on dominated terms in the general knapsack problem

E L S E V I E R Operations Research Letters 21 (1997) 31-37

On dominated terms in the general knapsack problem

Nan Zhu, Kevin Broughan* Department of Mathematics, University of Waikato, Private Bag 3105, Hamilton, New Zealand

Received 1 July 1995; revised 1 February 1997

Abstract

A necessary and sufficient condition for the identification of dominated terms in a general knapsack problem is derived. By general, we mean a knapsack problem which is unbounded, equality constrained and which has a parametric right-hand side. The given condition yields recently published results in the literature. A report on computational experiments for large-scale knapsack problems, demonstrating the effectiveness of this approach, is included. © 1997 Published by Elsevier Science B.V.

Keywords: Knapsack problem; Dominated term; Representation

1. Introduction

In Integer Programming (IP), preprocessing (in- cluding eliminating redundancies, fixing variables, tightening constraints, adding logical inequalities, and so on) can be regarded as an important phase between formulation and algorithm application, especially for large-scale real-world problems, see [5]. In this paper, we use the idea of fixing variables to reformulate a knapsack problem (KP) which is a NP-hard and classical problem in IP (see [3]). First, we give:

Definition I. The general knapsack problem (called KP(1)) is defined in the form:

F(b) = max ~ cjxj jeN

* Corresponding author.

subject to ~, ajxj = b, j~N

xj/> 0 integer, V j ~ N , (1.1)

where {a j} are given positive integers, {c j} are given arbitrary integers, b is a nonnegative integer parameter, j C N = {1,2 . . . . ,n} with n/> 2.

If {cj} ( j ~ N \ { n } ) are given positive integers, c. = 0 and a. = 1, then the problem (1.1) is an ordinary K P (called KP(2)):

F ( b ) = m a x ~, cjxj j~N\{n}

subject to ~, ajxj + x . = b, j~N\{n}

xj/> 0 integer, V j ~ N , (1.2)

where x, is a slack variable. By a given K P we mean that the KP has a fixed

b = bo. For a given KP(2), some results and algorithms can be seen in [3]. For a given KP(1), an

0167-6377/97/$17.00 © 1997 Published by Elsevier Science B.V. All rights reserved PII S0 1 67-6377(97)0001 8-7

32 N. Zhu, K. Broughan / Operations Research Letters 21 (1997) 31-37

enumeration algorithm is obtained by Yanasse and Soma [7]. For KP(2), an elimination criterion and a corresponding algorithm for eliminating dominated item types are given by [3]. Further work on dominance relations for KP(2) has been done by Dudzifiski [2], and by Pisinger [6]. 1 A similar result to Theorem 3.2 of [3] for KP(2) is provided by Babayev and Mardanov [-1], and that result is generalized and extended for the multi-dimensional KP and mixed IP problems.

In this paper, we are interested in discussing how to identify and eliminate dominated terms in KP before any of the known solution methods for a given KP is used. Our purpose is to establish a basis for discussing dominance relations in KP. The concept of the dominated term of KP is defined in Section 2. A necessary and sufficient condition for the identification of these dominated terms is provided. Several elimination criteria are set up, and recently published results in the literature can be considered as the special cases of our basic result. Computational experiments for large-scale problems are reported in Section 3.

2. Main results

KP(1) is in general harder to solve than KP(2) since in KP(1), there may not exist a feasible solution for some values orb. For a given KP(1), we call the term with index k e n (term k for short) the redundant term if the optimal solution value F(bo) (if it exists) does not change when k is removed from N. For a redundant term k, we can fix xk = 0 in the original problem. In KP(1), there may exist some terms that can be removed for all nonnegative integer values of parameter b without altering the optimal solution value F(b) (if it exists). Our work is to identify and try to eliminate this kind of redundant terms.

We first give two definitions.

Definition 2. A positive integer a k has a representation by the positive integers {a j} (j e N\{k})if there

1 After the work on this paper was finished, we became aware of Pisinger's work [6].

exist nonnegative integers {lj}, such that ak = Y.~ N,,{k} ljaj.

Definition 3. In KP(1), term k is dominated if the optimal solution value F(b) (if it exists) does not change when k is removed from N.

We provide a basic result.

Theorem 1. In KP(1), term k(k E N) is dominated if and only if there exist nonneotive integers {lj}, such that

ak= ~ ljaj and Ck<~ ~, IjCj. (2.1) j6N\{k] jcN\,{k}

Proof. Here we only give the proof of the necessary condition.

