OPTIMAL RESOURCE ALLOCATION AMONG AREAS OF EQUAL VALUE
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1 80 NZOR volume 4 number 2 July 1976 OPTIMAL RESOURCE ALLOCATION AMONG AREAS OF EQUAL VALUE John S. Cr o u c h e r* Macquarie Un i v e r s i t y, Sydney S u m m a r y This p a p e r c o n s i d e r s a two s i d e d r e s o u r c e a l l o c a t i o n g a m e in w h i c h b o t h p l a y e r s have a f i x e d n u m b e r of i n d i v i s i b l e u n i t s w h i c h m u s t be d i s t r i b u t e d o v e r a r e a s o f equal value. A r e s t r i c t i o n f o r b o t h sides is that there is only a l i m i t e d n u m b e r o f c h o i c e s f o r the n u m b e r of units w h i c h may be a s s i g n e d to a n y g i v e n target. A m e t h o d o f s o l u t i o n is p r e s e n t e d f o r the case in w h i c h b o t h sides have up to three such c h o i c e s, a n d a n e x a m p l e is p r o v i d e d to i l l u s t r a t e the theory. 1. Introduction For the past twenty years the subject of resource allocation has been widely researched, and numerous types of mathematical models have been employed in many fields. For example, the situation described in this paper may be regarded as one applicable to economics or accounting, where there are two firms in competition which operate on fixed budgets of $X and $Y respectively. These budgets are to be invested in an area which has been subdivided into k regions such that the potential profits to be gained from each region are equal. The problem for each firm is to allocate their budgets without knowledge of the other's allocation, and in doing so maximise their expected return. The reader will be able to visualise similar applications of the theory, but for purposes of the discussion here, however, the situation will be considered as one involving attack and defence of equally valued targets. The problem discussed is a complex one of determining optimal * Manuscript received August 26, 1975: revised Sept. 29, 1975.
2 81 offensive and defensive strategies in the (game-theoretic) ease when each side knows the others total stockpile size, but not his target allocation. Previous research in this area includes a model by Dresher I 4 J in which the defensive stockpile does not exceed that of the attacker and the targets are of unequal value. In this case an artificial payoff function is used, and the interesting result is that the optimal offensive strategy is to attack a single target with the entire stockpile. Cooper and Restropo [ 1] have used a similar model, although the optimal strategies have only been determined for special values of the parameters. Game theoretic models in which the units are assumed to be divisible may be found in Croucher [ 2] and Danskin [ 3]. 2. Formulation Suppose that there are k targets, where each target has a numerical value (to both sides) of v. The attacker has X units and the defender Y units, and these units may be allocated among the k targets. It is assumed that there is only a limited number of choices which can be made for the number of units allocated by either side to any one target. set Define a for the possible number of attacking units which may be used by Sv = {a.,...,a } where a, < a_ <... < a. A ' X 1 m 1 2 m strategy for the attacker corresponds to the selection of a vector r of real numbers such that r = {r.,...,r } where r l m i represents the fraction of the k targets to be attacked by a. units, l It is convenient to formulate the problem in terms of parameters A and D, defined by A = X/k and D = Y/k. Thus A and D represent the "average number of attacking and d e fensive units per target, respectively. In order for the problem to have a non-trivial solution for the attacker, it will be assumed that
3 82 since otherwise each target could be attacked with am units. By definition of r, the following conditions must hold: (2) E a.r. = A, i = l 1 m (3) E r. = 1, i = l 1 (4) r i > 0, i = 1,...,m. Similarly, define a set of Sy for the choice of the number of defensive units by Sy = {d^,...,dn } where d^ < d 2 <... < dn< A strategy for the defender is the selection of a vector s of real numbers such that s = {s...s } where s. represents the fraction of the k I n j targets defended by dj units. The following restrictions apply: (5) I d s - D < d J =1 J J (6) Z s. = 1 j = l J (7) Sj > 0, j = 1,...,n. Let f ^ - = f(a^,dj) be a general damage function representing the probability that a^ attacking units can destroy a target defended by d^ defensive units. If it is assumed that the values gained at each target are additive, and that both sides allocate their entire resources, then the expected p a y off (number of targets destroyed) to the attacker is given by the bilinear function m n (8) F (r,s ) = kv Z Z f., r s. ~ ~ i=l j=l 13 1 J
