Operations Research Proceedings 1991, Springer-Verlag Berlin Heidelberg (1992) 439 444 About the gap between the optimal values of the integer and continuous relaxation one-dimensional cutting stock problem Guntram Scheithauer and Johannes Terno Dresden University of Technology Abstract: The purpose of this paper is to show that the gap is possibly smaller than 2. Some helpful results are summarized. Zusammenfassung: Es werden Ergebnisse vorgestellt und diskutiert, die bei der Untersuchung der maximalen Differenz zwischen den Optimalwerten des ganzzahligen und stetigen eindimensionalen Zuschnittproblems erhalten wurden. 1 Introduction The well-known one-dimensional cutting stock problem arises when larger stock length are to cut into shorter piece length. The standard linear programming formulation given by Gilmore and Gomory (1961) is the following. Let be given the stock length L and the piece lengths l i, i = 1,..., m, not larger than L, and order quantities b i of piece i, i = 1,..., m. A cutting pattern is a combination of pieces having a total length not larger than L. It can be represented by a nonnegative integer vector a j = (a 1j,..., a mj ) T. The objective is to minimize the number of stock lengths needed to fulfill the order demands. Hence, z(x) = e T x = min (1) Ax = b (2) x 0 (3) x integer (4) 1
where the cutting patterns a j, j = 1,..., n, are the columns of A and b = (b 1,..., b m ) T, e = (1,..., 1) T, x = (x 1,..., x n ) T. Relaxing the integrality constraint leads to a linear program having an optimal solution x c. x c is in general fractionally. Let x int denotes an optimal integer solution. Diegel (1988) suggests the following proposition: the gap between z(x int ) and z(x c ) is less than 1. Two examples and some motivations seem to verify this proposition. But Fieldhouse (1990) presents counterexamples with z(x int ) z(x c ) = 1.0333... In the paper we show for some special cases that z(x int ) z(x c ) + 1 (5) where a denotes the smallest integer not smaller than a. Furthermore a defines the largest integer not larger than a and {a} is the fractional part of a. 2 Results 2.1 Problemreduction The continuous solution x c can be dissected according to integer and fractional parts of its components, hence x c = x c + {x c }. With α = e T x c and β = e T {x c } one has z(x c ) = α + β. Rounding down yields residual order quantities b = b A x c. If there exists an integer solution x red for (1) - (4) with right hand side b fulfilling then z(x red ) β + 1 x c + x red is an integer solution of the initial problem which performs the assertion (5). Hence, it is only necessary to consider right hand sides b which yield to continuous solutions having components less than 1. 2
2.2 Fundamental statements According to the previous section we may assume in the following that the components of solutions of the relaxation problem (1) - (3) have values less than 1. For abbreviation we set = L l i, i = 1,..., m. (6) A cutting pattern a j = (a 1j,..., a mj ) T is called elementary if Let = { ki if i = j, 0 if i j. γ = m b i (7) be the objective function value using only elementary patterns. In general, the optimal function value β is less than γ, i.e. β γ. (8) Lemma 1 Let be given positive integers and b i, i = 1,..., m, with b i, i = 1,..., m,. Then there exist γ + 1 nonnegative integer vectors a j = (a 1j,..., a mj ) T with γ +1 m = b i, i = 1,..., m, 1, j = 1,..., γ + 1. (9) Proof: The proof is done by induction. The proposition is fulfilled for n = 1 with a 11 = b 1, a 12 = 0. Let n b i γ n =. (10) Case a): Let be k n+1 = k r for one r with 1 r n. Then one can define b (1) r = b r + b n+1. If b (1) r k r all is done, otherwise we have b (1) r 2k r and with b (2) r := b (1) r k r, b (2) r k r, we get n,i r b i + b(2) r = γ n+1 1. k r Since b i for all i r and b (2) r k r there exist γ n+1 nonnegative integer vectors a j and with a r,γn+1 +1 := k r, a i,γn+1 +1 := 0, i r, it follows the statement. 3
