CHAPTER 4 IDENTIFICATION OF REDUNDANCIES IN LINEAR PROGRAMMING MODELS

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1 CHAPTER 4 IDENTIFICATION OF REDUNDANCIES IN LINEAR PROGRAMMING MODELS 4.1 INTRODUCTION While formulating a linear programming model, systems analysts and researchers often tend to include inadvertently all possible constraints although some of them may not be binding at the optimal solution. Since only a relatively small proportion of the constraints is binding at the optimal solution for most of the linear programming problems, systems analysts and researchers are interested in developing techniques for identifying redundant and nonbinding constraints. A constraint that is satisfied in exact measure at some optimum solution is a binding constraint. A redundant constraint enters into an optimum with positive slack and a variable enters with zero value. Redundancies, if any, in the model will waste computational effort. A technique that makes use of the intercept matrix to enable one to easily identify redundancies without investing computational effort has been developed and described in this chapter Redundant Constraints The Linear Programming Model can be written, in matrix form, as Maximize Z = CX Subject to AX < P0, X > 0

2 55 where C, X, A and P0 are matrices of order lxn, nxl, mxn and mxl respectively. Definition Let A X < bj be the ith constraint of the LPP and let A~ X < b~, X > 0 be the set of constraints of LPP excluding the ith constraint. The ith constraint is redundant if and only if there exists no vector X such that A~ X < b~ and Ai X > bt. Geometrically, the constraint Ai X < fy is redundant if and only if the convex set described by A X < b, X > 0 is identical with the set defined by A~ X < b~, X > 0. Illustration Consider the Polyhedral set (Bazarra et al 1990) defined by the following inequalities : Xj + 2x2 < 8 (1) 2x1 + x2 < 10 (2) 3xj^ + 4x2 < 20 (3) 9XJ + 8x2 < 56 (4) Xj^ > 0 (5) to IV o (6) The intersection of these six halfspaces gives the shaded set of Figure 4.1. Clearly the set is a convex set. If the third and fourth inequalities are disregarded, the polyhedral set is not affected. Such

3 Figure 4.1 Polyhedral set 56

4 57 inequalities are called (geometrically) redundant or irrelevant to the polyhedral set. Transform the inequations (1) through (4) into equations by adding slack variables s4, s2, s3 and s4. x1 + 2x2 + Sj = 8 (10 2xx + x2 + s2 = 10 (20 3x4 + 4x2 + s3 = 20 (30 9x4 + 8x2 + s4 = 56 (40 xl> x2> si> s2» s3> s4-0 (50 In Tables 4.1(a) through 4.1(e) all possible basic feasible solutions are generated. Table 4.1:Basic Feasible Solutions (a) Solution related to vertex O Basic variables XT x2 Solution si s S s (b) Solution related to vertex A Basic variables x2 s2 Solution si 1.5 xi s s o cn 3

5 58 (c) Solution related to vertex B Basic variables si s2 Solution x X sa s (d) Solution related to vertex B Basic variables s2 s3 Solution x X S s (e) Solution related to vertex C Basic variables X1 si Solution x s s s Prom the Table 4.1(a)-(e), it is obvious that for any feasible solution the value of the slack variable corresponding to inequality (4) is positive i.e., s4 > 0. Alternatively noxex exists such that in the corresponding solution of the adjoined equation system (10 through (50 the value of s4 becomes zero i.e., s4 * 0 for any x e X. The slack variable s3 has the same property except for the extreme point B. By now we can say that a constraint is called

6 59 redundant if, after deleting it, set X remains the same. If a boundary hyperplane that corresponds to a redundant constraint has at least one point in common with X, then this constraint is called relative redundant. Otherwise it is called absolute redundant (Gal 1979) Various Possibilities of Redundant Constraints Let us illustrate the various possibilities. Referring to Figures 4.2(a) - (c), in each of these figures constraint 1 is redundant. In all figures, the triangle OAB forms the feasible region defined by the constraints Aj X < (i = 2,3,...m) (Boot 1962). 4.2 EARLIER RESEARCH CONTRIBUTIONS A Chronological Survey Since only a relatively small proportion of constraints is binding at the optimal solution for most linear programming problems, there is considerable interest in developing methods for identifying redundant constraints. The earlier methods proposed in this direction are discussed below. Dantzig (1955) has suggested that some constraints can be anticipated to be non-binding and (equivalently) that certain activities (variables) are also anticipated to be in the optimum solution. The slacks of the nonbinding constraints and also the essential variables may be brought into the basis. The constraints in which the slack variables are basic, can together with the other variables, be dropped from the problem. When the optimum solution is obtained, these assumptions can be checked, and, if they are violated, the constraints may be reintroduced and made to take more number of iterations. If the number of errors in anticipating nonbinding constraints is relatively small, greater savings are achieved.

