¼1
j¼1
xij P 0;
xij 2 R;
i 2 I;
j 2 J:
For the relaxed problem, an optimal solution is given by the following theorem.
ð19Þ
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^ of the relaxed problem ðRLP 1 Þ are given by ^ and an optimal solution x Theorem 2. An optimal value l C
^ ¼ PT l
k¼1 csðkÞ =aksðkÞ
^xisðiÞ ¼ PT
ð20Þ
;
C=aisðiÞ
k¼1 csðkÞ =aksðkÞ
;
i 2 I; ^xij ¼ 0; i 2 I; j 6¼ sðiÞ; j 2 J;
ð21Þ
where sðiÞ arg maxj2J faij =cj g. Proof. The replacement of qij ¼ cj xij changes the original problem ðRLP1 Þ to ( ) m T X m X aij X ^ ¼ max min l q q 6 C; qij P 0; i 2 I; j 2 J : fqij g i2I cj ij i¼1 j¼1 ij j¼1 P By additional substitutions of cij aij =cj and d^i mj¼1 qij , we can transfer the problem as follows: ( ) X m T m X X ^ ^ ¼ max min cij qij qij ; qij P 0; i 2 I; j 2 J ; d 6 C; d i ¼ i¼1 i fd^i ;qij g i2I j¼1 j¼1 ( ) X T ¼ max min cisðiÞ qisðiÞ d^ 6 C; d^i ¼ qisðiÞ P 0; i 2 I ; i¼1 i fd^i ;qisðiÞ g i2I ( ) X T ^ ^ ^ ¼ max min cisðiÞ d i d 6 C; d i P 0; i 2 I ; i2I i¼1 i fd^ i g ( ) T X ¼ max ll 6 cisðiÞ d^i ; i 2 I; d^i ¼ C; d^i P 0; i 2 I : fd^ i ;lg i¼1
ð22Þ
An optimal value for the last reduced problem is given by T X ^ ¼ cisðiÞ d^i ; i 2 I; l d^i ¼ C: i¼1
This simultaneous system of equations has a solution of C C ¼ PT ; 1=c c ksðkÞ k¼1 k¼1 sðkÞ =aksðkÞ
^ ¼ PT l
csðiÞ =aisðiÞ d^i ¼ C PT : k¼1 csðkÞ =aksðkÞ
Transformation (22) is possible by the setting of d^i ¼ qisðiÞ ; i 2 I; qij ¼ 0; j 6¼ sðiÞ; j 2 J. That confirms us that the theorem gives us an optimal solution f^xij g. h Using the theorem, we can propose a simple greedy algorithm as an approximation method. The first step is to derive an integer solution fxij g ¼ fb^xij cg by rounding off a solution f^xij g of the relaxed problem ðRLP1 Þ. Residual budget PP C i j cj xij is used for the purchase of the cheapest items with price c0 for each knapsack in K, which is defined in Step 2 of the advance evaluation method and the sequential increment method. An entire set of the greedy algorithm is as follows: Step 1. By Theorem 2, solve the relaxed problem P P ðRLP1 Þ to obtain P a real solution f^xij g. Round off the solution by fxij g ¼ fb^xij cg. Calculate C R ¼ C i2I j2J cj xij and li ¼ j2J aij xij ; i 2 I. Step 2. l ¼ mini2I li ; K ¼ fnjln ¼ lg. Step 3. If jKjc0 > C R , terminate with an approximation value l and an approximation solution fxij g. Otherwise, modify a tentative solution fxnj g for each n 2 K as follows: xnj0 ðnÞ ¼ xnj0 ðnÞ þ 1;
ln ¼ ln þ anj0 ðnÞ ;
where j0 ðnÞ is given by definition (7). Return to Step 2 after renewing C R ¼ C R jKjc0 .
