Tibor ILLèS. Katalin MèSZêROS

Transcription

1 Yugoslav Journal of Operations Research 11 (2001), Numer A NEW AND CONSRUCIVE PROOF OF WO ASIC RESULS OF LINEAR PROGRAMMING * ior ILLèS Department of Operations Research E tv s LorÃnd University of Sciences udapest, Hungary Katalin MèSZêROS Faculty of Economics, University of Novi Sad Suotica, Yugoslavia Astract: In is paper a new, elementary and constructive proof of Farkas' lemma is given. he asic idea of e proof is extended to derive e strong duality eorem of linear programming. Zhang's algorims used, in e proofs of Farkas' lemma and e strong duality eorem, are criss-cross type algorims, ut e pivot rules give more flexiility an e original criss-cross rule of. erlaky. he proof of e finiteness of e second algorim is technically more complicated an at for e original crisscross algorim. o of e algorims defined in is paper have all e nice eoretical characteristics of e criss-cross meod, i.e. ey solve e linear programming prolem in one phase; ey can e initialized wi any, not necessarily primal feasile asis, ases generated during e solution of e prolem, are not necessarily primal or dual feasile. Keywords: Farkas lemma, strong duality eorem, criss-cross type pivot rules. * 1991 Maematics Suject Classification. 90C05. he first auor was supported in part y e Hungarian Ministry of Education under e grant FKFP No. 0152/1999 and y e Hungarian National Research Fund under e grant OKA O he first version of is paper was prepared during e visit of e first auor to W rzurg University, Institute of Applied Maematics and Statistics, sponsored y a scholarship from e Deutscher Akademischer Austaushdienst, DAAD A/98/07349.

2 16. IllÕs, K. MÕszÃros / A New and Constructive Proof of wo asic Results 1. INRODUCION On e 150 anniversary of e ir of Gyula Farkas in 1997, S. Zhang [14] pulished two new and finite pivot algorims for solving linear programming prolems. Zhang's algorims are generalizations of erlaky's Criss-cross meod [11, 12, 13]. 1 Klafszky and erlaky [7,8] gave a constructive proof to e well-known lemma of Gy. Farkas [2, 3]. Using e first algorim (FILO/LOFI rule) of Zhang [14, 15] and e so-called orogonality eorem (see for instance [7, 8, 5, 6]) we give herein a constructive proof to Farkas' lemma in a similar way as Klafszky and erlaky did in eir papers [7, 8]. his kind of constructive proof can e extended to verify e well-known strong duality eorem. We use Zhang's second algorim wi e most-often selected rule [14, 15]. Our proofs of e finiteness of Zhang's algorims are simpler an e original one. Let A R m n n m, c,x R, y, R and I = { 1, 2,..., n}. Wiout loss of generality we may assume at e rank of A is m, us A has full row rank. Let denote e i row vector of e matrix A, while j m a R denotes e ( i ) a R n j column vector of it. In our paper e following form of e Farkas lemma is proved in Section 2. heorem 1.1. (Farkas' lemma) From e following two systems of linear inequalities exactly one is solvale: Ax = x 0 ( A 1 ) y y 0 = 1 A ( A2 ) Our second goal in is note is to give constructive proof for e strong duality eorem of e linear programming prolem (Section 3). Now, let us consider e primal and dual linear programming prolems in e following form minc x Ax = x 0 (P) max y y A c (D) Furermore, let P e e set of primal feasile solutions 2, namely P : = { x R A x = ) n 1 Zhang [14, 15] proved e finiteness of one of his algorim, following e steps of erlaky's original proof [11, 12]. 2 n n n he R is e positive orant, us R = { x R : x 0}.

