m13-19 - CHAPTER XIX LINEAR PROGRAMMING AND THE THEORY OF...

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Unformatted text preview: CHAPTER XIX LINEAR PROGRAMMING AND THE THEORY OF GAMES 1 BY DAVID GALE, HAROLD W. KUHN, AND ALBERT W. TUCKER 2 The basic “scalar” problem of linear programming is to maximize (or minimize) a linear function of several variables constrained by a system of linear inequalities [Dantzig, II]. A more general “vector” problem calls for maximizing (in a sense of partial order) a system of linear func- tions of several variables subject to a system of linear inequalities and, perhaps, linear equations [Koopmans, III]. The purpose of this chapter is to establish theorems of duality and existence for general “matrix” problems of linear programming which contain the “scalar” and “vector” problems as special cases, and to relate these general problems to the theory of zero—sum two-person games. 1. NOTATION AND INTRODUCTORY Lemmas Capital letters, A, B, C, etc., denote rectangular matrices; lower-case letters, 5, c, u, 1:, etc, denote vectors, regarded as one-column matrices; and Greek letters (lower case) 5, k denote scalars—all quantities being real. A prime is used to denote transposition: thus A’ denotes A trans- posed, and 5" denotes a one-row matrix obtained by transposng the vector b. The number of components of a vector or the numbers of rows and columns of a matrix are not specified, but of course there are some implicit relations: thus the product As: implies that the number of columns of A is the same as the number of components of 31:. Vector equations and inequalities are based on the following notation: u = 0 means that all components of u are zero; a a 0 means that no components of u are negative; u 2 0 means u g 0 with u = 0 excluded; u > 0 means that all components of u are positive. Other usages follow naturally: thus u < 0 means —u > 0, ul ; uz means ul — H2 2 0, etc. It should be noted that the inner product 1 This chapter was presented in a preliminary form by A. W. Tucker'at a meeting of the Econometric Society at Boulder, Colorado, September 2, 1949. 3 Under contracts with the Office of Naval Research. 317 318 n. GALE, H. w. KUHN, AND A. w. TUCKER [PART III b’u>0ifb20, u>{l; ofcourse, b’u20ifbgO, ug0. Matrix equations and inequalities use the same rules: thus A 2 D means A — D Z 0 (Le, each element of A —- D is nonnegative, and at least one element is positive). The following lemmas provide the basis for the theorems in this chapter. Lemma 1 expresses a fundamental property of homogeneous linear inequalities observed by H. Minkowski [1896, p. 45]. Lemma 2 is an immediate consequence of Lemma 1, and Lemma 3 is a generaliza- tion of Lemma 2. LEMMA 1: In order that a homgeneousltnear inequality Me .>__ 0 hold for all u satisfying a system of homogeneous linear inequalities A’n g 0, tt is necessary and sufictem that b = Ax for some :5 g 0. For proofs the reader is referred to J. Farkas [1901, pp. 54}, H. chl [1935 or 1950, Theorem 3], and in this volume David Gale [XVIL corollary to Theorem 2} and M. Gerstenhaber [XVIIL Theorem 11]. LEMMA 2: In order that b’u < 0for no u g 0 such that A’u ; 0, it ts necessary and suficient that As g b for some a g 0. a: PROOF: In Lemma 1 replace A by [A I] and a: by [t :l, where I de- notes an identity matrix. Then, in order that Ht; 2 0 hold for all a satisfying A'u g 0, n g 0, it is necessary and sufficient that b = Ax + t for some a g 0, t g 0. That is, in order that b’u < 0 for no u g 0 such that A’u ; 0, it is necessary and suflicient that An: g b for some a: g 0. LEMMA 3:11: order that Ru 3 OJ‘or no u g 0 such that A'u ; 0, it is necessary and sufficient that As g By for some a: g 0, y > 0. PROOF: To show that the :E, y-condition is implied by the era-condition, we proceed as follows. Let bk denote the lath column of the matrix B. Then the u—condition implies that bin < O for no u g 0 such that z A'u ; 0, —B’u ; 0. Hence,' substituting [A —B] for A and [ 1‘] yr: for as in Lemma 2, we have Axk — By;G g 6;, for some at a 0, ya: a 0. Then, summing for all columns of B, A(Zxk) — B(Eyk) g 26:5. But 26;; = Bj, Where 3' denotes a vector whose components are all 1’s. So 110321;) 5 EU + 2%). That is, since 2% g 0 and j + 2y}; 2 j > 0, we have Angy forsome x; 0,y>0. This shows that the x, y—condition is implied by the u—condition. CHAP. XIX] LINEAR PROGRAMMING AND THEORY or GAMES 319 To show that the x, y—condition implies the u-condition We assume, if possible, that B’ao S 0 for some no a 0 such that A’ao a 0. Then ufiAxgo>u$By forall x;O,y>0. But, by the x, y~condition, ufiAx g uéBy for some a: a 0,1; > 0. This contradiction shows that the denial of the a—condition implies the denial of the a, y~oondition. Therefore the 2:, y-condition implies the u—condition. This completes the proof of Lemma 3. 2. LINEAR PROGRAMMING PROBLEMS Two general dual problems of linear programming are stated below. Each is based on the same given information—three matrices, A, B, 0—— and in each a matrix D is to be determined. A matrix D having a certain property is said to be maximal or minimal (under partial ordering by the mice of matrix inequalities explained in Section 1) if no other matrix A possessing the property is such that A 2 D or A S D, respectively. PROBLEM i: To find a maximal matrix D having the property that (1) Ca: I»; By for some 3 g 0, y > 0 such that An: g By. PROBLEM 2: To find a minimal matrix D having the property that (2) Ha é U!) for some a g 0, v > 0 such that A’u 1; C'v. It will be shown (in Theorem 4) that there exists a matrix D providing solutions for both problems if the following existence conditions both hold: (3) Act: g By for some :t a 0, y > O, (4) A’a ; C’v for some a g 0, v > 0. It will also be shown (in Theorem 2) that Problem 1 admits a particular matrix D as solution if, and only if, Problem 2 also admits this D as a solution. If the matrix B consists of a single column, b, and the matrix C’ consists of a single row, c’, then D becomes a scalar, 6, and y and i: become positive scalars that may be eliminated by dividing through by them. In this case the two general matrix probiems reduce to the following two simple 320 D. GALE, H. W. KUHN, AND A. W.‘T‘UCKER [mm In scalar problems: (a) to find the ordinary maximum, 5, of the linear func- tion c’a: constrained by Am g b, m a 0, and (b) to find the ordinary minimum, 6, of the linear function We constrained by A’u ; c, u a 0. PROBLEM 15: To find a maximal sector 5 having the property that 6’2: 2 6 for some a g 0 such that Ax g b. PROBLEM 25: To find a minimal scalar 6 having the property that b’u g 6 for some u g 0 such that A’u ; c. The “diet problem” of Cornfield and Stigler [1945] furnishes a typical example of Problem 26; another, more specialized, example occurs in the “transportation problem” of Hitchcock [1941] and Koopmans [XIV]. Fundamental methods for attacking such scalar problems have been developed by Dantzig III, XXI, and XXIII]. The duality and existence theorems for Problems 15 and 26 are contained in the corollary to Theorem 2 (at the end of Section 3 of this chapter) and in the remark following the proof of Theorem 4 (in Section 4 of this chapter). If the matrix B consists of a single column, b, but 0 consists of more than one row, then D becomes a vector, d, and 3; becomes a positive scalar that may be eliminated by division. In this case the two general matrix problems reduce to the following vector problems. PROBLEM 1d: To find a maximal vector (1 having the property that Caz ; d for some a: g 0 such that Ax g b. PROBLEM 2d: To find a minimal vector d having the property that b’u g d’v for some u g 0, v > 0 such that A’u ; C’v. A representative vector problem is the “efficient poin ” problem of Koopmans [III] fr0m which the general matrix problems in this chapter have evolved. The following equations relate our notation to Keep- mans’ partitioning of his technology matrix A, commodity