4. Numerical examples Here we consider a concrete budgeting problem of procuring missiles from the cost-efficient point of view. Player A is going to hide a total amount V of valuable material in T ¼ 4 shelters. On the other hand, Player B is going to purchase
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m ¼ 5 types of missiles using budget C. The price cj of a type j of missile and an estimated SSKP of a j-type missile against Shelter i, bij , are illustrated in Table 2. A type j ¼ 1; . . . ; 4 of missile is specifically effective for the same number of shelter j. But a type-5 of missile is for general-purpose use and effective to every shelter to some extent. The general-purpose missile costs cheaper than special-purpose one. The prices of special-purpose missiles have an order of c1 6 c2 6 c3 6 c4 . Player A can observe a Player B’s assignment plan of missiles aiming at the shelters of Player A from some intelligence information and he can make a best hiding plan of his material into m shelters. Considering this situation comprehensively, Player B is requested to make a cost-efficient procurement plan of missiles, especially taking account of the efficient combination of special-purpose and general-purpose missiles. The value V = exp nC given by Eq. (5) is the minimax expected amount of Player A’s material surviving the Player B’s attack and 1= exp nC is the minimax ratio of the survived amount to the total, i.e. the minimax survivorship ratio, where nC is an optimal value of the minimax IKP ðP 1 Þ. Fig. 1 illustrates the survivorship ratios in percentage while changing an integer total budget C from 0 through 50. The curve has a step-function form as the knapsack function does. Because of the form, some ratios can be the same for different budgets, e.g. we can see the same ratios from C ¼ 26 through 40. Considering the sensitivity of the budget, Player B asks some contractors or makers of missiles to compare the following eight alternates besides of the basic plan of Table 2. 1. 1SU: Upgrade the basic specification of the special-purpose types of missile j ¼ 1; . . . ; 4 and increase their prices, i.e. improve ajj of the j-type missile and price up cj by one for j ¼ 1; . . . ; 4. 2. 1GU: Upgrade the basic specification of the general-purpose missile j ¼ 5 and increase its price, i.e. improve ai5 ; i ¼ 1; . . . ; 4 and price up c5 by one. 3. 2SU: Price up cj of the special-purpose missiles j ¼ 1; . . . ; 4 by two and improve their basic performances ajj more over Alternate 1SU. 4. 2GU: Price up c5 of the general-purpose missile by two and improve its performance ai5 ; i ¼ 1; . . . ; 4 more over 1GU. 5. 1SD: Price down cj of the special-purpose missiles j ¼ 1; . . . ; 4 by one and downgrade their performances ajj . 6. 1GD: Price down c5 of the general-purpose missile by one and downgrade its performance ai5 ; i ¼ 1; . . . ; 4. 7. 2SD: Price down cj of the special-purpose missiles j ¼ 1; . . . ; 4 by two and downgrade their performances ajj more over 1SD. 8. 2GD: Price down c5 of the general-purpose missile by two and downgrade its performance ai5 ; i ¼ 1; . . . ; 4 more over 1GD. The naming of the alternates has a rule as follows. Alphabet ‘S’ or ‘G’ stands for ‘‘special-purpose’’ or ‘‘general-purpose’’, respectively. ‘U’ or ‘D’ indicates ‘‘upgrade’’ or ‘‘downgrade’’. Number ‘1’ or ‘2’ preceding those alphabets represents the extent to which the missile is upgraded or downgraded, or the missile is priced up or down. Performance parameter a0ij corresponding to the basic SSKP bij in Table 2 is calculated by bij ¼ 1 expða0ij Þ. Other parameters ajj ; j ¼ 1; . . . ; 4, for Alternate 1SU and 2SU, and ai5 ; i 2 I for 1GU and 2GU are calculated by the following evaluation formulas. 1 1 1U 0 2U 1U aij ¼ aij 1 þ ; aij ¼ aij 1 þ : ð23Þ cj þ 1 cj þ 2 Similarly the performance parameters corresponding to downgraded alternates are evaluated by 1 1 0 2D 1D ¼ a 1 ¼ a 1 a1D ; a : ij ij ij ij cj 1 cj 2
ð24Þ
The SSKPs reduced from formulas (23) and (24) are shown in Table 3. For the upgraded alternates, more expensive missile of 2SU or 2GU is improved less than less expensive one of 1SU or 1GU, compared with the increment of price. For the Table 2 Single shot kill probabilities Shelters
1 2 3 4 ci
Types of missile 1
2
3
4
5
0.60 0.01 0.01 0.01 5
0.01 0.60 0.01 0.01 6
0.01 0.01 0.60 0.01 7
0.01 0.01 0.01 0.60 8
0.20 0.20 0.20 0.20 4
Ratios(%)
R. Hohzaki, S. Nagashima / European Journal of Operational Research 193 (2009) 238–249
247
100 90 80 70 60 50 40 30 20 10 0 0
5
10
15
20
25
30
35
40
45
50
C Fig. 1. The minimax expected survivorship ratio by basic-performance missiles.
