Linear Programming in Industry pp 65-89 | Cite as

# Computational Procedures for Solving Linear Programming Problems

## Abstract

As we have seen in Ch. II^{2}, the simplex procedure can be described as a systematic method of examining the set of basic feasible solutions, starting in an arbitrary initial basis of *m* variables (activities) where m is the number of linear restrictions. If the initial basic solution does not satisfy the simplex criterion, we move to a neighboring basis by replacing one of the basic variables, and so forth, until a basic feasible solution is attained in which all of the simplex coefficients are non-positive (in a minimization problem, non-negative). By the Fundamental Theorem, such a solution will be an optimal solution.

## Keywords

Basic Solution Linear Programming Problem Basic Variable Transportation Problem Slack Variable## Preview

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## References

- 1.Numerical Exercises for this chapter are given at the end of the book.Google Scholar
- 2.See also the Appendix, B.Google Scholar
- 1.Exercise: Assuming that (1) is a problem of optimal utilization of two machine capacities, what is the economic interpretation of this initial basic solution?Google Scholar
- 2.Exercise: Explain why a basis which gives a non-feasible solution cannot be used as a starting basis.Google Scholar
- 3.This is a purely empirical rule. Experience has shown that, if it is followed, the optimal solution will in most cases be reached in fewer steps than otherwise.Google Scholar
- 1.Exercise: Draw the geometric picture of the problem (1) and indicate the three extreme points (corners of the area of feasible solutions) which correspond to the three computational steps (3), (4), and (5).Google Scholar
- 1.Exercise: Why cannot the simplex calculations be confined to the Po column (i.e., the basic feasible solutions) and the zi — c5 row (the simplex coefficients), which contain all the information we are really interested in?Google Scholar
- 2.See Appendix, C, pp. 107 ff.Google Scholar
- 3.Exercise: Explain why (i)—(iii) must always hold.Google Scholar
- 1.Exercise: Carry out these calculations.Google Scholar
- 2.The validity of this result can easily be checked using the formulas of the simplex algorithm, cf. the Appendix, pp. 107 ff.Google Scholar
- 3.Exercise: Assuming that (1) is a problem of optimal utilization of machine capacities, indicate the economic interpretation of the simplex coefficient as computed by the formula (6).Google Scholar
- 4.Exercise: Can the slack variables be used as an initial basis if the right-hand sides are all negative?Google Scholar
- 1.It is not necessary to specify the numerical value of M in the computations, although in some cases it is convenient to set it equal to some very large number when the problem is solved electronically.Google Scholar
- 1.In minimization problems, non-positive.Google Scholar
- 2.Alternate optima may also occur without zero coefficients appearing in the zg — el row, namely, when the optimal basic solution is degenerate so that one or more basic variables are zero. These variables can be replaced in the basis by others without affecting the value of the preference function. However, such “alternate” optima are in effect identical. Cf. Ch. VI, E, below.Google Scholar
- 1.Exercise: Explain why this must be so, and give a geometrical illustration of a case involving only two structural variables.Google Scholar
- 2.Of course the second-best solution here means the second-best basic solution. Obviously the best and the second-best solution can be combined to give intermediate solutions in more than m variables.Google Scholar
- 3.Cf. pp. 47 ff.Google Scholar
- 1.Exercise: Fill in the blank spaces in the three tableaux.Google Scholar
- 2.Exercise: Does it follow from the reasoning above that Tableaux IIb and IIc represent the third-best and the fourth-best solutions?Google Scholar
- 1.Cf. the Numerical Exercises at the end of the book.Google Scholar
- 2.Including the initial tableau.Google Scholar
- 3.This happens twice during the solution of the ice-cream problem of Ch. III. All of the four artificial variables that constitute the initial basis are replaced, which requires 5 tableaux (i.e., 4 simplex tableaux in addition to the initial one), but the 2 detours require 2 additional iterations so that the problem is solved in 7 steps.Google Scholar
- 1.The ice-cream problem of Ch. III is a case in point. By inspection of the coefficients, the region within which the optimal solution was to be found could be narrowed down considerably, and the particular basic solution which was actually used by the company was known (and was within the region). The problem was easily reduced such as to involve only four equations, and the majority of the coefficients were zero. The first basic solution turned out to be optimal, whereas it would have taken seven iterations (simplex tableaux) to solve the problem starting from an initial basis of artifical variables.Google Scholar
