A Concise Pivoting-Based Algorithm for Linear Programming

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International Journal of Advances in Management Science (IJ-AMS) Volume 3 Issue 3, August 2014 DOI: 10.14355/ijams.2014.0303.02

A Concise Pivoting-Based Algorithm for Linear Programming Zhongzhen Zhang*1, Huayu Zhang2 School of Management, Wuhan University of Technology, Wuhan, China Institute of Applied Informatics and Formal Description Methods, Karlsruhe Institute of Technology, Germany

1 2

zhangzz321@126.com; 2huayu.zhang@kit.edu

*1

Abstract This paper presents a pivoting-based method named pivoting algorithm for linear programming. In contrast to the simplex method this one never requires additional variables added into the problem. For a problem of m general constraints in n variables the size of the table for computation is (m + 1) × (n + 1). Generally speaking it is much smaller than the table used by simplex method even though these two methods are equivalent to each other.

Keywords Linear Programming; Pivoting Operation; Positive Basic Cone

Introduction Linear programming (LP) is the fundamental branch of optimization and is widely used in economics, management and engineering. The most well-known method for solving LP is the simplex method.

However this method is not good enough. We see that when a LP problem is solved by simplex method it must be transformed into a standard form where every variables are nonnegative and other constraints are equalities. According to simplex method, if a problem has a general inequality constraint (containing at least two variables), it must be changed into equality by introducing a slack or surplus variable; if the lower bound of a variable is not zero, it has to be changed into zero; and if a variable is free, it is replaced with two nonnegative variables. Moreover artificial variables are often required so that the formal computing can be conducted. These preliminary procedures make the original problem to be a big one and increase the computational burden. The simplex method is a column-processing method where the entering and leaving vectors are columns of the coefficient matrix of equality constraint. It binds up all the general constraints and is difficult to find redundant constraints in the iteration process.

This paper presents another kind of pivoting-based method for solving LP. This method never requires additional variables. For a problem of m general

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constraints in n variables the size of the table for computation is just (m + 1) × (n + 1). It is a rowprocessing method where the entering and leaving vectors are coefficient vectors of constraints. When a problem of k independent equality constraints in n variables is solved by this method, all the k equalities are first deleted by pivoting operations to result in a problem which has only inequality constraints in n − k variables. And some redundant constraints may be easily detected and deleted in the iteration process.

This paper is organized as follows. In Section 1 we introduce a kind of pivoting operation. In Section 2 we present the pivoting algorithm for LP. In Section 3 we discuss the relationship between the pivoting algorithm and the simplex algorithm. In Section 4 we show how to obtain the optimal solution of a linear program by solving the dual of this problem. In Section 6 we give some concluding remarks.

Pivoting Operation for Linear Programming We formulate linear programming in this form min z = c x s.t. ai x = bi, i = 1,…,l, ai x ≥ bi, i = l+1,…,m.

(1)

Where c is an n-dimensional row vector, x is an unknown n-dimensional column vector, ai is an ndimensional row vector, and bi is a scalar, i = 1,…,m.

In our method the following concepts are employed.

A set of maximum linearly independent vectors of a1,…,am is called a basis of (1). Vectors in the basis are called basic otherwise called nonbasic. The equalities and inequalities associated with basic vectors are called basic, otherwise called nonbasic. The system formed by basic equalities and inequalities is called a basic system and whose solution set is called a basic cone. The solution to the system of equations that are corresponding to basic (in)equalities, i.e., the vertex of the basic cone, is called basic solution.

International Journal of Advances in Management Science (IJ-AMS) Volume 3 Issue 3, August 2014

For a basic cone of (1), let I0 and I1 be the index sets for basic equalities and basic inequalities respectively.

Suppose that c is a linear combination of basic vectors as following:

c=

∑ w0 j a j

j∈I 0 ∪ I1

If w0j ≥ 0 for every j∈I1, the basic cone is called positive basic cone.

=( wis / wrs )br + = ( wis / wrs )br +

∑

j∈I 0 ∪ I1 \{s}

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[ wij − ( wis / wrs ) wrj ]b j

∑ [ wij − ( wis / wrs ) wrj ]b j .

j∈I 0 ∪ I1

Multiply (3) through by x(1) on the right hand side to have

= ai x (1)

(1) = ∑ wij a j x

∑ wij b j

j∈I 0 ∪ I1

j∈I 0 ∪ I1

Therefore Geometrically, our method for solving (1) begins with ai x(2) – ai x(1) = ( wis / wrs )br − ∑ ( wis / wrs ) wrj b j a positive basic cone whose vertex is denoted by x(1). If j∈I 0 ∪ I1 x(1) is feasible, then it is optimal for (1) by Karush= ( wis / wrs )(br − ∑ wrj b j ) = (wis / wrs)(br − ar x(1)).