3 . IllÕs, K. MÕszÃros / A New and Constructive Proof of wo asic Results 17 and let e set of dual feasile solutions, D, e given as m D = { y R y A c}. heorem 1.2. (Strong duality eorem) From e following two statements exactly one holds: = (1) here exists xˆ P and ŷ D such at c xˆ yˆ. (2) P = 0/ or D = 0/. Let us introduce e (primal) pivot taleau for e (primal) linear programming prolem, as follows A c * where all e data related to e prolem are arranged. Under e assumption at matrix A has full row rank, ere exists an m m nonsingular sumatrix A of A. Let us interchange e columns of A to otain e following partition A = ( A, AN ), where e sumatrix A N contains ose columns of A which do not elong to A. Now e linear system A x = can e written as A x A x =, where we group + N N e unknowns in e same way as e columns of matrix A, namely x = x, x ). Similarly, we can reorder e components of e vector c as c = c, c ). ( N ( N Now we are ready to restate some well-known concepts of linear algera and linear programming such as asis, asic solution, feasile asic solution, optimal solution and orogonality. Definition Any m m nonsingular sumatrix A of A is called a asis he x = A, x N = 0 is a asic solution of A x = for a given A. 3. Variales grouped in x are called asic variales, while ose corresponding to x N are called nonasic variales If A 0 en we say at x, ) is a (primal) feasile solution and A is a (primal) feasile asis. 1 ) 5. he vector y = ( c A R A 1 A 6. If c c holds en ( x N m A is called a dual asic solution. is said to e a dual feasile asis. 7. he primal feasile solution x P is said to e an optimal solution of e primal prolem, if c x c x holds for all x P.

4 18. IllÕs, K. MÕszÃros / A New and Constructive Proof of wo asic Results 8. he dual feasile solution y D is said to e an optimal solution of e dual prolem, if is e following y y holds for all y D. he (primal) pivot taleau corresponding to e asis A for e LP prolem 3 c A 1 A c A 1 A c 1 A 1 A and let us introduce e following notations = A = A A, = A, c = c c A A, and ζ c. he set of asic indices corresponding to e asis A is denoted y, while e set of nonasic indices is denoted y N. rivially, I = N. We need e concept of orogonality among vectors. Definition 1.4. Let a = 0. k a, R en vectors a and are said to e orogonal, if Using e pivot taleau we can introduce e following n-dimensional (column) vectors: and t ( i) ( i) n = ( tk ) k= 1 t ik, if k N = 1, if k = i 0, if k, k i t t kj, if k = 1, if k = j 0, if k N k n j = ( t( j ) k ) k = 1, j ( i) where t, i is equal to e i row of, while t, j N is formed from e column of extended y an ( n m) - dimensional negative unity vector. 4 j j 3 For e system ( A 1 ) e pivot taleau corresponding to e asis of A is simpler as you may see: A 1 A 1 A

5 . IllÕs, K. MÕszÃros / A New and Constructive Proof of wo asic Results 19 he following useful oservation is called e orogonality eorem (see for instance [7, 8, 6]). Proposition 1.5. Let a linear system are ases of e linear system, en A x = e given. Furermore, A and A ( i) j (t ) = 0 for all i and for all j N, holds, where and are e index sets corresponding to e ases respectively. A and, A heorem 1.1 is proved in Section 2. First we define an algorim to solve e system A ) and prove its finiteness. he algorim eier solves A ) or gives a certificate for e nonexistence of a solution. In is second case, using elementary computations, we can compute e solution of system A ). he solvaility of e LP prolem (P) is discussed in Section 3. A pivot algorim is defined using e (mostoften-selected variale) pivot rule of Zhang, [14, 15]. he finiteness of is second algorim is proved. he strong duality eorem, heorem 1.2., is otained as an easy consequence of e finiteness of e algorim. o of e presented algorims have e general property of e criss-cross meod [4], namely at e system A ) is solved wiout introducing artificial variales and using e so-called first phase ojective function (or oer techniques like e ig-m meod, [9]). Consequently, we do not need two phases to solve prolem (P), ecause e algorim can e initiated y any (not necessarily primal feasile) asis. Our proofs are purely cominatorial, erefore e only information at is used is e sign of e entries of e pivot taleau. hus, we use e alinski-ucker [1] notation which is very convenient for our purposes. Positive, nonnegative, negative and nonpositive numers are denoted y +,,, signs, respectively. If an entry in e taleau is denoted y en ere is no information aout e sign of at entry. ( 2 2. PROOF OF HE FARKAS LEMMA First, let us deal wi e solution of e system ( A 1 ). We introduce e following mapping ur : I N0, and let u 0 = ( 0, 0,..., 0) and r, ur ( i) = u r-1 if e i ( i), oerwise variale moves in e r iteration 4 he vector t j is a column of e dual simplex taleau as it is defined in [10].