(=>) Let term k be dominated, and k be eliminated from N in KP(1). Suppose ak has no representation by {aj}jeN\{k}. When b = ak, F(ak) does not exist in the new KP(1). However, in the original problem, there is an optimal solution: x* = 1, and x* = 0 otherwise, with F(ak) = Ck. This contradicts Definition 3. Therefore, ak has at least one representation by other {a j}.

Now suppose that for this term k it is the case that if ak=YO~N\(k}ljaj with integers lj>.O (VjE N\{k}), it is always true that Ck > Y~j~N\(k~ ljcj. When b = ak, in the new KP(1),

F(ak) = m a x ~ ljcj& ~ l*cj. j~N\{k} jeN\{k}

However, in the original KP(1), F(a t )= Ck> ~j~U',{k} l 'c j, SO we arrive at a contradiction. There- fore, condition (2.1) is satisfied. []

Since a, = 1 and c, = 0 in KP(2), using Theorem 1 it is easy to obtain

Corollary 1. In KP(2), term k(k~ N\{n}) is dominated if and only if there exist nonnegative integers {lj}, such that

ak > ~ ljaj and ck <~ ~ ljcj, (2.2) .je N ',{k,n} j~ N\{k,n}

term n is dominated if and only if there exists at least one j e N \ { n } such that aj = 1.

N. Zhu, K. Broughan / Operations Research Letters 21 (1997) 31-37 33

The significance of Theorem 1 is that all the dominated terms in KP(1) are identified. Below, letting I i take suitable values, several concrete elimination criteria are set up based on Theorem 1 to try to eliminate these dominated terms.

We first have

Theorem 2. In KP(2),/f there are two distinct terms j and k (j, k E N \ { n } ) such that

lak/a~] cj >>- ck, (2.3)

then term k is dominated.

Proof. In KP(2), a, = 1 and c, = 0. There always exists a nonnegative integer rj < aj satisfying ak = Lak/ajJa; + ria,. Under condition (2.3), Ck ~< Lak/a;Jcj + rjc,. Thus, term k is dominated using Theorem 1 (lj = Lak/ajJ, I, = r;, and li = 0 otherwise). []

Theorem 2 above can be regarded as Theorem 3.2 in [3] or Theorem 1 in [1]. The O(n 2) elimination algorithm for dominated terms satisfying condition (2.3) (see Corollary 3.1 in [3]) is an important part of the new Martello-Toth algorithm in [3] for solving a large-scale KP(2). In [2], it is shown that the dominance relation (2.3) is a partial order lead- ing to a more efficient elimination algorithm for KP(2). In [6], using condition (2.3) (i.e. Definition 2 in [6]), a faster reduction method for KP(2) is discussed through sorting according to nonde- creasing weights {aj}j~u\~,~. Note that using Corol- lary 1, it is easy to yield an elimination criterion (i.e. Definition 1 in [6]) for KP(2).

Using Theorem 1, a new criterion for KP(2) is derived.

Theorem 3. In KP(2), /f there are three distinct terms i,.j and k (i,j, k ~ N \ {n}) such that

k (a, -- L ak/a; J a;)/ai J ci + L ak/a;J c; >>. ck, (2.4)

then term k is dominated.

Proof. From the proof of Theorem 2, we know ak = Lak /a jda j + rj, where 0 ~< rj = ak -- Lak /a jJa j . There always exists an integer r'i satisfying 0 <~ r'i < al such that rj = Lrj/aiJai + r'ia,. Hence, ak = l ak/aj la~ + I rj/ai ]ai + rla,. Letting lj = Lag/

ajJ, li = L rJai J, I, = r~ and lh = 0 otherwise, using condition (2.4) and Theorem 1, term k is dominated. []

Using Theorem 3, a corresponding O(n 3) elimination algorithm is easily devised. Theorems 2 and 3 are not suitable for handling KP(1) directly. In the following, we give some results for KP(1). We first provide the formulas:

Inequalities. Let cl and c 2 be real, a 1 and a 2 be positive, ll and 12 be nonnegative with at least one li > O. Then the following inequalities hold:

min c l c2 ~< ~<max , . (2.5) ~al ' lxal + 12a2

Proof. We can always let c l / a l >~ c2/a 2. Then m a x { c l / a x , c z / a 2 } = cl/a~ ~=h, rr f in{c l /a l ,c2 /a2} = c2 /a2gk . Hence h/> k. We have

c2 k(l lal -1- 12a2) 11cl + 12C2 l lhal + 12ka2

a2 11al + 12a2 llal + 12a2 llal + 12a2

h(llal + 12a2) Cl ~< - [ ] llal + 12a2 al"

Inequalities (2.5) can easily be extended by induc- tion to the general form:

General Inequalities. Let {cl} be real, {ai} be positive, {li} be nonnegative with at least one Ii > 0, ie { 1 . . . . . m} with m >>. 2. Then the following inequalities hold:

rain fc t l<~i<~ m ~l<~i-<mliai~ ~ l<~i~mmaX . (2.6)

In KP(1), assume that

cl /a l = m a x { c J a j } , (2.7) jEN

and if cl/al = ci/ai then al <<. ai ( i~N\ {1} ) . This kind of KP(1) is called KP(I'). When al = 1, KP(I') obviously has the optimal solution: x* = b, and x* = 0 otherwise, with F ( b ) = clb. Therefore, we always assume a~ > 1 in KP(I'). If there exists k ~ N \ { 1 } such that cl/al = Ck/ak and al = ak, it is


reasonable to assume always term k is dominated and is eliminated in KP(I'). Hence, we give

A Property of KP(I'). In KP(I') in which !f c l /a l = cl/ai then al < ai is set for ieS\{1}, term 1 is undominated.

Proof. Suppose term 1 is dominated, i.e. there exist integers lj>~0 for all j~N\{1} such that a~ = Y4~N',(~I ljaj and cl <~ Y~j~N,,~t l jcj. Obviously, at least one I t # 0, and the corresponding l~ = 0 if ca/al = ci/ai (i ~ N\{1}). Letting N' be the index set of all 1 t 4: 0, then all of the above i41N'. Using (2.6), it follows

Cl _ Cl ~jeN\,{ll l jc j _ ~ j s N ' ljCj

aa ~ j eN , {1} l j a j ~ ~jEN',,{]] l taj ~ j s N ' l j a j

..< max ~ c ~ & % , jEN' ( a j ) at,

where j l ~N'. Since c~/al > Cjl/aj,, we arrive at a contradiction. []

Theorem 4. In KP(1), i f there are two distinct terms j and k ( j , k ~ N \ { 1 } ) such that

a j - a k ( m o d a l ) , aj <~ ak and pj <~ p,; (2.8)

o r

ak =-- 0 (modal), (2.9)

where a l (> 1) satisfies ca~a1 = maxjeN{cj/aj} , and p i~=c la l - ciaa (V i~N\{1}), then term k is dominated.

Proof. Using condition (2.8), we have ak = laaa + aj, where integer la >t 0. Since pj <~ Pk, we h a v e c l a j -- c j a I ~ Clak -- Ckal = C l ( l l a 1 + at) - Ckal. Thus, Ck <~ llca + C t. Hence, term k is dom-

inated using Theorem 1 (l t -- 1). Using condition (2.9), we have ak = 11 ax , where

integer ll > 0. Since cl /al >7 c,/ak = c , / l la l , c, <~ llca. Hence, term k is dominated. []

which is a relaxation problem of KP, in which Pi can be considered as the coefficient of xl in the objective function of the group problem, see e.g., [4,8]. Using Theorem 4, an O(n 2) elimination algorithm for KP(1) is easily devised. Note that if using KP(I') in place of KP(1) in the theorem, more dominated terms may be identified since term 1 in KP(I') always is an undominated term. In addition, Theorem 4 can also be applied to KP(2).

Example 1. Reformulate a KP(2):

F(b) = max l lxa + 22x2 + 6x3 + 15x4 + 35x5

q- 3 x 6

subject to

4xl + 9X2 + 3X 3 -I- 6X4 + 15X5 + 2X6 + X7 = b,

x j ) 0 integer, V j e N ,

where c l / a 1 > cha t (Vj~N\{1}), and b is a nonnegative integer parameter.

Using Theorem 3 or 4, it follows that terms 2 and 5 are dominated. Thus, terms 2 and 5 can be eliminated from the problem before using a solution method. If we apply Theorem 2 to this example, only term 2 is identified as being dominated.

Using Theorem 1, it is easy to extend Theorem 4 to the following criterion:

Corollary 2. In KP(1), i f there are two distinct terms j and k (j, k e N \ { r } ) such that

ca t =-ak(modar) , ca t <. ak,

o r

and cqj ~ qk',

(2.10)

a k = O ( m o d a , ) , and O<~qk, (2.11)

where c is a given positive integer, given r ~ N such that ar > 1, and q i ~ c r a i - ciar ( V i e N \ { r } ) , then term k is dominated.

The use of the modular arithmetic method is linked with solving the group knapsack problem,

The proof of Corollary 2 is similar to the proof of Theorem 4, so is omitted.


