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Unformatted text preview: ming {GIL 3. :g 33] Methods of Integer Programming .. 245 :putational advantage and his version will be presented here. The third
algorithm applies to the pure zero—one problem. This is the socalled additive
algorithm due to E. Balas. An extension of the zeroone additive algorithm
to nonlinear binary problems will also be presented. astraint to 1thand side 8 31 CuttingPlane Algorithms The concept of the cutting plane will ﬁrst be illustrated by an example. rig for the approximation of a
fConsider the integer linear programming problem Section 1631.
maximize x0 = 7x1 + 9x2 —x1 + 3x2 5 6
7x1 + 352 S 35 x1, x2 nonnegative integers rized as he optimal continuous solution (ignoring the integrality condition) is
shown graphically in Figure 84. This is given by x0 = 63, x1 = 9/2, and rily for integer linear problems,5 {52 x 7/2: WhiCh is ﬂoniﬂtﬁgel‘. systematically adding special
represent necessary conditions:
. is gradually modiﬁed until its"E
1e integer conditions. The name
t the added “secondary” co _'
parts of the solution space that: (9/2, 7/2); x9 2 63
(4, 3);):0 E 55 Secondary constraintsm
l1\ \ itforward idea of enumerating
to develop “clever” tests that
ble integer explicitly but auto_
mplicitly. The most prominent.
:chnique. It also starts with th ..
oartitions” the solution space
sin no feasible integer points. _
:s when all the integer variables;
Lblcs simpliﬁes the search proi; Figure 8—1 The idea of the cutting plane algorithm is to change the convex set of the
solution space so that the appropriate extreme point becomes allinteger.
Such changes in the boundaries of the solution space should result still in a
convex set. Also this change should be made without “slicing off” any of
the feasible integer solutions of the original problem. Figure 8—1 shows how
two (arbitrarily selected) secondary constraints are added to the problem
 with the new extreme point (4, 3) giving the integer optimal solution. Notice
that the area sliced loll the original solution space (shaded area) does not
_' include any integer values. ' apply primarily to the linear,
)ped by R. E. Gomory, include
the pure integer problem, and,
or the mixed integer problem.
ally developed by A. H. Land.
odiﬁcation offers greater corn 'r 33] 7 Met. 246 Integer Programming [Ch. 8 The variables )6i (i = l,
_ablesw,(j:1,2,...,n
arranged as such for cc Consider the ith equa
value. The following analysis shows how the secondary constraints are developed
systematically for the pure and mixed integer problems. 8'311 The Fractional (Pure Integer) Algorithm
' xi = is: —
A basic requirement for the application of this algorithm is that all the z
coefﬁcients and the right—hand side constant of each constraint must be integer. For example, the constraint Any such equation will
coeﬂicients of the objec
also integer and the xor
convergence proof of ti
I Let xl+%XQ$'];2§ must be transformed to 6371 + 2x2 5 39 iwhereN = {a} is the larg
land 0 sfij < 1; that i
negative fraction. For e where no fractions are present. The latter is achieved by multiplying both
sides of the original constraint by the least commonmultiple of the denom
inators. The above requirement is imposed since, as will be shown later, the pure
integer algorithm does not differentiate between the regular and slack vari
ables of the problem in the sense that all variables must be integers. The
presence of fractional coelﬁcients in the constraints thus may not allow the
slack variables to assume integer values. In this case, the fractional algorithm
may indicate that no feasible solution exists, even though the problem may
have a feasible integer solution in terms of nonslack variables. (See Problem
8—7 for an illustration of this case.) The details of the algorithm will be discussed now. First, the problem is
solved as a regular linear programming problem; that is, disregarding the
integrality condition. If the optimal solution happens to be integer, there is
nothing more to be done. Otherwise, the secondary constraints which will
force the solution toward the integer solution are developed as follows. Let
the ﬁnal optimal tableau for the linear program be given by The source row thus yie 1‘;— In order for all the v
9f. the above equation 1
illust also be integer. Gi
2&1]?jo 2: 0. Consequ Solution This means f, w 2L1];
be integer, a necessary 0 ogramming [Ch. 1e secondary constraints are developed
integer problems. gorithm tion of this algorithm is that all the
constant of each constraint must be <2
x2_2 1x2 5 39 latter is achieved by multiplying both
least common multiple of the denom since, as will be shown later, the pure
ate between the regular and slack vari
._at all variables must be integers. The
:he constraints thus may not allow the
as. In this case, the fractional algorithm
1 exists, even though the problem may
ms of nonslack variables. (See Problem te discussed now. First, the problem is
ling problem; that is, disregarding the
solution happens to be integer, there is
:, the secondary constraints which will
solution are developed as follows. Let [I program be given by Wi w, wﬂ Solutlon :.§ 813] Methods of Integer Programming 247 ' The variables x, (i = l, 2, . . ., m) represent the basic variables while the vari—
f ables w, ( j m l, 2, . . ., n) are the nonbasic variables. These variables have been
. arranged as such for convenience. Consider the ith equation where the basic variable x, assumes a noninteger
 value. Ila
x, = 18, — Z 'aifwj, :8: noninteger (Source row)
i=1 I Any such equation will be referred to as a source row. Since in general the
coeﬂ‘icients of the objective function can be made integer, the variable x0 is
also integer and the xn—equation may be selected as a source row. Indeed, the convergence proof of the algorithm requires that x0 be integer.