6 20. IllÕs, K. MÕszÃros / A New and Constructive Proof of wo asic Results for r = 1, 2,..., k. It is easy to show at u r u r 1 and u r u r 1. he asic idea of e pivot rule is e following: from e infeasile variales choose e one to leave e current asis which entered most recently and from ose which are candidates to enter e asis choose e one which has left e asis most recently. Algorim 2.1. Let a asis r = 1. A for e system A ) e given wi e corresponding pivot taleau. Let Step 1. Let J : { i : x < 0}. = i If J = 0/ en e system A ) is solved, SOP Step 2. Let K : { j N : t < 0}. Step 3. Now, else let Jmax : = { j J : ur1 ( j) ur1 ( i), for all i J} choose an aritrary index k Jmax and go to Step 2. = kj If K = 0/ en e system A ) is solved, SOP else let K max : = { j K : ur1 ( j) ur1 ( i), for all i K}, choose an aritrary index l Kmax and go to Step 3. x k leaves and x l enters e current asis. Let us update e vector u as follows r, if i = k or i = l ur ( i) = ur 1 ( i), oerwise Increase e value of r y 1, namely r : = r + 1 and go to Step 1. Let us extend e definition of e vectors (i) t and t j to e column of, as well. From now on, we assume at e index (which elongs to e column vector ) is always in e set of N. Now we can apply e orogonality eorem for e matrix [ A, ] and en e vectors Lemma 2.2. Algorim 2.1 is finite. (i) t and j t ecome n tuples. Proof: y contradiction. Let us assume at e algorim is not finite. his means at ere exists (at least) one example in which e algorim is not finite, us it generates an infinite sequence of pivot taleaus, i.e. infinite sequence of ases. ut e numer of all possile ases for a given prolem (wi an m n matrix A ) is finite (at

7 . IllÕs, K. MÕszÃros / A New and Constructive Proof of wo asic Results 21 n most, erefore some of e ases should occur infinitely many times. 5 From m ose examples for which cycling occurs choose one wi e smallest possile size. For such prolems all e variales have to change eir asis status during a cycle. Let us consider e sequence of pivot taleaus generated y e algorim and let us denote y at which satisfies e following criteria: ere is a variale x q which changes its asic status for e first time; after is pivot taleau all e variales have changed eir asic status at least once. We have two choices: e variale x q eier enters or leaves e ases at e pivot taleau. It follows from our counter assumption at ere should e anoer asic taleau, such at e variale x q will change its asic status for e first time since. Let us analyze e (sign) structure of e pivot taleaus Case 1. Let us deal wi e case when e variale and enters follows. hen e sign structure of e pivot taleaus and. x q leaves e pivot taleau and are as xl Figure 1. (l) Using e orogonality eorem (Proposition 1.5) e vectors and t are orogonal. On e oer hand, ased on e pivot rule of Algorim 2.1 if t < 0, for some i N \{ q, } en ur 1 ( i) < ur1 ( q), means at e variale x i did not change its asic status since e pivot taleau, erefore i N, so t i = 0. From is oservation we may get at ( l) = l lq q 0 ( t ) + t > 0, ecause t = 1, < 0, t < 0 and t < 0. A contradiction is otained. q l lq li 5 his phenomena is known in e literature as cycling, see for instance [9, 10, 5, 6].