vector y, and price vector p, as regards primary and final commodities: A = Ami) b 3"; _7lpri: u = pprir C = Afinr d = yfim 1’ = Prin- The extension to include intermediate commodities is indicated at the end of Section 6 of this chapter. Of course, there are also vector problems, 1d' and 2d', that occur when the matrix C consists of a single row 6’ and 1) becomes a positive scalar that may be eliminated by division. CHAP. XIX] LINEAR PROGRAMMING AND THEORY or GAMES 321 3. DUALITY In preparation for the duality theorem (Theorem 2), we will now prove that the following new forms of Problems 1 and 2 are equivalent to the original forms. PROBLEM I (new form): To a matrix D having both of the folloun‘ng properties: (1) 01: ; Dy for some a: g 0, y > 0 such that Ax g By, (2*) Ca: 2 Dy for no a: g 0, y a 0 such that An: g By. PROBLEM 2 (new form): To find a metric: D having both of the following properties: (2) Ha g 19’!) for some u g 0, v > 0 such that A’a ; 0’1), (1*) Eu 5 Us for no u g 0, e g 0 such that A'u ; C'v. Properties (1) and (2) occur also in the original statements of Prob- lems 1 and 2. The new properties (2*) and (1*) are so denoted because they are equivalent to (2) and (1), respectively, as will be shown in the course of the proof of Theorem 2. It is to be remarked that a matrix D having both properties (1) and (2*) must produce equality, Ca: = Dy, in property (1), and similarly that a matrix D having both properties (2) and (1*) must produce equality, B’u = Us, in property (2). HEOREM 1'. The new forms of Problems 1 and 2 are equivalent to the original forms. PROOF: To show that a solution D for the new Problem 1 is maximal as regards matrices having property (1), let us assume, if possible, that there is a matrix A 2 D having property (1). That is, Cw ; Ay for some a: g 0, y > 0 such that As; g By. Then 02: g Ag; 2 Dy for the same a and y—thereby contradicting prop- erty (2*) possessed by D as a solution for the new Problem 1. Conse- quently, D is maximal as regards matrices having property (1). A simi- lar argument shows that a solution D for the new Problem 2 is minimal as regards matrices having property (2). To show that a solution D for the original Problem 1 possesses prop- erty (2*), let us assume, if possible, that 0220 Z Dyo for some 2:0 2 0, yo 2 0 such that Are g Byu. 322 D. GALE, H. W. KUHN, AND A. W. TUCKER [PART III Adding this to (1), we get 0(x+:co)ZD(y+yo) forsome $+$o a 0,y+yo?0 such that A(x + 2:0) g B(y + yo). In the system of inequalities 0(1 + 9:0) 2 D(y + yo) there must be at least one individual inequality containing >, and so any element in the corresponding row of D may be increased slightly without disturbing the inequality. Then D is not maximal as regards matrices having prop- erty (1)—thereby contradicting the hypothesis that D is a solution for the original Problem 1. Hence D must possoss property (2 *). A similar argument shows that a solution, D, for the original Problem 1 possesses property (1*). This completes the proof of Theorem 1. THEOREM 2 (duality theorem): A matrix D is a solution for Problem 1 if, and only if, it is a solution for Problem 2. A PROOF: It follows directly from Lemma 3, by substituting[ C] for B u, A, [ D] for B, and i: ] for u, that a matrix D has property (1) if, and — u only if, it has property (1*). Then, replacing A, B, C, D, 21,11, u, v in (1) and (1*) by —-A’, --C’, —B’, —D’, u, v, 1:, 3;, respectively, it folloWs that a matrix D has property (2) if, and only if, it has property (2*). In face of Theorem 1, this completes the proof of Theorem 2. COROLLARY: Problems 15 and 25 have a unique common solution, 6, or else no solution at all. PROOF: Mom Theorem 2 it follows that both problems have a com- mon solution 6 if either admits 5 as a solution. Suppose that 51 provides another solution for either problem. Then, by Theorem 2, 51 provides also a solution for the dual problem. Clearly, 51 cannot exceed 6 due to the maximal property of 5, nor can 6 exceed 61 due to the maximal property of 51. So 51 = 6, which completes the corollary. 