Table 3 Single shot kill probabilities for alternates
b11
b22
b33
b44
bi5 ; i 2 I
0.705 – 0.657 – 0.600 0.497 – 0.368 –
0.692 – 0.649 – 0.600 0.520 – 0.423 –
0.682 – 0.643 – 0.600 0.534 – 0.457 –
0.674 – 0.639 – 0.600 0.544 – 0.480 –
– 0.268 – 0.235 0.200 – 0.138 – 0.072
Ratio(%)
2SU 2GU 1SU 1GU Basic plan 1SD 1GD 2SD 2GD
SSKPs
100 90 80 70 60 50 40 30 20 10 0
1SU 1GU 2SU 2GU
0
5
10
15
20
25
30
35
40
45
50
C Fig. 2. The minimax expected survivorship ratio by upgraded plans.
Ratio(%)
Alternates
100 90 80 70 60 50 40 30 20 10 0
1SD 1GD 2SD 2GD
0
5
10
15
20
25
30
35
40
45
50
C Fig. 3. The minimax expected survivorship ratio by degraded plans.
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R. Hohzaki, S. Nagashima / European Journal of Operational Research 193 (2009) 238–249
Ratio(%)
downgraded alternates, a cheaper missile of 2SD or 2GD is more downgraded than a missile of 1SD or 1GD, compared with the decrement of price. Fig. 2 illustrates the minimax expected survivorship ratios for the upgraded alternates and Fig. 3 for the downgraded alternates. The ratios for all alternates, including the basic plan, are shown in Fig. 4. We can improve the sensitivity of the survivorship ratio to budget C by comparing several alternates. The improvement of the sensitivity enables us to budget for the construction of weapons with elaborate consideration from the cost-efficient point of view. In Table 4, each small table has some information about an optimal solution for a given budget C: the best alternate, the minimax expected survivorship ratio (%) and an optimal assignment of missiles to facilities fxij ; i 2 I; j 2 Jg, in detail. In
100 90 80 70 60 50 40 30 20 10 0
0
5
10
15
20
25
30
35
40
45
50
C Fig. 4. The minimax expected survivorship ratio by all plans.
Table 4 Optimal assignment of missiles to shelters Shelters
Types of missile
C = 8–11
2GD
1 2 3 4
0 0 0 0
C = 18–20
2SD
1 2 3 4
1 0 0 0
C = 26–29
Basic
1 2 3 4
1 0 0 0
C = 38–41
2SU
1 2 3 4
1 0 0 0
C = 44–47
1SD
1 2 3 4
2 0 0 0
1
2
3
4
0 0 0 0
0 0 0 0
1 1 1 1
63.3 0 1 0 0
0 0 1 0
0 0 0 1
0 0 0 0
40.0 0 1 0 0
0 0 1 0
0 0 0 1
0 0 0 0
31.8 0 1 0 0
0 0 1 0
0 2 0 0
0 0 2 0
0 0 0 1
0 0 0 1
25.3 0 0 0 2
Types of missile
C = 12–14
1GD
1 2 3 4
0 0 0 0
C = 21
2SD
1 2 3 4
2 0 0 0
C = 30–33
1SU
1 2 3 4
1 0 0 0
C = 42
2SU
1 2 3 4
1 0 0 0
C = 48–50
1SD
1 2 3 4
3 0 0 0
5
92.8 0 0 0 0
Shelters
0 0 0 0
1
2
3
4
0 0 0 0
0 0 0 0
1 1 1 1
57.7 0 1 0 0
0 0 1 0
0 0 0 1
0 0 0 0
36.1 0 1 0 0
0 0 1 0
0 0 0 1
0 0 0 0
30.8 0 1 0 0
0 0 1 0
0 2 0 0
0 0 2 0
0 0 0 1
0 0 1 1
23.1 0 0 0 2
Types of missile
C = 15–17
2SD
1 2 3 4
1 0 0 0
C = 22–25
1SD
1 2 3 4
1 0 0 0
C = 34–37
2SU
1 2 3 4
1 0 0 0
C = 43
2SD
1 2 3 4
3 0 0 0
5
86.2 0 0 0 0
Shelters
0 0 0 0
1
2
3
4
5
80.0 0 1 0 0
0 0 0 0
0 0 0 0
0 0 1 1
50.3 0 1 0 0
0 0 1 0
0 0 0 1
0 0 0 0
32.6 0 1 0 0
0 0 1 0
0 0 0 1
0 0 0 0
29.5 0 3 0 0
0 0 2 0
0 0 0 2
0 0 0 0
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each small table, the range of budget C, the best alternate and the minimax ratio are written in the first row. In below rows, an optimal assignment of missiles are tabled. We can analyze the results as follows. (1) 0 6 C 6 7: For such low budgets as C < Tc0 ¼ 8, Player B cannot construct enough missile to be assigned to all shelters of Player A. (2) 8 6 C 6 11: Player B cannot afford to purchase any expensive special-purpose missile and comes to buy some general-purpose missiles by the most degraded plan 2GD because the missiles are comparatively cheaper. (3) 12 6 C 6 14: As similar to the preceding case, Player B buys only general-purpose missiles but he use a moderately degraded plan 1GD. (4) 15 6 C 6 17: By Plan 2SD, Player B constructs special-purpose missiles j ¼ 1; 2, which are assigned to Shelter i ¼ 1; 2, because their