Example 2. Reformulate a KP(1):

F(b) = max 49xl + l l x 2 q- 22X3 + 6X 4 q- 15X5

q- 35X 6 d- 3X 7

subject to

16xl + 4x2 + 9X3 q- 3X4 -I- 6X5 q- 15X6 -t- 2X7 = b,

x i~>0 integer, V j • N ,

where cl/al > c/aj (VjeN\{1}), and b is a nonnegative integer parameter.

For this equality constrained problem (F(1) does not exist), Theorems 2 and 3 cannot be applied, and Theorem 4 has no effect. Letting c = 1 and choos- ing r = 2 (or 3 or 5), term 6 is identified as being dominated using Corollary 2.

Note that before using elimination results of this paper for a given KP(1), knowing whether F(bo) exists is not required since the original problem is equivalent to the new problem obtained using the elimination.

In the following, we add a condition to KP(1) by assuming

cl/al >t c2 /a2 ~ "" ~ c~/a~. (2.12)

This kind of KP(1) is called KP(I"). Using Theorem 1 and formulas (2.6), we give

Theorem 5. In KP(I"). ifak(k • N \ ( 1 }) has a representation by a~, ... ,ak-1, then term k is dominated.

Proof. Let ak = E1 <~j<<.k--1 l jaj, where {li} are nonnegative integers. Using (2.6), we have

C k < C k - l _ min {c-~j}~ ~l<~j<~k-lljcj ak Ok-1 l~j<~k-I ~.l<~j<<.k-lljaj

_ ~ , l ~ j <. k - ll;cj

ak

Hence, we have Ck <<. ~1 <,j <. k -- 1 ljc;. By Theorem 1, term k is dominated. []

Theorem 1 is easily generalized for a mixed integer KP.

Theorem 6. In a mixed integer KP, where the form looks like KP(1) but given aj > 0, given cj real, (k/ j • N), parameter b >t 0, integer xj, >10 (V j l • N1 ___a

{1 . . . . ,p}), and real xj~ >~ 0 ( 'V'j2•N2 ~ {p + 1 . . . . . n}), term k(k • N 1 ) is dominated if and only if there exist integer ljl ~ 0 (V j l • Nl\{k}), and real l j2/> 0 (V j2 • N2), such that

ak= ~. ljaj and Ck<~ ~ ljcj. (2.13) j~N\(k} jeN\{k}

The proof of the theorem is similar to the proof of Theorem 1, so is omitted. Based on Theorem 6, effective elimination criteria and efficient algorithms can be developed. Note that in the special case: p = n, Theorem 6 can be considered as The- orem 1. Ifp = 0, it is easy to prove by aj = (aj/al)al and cj <~ (aflal)cl, Vj • N\{ 1 }, that except for term 1, (assuming cl/al = maxj~N{cflaj}), all terms can be regarded as being dominated. Therefore, this kind of KP, that is a simple Linear Programming problem, has the optimal solution: x* = b/al, and x* = 0 otherwise, with F(b) = cl b/a~.

3. Computational experiments

In this section, we analyse experimental results only using Theorem 2 (TH2) and Theorem 4 (TH4) of this paper to identify and eliminate dominated terms of KP. The computational language used was FORTRAN. All runs have been executed on a SparcStation 2 Workstation.

Three types of test problems with integer data were investigated: first was KP(1), second KP(2), and finally the (unbounded) subset-sum problem (SSP) sometimes called the value independent problem (see [3, 4]). A SPP can be considered as a completely correlated KP(2): cj = aj.

For KP(1), we considered the data set {(c j, aj)}j~N, where all cj were distributed uniformly random in [I, u], and corresponding aj uniformly random in [10,u], where u = 1000 and 10000, respectively. For KP(2), we considered the data set {(cj, aj)}j~NU {(0, 1)}j=n+l, where the {(cj, aj)}j~N were produced as for KP(1) above. For SSP, we considered the data set ((aj, aj)}j~NU ((0, 1)}j=n+ 1, where all a i were produced as for KP(1) above.

Let r~l, ~i2, ti3 denote the mean number of re- maining integer variables (corresponding to pos- sibly undominated terms, for KP(2) and SSP term