4 8 3 Hence the attacker chooses a point given by r = (r, r ) from a convex set G in Rm, subject to (2), (3), and (4), and the defender chooses a point s = (s,...,s ) from a convex set H in Rn, subject to (5), (6), and (7). From the minimax theorem for bilinear forms over convex sets, there exist optimal strategies for the players. The r, s are chosen such that inin F(r, sj = max F(r, s ) = value of game, sell ' reg 3. Special Case: m=n=3 In this section it is assumed that the values of X,Y,k are arbitrary fixed positive integers, v is a real number and fjj represents an arbitrary payoff function. From (8), the payoff function may be written 3 3 (9) F ( r,s ) = kv il 1 ^ fij r isj * Inequalities (1) and (5) imply that 0 < A < a^ and 0 < D < dj. From (2) and (3) it can be shown that (10) r 1 = [ (a,-a2) r 3 + (a2-a) ]/ (a2- a ^, (11) r2 = I (A-aj) + ( a ^ a ^ r 3J/ (a2-aj), and from (5) and (6) (12) Sj = l (d 3-d,) s^ * (d,-d))/ (d2-d1 ), (13) s 2 = [ (D-d1 ( (d1-d,) s 3]/(d2-d1). The expected payoff may now be written as a bilinear function F* of the variables r^.s^. From equations i) to (13),
5 84 (14) F* (r3,s3) = kv[c1r3s3 + c2r3 + c3s3 + c4l / ( a ^ a ^ (d^dj) The c. are constants given by (15) C 1 = Ml p (16) c 2 = ~ 4 p* (17) C3 = *1 p* T ^2 * T? 2 T v 3» and (18) C4 I * c p* T y 3 where the are vectors defined below, P represents the 3x3 matrix whose entries are the fi, and P* is the same matrix as P except that the entry f33 = 0: Vi = (d3-d2, d1 -d3, d2 -d1), v 2 - (a3-a2, ar a3, a ^ a ^, \?3 ~ (a2-a, A -a ^» } and v4 = (d2-d, D- d ^, d 2-d1). (where the superscript T denotes vector tranpose). As ^rom anci it: f llows that (19) max (0,(A-a2)/(a3-a2)) < r3 <_ (A-a^)/ (a3*a^)) Also, s1,s2 0 implies that (20) max (0,(D-d2)/ (d3-d2) s3 (D-d1)/(dj-d^) The left hand sides of inequalities (19) and (20) are determined by the values of A and D. Consider the ordered pair (A,D) as belonging to one of four regions (i = 1,...,4)
6 85 as shown below in Figure 1. D Figure 1 R 2 R4 R 1 Once the region is determined for the given pair (A,D), inequalities (19) and (20) are fixed. For each of the R-, we can graph the corresponding areas for mapped out by (19) and (20). Consider the feasible area for the location of the ordered pair (r^.s^), and divide this into four areas, A.,, Aj, A ^, as shown in Figure 2. Figure 2 (D-d1)/(d^'djI (D-d2)/(d3-d2J * r. (A-a2)/(a3-a2) (A-ap/fa^-aj) Table 1 shows the feasible area for (r^.s^) for the corresponding region containing (A,D).
7 86 Table 1 Corresponding R^ A^ and A-, R 2 A 2 R 3 A., and A 3 R4 A 1, A-,, A, and A / Since F* is bilinear in (r^.s-) it will have a saddle point. By definition, a point (xq,yo ) will be a saddle point of F* if it falls in a feasible area, and at r^ = x q, s^ = y : (21) = = 0, Sr3 9s3 32F* 82F* 92F* 32F* n and A = ^r < 9r3 8s 3 3r3 3s3 3s3 3r3 It is easy to verify that A < 0 for all parameter values, and (14) and (21) imply that (22) x Q = -c3/ c 1, y Q = -c2/c l. The steps for determining the appropriate saddle point (and hence the optimal strategies for both sides) are outlined b e l o w : 1. From (15), (16), (17), (22) the values of x0 >Y0 may be found and written as linear functions of A and D, respectively, i.e., (23) Xo = 'C 3/ C l = Y! A + Y 2. ^o = "c 2/c l = y 3 D + y 4 where the may be obtained from (15j, (16), and (1 J.
8 Consider a particular region. The corresponding area may be found from Figure 2 and Table 1, and will consist of a set of inequalities of the type: (24) ct1 A + a, < r. < a. A + a j ^ 4 B1 n «6, < s3 < B 3 D By combining equations (25) with the inequalities (24), a set of inequalities can be found for A and D such that (xq,y0 ) lies in the area corresponding to R ^. Let these inequalities map out the sec S^. If the ordered pair (A,D) for the given problem lies in S^, then (x,y ) is the saddle point. 4. If (A,D) lies in the region R. (R j n S ^ ), then the saddle point lies at one of the four corner points of the area corresponding to R ^. In this case the saddle point may be found by searching among the four possibilities, and using the definition that (x,y ) is a saddle o o o point if (25) F*(r-,y ) < F*(x,y ) < F* (x,st) v. i \ o o o 3 for all r,,s- in the region R.~(R. n S ). 3 ^ 6 l v l l 5. Co to step 2 and repeat steps 2 to 4 for each region R^. 6. The value of the game and the optimal values of the r., s can be determined from equations (9) to (13). 4. Numerical Example The solution for the above problem will now be determined for arbitrary values of A,D,k,v, specific sets S^,Sy and payoff function F *. Consider the following damage function
9 88 f f i d j j - f4j i - d. (d./a.) J if 0 < d. < a. y i ~ J ~ i if 0 _< a i < dj and sets Sx = {0, 2, 3}, Sy = {0, 2, 3). These sets imply that if a particular target is to be attacked or defended, then either 2 or 3 units may be used only. From equations (10) to (18), the expected payoff F* may be written in terms of r 3,s3 as follows: F*(r3,s3)=kv[ -141 r 3s As3+(65D-54)r3]/108+kvA(2-D)/4 where max (0,A-2) _< r 3 A/3, and max (0,D-2) <_ s 3 D/3. The regions and corresponding areas A^ are given in Table 2. From (22) xq = 9 A / 4 7, y Q = (65D-54)/ Table 2 Region Area Interval Interval Interval Interval for A for D for r 3 for s 3 R1 R2 R 3 R 4 [2,3) (0,2] [A-2.A/3] [ 0,D/3] [2,3) 12,3) [A-2.A/3] [D-2,D/3] (0,2] [2,3) I 0,A/3] [D-2,D/3] (0,2] (0,2] I 0,A/3] [ 0,D/3] Each of the four regions can now be examined to determine the appropriate saddle point. Region R ^ : The corresponding area from Table 2 is defined by A - 2 _< r 3 _< A/3 and 0 <_ s 3 D/3. Upon substituting the values of x0>y0 respectively, into these inequalities, we find that S ± is the set S 1 = { (A,D) 12 < A <_ 47/19, 54/65 < D < 2}. Hence if (A,D) is in S^, then (x 0»y0 ) is a saddle point.
