On the Chvatál-Complexity of Knapsack Problems

Transcription

1 R u t c o r Research R e o r t On the Chvatál-Comlexity of Knasack Problems Gergely Kovács a Béla Vizvári b RRR 5-08, October 008 RUTCOR Rutgers Center for Oerations Research Rutgers University 640 Bartholomew Road Piscataway, New Jersey Telehone: Telefax: rrr@rutcorrutgersedu htt://rutcorrutgersedu/ rrr a College for Modern Business Studies, Tatabánya, Stúdium tér, H-800 Hungary b RUTCOR, Rutgers University, bvizvari@rutcorrutgersedu, and Deartment of Industrial Engineering, Eastern Mediterranean University, Famagusta, Mersin 0, Turkey, belavizvari@emuedutr, and Deartment of Oerations Research, Eötvös Loránd University of Budaest, H-088 Budaest, Pázmány Péter sétány /C, H-7 Hungary, vizvari@matheltehu

2 Rutcor Research Reort RRR 5-08, October 008 On the Chvatál-Comlexity of Knasack Problems Gergely Kovács Béla Vizvári Abstract There is a famous result of Chvatál giving a theoretical iterative rocedure, which determines the integer hull of a olyhedral set It starts from the olyhedral set itself and in each iteration new cuts are introduced This result is considered the theory of the Gomory method of integer rogramming The simlest integer rogramming roblem is the knasack roblem The number of iterations in Chvatáls rocedure is investigated in the case of some secial binary knasack roblems

3 RRR 5-08 Page Introduction The integer hull of a olyhedral set is the convex hull of the integer oints of the set If a olyhedral set is equal to its integer hull then any linear integer rogramming and linear rogramming roblems having the same objective function have at least one common otimal solution This is the underlying idea of the cutting lane methods [] introduced a concet of generating cuts such that the cuts do not eliminate any integer solution Assume that the olyhedral set is defined by the inequality system: Ax b, () where A is an m n matrix, b and x are vectors of m and n dimensions, resectively Assume that λ is an m dimensional nonnegative vector such that the vector A T λ is integer Then all integer vectors x of the olyhedral set must satisfy the inequality λ T Ax λ T b () In general () is a valid cut of the integer hull Furthermore if λ T b is non-integer then it will cut off a art of the olyhedral set It can be roven, see also [3], that: there are only finite many significantly different Chvatál cuts of tye () and if the Chvatál cuts added to the set of inequalities () and in this way a new the olyhedral set is defined, and the whole rocedure is reeated, then after finite many iterations the olyhedral set becomes equal to the integer hull In this aer the number of iterations is called Chvatál rank From ractical oint of view it is imortant that how large can it be To our best knowledge this roblem has not been analyzed The reason is that the integer rogramming roblems are NP-comlete, therefore the number of iterations should be great in general On the other hand imortant subclasses can be those ones, where the number of iterations is small, eg it is only Such a result can be hoed only in the case of roblems, which have a simle structure Thus one of the first candidates are secial structured knasack roblems In this aer binary knasack roblems are investigated The following notations and assumtions are used Let n be the number of variables, ie the dimension of the roblem All variables are binary variables The knasack constraint is given in the form: where a, a,, a n are ositive integers Furthermore a x + a x + + a n x n b, (3) a a a n (4) It follows from the binary roerty of the variables that the values of all comonents of an element of the integer hull are between 0 and Therefore the initial olyhedral set is defined by the following system of inequalities, where the index of the inequality is used in the further analysis as it is indicated here:

4 Page RRR 5-08 index Right-Hand Side Left-Hand Side 0 a x + a x + + a n x n b x x n x n n + x 0 n + x 0 n x n 0 Using the same index set the multiliers of the inequalities of this original constraint set are denoted by λ 0,, λ n In general the set of integer feasible solutions is emty if and only if b < 0 inequality is required, which cannot be satisfied It is obtained be the multiliers λ 0 =, λ = λ = λ n = 0, λ n+ = a, λ n+ = a, λ n = a n Then an as then the resulted inequality has the classical form showing the unsatisfiability of the system: Therefore it is assumed further on that b 0 0 = 0x + 0x + + 0x n b < 0 In section it is shown that u to 3 dimensions all binary knasack roblems have Chvatál rank at most An examle having a higher rank is rovided in section 3 in dimension 4 This examle is generalized and analyzed based on a result of [] in section 4 The Case of Dimensions,, and 3 It follows from the assumtions that in the case of n = there are only two ossible sets of integer feasible solutions: {0, }, and {0} For the first one the existing inequalities 0 x are determining the integer hull In the second case the integer hull is the 0 oint itself It is obtained from 0 x, and its counter art generated by λ 0 = /a, λ = λ = 0 Thus the Chvatál rank is either 0 or Assume that n = The it follows from (4) that the left-hand side of (3) is decreasing for the following sequence of binary vectors: (, ), (0, ), (, 0), (0, 0) Thus if a set of integer feasible solution consists of k ( k 4) vectors then these vectors are the last k vectors of the sequence Hence it is not difficult to see that the Chvatál rank is again either 0 or If n = 3 then the maximal elements of the feasible solutions belong to one of the cases of the table below: case maximal vectors inequalities of the feasible set emty 0 (0, 0, 0) y i 0 3 (, 0, 0) y 0, y (, 0, 0), (0,, 0) y + y, y (,, 0) y (, 0, 0), (0,, 0), (0, 0, ) y + y + y 3 7 (0, 0, ), (,, 0) y + y 3, y + y 3 8 (,, 0), (, 0, ) y + y 3 9 (,, 0), (, 0, ), (0,, ) y + y + y 3 0 (,, ) emty

5 RRR 5-08 Page 3 Table The non-trivial inequalities of the integer hall of the feasible set of knasack roblems in the 3-dimensional case The knasack roblem has a Chvatál rank at most in all of the cases The statement can be shown in a trivial way for cases,, 3, 5, 0 E g in case 5 the multiliers are: λ 0 = a 3, λ = 0, λ = 0, λ 3 = 0, λ 4 = a a 3, λ 5 = a a 3, λ 6 = 0 The other cases can be solved as follows Case 4: The inequality of y 3 0 as generated for case 5 and another cut with λ 0 = b, λ = a b, λ = a b, λ 3 = 0, λ 4 = 0, λ 5 = 0, λ 6 = a3 b Case 8: λ 0 = b, λ = 0, λ = a b, λ 3 = a3 b, λ 4 = a b, λ 5 = 0, λ 6 = 0 Case 9: λ 0 = b, λ = a b, λ = a b, λ 3 = a3 b, λ 4 = 0, λ 5 = 0, λ 6 = 0 Case 7: The cut of case 8 and another cut with λ 0 = b, λ = a b, λ = 0, λ 3 = a3 b, λ 4 = 0, λ 5 = a b, λ 6 = 0 Case 6: λ 0 = a, λ = a a, λ = 0, λ 3 = 0, λ 4 = 0, λ 5 = 0, λ 6 = a3 a 3 A Counterexamle in Dimension 4 The 4-dimensional case is the first one, where there is a knasack roblem with an integer hull having a Chavtál rank higher than Here is the examle: a =, a =, a 3 = 4, = 30 and b = 53 It is easy to see that the maximal feasible solutions are the vectors (,,, 0), (0, 0,, ), (0,, 0, ), (, 0, 0, ) The hyerlane y + y + y 3 + y 4 = 3 contains all of these maximal feasible oints Therefore y + y + y 3 + y 4 3 (5) is a valid cut of the integer hull If the cut could be generated then there were nonnegative λ s satisfying the following conditions: λ 0 + λ λ 5 = λ 0 + λ λ 6 = 4λ 0 + λ 3 λ 7 = 30λ 0 + λ 4 λ 8 = 53λ 0 + λ + λ + λ 3 + λ 4 < 4 The system has a feasible solution if and only if the otimal value of the linear rogramming rogram: min 53λ 0 + λ + λ + λ 3 + λ 4 λ 0 + λ λ 5 = λ 0 + λ λ 6 = 4λ 0 + λ 3 λ 7 = 30λ 0 + λ 4 λ 8 = λ 0, λ 8 0