7 60 3 (c) Figure 4.2 Various Possibilities of Redundant Constraint

8 61 If the variables are known to be present in the optimum solution, then no additional iterations need be made. Boot (1962) has presented a method for identifying redundant constraints before beginning to solve a problem. In this method, only one constraint is checked for redundancy at a time. The method is as follows. First, establish that the convex set described by the problem is not empty. Then, to test whether a constraint is redundant or not, assume that the constraint, Aj X < bj, is violated, that is, ApC = bj + e, where e > 0 is a small but finite number. Use this relationship to eliminate one of the variables one of the variables from all of the other constraints, such that the eliminated variable is non-negative. If the resulting convex set is non-empty, then the constraint is not redundant. If the set is empty, the constraint is redundant. Each constraint may be so checked, the redundant ones are being discarded as they are found. However, the method can require considerable computation, since a feasible solution must be found to show that a convex set is non-empty. Thompson et al (1966) have proposed a method for identifying redundant constraints prior to the start of solving the problem. It is suggested that a constraint A X < fy, is E~ is redundant if AE_1 has nonnegative components where E is the set of inequalities whose slacks are not basic at a basic feasible solution X (i.e.) E : Ae X* = Be j ( E~ : Ae~ X < BE~, E~ is the complement of E. Ae is a square (nxn) nonsingular submatrix of A. Mattheiss (1973) has constructed a theorem for identifying the redundant constraints of a linear programming problem. This theorem states that a constraint is redundant for the feasible region of Linear Programming if and only if its associated slack variable is in the basis of every primary subsystem of the linear program.

9 62 Brearley et al (1975) have suggested a procedure to find out the redundant constraints and to fix variables at their bounds. The procedure for identifying redundant constraints is as follows: Step 1. Compute the upper and lower row bounds of the constraints. The ith constraint upper and lower bounds are Ui = I jepi ay ) + X ay lj and jenj Li = I ajj lj + X aij Uj respectively jenj jepj where Pi SZ (j : aij > 0} Ni = (j : aij < 0} and lj is the lower limit of Xj and Uj is upper limit of Xj, j = 1,2,...n tvi Step 2. i. The i upper bound constraint is redundant if Uj < fy ii. The i lower bound constraint is redundant if Lj > bj. Tomlin and Welch (1986) and Bixby and Wagner (1987) have presented an algorithm for identifying duplicate rows in an LP matrix, i.e. that is rows which are identical except for a scaler multiplier. Many researchers like Ye (1990), Mitchel (1986), Goffin et al (1990), Tone (1991), Den Hertog et al (1992) and Imbert et al (1996) have proposed a strategy for reducing the computational effort in this direction and solving Linear Programming Problems. In 1997, Gondzio discussed the presolve procedure of detection and removal of different linear dependencies of rows and columns in a constraint matrix.

10 63 This work addresses a different strategy for identifying redundancies in LPP using the intercept matrix. 4.3 DEDUCTING REDUNDANT CONSTRAINTS The methods so far proposed for the identification of redundant constraint could not be implemented as it required excessive computational efforts. The computational time is one of the critical factors for real time applications in the solving of large scale problems. Thus there is a need to develop an efficient algorithm to identify the redundancies prior to solving the problem. An attempt is made in this section to use the intercept matrix to identify redundant constraints without investing any computational effort Method for Identifying Redundant Constraints If a slack variable is in the optimal basis, the corresponding constraint will be redundant. The technique proposed here predicts a set of slack variables in the optimal basis prior to solving the problem. The procedure is the outcome of an indepth study of the theorem of Mattheiss (1973). Let us consider the linear programming problem which has m constraints and n variables. Maximize Z = CX subject to AX < P0, X > 0 Step 0 : Let I be the set of subscripts associated with the initial basic variables (slack variables). Initially let that set be I = {1,2,...m}. Let J be the set of subscripts associated with the initial decision variables. Initially let that set be J = {l,2,..,n}.