4. EXISTENCE In preparation for the existence theorems (Theorems 4 and 5) we introduce a third problem based on the same data as Problems 1 and 2 and employing jointly the two properties involved in the original forms of Problems 1 and 2. PROBLEM 3: To find a matrix D that has both the following properties: (1) 02: ; Dy for some 3 a 0, y > 0 such that As; g By; (2) 3’1: g Us for some u g 0, v > 0 Such that A’u ; C’v. CHAP. XIX] LINEAR PROGRAMMING AND THEORY OF GAMES 323 A problem of this symmetric sort was formulated by von Neumann [1947] for the case in which I) reduces to a scalar 5, corresponding to Problems 15 and 26. THEOREM 3: A mm D 2's a“. solution for Problem 1 or 2 if, and only if, it is a solution for Problem 3. PROOF: It is an immediate consequence of the equivalence Of prop- erties (1) and (1*), and of properties (2) and (2*), established in the proof Of Theorem 2, that a matrix D has properties (1) and (2*) or (1*) and (2) if, and only if, it has properties (1) and (2); and of course, by Theorem 1, a matrix D has properties (1) and (2*) or (1*) and (2) if, and only if, it is a solution for the original Problem 1 or 2. This com- pletes the obvious proof. Remark: Problem 3 is not changed if the leading inequalities in prop- erties (1) and (2) are made equalities: 02: = Dy and Ho = D’v. This follows from the obvious facts (pointed out in sentences just preceding Theorem 1) that a matrix D having properties (1) and (2*) must give Co: = Dy and that a matrix D having properties (2) and (1*) must give B’u = 170. THEOREM 4 (existence theorem): There exists a solution, D, for Prob- lem 3, and so for Problems 1 and 2 also, if, and only if, the following exist- ence conditions are both satisfied: (3) Ax g By for some a: a 0,3; > 0, (4) A’u ; 0’2; for some u g 0, v > 0. PROOF: Let b = Bye, and c = 0%, where yo and so are the values of y and v in any particular set Of as, y and u, I) that satisfy the existence conditions (3) and (4). Then (3) and (4) imply that (36) An: g b for some 3 g 0, (46) A’u ; c for some u g 0. [These two conditions are denoted by (36) and (45) because they are the counterparts of (3) and (4) for the scalar problems, 15 and 25.] By Lemma 2, (35) and (46) are equivalent to (35*) b’u < O for no u g 0 such that A’u ; O, (45*) 6’2: > 0 for no 5 a 0 such that As :5; 0, where in the ease of (46) and (45*) we must replace A, b, u, x in Lemma 2 by -A', —c, 9:, u, respectively. 324 n. GALE, H.‘ w. KUI—IN, AND A. W. TUCKER [PART III The inequality b’u g c’m holds for all 7k 2 D, u g 0, x g 0 such that Act g Rb, A’a ; M. For, if A > O, we have b’u ; h‘lu'Ax ; c’x, and, if A = 0, we have b’u ; 0 ; c’r, by (36*) and (45*). Consequently, 0 ’ A x b u < 0 for no 'u g 0 —c a; x —c' k such that 0 A a g 0. -A' 0 2: So, by Lemma ‘2, b’ ——c’ O 0 A [no]§ b forsome [no];(). :30 1170 —A’ 0 -c Multiplying these out, We get b’uo g c’azo, Azo g b, A’uo g c -for some no a 0, 2:0 2 0. But b’up g. ugAxo _2_ deg, so (5) b’uo = its/1:50 = do). That is, replacing b and c by Ego and 0%, we have 11433210 = utAwo = vtho- Let c ’B h 'B c " D = 3,0110 or 1:0 + 370.7 quIo 90h .7 310 h and j denoting vectors all of whose components are 1’s. Then, in either case, according as 11434430 75 0 or = 0, Dye = 0x0, and 93D = ufiB. This means that our D has properties (I) and (2) for the go, :20 taken initially and the me, no arising in the course of the argument (see remark below). Consequently, D is a solution for Problem 3—and so, by Theorem 3, for Problems 1 and 2 also. Conversely, it is obvious that (3) and (4) must hold if there exists a D having properties (1) and (2). This completes the proof of Theorem 4. ems. xxx] LINEAR PROGRAMMING AND THEORY on GAMES 325 Remark: It is to be noted that the gist of the above proof—namely, the part from conditions (35) and (45) to equation (Ed—amounts to showing that Problems 16 and 25 have a common solution, 6 = b’uo = uLAxo = c’xo, when (36) and (46) both hold. THEOREM 5 (existence theorem) : A solution, D, exists for Problem 1 if, and only if, the following existence conditions both hold: (3) As g By for some a: g 0, y > 0, (4*) Cr: 2 0 for no x a 0 such that Ax g 0. Similarly, a solution, D, exists for Problem 2 if, and only if, the following existence conditions both hold: (4) A’u g C’o for some n g 0, o > 0, (3*) 3’11. 