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Gomory Notes from Taha 10-12

Gomory Notes from Taha 10-12 - ming{GIL 3:g 3-3 Methods of...

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Unformatted text preview: ming {GIL 3. :g 3-3] 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 so-called additive algorithm due to E. Balas. An extension of the zero-one additive algorithm to nonlinear binary problems will also be presented. astraint to 1t-hand side 8- 3-1 Cutting-Plane 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 16-3-1. 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 pro-i; 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 all-integer. 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 3-3] 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. 8'3-1-1 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 xo-r 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 right-hand 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 iii-vi 2 0. Consequently, fi — jifnwj 5ft This means f, — 2;; fgjwj < 1 because f; < I. Since the left-hand 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 so-called 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 4-5) 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 § 8-3] 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 primal-fez which are dual—feasible. Th putationally promising, hovs :) Example 8’3—1 Consider the problem wl Section 8-3-1. 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 § 8-3] 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 “all-integer . I . integer” algorithm. The algorithm starts with an initial all-integer 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 cutting-plane 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 primal-feasible as compared with Gomory’s algorithms, i which are dual-feasible. 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 8-3—1 Consider the problem which was solved graphically at the beginning of a Section 8-3-1. 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 avg-equation. This gives 'antages : automatic calculations may yield hough this can be overcome by Integer Programming § 8-3] 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 _ § 8-3] 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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