Kuhn-Tucker (KKT) conditions. Otherwise there exists j∈I 0 ∪ I1 a violating constraint say ar x ≥ br such that ar x(1) < br. If Rewrite it as the half space ar x ≥ br is inconsistent with the cone, there is no feasible solution. Otherwise x(1) is projected ai x(2) − bi = ai x(1) − bi − (wis /wrs)(ar x(1) − br). along a particular edge of the cone onto the boundary Letting σi = aix(1) − bi, σr = arx(1) − br and σi’ = ai x(2) − bi, we ar x = br of the half space ar x ≥ br to constitute a new have positive basic cone with a value of the objective function not less than cx(1). Then the same process σi’= σi − (wis /wrs)σr，i∈I3∪I2\{r}. repeats.

Similarly, from (5) and (2) we have Algebraically, the iteration is accomplished by the c x(2) − c x(1) = (w0s / wrs) (br − ar x(1)) following pivoting operation. or For a basic cone of (1) with vertex x(1), let I0, I1, I2 and I3 f0’ = f0 − (w0s / wrs)σr be the index sets for basic equalities, basic inequalities, nonbasic inequalities and nonbasic equalities where f0 = c x(1) and f0’ = c x(2). respectively. Suppose that On the other hand from (4) we can obtain (2) c = ∑ w0 j a j , j∈I 0 ∪ I1 as x(2) − bs = − (1/wrs)(ar x(1) − br).

ai =

∑ wij a j , i∈I3∪I2.

j∈I 0 ∪ I1

(3)

If wrs ≠ 0 for r ∈I3∪I2 and s∈I1, we solve as from the rth expression of (3) to have (4) as = (1 / wrs )ar + ∑ (− wrj / wrs )a j , j∈I 0 ∪ I1 \{ s}

Substitute it into (2) and other expressions of (3) to have c = (w0s / wrs)ar + ∑ [ w0 j − ( w0 s / wrs ) wrj ]a j , (5) j∈I 0 ∪ I1 \{s}

ai = ( wis / wrs )ar +

∑

j∈I 0 ∪ I1 \{s}

[ wij − ( wis / wrs ) wrj ]a j ,

i∈I3∪I2\{r}.

(6)

Multiply (6) through by x(2) on the right hand side to have

ai x = ( wis / wrs )ar x

(2)

+

∑

j∈I 0 ∪ I1 \{s}

σs’ = − σr /wrs.

The above operational process is called a pivoting (operation), wrs is called pivot, the row of wrs is called pivot row and the column of wrs is called pivot column. We say ar enters and as leaves the basis. The exchange of these two vectors is denoted by ar ↔ as.

The pivoting process is simply represented by tables 1 and 2.

TABLE 1. INITIAL TABLE

Now we have a new basic cone whose index set is {r}∪I0∪I1\{s} and whose vertex is denoted by x(2).

(2)

Let σs’ =as x(2) − bs to have

[ wij − ( wis / wrs ) wrj ]a j x

(2)

c ar ai

as w0s wrs* wis

aj w0j wrj wij

f0

σr σi

TABLE 2. RESULT OF THE PIVOTING

c as ai

ar w0s/wrs 1/wrs wis /wrs

aj w0j’ −wrj /wrs wij’

f0’ −σr /wrs σi’

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where w0j’= w0j – (w0s / wrs)wrj, j∈I0∪I1\{s}; wij’= wij – (wis / wrs)wrj, i∈I3∪I2\{r}, j∈I0∪I1\{s}; σi’= σi – (wis / wrs)σr, i∈I3∪I2\{r}, f0’ = f0 – (w0s / wrs)σr.

The matrix (wij’) for i∈I3∪I2\{r} and j∈I0∪I1\{s} is called residual matrix with respect to wrs.

For Table 1, σi = ai x(1) − bi is called the deviation of the nonbasic vector ai or the associated constraint with respect to x(1) and w0j is called the (reduced) cost of the basic vector aj or the associated constraint. If

w0s / wrs = min{w0j/wrj | wrj > 0, j∈I1} which ensures the new basic cone to be positive as well. Because a pivoting on wrs yields c = (w0s / wrs)ar +

∑

j∈I 0 ∪ I1 \{s}

[ w0 j − ( w0 s / wrs ) wrj ]a j .

in which all the coefficients associated with the index set {r}∪I1\{s} are nonnegative.