5 0 for no u a 0 such that A’u ; O. PROOF: By Lemma. 3, conditions (3*) and (3) are equivalent. Like- wise, replacing A, B, u, z, y in Lemma. 3 by —A’, —C’, x, u, v, we see that (4*) and (4) are equivalent. Hence (3) and (4*) or (4) and (3*) hold if, and only if, (3) and (4) hold. And, by Theorem 4, a, Solution, D, exists for Problems 1 or 2 if, and only if, (3) and (4) hold. This completes the proof of Theorem 5. ‘ Remarks: It is to be noted that each of the four existence conditions (3), (4), (3*), (4*) is necessary and sulficient that there exist a. matrix D having the corresponding one of the four properties (1), (2), (1*), (2*). Thus (3) or (4) is implied by the existence of a matrix D having property (1) or (2) ; and conversely, if (3) or (4) holds, we can construct a. matrix D having property (1) or (2) merely by taking large enough negative or positive elements, respectively. The equivalence of (1) to (1*), etc., then shows that (3*) or (4*) is necessary and sufficient for the existence of a matrix D having property (1*) or (2*), respectively. It is to be noted also that the existence conditions (3), (3*), (4), (4*) can be interpreted in terms of special “null” problems, 1d’, 2d’ and 2d, 1d, in which 0’ = 0 and b = 0, respectively. For, with C' = o’ = 0, property (1) or (1 *) is held by D = d’ = 0 if, and only if, condition (3) or 3*) holds, while property (2*) or (1) is held trivially; and, with B = b = 0, property (2) or (2*) is held by D = d = 0 if, and only if, condition (4) or (4*) holds, while property (1*) or (2) is held trivially. Hence the special “null” problem, 1d’, 2d’, 2d, or 1d, admits a. null solu- tion (d’ = 0 or d = 0) if, and only if, the corresponding existence condi- tion (3), (3*), (4), or (4*) holds. 326 n. GALE, H. w. KUHN, AND Afiw. TUCKER [PART III 5. PROGRAMMING AND GAMES Let A be the “payo ” matrix of a zero—sum two-person game [von Neumann and Morgenstern, 1944, Chapter III]. Then, to solve the game, we must find the value, A, of the game and optimal (or good) mixed strategies, u and 2:, characterized by the following relations: A’ughi, ago, g’u= 1, Axglg,‘ :rgD, i’a:=1, Where g and 2' are vectors whose components are all 1’s. The fact that such ii, a, :1: always exist—the main theorem for zero-sum two-person games—can be established as a by-product of Theorem 4. To this end, assume that A > G—not an essential restriction, since the same arbitrary constant K can be added to all the elements of a game matrix without affecting the game (except to increase the value of the game by x). Then x must be positive (if it exists), and the relations above can be divided throughout by h. The divided relations may be rewritten in reverse order, as follows: (1a) 11’s: = 5 for some a: a 0 such that As g 9, (2a) g’u = 6 for some u g 0 such that A’u ; t; where now 5, 2:, u replace the previous l/k, x/h, u/k. This amounts to Problem 3 for the pecial scalar case A > 0, B = g, C = t", D = 5. (See remark preceding Theorem 4 concerning the use of equations involv- ing 6 rather than inequalities.) By Theorem 4 this scalar problem has a solution, 6, because the existence conditions, Asgg forsome 35:0; A’ugi forsome ago, are easily satisfied by taking 2: = 0 and a sufficiently large. We carry the solution back to the initial game relations by dividing (1a) and (2a) throughout by 5, which is clearly positive—and unique, by the argument of the corollary to Theorem 2. Henco we conclude that the game with payoff matrix A has a unique value, A = 1/3, and at least one pair of optimal mixed strategies, at and :t. Suchreduction