11 64 Step 1 : Construct an intercept matrix "0" using the following relationship 0ji =(P0)i/aij ; ay > 0 for je J, iel. Step 2 : Determine the promising variables making use of the following i. Calculate Zj-Cj = CgB'1 Pj-Cj for all nonbasic variables. ii. Let pj = min {0-}, for je J iel iii. Compute Zj'-Cj' = Pj(zj-Cj) for jej Step 3 i. Let zk' - ck = min {Zj'-Cj'} jej ii. iii. Take away the element k from the set J, i.e., J = J-{k} If zk'-ck' > 0, then the problem has no redundant constraint stop. Otherwise, iv. Let 0kj = min {0M} = Pk iel V. Take away the element from the set I, i.e, I = I-U} and vi. Find p such that min {0p$} = Pp for pej. If so, take away such p elements from the set J, i.e., J =J-{pl. Step 4 If J = {<)>}, then go to Step 5. Otherwise, go to Step 3. Step 5 If I = {())}, then the problem has no redundant constraint, stop. Otherwise, the constraints whose intercepts 8ji * max {Bj}, iel, are redundant. Stop j=l,2,...n

12 Illustration of the Method This section illustrates with examples the working of the method for identifying Redundant constraints in a step by step manner. Example 1 Maximize Z = 3XJ + 4x2 subject to Xj + 3x2 < 15 2xj + x2 < 10 2xj^ + 3x2 < 18 xl + x2 * 7 Solution Step 0 : 4xj + 5x2 < 40 x1} x2 > 0 I = {1,2,3,4,51; J = U,21 Steps 1 & 2 : The intercept matrix is Basic variables Decision variables S1 s2 s3 s4 s5 Zj-Cj p, f ^ f Zj-Cj xi X Step 3 Iteration No. k J l I J 1 2 {11 1 {2,3,4,51 { {<t»l 2 {3,4,5} {0}

13 66 Steps 4 & 5 : J = {<>} The constraints 3,4 and 5 are redundant. The optimal solution is Z = 25; Xj = 3, x2 = 4, s3 = 0, s4 = 0, s5 = 8 Example 2 Maximise Z = 2x1 + x2 subject to the constraints xl + x2 < 1 Xj + 2x2 < 4 x1? x2 > 0 Solution I = {1,2} ; J = {1,2} The intercept matrix is Decision variables Basic variables Si s2 Zj-Cj zrci' xi x The 2nd constraint is redundant. The optimal solution is Z = 2; Xj_ = 1, s2 = 3

14 67 Example 3 Maximize Z = 61xx + 209x x3 + 33x x x x xg+ 12x x10 subject to the constraints 16X-L + 25x2 + 22x3 + 4x4 + 9x5 + 8x6 + llx7 + 29x8 + 20x9 + 22x10 <11 5xj + 22x2 + 15x3 + 30x4 + 24x5 + 15x6 + 14x7 + 28xg + 31x9 + 25x10 < 53 22x1 + 17x2 + 9x3 + 32x4 + 26x5 + 20x6 +16x7 + 16xg + 26x9 + 24x10 < 50 14xj + 9x2 + 32x3 + 22x4 + 30x5 + 18x6 + 18x7 + 32xg + 15x9 + x10 < 40 32xj + 30x2 + 10x3 + 30x4 + 7xg + 29xg + 15x7 + xg + 19x9 + 26x10 < 4 12xj + 4x2 + 30x3 + llx4 + 23x5 + 29xg + 8x7 + 2xg + 0x9 + 23x10 <31 22xj + 23x2 + 26x3 + 13x4 + 6x5 + 13xg + 32x7 + llxg + 8x9 + 5xlg < 39 Xj > 0, j = 1,2,3, Solution I = {1,2,3,4,5,6,71 J = {1,2,3,4,5,6,7,8,9,10} The intercept matrix is given by Decision variables Basic varia jles S1 s2 s3 s4 sf? sfi s7 zrei Pi zrci' x, x X, X Xr XR x XR x x

