Maintaining the All-Integer Property of an ILP When Using Cuts in Rational DataAuthor(s): Søren HolmSource: Operations Research, Vol. 30, No. 1 (Jan. - Feb., 1982), pp. 208-209Published by: INFORMSStable URL: http://www.jstor.org/stable/170321 .

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Maintaining the All-integer Property of an ILP when Using Cuts in Rational Data

S0REN HOLM Odense University, Odense, Denmark

(Received September 1980; accepted January 1981)

If all data for an ILP is integer and a cut is given in rational data, then there exists a transformation of the cut such that all the coefficients and the right hand side constant are integers, thus requiring the slack variable also to be an integer. By repeating this transformation inductively an all-integer problem remains all-integer. As an application of this result, it is shown how to solve an ILP by using transformed strong MILP cuts. As these cuts are superior to fractional cuts, a more efficient algorithm results.

(1 ONSIDER the integer linear program (ILP)

max cx

s.t. Ax= b

x > 0 integer

where all the parameters are assumed integers. Given a cutting-plane, in tableau variables,

EjE-RdjX; I dO (do>O) (1)

and assuming the cut is in rational data, we may write dj = e1/E for integer ei and a fixed integer denominator E>- 1 so that (1) is equivalent to

ZjeR eXj >- e0. (2)

Since all tableau variables are integers, the slack variable s for (2) is integer, so we may append a new row

s = -eO + EjE>R ejXj

with a new integer variable s - 0. By repreating this process inductively, an all-integer problem remains all-integer. As an application of this observation, consider the MILP cut (Gomory [1963]) used on the ILP.

Subject classification: 628 solving ILP using MILP-cuts.

Operations Research Vol. 30, No. 1, January-February 1982

0030-364X/82/3001-0208 $01.25 ( 1982 Operations Research Society of America

208

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Holm 209

Here do = fo and

if fjfo (iE R-) J fio(i - MA - fo) if j > fo (i E R+)

where the fs are the fractional values of the source row. Salkin [1971] has shown that this cut is superior to Gomory's [1960]

fractional cut, but as Dantzig [1963] has pointed out its slack variable is in general not required to be integer, so that one has a MILP on later iterations. However, from the above observation, as the cut is in rational data there exists E such that all the coefficients and the slack variable in the transformed cut will be integers. In particular, let the absolute value of the determinant for the corresponding basis be I det B I = D, so f. = g./D, with g. and D integers. Let E = D(D - go) then eo = (D - go)go and

f (D-go)gj for jE R eJ (D-gj)go for j E R +

so the transformed MILP cut with integer data will be

(D - go) ZjeR- gjXj + go EjeR+ (D - gj)xj - s = (D - go)go.

By repeating this process of appending transformed MILP cuts, the all- integer problem remains all-integer and results in a more efficient algo- rithm.

ACKNOWLEDGMENT

The author wishes to thank the referee for helpful comments. This research has in part been funded by the Danish Social Sciences Research Council.

REFERENCES

DANTZIG, G. B. 1963. Linear Programming and Extensions. Princeton University Press, Princeton, N.J.

GOMORY, R. E. 1960. An Algorithm for the Mixed Integer Problem, P-1885, The Rand Corporation, Santa Monica, Calif. (February).

GOMORY, R. E. 1963. An Algorithm for Integer Solution to Linear Programs, in Recent Advances in Mathematical Programming, pp. 269-302, R. L. Graves and P. Wolfe (eds.). McGraw-Hill, New York.

SALKIN, H. M. 1971. A Note on Gomory Fractional Cuts. Opns. Res. 19, 1538- 1541.

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