of games to program- ming problems is treated in this volume by Dantzig [XX] and Dorfman [XXII]. It will now be shown that Problems 1 and 2, in full generality, are related through Problem 3 to a zero-sum two-person game. CHAP. XIX} LINEAR PROGRAMMING AND THEORY or GAMES 327 THEOREM 6: A matrix D is a solution for Problem 1 or 2 if, and only if, the game with the payofi‘ matrix L23 ‘3] has value zero and optimal mixed strategies H [ll 2: ’ y suchlhalo>0tmdy>0. PROOF: Substituting [ A ‘3] [“l ["l [g] [1'] _C D I v i y l 5 j i for A, u, :5, g, i, respectively, in the basic relations for a zero-sum two- person game stated at the beginning of this section (9, h, i, j being vectors whose components are all 1’s), and requiring 7x = 0, v > 0, y > 0, we get ' A’u ; C’v, Hit 23 D’v, u _2_ 0, o > 0, g’u + 11'?) = 1; Angy, ngDy, $20, y>0, i’x+j’y=1. But these amount to properties (2) and (1) of Problem 3, coupled with the “normalizations” g’u + M: = 1 and i’a: + j’y = l, which can always be achieved in Problem 3, because the inequalities v > 0 and y > 0 assure that (2) and (1) can be divided by g’u + [W and i’x + j’y, respectively. Therefore Theorem 6 is a direct consequence of Theorem 3. This completes the proof. One further theorem relating linear programming to games is‘ stated below. It follows out an ingenious idea of Dantzig [XX] and Brown [XXIV]. There does not seem to be any natural generalization for Problems 1 and 2. THEOREM 7: A solution, 6, exists for Problems 16 or 25 if, and only if, the symmetric game with the payofl' matrix 0 A —b ———A’ 0 o b’ —c’ 0 has an optimal mixed "strategy whose last comth is positive. 328 D. GALE, H. W. KUHN, AND A. w. TUCKER {PART III PROOF: We will not give the proof explicitly, but it is contained in the proof of Theorem 4. (See the remark at the end of Theorem 4.) Remark: Theorems 6 and 7 do not exclude necessarily the possibility that there also exist optimal mixed strategies lacking the specified posi- tiveness. Thus the symmetric game above may also possess an optimal mixed strategy, a x a 0 even when Problems 15 and 26 have a solution, 3. In this particular event, b’u = c’x = 0 due to conditions (35*) and (45*). 6. PROBLEMS warn Cons'rnsmrr EQUATIONS The following dual problems present themselves when a system of equations, Ex = Fy, is added to the constraints Ax g By, a: g 0, y > 0 in Problem 1. PROBLEM 4: To find a maximal matrix D having the property that Cm gDyfmsmxg0,y>OsuchthatAz gBy,E:t=Fy. PROBLEM 5: To find a minimal matrix D having the property that B’u +F’wgD’vfor some ago, o>0, w, such that A’u+E’w§C’o, the vector to being unrestricted in sign. These problems can be regarded as arising from Problems 1 and 2 by A B u substituting E for A, F for B, and ml for u. Then to = "E —F 102 ml -— w; is a Vector whose components take all values, unrestricted in sign, as the vectors ml and wz vary subject to the constraints w, g 0 and wz ; 0. Conversely, any vector to can be expressed as the differ- ence wl -- 1:22 of two vectors ; 0, say, by taking 2101 = l 'w I + w, and 21.02 = I w] — w, where I w] is the vector whose components are the absolute values of the components of it. There are exactanalogues of Theorems 1-7 for these two problems, which the reader may easily formulate for himself. If the matrices B and F consist of single columns, b and I, then 1) becomes a vector at, and y becomes a scalar that may be eliminated by CHAP. XIX] LINEAR PROGRAMMING AND THEORY OF GAMES 329 division. In this case the general problems, 4 and 5, reduce to vector problems that bear on Koopmans’ treatment of “efficient points” in the presence of intermediate commodities [III]. To cover this extension the following line should be added to the table of corresponding notations near the end of Section 2: E = ill-int, I f = 0; w = pint.- ...
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m13-19 - CHAPTER XIX LINEAR PROGRAMMING AND THE THEORY OF...

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