15 68 The constraints 2,3,4,6 and 7 are redundant. The optimal solution is Z = ; x5 = 0.541, x8 = 0.211, s2 = 34.09, s3 = 32.55, s4 = 17, s6 = 18.13, s7 = Computational Experience This section discusses the efficiency of the method and concludes with the presentation of the observations made. The efficiency of the algorithm is tested by solving LPP before and after the model reduction. Table 4.2 provides a comparison of the computational results. These results show that the proposed algorithm is useful to identify redundant constraints in a given LP problem and it also reduces the computational effort and memory requirements. For this purpose, the author used the problems of types employed by Kuhn and Quandt (1962). These problems have the canonical form with Cj = 1 for j = 1... n, fy = 500 for i = l,2...m and a^ is generated uniformly within the interval (0,100) for i = 1,2... m, j = 1,2... n. Table 4.2: Comparison of Computational Efforts Required with and without Redundant Constraints SI. No. Size of the Problem No.of Constraints No.of Variables No.of redundant constraints No.of multiplications/divisions to solve LPP With redundant Without redundant O o

16 IDENTIFYING REDUNDANT VARIABLES A variable which has zero value in every optimal solution is redundant. Thompson et al (1966) have proposed a theorem for declaring redundant variables in linear programming problems, which states that "if at any iteration of the simplex method solution, a variable is not profitable with respect to a profit function (i.e., Zj-Cj > 0), and has only positive coefficients in the constraint equations, then there exists an optimum solution that does not contain that variable". For example, consider the following linear programming problem Maximize Z = 4xx + 3x2 + 5x3 + 2x4 + 5x5 subject to the constraints 3X]^ + 2x2 - x3-2x4 + 4x5 < 1 2xj + x2 + 3x3 + x4 + 2x5 < 1 Xi, x2, x3, x4, x5 > 0 The simplex tables of the linear programming problem are Basic X1 x2 x3 x4 x5 x6 x7 Solution x X? Z x x z

17 70 Basic xl x2 x3 x4 x5 x6 x7 Solution xf> x Z xf> x Z x x Z When the theorem of Gerald L. Thompson et al (1966) is applied the variables Xj_, x3 become redundant. The present work has suggested a new approach for identifying the redundant variables in linear programming problems without solving the problem Algorithm for Identifying Redundant Variables in the Primal Model Consider the linear programming model which has m resource constraints and n decision variables. Maximize Z = CX subject to the constraints AX < P0, X > 0

18 71 Step 0 : Let I be the set of subscripts associated with the initial basic variables. Let that initial set be I = {1, 2,... m}. Let J be the set of subscripts associated with the decision variables. Let that initial set be J = {1, 2,... n}. Step 1 Construct the intercept matrix 0 using the following relationship 0ji = (PoV^ji aij > 0 for Je ie 1 Step 2 Scan the 0 matrix row-wise and identify the minimum intercept in each row. m Let pj = min {0^}, je J i=l Step 3 Check whether more than one j s (je J) have the minimum intercept value in the same column of the 0 matrix. If yes, then go to Step 4. Otherwise, the model has no redundant variables. Stop. Step 4 Let the number of minimum values of the intercepts in the 0 JLl. matrix lie in the i column. i. Identify the maximum of the intercept in the ith column. ii. iii. iv. The decision variable(s) not corresponding to this maximum intercept bring(s) out redundant variable(s). Let the set of subscripts corresponding to the basic and redundant variables be Jj Set J = J - Jl Step 5 Go to Step 3

19 Illustration The following examples illustrate the steps of the algorithm. Example 1 : Consider the example given in Section 4.4 Step 0:1 = {1,2}; J = {1,2,3,4,5} Steps 1 & 2 : The intercept matrix 0 is Decision variables Basic variables sl xi X X x4-1 1 X s2 ft Steps 3 & 4 : Xj_, x3, x5 are redundant variables. The optimal solution is Z = 2.75; x2 = 0.75, x4 = 0.25 Example 2 : Maximize Z = 6x4 + 10x2 + 13x3 Subject to 0.5XJ + 2x2 + x3 < 24 x4 + 2x2 + 4x3 < 60 xi> x2> x3-0

20 73 Step 0:1 = {1,2} J = (1,2,3} Steps 1 & 2 : The intercept matrix 0 is Basic variables Decision variables sl s2 Pj xi x x Steps 3 & 4 : x2 is a redundant variable. The optimal solution is Z = 294; Xj = 36; x3 = Computational Results Thomson (1966) has attempted to identify the number of redundant variables in the primal model after generating the Simplex table. His algorithm could identify only a few redundant variables after investing some computational effort. But the algorithm developed in this thesis could identify more redundant variables than Thomson s without wasting computing time. The algorithm developed in this thesis reduces the size of the LP models leading to the reduction of computational effort. The presence of redundant rows and columns permits more variables to pop in and out of the basis. The elimination of redundant constraints and variables abinitio curtails this tendency of popping variables to the least minimum possible. The developed algorithm not only saves memory and computing time due to the elimination of rows and columns but also minimizes the popping variables.