6 Page 4 RRR 5-08 is less than 4 On the other hand the otimal solution is: λ 0 = 5, λ = λ = 5, λ 3 = λ 4 = λ 5 = λ 6 = λ 7 = λ 8 = 0 The otimal objective function value is 4 In general there are 8 different sets of maximal feasible solutions in dimension 4 if inequality (4) is satisfied 0 out of 8 are equivalent to the 0 cases of the 3-dimension with the modification that the last comonent of all maximal feasible solutions is 0 Here are the further 8 cases: 5, (,0,0,0), (0,,0,0), (0,0,,0), (0,0,0,) y + y + y 3 + y 4 (0,0,,0), (0,0,0,), (,,0,0) y + y 3 + y 4, y + y 3 + y 4 3 (0,0,0,), (,,0,0), (,0,,0) y + y + y 3 + y 4, y + y 3 + y 4 4 (0,0,0,), (,,0,0), (,0,,0), (0,,,0) y + y + y 3 + y 4 5 (0,0,0,), (,,,0) y + y 4, y + y 4, y 3 + y 4 6 (,,0,0), (,0,,0), (,0,0,) y + y 3 + y 4 7 (,,0,0), (,0,,0), (0,,,0), (,0,0,) y + y + y 3 + y 4, y + y 4, y 3 + y 4 8 (,0,0,), (,,,0) y + y 4, y 3 + y 4 9 (,,0,0), (,0,,0), (0,,,0), (,0,0,), (0,,0,) y + y + y 3 + y 4, y 3 + y 4 0 (,0,0,), (0,,0,), (,,,0) y + y + y 4, y 3 + y 4 (,,,0), (,,0,) y 3 + y 4 (,,0,0), (,0,,0), (0,,,0) and y + y + y 3 + y 4 (,0,0,), (0,,0,), (0,0,,) 3 (,0,0,), (0,,0,), (0,0,,), (,,,0) y + y + y 3 + y (0,0,,), (,,0,) y + y 3 + y 4, y + y 3 + y 4 5 (0,,0,), (,,,0), (,0,,) y + y + y 4, y + y 3 + y 4 6 (,,,0), (,,0,), (,0,,) y + y 3 + y 4 7 (,,,0), (,,0,), (,0,,), (0,,,) y + y + y 3 + y (,,,) 0 y i ı =,, 3, 4 Table The non-trivial inequalities of the integer hull of the feasible set of knasack roblems in the 4-dimensional case As it can be seen from Table the above examle belongs to case 3 All other cases have Chvatál rank The roofs are similar to the 3-dimensional case, the details remain to the reader Even case 3 has Chvatál rank for certain coefficients Theorem 3 The Chvatál rank of case 3 is higher than if and only if a + a + a 3 b (6) a 3 + b (7) a + a + > b (8) a 3 < (9) a + a + a 3 + b (0) Proof We obtain the first set of conditions from the fact that the oints (,,, 0), (0, 0,, ), (0,, 0, ), (, 0, 0, ) are feasible The first two oints imly the inequalities (6) and (7) It follows from (4) that the similar inequalities of the third and fourth oints are consequences of (7) Any oint, which is greater than at least one of these four oints, must be infeasible The minimal such oints are (, 0,, ), (0,,, ), (,, 0, ) It follows from (4) that the last one gives the minimal left-hand side Thus it is enough to claim only inequality (8)

7 RRR 5-08 Page 5 Case I If a4 a 3 then let λ 0 = a 3, λ = a a 3, λ = a a 3, λ 3 = 0, λ 4 = a4 λ 5 = λ 6 = λ 7 = λ 8 = 0 Then the coefficients of the left-hand side are,,, The value of the right-hand side is ( a ) ( + a ) ( + a ) 4 + b = b a a + 4 () a 3 a 3 a 3 a 3 a 3 It follows from (8) that it is less than 4, ie it can be rounded down to 3 Thus (9) is a necessary condition of having a Chvatál rank greater than one Case II If a 3 < a4 then let λ 0 =, λ = a, λ λ 4 = λ 5 = λ 6 = λ 7 = λ 8 = 0 = a, λ 3 a 3, = a3, Then the coefficients of the left-hand side are,,, The value of the right hand side is ( b + a ) ( + a ) ( + a ) 3 = b a a a () This value is less than 4 if and only if b a a a 3 <, (3) what is equivalent to (0) Thus (6)-(0) are necessary conditions of having a Chvatál rank greater than It is easy to see that in any system of the λ multiliers giving a minimal right-hand side both λ i and λ 4+i can not be ositive (i =,, 3, 4) Thus the otimal choice of λ i (i =,, 3, 4) if λ 0 is fixed, λ i = max{ λ 0 a i, 0}, if i =,, 3 and λ 4 = max{ λ 0, 0} Thus the right-hand side in an otimal solution is λ 0 b + max{ λ 0 a, 0} + max{ λ 0 a, 0}+ Its value is equal to +max{ λ 0 a 3, 0} + max{ λ 0, 0} (λ 0 a + max{ λ 0 a, 0}) + (λ 0 a + max{ λ 0 a, 0}) + ( + (λ 0 a 3 + max{ λ 0 a 3, 0}) + λ 0 + max{ λ a ) 4 0, 0} + +λ 0 (b (a + a + a 3 + )) Here the first three terms are at least and the fourth term is at least +max{ λ 0, 0} Thus the sum of the first four terms is at least 4 If (0) holds then the total value of the right-hand side can not be less than 4, ie the cut (5) can not be generated in the first Chvatál iteration QED