Also we have cx(2) – cx(1) = (w0s / wrs)(br – ar x(1)) ≥ 0.

w0j ≥ 0, j∈I1 σi = ai x(1) − bi = 0, i∈I3 σi = ai x(1) − bi ≥ 0, i∈I2 then x(1) is the optimal solution. For an r∈I2 if σr = ar x(1) − br< 0, we say ar x ≥ br to be a violating inequality against x(1) and say |ar x(1) − br| / || ar ||2 to be the distance from x(1) to ar x ≥ br.

Pivoting Algorithm for Linear Programming Principle of Pivoting Algorithm As mentioned above the algorithm for solving (1) begins with a positive basic cone whose vertex is supposed to be x(1) and the index sets for basic equalities, basic inequalities, nonbasic inequalities and nonbasic equalities are I0, I1, I2 and I3 respectively.

For a nonbasic inequality ar x ≥ br (r ∈ I2), if ar x(1) < br, i.e. the deviation σr = ar x(1) − br < 0, then ar is a candidate to enter the basis. Suppose

ar =

If σr = ar x(1) − br < 0 and ar is entering the basis, we will determine a pivot by the minimal ratio

∑ wrj a j

j∈I 0 ∪ I1

as before. If wrj ≤ 0 for any j∈I1, then (1) has no feasible solution. The reason is as follows.

where x(2) is the vertex of the new positive basic cone.

It implies that the value of the objective function is non-decreasing and is increasing if the cost w0s of the leaving vector is positive.

Now suppose that a nonbasic equality ar x = br (r ∈ I3) is entering the positive basic cone with vertex x(1).

Since ar x = br is equivalent to two inequalities ar x ≥ br and −ar x ≥ −br, if ar x(1) < br and wrj ≤ 0 for any j∈I1, or ar x(1) > br and wrj ≥ 0 for any j∈I1, then (1) has no feasible solutions. Otherwise we determine the pivot wrs as follows.

If σr = ar x(1) − bi < 0, wrs satisfies w0s / wrs = min{w0j / wrj| wrj > 0, j∈I1}, and if σr = ar x(1) − bi > 0, wrs satisfies w0s / wrs = max{w0j / wrj| wrj < 0, j∈I1}.

If σr = ar x(1) − bi = 0 and wrj = 0 for any j∈I1, then ar x = br is a redundant equality constraint. Because in this case ar is a linear combination of coefficient vectors of the system of equations aj x = bj (j∈I0) and br is the same linear combination of bj (j∈I0).

Since −ar = −∑ j∈I 0 ∪ I1 wrj a j where −wrj ≥ 0 for any j∈I1,

Computational Steps of Pivoting Algorithm

by Farkas lemma (Bazarra & Shetty,1979), any solution x of the system of inequalities

With above explanation we formally state the algorithm for (1) as follows.

aj x = bj, j∈I0; aj x ≥ bj, j∈I1

Algorithm 1. Pivoting algorithm for (1).

satisfies −ar x ≥ −ar x therefore satisfies ar x ≤ ar x < br and not satisfies ar x ≥ br. Hence (1) has no feasible solution.

(1)

(1)

Suppose that c = ∑ w0 j a j j∈I 0 ∪ I1

where w0j ≥ 0 for any j∈I1.

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Step 1. Initial step. Let c = (c1,…,cn), x = (x1,…,xn)T, ej be the jth row of the identity matrix of order n and M be a number large enough. For j = 1,…,n, if cj ≥ 0 and there is a constraint in the form of xj ≥ lj, let xj ≥ lj be a basic inequality otherwise let xj ≥ −M be a basic inequality, that is ej is a basic vector and xj(0) = lj or −M; if cj < 0 and there is a constraint in the form of xj ≤ uj, then let −xj ≥ −uj be a basic inequality otherwise let −xj ≥ −M be a basic inequality, that is −ej is a basic vector and xj(0) = uj

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or M. Where xi ≥ −M and −xi ≥ −M are called artificial inequalities. Other constraints are nonbasic whose deviations are σi = ai x(0) − bi, i = 1, … ,m, where x(0) = (x1(0), …, xn(0))T. Thus the initial table is constructed as shown by Table 1 where the index sets for basic equalities, basic inequalities, nonbasic inequalities and nonbasic equalities are I0, I1, I2 and I3 respectively.