21 74 The Table 4.3 shows the computational effort required for solving problems with and without redundancies. Table 4.3 : Comparison of Computational Efforts Required with and without Redundancies Size of the problem No.of redundancies No.of iterations required m n Constraints Variables With redundancies Without redundancies REDUNDANCY RELATIONSHIP BETWEEN PRIMAL AND DUAL MODELS Redundant resource constraints and redundant variables in a LP model not only occupy more storage in a computer but also consume more computational time. The algorithm presented in this chapter avoids the wastage of storage and improves the computing time by removing redundancies. The intercept matrix is used for identifying both the redundancies in one stroke. The intercept matrix is used for identifying both the redundancies for a given model. The twin properties between the two models should hold good i.e. for every primal redundant constraint there exists a dual redundant variable and vice versa. Any redundant primal constraint should have a non-negative value for the corresponding surplus variable in the optimal solution. According to the complementary slackness theorem the corresponding dual variable should be zero. Conversely any redundant dual constraint should have a corresponding redundant primal variable. There is row(s) reduction and column(s) reduction resulting in an overall model reduction.

22 Lemma on Redundant Inequalities (Chames et al 1962) Consider the pair of linear programming problems written in the form Maximize X + C'2 Y Subject to A X + BY < bj D X + EY < b2 X, Y>0 and Minimize bjw' + b2u' Subject to AW' + DU' > BW' + EU' > C'2 W', U' > 0 where A,B,D,E represent a four-part partition of a standard linear programming constraint matrix, bj and b2 form a corresponding partition of the usual b or stipulations vector, and C'i, C'2 form a partitioning of the coefficients in the functional vector C. Lemma Part (a) If the constraints AX + BY < bj are redundant in an optimum solution to the maximizing problem then there will exist solutions W', U' to the $ a e * minimization problem with bjw ' = 0, AW ' = BW ' = 0.

23 76 Part (b) If the constraints W'A + U'D > C\ are redundant in an optimum solution to the minimizing problem then there will be solutions X, Y to the maximising problem with C\ X = 0, AX = 0, DX = Illustrative Examples The redundancy relationship between the primal and the dual model is explained using the following examples. Example 1 Consider the following primal and dual problems. Primal: Maximize 4x4 + 3x2 subject to the constraints x4 + 2x2 < 2 Xj - 2x2 < 3 2xj + 3x2 < 5 x4 + x2 < 2 3x^ "f* x2 3 Xi, x2 > 0 Dual : Minimize 2yx + 3y2 + 5y3 + 2y4 + 3y5 subject to the constraints y4 + y2 + 2y3 + y4 + 3y5 > 4 2yx - 2y2 + 3y3 + y4 + y5 > 3 yi» y2> ys» ys - -

24 77 The primal constraints 2,3 and 4 are identified as redundants by the proposed algorithm. The optimal solution to the primal is = 0.8, x2 = 0.6, s2 = 3.4, s3 = 1.6, s4 = 0.6 with objective value 5. Utilizing the theorem of complementary slackness, we can conclude that the dual variables y2, y3 and y4 corresponding to primal constraints 2,3 and 4 are redundant in the dual model. Example 2 Consider the example given in Section 4.4. The primal variables x4, x3 and x5 are identified as redundants by the proposed algorithm. Utilizing the theorem of complementary slackness, we conclude that the dual constraints 1, 3 and 5 corresponding to the primal variables x4, x3 and x5 are redundant in the dual model. 4.6 CONCLUSION Simple heuristic algorithms have been presented for identifying the redundant constraints and variables, if any, in the linear programming models apriori to the start of the solution process with the help of the intercept matrix. The redundant constraints and the variables in the model are then eliminated, and the resulting model is solved by using the Multiplex Algorithm in order to establish the validity of the above algorithms. A significant reduction in the computational effort of the order 5% to 40% is thereby achieved.