8 Page 6 RRR A Secial Class of Knasack Problem [] has given a comlete descrition of the integer hull of a secial binary knasack roblem The inequality of the roblem is x + x + + x m + x m+ + x m+ + + x m+m b, (4) where m, m,, b are ositive integers The set of binary feasible solutions of (4) is denoted by F Let I {,,, m + m } be an arbitrary index set The sum of variables belonging to that set is x(i) Let T = {m +,, m + m } and S {,,, m } Denote the number of element of S by s Finally let q be a ositive integer such that q < h(s, q) = max{x(s) + qx(t ) : x F } The value of h(s, q) is given by the formula { b if s b h(s, q) = max{s + q b s, b ( q) b s } if b > s { } b Among the last m variables at most l max = min m, can have value in any feasible solution The main result of [] is the following: Theorem 4 (A) The integer hull of the knasack roblem is described by the following system of inequalities: (4), x(t ) l max, x(s) + qx(t ) h(s, q), S : S {,,, m } and q : q <, 0 x i, i {,,, m + m } (B) The inequality x(s) + qx(t ) h(s, q) defines a facet of the integer hull if and only if (s > q or s = q = ) and s {q + b, q + b,, q + b l max } 4 The case m = First we shall discuss the case m = in a detailed way We assume that b otherwise the roblem is trivial Then l max = It follows from Theorem 4 that for each q > the only value of s giving a facet of the integer hull is Hence it follows that h(s, q) = s s = q + b (5) The first question to be investigated is when has a facet with arameters satisfying (5) a Chvatál rank equal to For the sake of simlicity we assume that S = {,, s} The best cut of this tye what can be achieved in the first Chvatál iteration is described by the following linear rogram:

9 RRR 5-08 Page 7 min bλ 0 + λ + λ + + λ m+ λ 0 + λ λ m+ = λ 0 + λ s λ m+s+ = λ 0 + λ s+ λ m+s+ = 0 λ 0 + λ m λ m+ = 0 λ 0 + λ m+ λ m+ = q λ 0, λ m+ 0 This linear rogramming roblem will be analyzed by using simlex method If the simlex method detects an otimal solution with an objective function value at least s+ then the Chvatál rank is at least These cases will be searated when we follow the ath of otimization In what follows the negative of the objective function is maximized instead of the original objective function because of some technical reasons It is easy to see that the variables λ, λ m+ form a feasible basis simlex tableau is this, where all missing elements are 0: The aroriate λ 0 λ λ s λ m+ λ m+ λ m+ RHS λ λ s λ s+ 0 λ m 0 λ m+ q OBF b m q s If b m + then the integer hull is the comlete unit cube and the roblem is trivial In that case the solution rovided by the simlex tableau is otimal, ie the inequalities x(s) + qx(t ) h(s, q) are not facet defining, but on the other hand the Chvatál rank is 0 as the olyhedral set is the unit cube If m + > b then the only candidate to enter the basis is variable λ 0 If m > s then any of the variables λ s+,, λ m may leave the basis Assume that λ s+ leaves the basis Then the new simlex tableau is this: λ 0 λ λ s λ s+ λ m+ λ m+ λ m+ RHS λ 0 λ s 0 λ 0 0 λ s+ 0 0 λ m 0 0 λ m+ 0 q OBF b + m b + m q s