Table 1

The effectiveness of T H 2

T H 2 C P U Time

u n til ?12 ?/3 "~1 $2 $3

1000

10000

500 3.48 167.04 - - 0.011 0.323

1000 3.64 185.32 ..... 0 .024 0.772

2000 --- 3.64 184.04 0.050 1.672

5000 3.82 176.58 - - 0 .140 4.312

10 000 3.84 176.00 - - 0 .307 8.874

50 000 3.84 176.00 1.890 45.965

500 .... 3.04 390.10 - - 0 .010 0.705

1000 - 3.00 632.10 0.022 2.356

2000 3.10 926.36 - - 0 .045 7.097

5000 - - 3.36 1249.72 - - 0 .126 25.830

10 000 - - 3.54 1344.36 - - 0 .275 62.871

50 000 3.94 1239.18 - 1.671 353.488

Table 2

The effectiveness of T H 4

/~/ ?1

T H 4 C P U Time

nl n2 n3 $1 $2 $3

1000

10000

500 14.78 14.82

1000 12.36 12.40

2000 11.32 11.36

5000 10.72 10.90

10000 10.24 10.48

5 0 0 0 0 10.00 10.44

500 43.04 43.04

1000 23.50 23.50

2000 19.04 19.04

5000 14.32 14.34

10000 12.22 12.28

5 0 0 0 0 10.62 10.92

12.72 0.052 0.042 0.037

11.50 0.085 0.084 0.085

11.18 0.169 0.174 0.171

11.02 0.473 0.479 0.378

11.00 1.054 1.068 0 .820

11.00 5.941 5.978 4 .000

32.26 0.090 0.091 0 .074

19.30 0.157 0.140 0 .134

15.66 0.306 0.261 0 .214

12.72 0.696 0.601 0.520

11.70 1.240 1.246 1.074

11.00 6.281 6.253 5.979

n + 1 is included since a t ~> 10. From Corollary 1, term n + 1 (Cn+ 1 = 0 , an+l = 1) always is undominated), computed over 50 problem instances, after using TH2 and TH4, respectively, in the three type test problems: KP(1), KP(2) and SSP.

For each type of the problem and specified number n of integer variables, Tables 1 and 2 present the ~ and corresponding average running time g~ (sec- onds) using TH2 and TH4, respectively. The data was not sorted when using TH2. When using TH4,

the sorting time for (cl, al), that satisfied condition (2.7) (for SSP, it was chosen al <~ a j), was included in g~ of Table 2.

Note that from theoretical point of view (i.e. using the simple criterion: in a KP, if aj = ak and cj >~ ck, j ~ k, then term k is dominated), except term n + l there are at most min {n, 991}, min{n,9991} undominated terms in every problem instance wherein u = 1000, u = 10000, respectively.


The total number of problem instances that we discussed (using TH2 and TH4) was 1200 + 1800 = 3000. The experimental results showed that the identifying effect of TH2 and TH4 was different to the same problem.

For KP(2), TH2 worked better than TH4 (some experimental results can also be seen in [2, 3, 6]), but TH2 cannot handle KP(1). For the test problem of the SSP type, the number of possible undominated terms after using TH2 was at least 12 times greater than that after using TH4, and was computationally expensive. The effectiveness of TH2 was more sensitive than that of TH4 to the relation between cj and aj (see also I-2, 3, 6]). In TH2, increasing the correlation relation weakened the effectiveness of the elimination criterion, i.e. increased the number of possible undominated terms.

Using TH4 in KP(1), KP(2) and SSP, respectively, the numbers of possible undominated terms of the problems were small. Increasing the number, n, of variables of KP(1) increased the number of dominated terms that were eliminated. When n = 50 000, on the average 99.98 % of integer variables in KP(1), wherein u = 1000 or u = 10000, were fixed at the value of zero before using any of the known K P algorithms. Since the modular arithmetic method is used, the effectiveness of TH4 might depend on the relative sizes of al and n in a problem: the smaller the value ofa~ and larger the

value of n (especially for a large-scale problem with a~ << n), the greater the number of dominated terms which are eliminated.

Acknowledgements

The authors would like to thank Prof. G.L. Nem- hauser and a referee for their suggestions and com- ments.

References

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[2] K. Dudzifiski, A note on dominance relation in unbounded knapsack problems, Oper. Res. Lett. 10 (1991) 417419.

I-3] S. Martello, P. Toth, Knapsack Problems: Algorithms and Computer Implementations, Wiley, Chichester, 1990.

[4] G.L. Nemhauser, L.A. Wolsey, Integer and Combinatorial Optimization, Wiley, New York, 1988.

[5] G.L. Nemhauser, The age of optimization: solving large- scale real-world problems, Oper. Res. 42 (1994) 5-13.

[6] D. Pisinger, Dominance relations in unbounded knapsack problems, Technical Report No. 94/33, DIKU, Copen- hagen, Denmark.

[7] H.H. Yanasse, N.Y. Soma, A new enumeration scheme for the knapsack problem, Discrete Appl. Math. 18 (1987) 235-245.

[8] N. Zhu, On the relationship between the knapsack problem and the group knapsack problem, Technical Report No. 26, ISOR, Wellington, NZ, March, 1993.

on dominated terms in the general knapsack problem

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