- 1.This procedure is somewhat reminiscent of the method we used in the ice-cream problem of Ch. III to narrow down the region within which the optimal solution was to be sought. The ingredients were arranged in groups, but a complete partitioning was not possible except for ingredients 12 and 13, whose quantities could then be determined independently of the first 11 ingredients.Google Scholar
- 2.Reproduced by permission from K. EIBEMANN (1954), published by International Business Machines Corporation, U.S.A.Google Scholar
- 1.This new variable should not be confused with the slack variable already present in the fifth equation.Google Scholar
- 2.Exercise: Solve the problem as indicated.Google Scholar
- 3.A degenerate solution in k positive variables implies that a system of m linear equations in k variables (m> k) has a solution; clearly this is an exceptional case. The attention given to the problem of degeneracy in the literature is quite out of proportion to the actual frequency of degenerate solutions in practical problems, as well as to the relatively insignificant computational difficulties caused by degeneracy. Except for the rather special transportation problem, degenerate solutions do not occur very frequently in practice. (However, two of the concrete problems in Ch. V had degenerate basic solutions.)Google Scholar
- 4.Exercise: Construct a geometric example (in two structural variables) in which one of the basic feasible solutions is degenerate.Google Scholar
- 1.The existence of a degenerate solution was obvious from the outset; since Po is proportional to P2 in the first tableau, the restrictions will be satisfied by a positive solution in x2 only.Google Scholar
- 2.Cf. Ch. II, p. 14, and the Appendix, B. The point is that, even though the sign of the simplex coefficient zl — ci in Tableau III indicates that the solution could be improved if xi could be made positive, this is not feasible because x4, which is zero in the basic solution, would then become negative.Google Scholar
- 3.Exercise: Why can cycling occur only in the case of degeneracy?Google Scholar
- 4.One of the very few examples that exist is constructed by BEALE, cf. E. M. L. BEALE (1955) or E. O. HEADY and W. CANDLER (1958), pp. 146 ff. Another example is found in N. NIELSEN (1956).Google Scholar
- 1.By “smallest” is meant algebraically smallest; thus, for example, —2 is smaller than 1, and —3 is smaller than —2.Google Scholar
- 2.Exercise: Explain why.Google Scholar
- 3.The rule, which is due to A CHARNES, is derived from a modified simplex procedure which makes it possible to steer clear of degenerate solutions by slightly distorting the problem. In the initial tableau, each element xt of the Po column is replaced by the polynomialGoogle Scholar
- 1.Instead of writing x13 = 0, we might have gone round by x22; similarly, x25 might have been chosen as a basic variable instead of x34. Indeed, any two blank entries in the table (i.e., zero variables) may formally be considered as representing basic variables together with the five variables that are positive in the basic solution. Cf. Ch. VI, E, above.Google Scholar
- 2.See A. CHARNES and W.W. COOPER (1954).Google Scholar
- 1.Exercise: Why could we not have adjusted x22 or x24 instead of X23?Google Scholar
- 2.We cannot choose to adjust x33 instead of x13; we are examining the effects on basic variables only.Google Scholar
- 1.If some of the simplex coefficients in the tableau had been zero, the solution would still have been optimal, but not unique.Google Scholar
- 2.Exercise: Solve the problem in five steps by the general simplex method, omitting the equation referring to the third origin, and using the same initial basis as above. Show that the calculations correspond to those above, step for step and coefficient for coefficient.Google Scholar
- 3.Following the shortest possible path does not imply that the direction is changed as soon as a circle is reached. For example, in calculating z2i — cal in the first tableau above, the circled value of xis had to be skipped over. (Exercise: Explain why.)Google Scholar
- 4.These circles may be thought of as stepping stones used in exploring the vicinity of xii, hence the method is known as the “stepping stone method”.Google Scholar
- 5.Exercise: Explain why.Google Scholar
- 1.Cf. S. VAJDA (1958), p. 4. For further computational short cuts see op. cit., Chs. II and X.Google Scholar
- 2.Exercise: Solve the transportation problem of Ch. V, D, by the stepping stone method, using this device to find an initial basis.Google Scholar
- 3.Cf. A. CHARNES and W. W. COOPER, op. cit., pp. 60 ff.Google Scholar
- 4.The formal characteristics which such problems must have, and some of the devices by which they are converted into transportation problems, are described in A. HENDERSON and R. SCHLAIFER, op. cit., pp. 98 f.Google Scholar