10 Page 8 RRR 5-08 There is only one otential candidate to enter the basis as b+ m = s+ q+ m = s m + q 0 as m s and q The only case when the current basis is otimal, is m =, q = It imlies that b + = m + Then the maximal feasible solutions are the following vectors: (a) the first m comonents are, and the last comonent is zero, (b) the last comonent is and b = m comonents among the first m are, ie the maximal feasible binary solutions are such that they have only zero comonent, their all other comonents are Thus this case is a generalization of case 7 of Table Hence the only inequality what must be generated to obtain the integer hull is m + j= It can be generated by the following weights: x j m λ 0 =, λ = = λ m =, λ m + = = λ m+ = 0 Then all coefficients of the left-hand side are in the generated inequality and the right-hand side is which can be truncated to m b + m = m +, Now assume that m + > b In general the inequality b + u m < 0 is a consequence of (5) if and only if u < m s + q After the entering of variable λ m+s+ into the basis the next candidate will be variable λ m+s+3 with reduced cost b + m The simlex method is iterating in this way until all of the variables from λ m+s+ to λ m enter the basis At this moment the simlex tableau is as follows: λ 0 λ λ s λ m λ m+ λ m+ λ m+ RHS λ 0 λ s 0 λ 0 0 λ m+s+ 0 0 λ m 0 0 λ m+ 0 q OBF q q q s The next variable to enter the basis is λ m+ As q < variable λ m+ is leaving the basis The new simlex tableau is this: λ 0 λ λ s λ m+ λ m+ RHS λ 0 q λ s 0 q λ 0 q λ m+s+ 0 λ m 0 OBF q 0 0 q q q q q s

11 RRR 5-08 Page 9 This is the otimal simlex tableau and the otimal objective function value is q + q + s Thus the Chvatál rank of the facet defining cut is if and only if q This is equivalent to the inequality + q + s < h(s, q) + = s + q q + > 0 (6) If m + > b and s = m then the only candidate to enter the basis is again variable λ 0 Then according to the ratio test only λ m+ may leave the basis After ivoting the new simlex tableau is just the same as the otimal tableau of the revious case: λ 0 λ λ m λ m+ λ m+ RHS λ 0 q λ m 0 q λ 0 q OBF q q q q s Hence the Chvatál rank is if and only if (6) holds Thus the following lemma has been roven: Lemm Let m,, and b be ositive integers such that m + > b + Then the Chvatál rank of the integer hull of the set { x IR m + x + + x m + x m+ b; 0 x i, i =,, m } (7) is if and only if a ositive integer q with q < exists such that (6) holds If is fixed then the left-hand side of (6) can be considered as a single variable quadratic function of q, say f(q) The roots of f(q) are: q = + 4, q = 4 Hence if = 3 then the equation f(q) = 0 has only comlex roots, ie f(q) lies comletely above the horizontal axis Thus for = 3 the Chvatál rank of the integer hull is indeendently from the value of m It is easy to see that if 4 then the roots have the following roerties: q, q Inequality (6) is satisfied if q = or q = and it is not satisfied for any integer between and For examle if 4 then the cuts of tye x i + + x is + x m+ with s = b +, have a Chvatál rank at least This fact leads to the following theorem Theorem 4 Let m,, and b be ositive integers such that m + > b + and 4 Then the Chvatál rank of the integer hull of the set (7) is at least The main content of the theorem is that although the set defined in (7) has one of the simlest definitions among the sets of binary vectors, its Chvatál rank is still large s