10 89 Suppose (A,D) belongs to the set ^ ' ( ^ n Sj). Then the saddle point lies at one of the four corner points of (Aj U A 2), i.e., (A-2,0), (A/3,0), (A-2,D/3), (A/3,D/3). Using (25), it can easily be shown that (i) If (A,D) is in the set { (A,D) 2<A<47/19, 0<D<54/65}, then F*(r3,0) < F*(A-2,0) <_ F*(A-2,s3). Hence if (A,D) lies in this set, then (A-2,0) is the saddle point. (ii) If (A,D) is in the set { (A,D) 47/19 <_ A <_ 3, 0 < D <_ 2, then F*(r3,D/3) < F*(A-2,D/3) <_ F*(A-2,Sj). Hence if (A,D) lies in this set, then (A-2,D/3) is the saddle point. Region R 2 : In this case S 2 is the set {(A,D) 2 <_ A <_ 47/19, 2 <_ D < 3}. If (A,D) lies in this set then (xq,yo ) is the saddle point. The set R 2~(S2 n R 2) is { (A,D) 147/19 <_ A < 3, 2 _< D < 3. If (A,D) is in this set then F*(.r3,D/3) < F* (A- 2,D/3) ^ F * ( A - 2, s 3) and hence (A-2,D/3) is the saddle point. 7 Region R 3 : In this case the set Sj is R, and hence (x 0»y0 ) is a saddle point for all (A,D) in Region R ^ : In this case S^ is the set {(A,D) 0 < A _< 2, 54/65 _< D <_ 2}. If (A,D) lies in this set then (x0 >y0 ) is the saddle point. The set R^'fR^ n S^) is { (A,D) 0 < A <_ 2, 0 < D <_ 54/65. If (A,D) is in this set then (0,0) is a saddle point, since
11 90 h-* -fc* lo\ Ol tsi M -P* *o (0.2] o K-*j^ > (0,3) o O'! Ln Cn -P* o O'! Ln Cn -P* OnIcn Lnj 04 a o O M NJ > 5 > 4 h-» 3-A 3-A M > VOX i rsj > 1 INJ > fo O s 55 OJ I-* ho t3 h-1 1 M O h-* P* Ol M- a (/i H* O H a H O *0 04 a (/> K) Oj U o o (65D-54) H-» -P* h-± </> 04 < 04 Lri I o ** < ro to a v / Akv(2-D) 4 h-* ID > *- > * < 4^ OJ O <
12 91 F * ( r 3,0) < F*(r3,s3) < F*(0,s3) This completes the solution (in terms of r3,s3) for all possible values of the parameters A and D. A summary of the optimal strategies, as well as the value of the game, are given in Table 3. As the targets are of equal value, the specific targets to be attacked and defended can be chosen in an arbitrary fashion. 5. Remarks This paper has examined the optimal attack and defence strategies for a two sided resource allocation problem in which units are indivisible and there is a limited choice for the number of units which may be assigned to any one target. It should be noted that the optimal strategies obtained may not always yield an integral number of targets. Although the solution is correct for all values of k, it is especially appropriate for large values. In particular, if the problem is such that an integral solution is required, a rounding pro? cedure on the continuous solution may be used. Even for the case of equally valued targets, models such as the one discussed may be quite difficult to analyse. The results thus far achieved in this area, however, have been far from satisfactory, and there remains a great deal more research to be don e. REFERENCES [1] Cooper, J.N. and R.A. Restropo, "Some Problems of Attack and Defence11, Soc. Indus. Appl. Math. Rev., 9, 1967, pp [2] Croucher, J.S., "Application of the Fundamental Theorem of Games to an Example Concerning Anti-Ballistic Missile Defence, Naval Res. Log. Quart., Vol. 22, No. 1, 1975, pp [3] Danskin, J.M., The Theory of M a x - M i n, Springer-Verlag, New York, [4] Dresher, M., Games and Strategy: Theory and Applications, Prentice-Hal1^ 1961.
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