Case b): Because of case a) we may assume 1 k 1 < k 2 <... < k n+1. Hence k n+1 n + 1. Because of γ n γ n+1 there exist γ n+1 + 1 nonnegative integer n-dimensional vectors a j = (a 1j,..., a nj ) T with and n = b i, i = 1,..., n, 1, j = 1,..., γ n+1 + 1. Now we consider these vectors as (n + 1)-dimensional vectors setting Hence According to (10) we have a n+1,j := (1 n )k n+1. (11) 1 1 n+1 < 1, j = 1,..., γ n+1 + 1. k n+1 ) n b i b n+1 k n+1 (γ n+1. Hence, b n+1 a n+1,j k n+1 γ n+1 1 n b i a n+1,j k n+1 = k n+1 γ n+1 1 n a n+1,j 1 k n+1 n+1 = k n+1 γ n+1 ( < k n+1 (γ n+1 (γ n+1 + 1) 1 1 )) k n+1 ( = k n+1 1 + γ n+1 + 1 ) k n+1 k n+1 1. Because of the integrality of b n+1 and the a n+1,j it follows a n+1,j b n+1 4
and we can choose suitable nonnegative integers a n+1,j, j = 1,..., γ n+1 + 1, with This completes the proof. a n+1,j a n+1,j and a n+1,j = b n+1. Lemma 2 Let be given positive integers and b i, i = 1,..., m. Then there exist γ + 1 nonnegative integer vectors a j = (a 1j,..., a mj ) T with γ +1 m = b i, i = 1,..., m, 1, j = 1,..., γ + 1. Proof: Let α i := b i and β i := b i α i. It holds γ = m b i = m m α i + β i =: Γ 1 + Γ 2 Because of Lemma 1 there exist Γ 2 + 1 nonnegative integer vectors a j with and m Γ 2 +1 = β i, i = 1,..., m, 1, j = 1,..., Γ 2 + 1. Using α i times the corresponding elementary patterns the assertion is proved. Theorem 1 Let β be the optimal function value of problem (1) - (3) and γ be defined according to (7). If β = γ then there exists an integer solution of problem (1) - (4) with objective function value not larger than β + 1. Proof: Because of Lemma 1 resp. 2 there exist γ + 1 = β + 1 nonnegative integer vectors a j which fulfill condition (2). Since the a j represent feasible cutting patterns the assertion is proved. 5
2.3 Generalizations Let κ := j sign(x j ). If x corresponds to a continuous solution then obviously κ m. Lemma 3 If β = 1 or β κ 1 then the assertion (5) is fulfilled. Proof: For β 1 j a j {x j } is by itself a cutting pattern. In the other case we get κ patterns by rounding up. The difference m 1 seems to be an essential characterization for a cutting pattern and also for instances of cutting problems. Generally holds Lemma 4 If γ β = i i / 1 1 x j j j x j = j x j ( i ) 1 for all columns j then (5) is fulfilled. (12) Proof: Formula (12) yields γ β 0. Together with (8) the premises of the theorem are fulfilled. Remark: In general, there is max { i : i l i L, a j = (a 1j,..., a mj ) Z m + } 1.7. This follows from a Lemma given in Coffman et al. (1980). In the following an example with β < γ is described. i l i b i a 1 a 2 a 3 1 15 2 1 1 1 0 Let L = 35. 2 10 3 1 2 0 0 3 6 5 6 0 3 5 With the continuous solution x 1 = 0.5, x 2 = 0.5, x 3 = 0.9 one has β = 1.9 < 2 < γ = 61 30. It is to denote that there exists optimal integer solutions with only 2 patterns. It is remarkable that all patterns in the example with L i l i > l 3 have the property i 1. 6
2.4 The largest gap found The following example enlarges the gap given by Fieldhouse(1990). i l i b i 1 44 3 2 Let L = 132. 2 33 4 3 3 12 11 6 Here we have β = γ = 2 3 + 3 4 + 6 11 = 259 132 = 1.962121... There exists no integer solution with 2 patterns. Hence, the gap equals 1.0378.... The same gap arises if the pieces with length 33 are replaced by pieces of lengths 66 and 33 each having order quantity 1. References [1] E.G.Coffmann,jr., M.R.Garey, D.S.Johnson, R.E.Targon, Performance bounds for level oriented two-dimensional packing algorithms, SIAM J.Comput. 9 (1980)p.808-826. [2] A.Diegel, Integer LP solution for large trim problems, Working Paper 1988 University of Natal, South Africa. [3] M.Fieldhouse, The duality gap in trim problems, SICUP-Bulletin No. 5,1990. [4] P.C.Gilmore, R.E.Gomory, A linear Programming approach to the cutting stock problem, Op. Res. 9(1961)p.849-859. [5] J.Terno, R.Lindemann, G.Scheithauer, Zuschnittprobleme und ihre praktische Lösung, Verlag Harry Deutsch Thun und Frankfurt/Main 1987. 7