8 22. IllÕs, K. MÕszÃros / A New and Constructive Proof of wo asic Results Case 2. Let us assume at e variale leaves at. x q enters e asis at pivot taleau and xl Figure 2. aking into consideration e sign structure of ese taleaus, e pivot rule of Algorim 2.1. and using e orogonality eorem (Proposition 1.5), as in e previous case, we can show at o and cannot occur in e sequence of pivot taleaus generated y Algorim 2.1. herefore Algorim 2.1. is not cycling. Now, we are ready to prove e Farkas' Lemma. Proof of heorem 1.1: Let us assume at o A ) and A ) have a solution, en from ( 2 A x = it follows at y A x = y. aking into consideration at y A 0 and x 0 it follows at 0 y A x = y = 1, ecause y = 1 solution. holds. his is a contradiction, us o systems cannot have a We need to show at one of e systems is solvale. Let us apply Algorim 2.1. to system A ). According to Lemma 2.2. Algorim 2.1. is finite, erefore it eier stops in Step 1 wi J = 0/, which means at a (asic) feasile solution of e system A ) is found, or it reports at K = 0/ (Step 2), us we otain a pivot taleau (k) such at t 0 inequalities and < 0. 6 Now it is ovious at e following system of k 6 his is known from e literature as primal infeasiility criteria, see for instance [9, 10, 5, 6].

9 . IllÕs, K. MÕszÃros / A New and Constructive Proof of wo asic Results 23 ( t ( k) ) x =, k x 0 (1) has no solution. herefore e system ( A 1 ) cannot have a solution. Using e corresponding asis we can compute a solution 7 of e system ( A 2 ) as 1 (( k) A 1 ) k y = e, where k m e R is e k unity vector. 3. PROOF OF HE SRONG DUALIY HEOREM Let us consider e primal linear programming prolem, (P). Let us introduce e following mapping v r : I N0, and let v 0 = ( 0, 0,..., 0), furermore v = ( ) +, ( ) 1 i 1 if e i v i r r vr 1( i), oerwise. Vector status until e end of e Algorim 3.1. variale moves in e r iteration v r counts how many times e variales have changed eir asic r iteration. Let a asis A of e primal linear programming prolem (P) e given wi e corresponding pivot taleau and let r = 1. Step 1. Let J : { i I : x < 0 or < 0}. = i c i If J = 0/ en e current asic solution is optimal, SOP, Step 2. (a) Primal iteration: else let Jmax : = { j J : vr1 ( j) vr1 ( i), i J} and choose an aritrary index k Jmax. k N. Define e set of indices K { i : t > 0}. P = ik If K = 0/ en D = 0/, ere is no dual feasile solution, SOP, p else let K : { j K : v ( j) v ( i), i K } P, max = P r1 r1 P 7 Using e orogonality eorem (Proposition 1.5) it is easy to check at if e current pivot taleau is denoted y and e corresponding asis y A en tkj = (z k) a j and k = ( zk), 1 m where z k = (( ek) A ) and ek R is e k unity vector. See for instance [7, 8, 5, 6].

10 24. IllÕs, K. MÕszÃros / A New and Constructive Proof of wo asic Results Now, and choose an aritrary index l K P, max. x l leaves e asis, while x k enters it, and vr 1( i) + 1, if i = k or i = l vr ( i) = vr 1( i), oerwise. Increase e value of r y 1 and go to Step 1. () Dual iteration: k. Define e set of indices K : { i N : t < 0}. D = ki If K = 0/ en P = 0/, ere is no primal feasile solution, SOP, Now, D else let K : { j K : v ( j) v ( i), i K } D, max = D r1 r1 D and choose an aritrary index l K D, max. x k leaves e asis, while x l enters it, and vr 1( i) + 1, if i = k or i = l vr ( i) = vr 1( i), oerwise. Increase e value of r y 1 and go to Step 1. he most often selected infeasile variale is chosen y e pivot rule of e algorim in Step 1. Using exactly e same rule in Step 2 from e candidate variales, e most-often selected is chosen again. If we have more an one candidate eier in Step 1 (elements of J max ) or in Step 2 (in case (a) e elements of K P, max and in case () e elements of K D, max ) en we may choose from em aritrarily. he finiteness of e algorim will e proved using e orogonality eorem (Proposition 1.5). In e case of linear programming, e vectors t ( i) R n+1 and t j R n+1, furermore space of e following matrix (i) t elongs to e row space, while t j elongs to e null A c. 0 From now on we assume at e index c (which elongs to e row vector c ) is always in e set of. Lemma 3.2. he Algorim 3.1. is finite. Proof: he proof of is lemma is very similar to e proof of Lemma 2.2. Let us assume to e contrary, at e Algorim 3.1. is not finite. ut e numer of possile ases is finite, erefore at least one asis should e repeated infinitely many times. hus cycling must occur. From ose examples where cycling occurs choose one wi e smallest size, which means all e variales enter and leave e asis during a cycle.