prices are comparatively cheaper. To each of Shelter i ¼ 3; 4, he assigns a general-purpose missile with basic performance. (5) 18 6 C 6 21: Player B has enough budget to assign special-purpose missiles to all shelters although he uses the most degraded plan 2SD. (6) 22 6 C 6 25: Player B uses a similar assignment to the preceding case but he adopts the moderately degraded plan 1SD. (7) 26 6 C 6 29: As similar to the preceding case, Player B assigns a special-purpose missile with basic performance to each shelter. He can reach the same effectiveness by Alternate 1GU, 2GU, 1GD or 2GD. (8) 30 6 C 6 33: Player B uses the upgraded plan 1SU of assigning only special-purpose missiles. (9) 34 6 C 6 42: Player B uses the most upgraded plan 2SU and assigns the highest-quality missiles to every shelter. He complements an excess budget by purchasing some general-purpose missiles. (10) C ¼ 43: Player B purchases several special-purpose missiles for every shelter by the degraded plan 2SD. The assignment of multiple special-purpose missiles is effective for shelters, even though the missiles are downgraded. (11) 43 6 C: Player B still focuses on the multiple assignment of special-purpose missiles but he uses the moderately degraded plan 1SD.
5. Conclusions In this paper, we deal with a Stackelberg equilibrium for a missile procurement problem. The problem can be regarded as a distribution game of general discrete resources with a leader and a follower. This type of problem has a wide variety of applications in the fields of the military or international affairs, as mentioned in Introduction. As we did in this study, there are many needs for the extension of classical resource allocation problems (RAPs) to games. Especially, it is more difficult to propose practical solution methods for the discrete optimization problem of the RAP than other types of the RAP. Therefore, we cannot avoid the development of practical algorithms for the discrete RAP formulated in the game. This paper is a trial for the development. Although for the integer programming problem formulated in this paper, its objective function is not separable for variables, we can transform the problem into a maximin integer knapsack problem (a maximin IKP) equivalently. This paper utilizes mathematical theory known about the IKP to propose some solution methods by incorporating the theory. One of main purposes of this paper is to propose methods for an equilibrium of a resource allocation game with discrete resources. To improve algorithms for the maximin-type of IKP further or develop more efficient algorithms from a computational point of view is another. This would be a future study for us. We deal with a Stackelberg-type of game in this paper but we leave the extension of it to a standard game with simultaneous decisions of players as our future work. References [1] P.C. Gilmore, R.E. Gomory, The theory and computation of Knapsack function, Operations Research 14 (1966) 1045–1074. [2] R.S. Garfinkel, G.L. Nemhauser, Integer Programming, John Wiley & Sons, 1972. [3] R. Hohzaki, K. Iida, An integer resource allocation problem with cost constraint, Journal of the Operations Research Society of Japan 41 (1998) 470– 482. [4] F. Lemus, K.H. David, An optimal allocation of different weapons to a target complex, Operations Research 11 (1963) 787–794. [5] A.S. Manne, A target-assignment problem, Operations Research 7 (1959) 258–260. [6] R.H. Nickel, M. Mangel, Weapon acquisition with target uncertainty, Naval Research Logistics 32 (1985) 567–588. [7] J.B. Kadane, Discrete search and the Neyman–Pearson lemma, Journal of Mathematical Analysis and Applications 22 (1968) 156–171. [8] T. Ibaraki, N. Katoh, Resource Allocation Problem, The MIT Press, London, 1988.