12 Page 0 RRR An uer bound of the Chvatál rank In this subsection an inductive rocedure is given to generate all of the facet defining cuts To any induction two things have to be shown: the iterative ste and the fact that the conditions of the start of the iterative rocedure, are satisfied We shall discuss the inductive algorithm in this order First assume that the facet defining cuts for q = i and q = i, where i <, are exiting and have been already generated We shall show that in this osition the facet defining inequalities for q = i + and q = i can be generated in one Chvatál iteration Case q = i + We got a facet defining inequality if s = b + i + For the sake of simlicity assume that S = {,, s} According to the assumtion we have all inequalities with s 0 = s and q 0 = q The set S has s subsets with s 0 number of elements We shall consider the aroriate s inequalities with arameters s 0 and q 0 All of these inequalities will have the same weight, say u, in the generation of the required inequality Only one inequality will have a weight different from 0 among the inequalities belonging to the arameters s = b i, q = i Assume that the aroriate set is S = {,, b + i} The weight of this inequality is denoted by v Finally the weights of the inequalities x j 0 (j = s +,, b i) are again v The value of u and v will be chosen such that the coefficients of the left-hand side of required inequality are obtained The coefficient is obtained from the s of s inequalities all having the weight u and from the of the last inequality with weight v Thus the equation (s )u + v = (8) must hold The coefficient of x m+ is i + in the required inequality In the same osition the coefficients of the inequalities used in the generation are i and i, resectively Hence we obtain the equation The solution of the system (8)-(9): u = siu + ( i)v = i + (9) i (s )( i) i, v = (s )u = s i (s )( i) i (0) It is easy to see that both u and v are nonnegative These multiliers give a Chvatál rank if and only if us(s ) + v(b i) < s + () By substituting (8) the left-hand side of () can be reformulated as follows: us(s ) + v(b i) = us(s ) + ( (s )u)(b i) = u(s )(s b + i) + b i Hence () is equivalent to u(s )(s b + i) + b i s = (u(s ) )(s b + i) = v(b s i) < As s = b + i + the equation b s i = i holds Thus v(b s i) = v( i ) = s i ( i ) < () (s )( i) i The value of s is at least as b > and i From the assumtion that i < it follows that denominator is ositive Then () holds if and only if s si s i + i + i + i + < s si + i i (3)

13 RRR 5-08 Page It can be reformulated as which is true as s i = b 0 < i( i ) + s i, (4) Case q = i The inequality to be generated has the arameters s = b i and q = i For the sake of simlicity it is assumed that S = {,, b i } The airs of arameters used in the revious case, ie s 0 = b + i, q 0 = i and s = b i, q = i, are used again The generation is made by the following construction The facet defining inequalities with s 0 and q 0 such that the index set of their s is a subset of S, are used with the same weight, say u The facet defining inequality with arameters s, q and S = {,, b i} and the inequality x b i 0 have the weight v Each of the required inequality is in ( b i b + i ) inequalities of the used ones of tye s 0, q 0 Hence the equation ( ) b i u + v = (5) b + i is obtained The number of inequalities of the first tye is ( ) b i b + i in the construction Thus to get the coefficient of x m+ the equation ( ) b i u + ( i)v = i (6) b + i must be satisfied The solution of the (5)-(6) system is u = It is easy to see that ( b i b +i ) ( ), i b i b +i i i b i b +i i > 0 v = i b i (7) b +ii Hence it can be roven that both u and v are nonnegative The right-hand side of the generated inequality is ( ) b i RHS = (b + i)u + (b i)v b + i The construction gives the required Chvatál cut if and only if RHS < b i Let ( ) b i A = and B = b + i i b i b +i i Then RHS = B A Ab i (b + i) + (b i) B(b i) = b i B < b i b + i

14 Page RRR 5-08 What is remained to show is that the algorithm can be started If q = then s = b + and from the assumtion m + > b it follows that m s If q = then s = b, which can be greater than m In that case the inequality x + + x m + ( )x m+ b (8) is equivalent to (7) (8) can be generated by the following multiliers: λ 0 =, λ = λ = = λ m =, λ m + = = λ m+ = 0 The same rocedure can be reeated until the aroriate s will not exceed m These results can be summarize in the following statement Theorem 43 Let m,, and b be ositive integers such that m + > b + and 4 Then the Chvatál rank of the integer hull of the set (7) is at most 4 The case m > It can be shown that the facet defining inequality has Chvatál rank if and only if (6) holds, in a similar way by analyzing the simlex method The details remain to the reader References [] Chvatál, V, Edmonds olytoes and a hierarchy of combinatorial roblems, Discrete Math, 4(973), [] Dahl, G, Foldnes, N, Comlete descrition of a class of knasack olytoes, Oerations Research Letters, 3(003), [3] Schrijver, A, The Theory of Linear and Integer Programming, J Wiley & Sons, Chichester, 986