11 . IllÕs, K. MÕszÃros / A New and Constructive Proof of wo asic Results 25 Let x q e e variale which moves last and A e first asis when changes its asic status. (Wiout loss of generality we may assume at asis A.) Let us denote y A at asis when x q enters at x q moves for e second time. We assume at after e asis A, all e variales have changed eir asic status at least once. It may happen at anoer variale x w, togeer wi x q, changes its asic status at A for e first time. We now have e following cases: + * xw (3a) (3) Figure 3: At primal iteration x q enters e asis and (a) change its asic status in Step 1.; () o x q is e only candidate to x q and x w change eir asic status for e first time If 0 en (3a) and (3) are equivalent taleaus. w xw (4a) xw * (4) Figure 4: At dual iteration x q enters e asis: (a) x q has een selected uniquely; () o x q and x w change eir asic status for e first time In (4a) and (4) e sign structure of e row of asis when e variale x w is e same. A is e x leaves e asis for e first time. We have two cases: q leaves e asis eier in Step 2 (a), primal iteration, or in Step 2 (), dual iteration.

12 26. IllÕs, K. MÕszÃros / A New and Constructive Proof of wo asic Results xs (5a) (5) Figure 5: he variale x q leaves e asis A at primal (5a) or at dual (5) iteration If t > 0 and i \{ q} (see Fig. 5, part (5a)) en, according to e pivot rule is of Algorim 3.1, vr 1 ( i) = vr1 ( q) = 1, us i. Similarly, if t ci < 0 ( t j < 0), where i N ( j \{ q}) en i N ( j ) holds ecause v ( i) v ( q) = ( v ( j = = v r ( q) 1) using e case (5). 1 = r 1 = r1 1 r1 ) Now we have e following four possile cases, namely: e variale e asis A at a) primal iteration and leaves e asis A at primal iteration; ) primal iteration and leaves e asis A at dual iteration; c) dual iteration and leaves e asis A at primal iteration; d) dual iteration and leaves e asis A at dual iteration. x q enters Let us deal wi e case (a), where we have e sign structures given at (3a) (c) (or (3)) and (5a). ecause is e same vector for o (3a) and (3) we do not need to separate ese two sucases. From e pivot taleau shown on (5a) we use e vector t. We know at (t ) t = 0, according to e orogonality eorem s s (Proposition 1.5). aking into consideration e signs of e entries in especially if t is > 0, i \{ q} en i, us t = 0, we have = s cq qs cc cs 0 ( ) t t + < 0. ci (c) and t s,