Let :8: m [55} +ﬁ
(xi = [mil +ij whereN = [a] is the largest integersuch thatN 5 a. It follows thatO <1", < l
and 0 s f”. < 1; that is, f, is a strictly positive fraction and f” is a non
negative fraction. For example, ' a la] f=a [ad 1;,— 1 1/2
— 2t —3 2/3
W 1 —l 0 —2/5 21 3/5 The source row thus yields ﬂ _ :1":jo = xi — {131} + Z: fail“); In order for all the variables x, and w, to be integer, the righthand side
of the above equation must be integer. This implies that the left—hand side
must‘also be integer. Given ﬂ} 2 O and w, 2 0 for all i and j, it follows that L1 iiivi 2 0. Consequently, fi — jifnwj 5ft This means f, — 2;; fgjwj < 1 because f; < I. Since the lefthand side must  be integer, a necessary condition for satisfying integrality becomes f, — quwj .<_ 0
i=1 248 Integer Programming [Ch]. 8 The last constraint can be put in the form ﬂ Si = Z ﬂjwj e f, (Fractional cat) i=1
where S, is a nonnegative slack variable which by deﬁnition must be an
integer. This constraint equation deﬁnes the socalled fractional cut. From
the last tableau, w, 2 O and thus S, = mﬁ, which is infeasible. This means
that the new constraint is not satisﬁed by the given solution. The dual simplex
method (Section 45) can then be used to clear this infeasibility, which is
equivalent to cutting off the solution space toward the optimal integer solution. .
The new tableau after adding the fractional cut will thus become Basic x1 xi xm wl w, wniS, Solution If the new solution (after applying dual simplex method) is integer, the process
ends. Otherwise, a new fractional cut is constructed from the resulting tableau
and the dual simplex method is used again to clear the infeasibility. This
procedure is repeated until an integer solution is achieved. However, if at any
iteration the dual simplex algorithm indicates that no feasible solution exists,
the problem has no feasible integer solution. The algorithm is referred to as the “fractional” method because all the
nonzero coefﬁcients of the generated out are less than one. The above algorithm may indicate, at ﬁrst thought, that the 'size of the
simplex tableau can become very large as new cuts are augmented to
the problem. This is not true. In fact, the total number of constraints in the
augmented problem cannot exceed the number of variables in the original
problem; that is, (m + n). This follows since if the augmented problem
includes more than (in + n) constraints, one or more of the slack variables
S, associated with the fractional cuts must become basic. In this case the
associated equations become redundant and may be dropped from the tableau completely.
The fractional algorithm has two disadvantages: 1. The round—off errors that evolve in automatic calculations may yield
the wrong optimal integer solution. Although this can be overcome by § 83] Method. scparately storing the hunt
fractions (hence avoiding d
numbers may exceed the av 2. The solution of the p
integer solution can be obta
This means that there will
calculations are stopped pr:
(integer) solution. The ﬁrst difﬁculty was 0
integer” algorithm. The ah
(that is, all the coefﬁcients
dual simplex algorithm. A :
such that its addition to th
coeﬁicients. However, the f:
integer optimal solution is r The second difﬁculty was c
that start integer and feasib.
feasibie and integral until tl
this algorithm is primalfez
which are dual—feasible. Th
putationally promising, hovs :) Example 8’3—1
Consider the problem wl
Section 831. The optimal c Since this solution is nor
tableau. Generally, any of
honinteger solution can be 5
0f thumb, one usually cho
Smce both equations in th;
,1 =ﬁ2 : 1/2, either one the a imming [051. 249 § 83] Methods of Integer Programming
'm separately storing the numerators and the denominators of the different
fractions (hence avoiding decimal calculations), the sizes of the associated
numbers may exceed the available capacity of the computer. 2. The solution of the problem remains infeasible in the sense that no
integer solution can be obtained until the optimal integer solution is reached.