13 . IllÕs, K. MÕszÃros / A New and Constructive Proof of wo asic Results 27 he last (strict) inequality holds ecause t < 0, t < 0, = 1 and t > 0. hus we cq cs cc have otained a contradiction, namely (3a) (or (3)) cannot occur in e sequence of pivot taleaus produced y e algorim togeer wi e taleau shown in (5a). Case (). Let us consider e vectors (c) and from e pivot taleau (3a), while e (c) vectors t and t are from e pivot taleau (5). Applying e orogonality eorem (Proposition 1.5) twice and summing e terms we get ( c) + ( t ) t ( t ) t previous expression in more detail, us = 0. Using e remark given after Figure 5, we can compute e qs 0 = ( ) t + ( ) + t + t cc c c = ζ ζ + t cq q cq q + cc c + ζ ζ > 0, + c + t cq q (2) ecause t = = 1, = t = 1, = t = 0, = ζ, = ζ, < 0 and t < 0. cc cc q cq c cq cq herefore o pivot taleaus (3a) and (5) cannot occur. Now we need to pay more attention to e case when pivot taleaus (3) and (5) are considered. In expression (2) e term t appears. Unfortunately, we have no cw tw information aout e sign of e element w (see Fig. (3)). he element w can e o negative and nonnegative. Furermore, it may happen at e element t cw is negative or nonnegative. herefore we have four sucases depending on e sign of w and t cw. If t 0 and t 0 en e proof goes along e same lines as for e w taleaus (3a) and (5). cw If t cw < 0 en let us consider e vectors q and orogonality eorem (Proposition 1.5) we have ( t ) = 0. hen = q cw wq cq qq c q cw wq 0 (t ) + t + = t < 0 q q (c) t. Using e where t = 0 and t = 0, ecause q and N, furermore t < 0 and t wq cq q > 0. hus we have a contradiction in is sucase. Let us now consider e sucase when t < 0 and t 0. In is situation cycling may occur in two different ways: (i) if e variales asic status at e asis enters a asis coming after w A, or (ii) if e variale A. cw x q and x q leaves e asis cw x w change eir A and xw

14 28. IllÕs, K. MÕszÃros / A New and Constructive Proof of wo asic Results Let us analyze first e case (i). Now, e vectors he sign structure of (q) t has e following properties: (q) t and are considered. t < 0, t < 0 and if t < 0 and i w en i N, namely t = 0. q qw qi From e orogonality eorem (Proposition 1.5) we know at ( t ) t = 0 and using e previous information we may compute in more details as follows ( q) = qw w q qq q qw w q 0 ( t ) + t + t = t t > 0, i ( q) ecause t < 0, < 0, < 0, = 1, = 1 and t = 0. hus we have otained qw w q qq contradiction once more. Now, we need to analyze e case (ii), us we take into consideration e first asis A after A such at x w enters e asis and x q is a nonasic variale at A 8. he sign structure of A is q x w and ecause Furermore, if q N en according to e pivot rule we have t 0, < 0. < 0, i w en i N, erefore t ci = q cw wq cc cq 0 ( t ) t + < 0, since t = 1 and t < 0. hus a contradiction is otained. cc cq cq cw (a). After is complicated case let us analyze (c) and (d), which are similar to case In case (c), we consider e vectors (w) and t s. Using e orogonality eorem (Proposition 1.5) and e sign structure of e vectors we have 8 he existence of such asis A is necessary to get a cycle, ecause at A e variale xw was in e asis, while x q was out of e asis.

15 . IllÕs, K. MÕszÃros / A New and Constructive Proof of wo asic Results 29 ( w) = s wq qs w s wc cs wq qs 0 ( ) t + t + t = < 0, where t = = 0, < 0 and t > 0. hus a contradiction is otained. s wc wq In e last case, (d), we consider e vectors orogonality of e vectors (proved in Proposition 1.5) we get ( w) = wq q w wq q w 0 ( ) t + t = t > 0, qs (w) and t and instead of e since t < 0, < 0 and t < 0. his contradiction shows at case (d) cannot occur, as well. wq q w his completes e proof, ecause none of e possile cases can occur. Now we are ready to prove e strong duality eorem. Proof of e strong duality eorem (heorem 1.2): he two statements of e eorem exclude each oer. Let us apply Algorim 3.1. for e linear programming prolem (P). he algorim terminates wi one of e following cases: 1. Variale x k leaves e current asis and we cannot choose any variale to enter e asis (Algorim 3.1, Step 2 ()). hen we have x < 0 and t 0, for all i I us P = 0/. ki k = k 2. Variale x k enters e current asis and we cannot choose any variale to leave e asis (Algorim 3.1, Step 2 (a)). hen we have c < 0 and t 0, for all i us D = 0/. 3. According to e pivot rule of Algorim 3.1. we cannot choose a variale eier to leave or to enter e current asis, us x 0 and c 0, for all i I (Step 1., J = 0/ ). herefore e current asis is optimal, so an optimal solution is found to e primal prolem. It is ovious at if 1 or 2 occurs en e statement (2) of e strong duality eorem is otained. For is, we only need to show at statement 1 and 2 are true. Statement 1 is proved during e verification of e Farkas Lemma (see (1)). Statement 2 9 can e verified as follows. Let us assume to e contrary, at ere exists a dual feasile asis ut A for which t 0 has to e orogonal to e vector t 0, (c) = k cc ck c k ck 0 ( t ) t + t = < 0, i k i ik k 9 Known in e literature as dual infeasiility criteria, see for instance [9, 10, 5, 6].