This means that there will be no “good” integer solution in store if the calculations are stopped prematurely prior to the attainment of the optimal . :
(integer) solution. , Fractional cut) which by deﬁnition must be a
the so—called ﬁactfonal cut. From..
ﬁ, which is infeasible. This mean
to given solution. The dual simplex }
3 clear this infeasibility, which i d {11 o timal inte e . .
pace towar e p g The ﬁrst difﬁculty was overcome by the development of an “allinteger . I . integer” algorithm. The algorithm starts with an initial allinteger tableau
 (that is, all the coefficients are integers) suitable for the application of the .
dual simplex algorithm. A special additional constraint is then constructed !
such that its addition to the tableau will preserve the integrality of all the ‘
coeﬁicients. However, the fact that the solution remains infeasible until the g l
_ integer optimal solution is reached still presents a disadvantage. '
The second diﬁiculty was considered by developing cuttingplane algorithms
that start integer and feasible but nonoptimal. The iterations continue to be
feasible and integral until the optimum solution is reached. In this respect, ;.
this algorithm is primalfeasible as compared with Gomory’s algorithms, i
which are dualfeasible. The primal algorithms do not appear to be com .1]
putationally promising, however. inal cut will thus become plex method) is integer, the process___
istructed from the resulting tableau
ain to clear the infeasibility. Thi
ion is achieved. However, if at any
tes that no feasible solution exists
in. actional” method because all the I» Example 83—1
Consider the problem which was solved graphically at the beginning of a
Section 831. The optimal continuous solution is given by ‘ 5 Solution 1. : re less than one. 0 0 28/11 ' 15/11 63 i ' !
ﬁrst thought, that the 'size of the 7/22 1/22 7/2 ' i
as new cuts are augmented to 0 _1/22 3/22 9/2 total number of constraints in the
.mber of variables in the original
Since if the augmented problem
me or more of the slack variables
st become basic. In this case the _
and may be dropped from the; Since this solution is noninteger, a fractional cut must be added to the
tableau. Generally, any of the constraint equations corresponding to a
noninteger solution can be selected to generate the cut. However, as a rule
of thumb, one usually chooses the equation corresponding to max, {)2}.
'Since both equations in this problem have the same value of j}, that is,
f, = f2 = 1/2, either one may be used. Consider the avgequation. This gives 'antages : automatic calculations may yield
hough this can be overcome by Integer Programming § 83] Methods of In 1 1 Hence, the corresponding fractional cut is given by
7 I l ‘nh—nht 2 or The dual simplex‘method now yiel 5'1 This gives the new tableau ' which gives the optimal integer sol:
_ The reader can verify graphical
“cuts” the solution space as desire The dual simplex method yields . '
' :_can be expressed In terms of x1 an
: stitution as follows. Basic 1/7
1/7 — 1/7
—22/7 Since the solution is still noninteger, a new cut is constructed. The x1
equation is written as 6 4
x1+ (0+%)x4+(—1+7)S1:(4+7) which gives the cut 1 6 4
Sg~7x4e~7S1=“7 Adding this constraint to the last tableau, one gets Figure Sal shows that the addition 0
_. W(optima1) extreme point (4, 3). .. Strength of the Pl‘actfonal Cut '__The above development indicates
Cut depends directly on the “source:
.1iTerent inequality cuts may be gen
The question naturally arises: Which
bf: measured in terms of how deep the rig [Ch_ 3 1 _ § 83] Methods of Integer Programming .. 251 '. The dual simpiex method now yields I which gives the optimal integer solution x0 : 55, x; : 4, x2 = 3.
I The reader can verify graphicaily that the addition of the above cuts
“cuts” the solution space as desired (see Figure 84). The ﬁrst cut 7 1 ]
81—2—2x3_ﬁx4=—§  can be expressed in terms of x1 and x2 only by using the appropriate sub
stitution as follows.  s —l(6+x m3x)—i(35—7x —x)=—1 1 22 1 2 22 1 2 2 or S1+X2=3 which is equivalent to
cut is constructed. The x; 2w) Similarly, for the second cut 1 6 4
S2—?X4¥§SI:'_77" the equivalent constraint in terms of x1 and x2 is
X; + x2 5 7 Figure 8—1 shows that the addition of these two constraints will result in the
new (optimai) extreme point (4, 3). « Strength of the Fractional Cut The above development indicates that the speciﬁc inequaiity deﬁning a cut depends directly on the “source row” from which it is generated. Thus, ' different inequality cuts may be generated from the same simplex tableau.
The question naturaiiy arises: Which out is the “strongest” ?Strength could
be measured in terms of how deep the inequality cuts into the sohition space. ...
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