16 30. IllÕs, K. MÕszÃros / A New and Constructive Proof of wo asic Results k cc ecause t = 0, t = 1 and t < 0, gives a contradiction. ck Statement 3 is true ecause if we denote e asic feasile solution, produced y e algorim, y ˆ 1 x ( A, 0) where 1 A 0, and ˆ 1 y = ( c A ) en c xˆ = c 1 = A = yˆ. Now applying e weak duality eorem of linear programming [9, 10, 5, 6], we may show e optimality of xˆ and ŷ, us statement (1) of e strong duality eorem is otained if Algorim 3.1. stops in Step 1. his completes e proof of e strong duality eorem. Acknowledgement. he auors are grateful to Filiz Erilen for reading an earlier version of is paper and making useful comments at improved e presentation of is paper. REFERENCES [1] alinski, M.L., and ucker, A.W., "Duality eorem of linear programs: A constructive approach wi applications", SIAM Review, 11 (1969) [2] Farkas, Gy., "Applications of Fourier s maematical principle", Maematikai Òs ermòszettudomãnyi èrtesú, 12 (1894) (in Hungarian). [3] Farkas, J., "heorie der einfachen ungleichungen", Journal f r dei Reine und Angewandte Maematik, 124 (1901) [4] Fakuda, K, and erlaky,, "Criss-cross meod: A fresh view on pivot algorims", Maematical Programming, 79 (1997) [5] IllÒs,., Pivot Algorims of Linear Programming, E tv s LorÃnd, University of Sciences, Faculty of Natural Sciences, udapest, Lecture Notes, 1998 (in Hungarian). [6] IllÒs., Linear Programming, Eastern Mediterranean University, Faculty of Arts and Sciences, Famagusta, Lecture Notes, [7] Klafszky, E., and erlaky,., "Application of pivot technique in proving some asic eorems of linear algera", Alkamozott Matematikai Lapok, 14 (1989) (in Hungarian). [8] Klafszky, E., and erlaky,., "A role of pivoting in proving some fundamental eorems of linear algera", Linear Algera and its Application, 151 (1991) [9] Murty, K.G., Linear Programming, John Wiley & Sons, [10] PrÒkopa, A., Linear Programming, JÃnos olyai Maematical Society, udapest, 1968 (in Hungarian). [11] erlaky,., "A new, finite criss-cross meod for solving linear programming prolems", Alkamozott Matematikai Lapok, 10 (1983) (in Hungarian). [12] erlaky,., "A convergent Criss-cross meod", Ma. Oper. und Stat. Ser. Optimization, 16 (1985) [13] erlaky,., "A finite criss-cross meod and its applications", MA SZAKI anulmãnyok, udapest, 179 (1986) (in Hungarian). [14] Zhang, S.A., "A new variant criss-cross pivot algorims for linear programming ", echnical Report 9707/A, Econometric Institute, Erasmus University Rotterdam, [15] Zhang, S.A., "A new variant criss-cross pivot algorims for linear programming ", European Journal of Operations Research, 116 (1999)