Step 2. Preprossing, put nonbasic equalities into the basic system as many as possible. (i) If I3 = Ø, goto Step 3. Otherwise for an r∈I3, when the deviation of ar is negative, positive or zero, goto (ii), (iii) and (iv) respspectively. (ii) (a) If wrj ≤ 0 for any j∈I1, (1) has no feasible solution, stop. Otherwise (b) select a basic inequality as x ≥ bs such that w0s / wrs = min{w0j / wrj : wrj > 0, j∈I1}.

Carry out a pivoting on wrs, let I3 := I3\{r}, I0 := I0∪{r}, I1 := I1\{s}, I2 := I2∪{s}, return to (i). (iii) (a) If wrj ≥ 0 for any j∈I1, (1) has no feasible solution, stop. Otherwise (b) select a basic inequality as x ≥ bs such that w0s / wrs = max{w0j / wr : wrj < 0, j∈I1}.

Carry out a pivoting on wrs, let I3 := I3\{r}, I0 := I0∪{r}, I1:=I1\{s}, I2 := I2∪{s}, return to (i). (iv) If there is wrj > 0 for an j∈I1, go to (ii) (b); otherwise if there is wrj < 0 for an j∈I1, go to (iii) (b); otherwise let I3 := I3\{r} and return to (i).

Step 3. Main iterations, exchange inequalities and basic inequalities.

nonbasic

(i) If all the deviations of nonbasic vectors are nonnegative, the current basic solution is optimal, stop.

Otherwise (ii) select a nonbasic vector ar (r∈I2) with a negative deviation to enter the basis. If wrj ≤ 0 for any j∈I1, there is no feasible solutions, stop. Otherwise let a basic inequality as x ≥ bs leave the basis that satisfies w0s / wrs = min{w0j / wrj | wrj > 0, j∈I1}.

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are formally written as ai x ≥ bi and −ai x ≥ −bi’ in our method. Suppose that the basic solution is x , then deviations of ai and −ai are ai x − bi and −ai x + bi’ respectively. The sum of these two deviations is bi’ − bi.

If one of ai and −ai is basic, the deviation of the other one is bi’ − bi > 0 therefore can be ignored. If both ai and −ai are nonbasic, since the coefficients in their expressions in terms of basic vectors are just opposite in signs, these two vectors can share one row in a table.

The purpose to transfer nonbasic equalities into the basis in the preprocessing stage is to make the basic solution satisfy all the equality constraints. Once an equality enters the basic system it never leaves the basis. For this we never choose a pivot in the column of basic equalities. Therefore the columns of basic equalities can be eliminated.

Reamrk 2. The identification of redundant inequality constraints If the deviation of a nonbasic vector ar is nonnegative and wrj ≥ 0 for any j∈I1, then ar x ≥ br is redundant by Farkas lemma where I1 is the index set of basic inequalities. Similarly, if the deviation of ar is nonpositive and there is only one s∈I1 such that wrs > 0, then as x ≥ bs is redundant, because a pivoting on wrs yields the above case.

Reamrk 3. The obtainment of basic solution Suppose x = ( x1 ,…, xn )T is the basic solution. If xj ≥ lj or

−M is basic, x j = lj or −M, i.e., x j equals to the lower bound of xj; if −xj ≥ −uj or −M is basic, then x j = uj or M, i.e., x j equals to the upper bound of xj. While xj ≥ lj or

−M is nonbasic, since its deviation is σj = ej x − li = x j − lj or x j + M, hence x j = σj + lj or σj − M, i.e., x j equals to the deviation of ej plus the lower bound of xj; if −ej is nonbasic, we see that x j equals to the upper bound of xj minus the deviation of −ej. For the final table, only when every component of the basic solution x is finite, i.e. not equal to −M or M, x is the optimal solution, otherwise no optimal solution for (1).

Reamrk 4. Rules for the iteration

Reamrk 1. The simplification of the table

Like simplex method, there are many rules to select entering and leaving vectors for iterations. Suppose I1 and I2 are index sets for basic inequalities and nonbasic inequalities respectively and σi is the deviation of ai x ≥ bi with respect to the current basic solution. Three rules for iterations are given below.

Sometimes a linear program contains constraints in the form bi’ ≥ ai x ≥ bi, especially in the form ui ≥ xi ≥ li, which

Rule 1. (The smallest deviation rule) Among nonbasic inequalities with negative deviations, select a nonbasic

Carry out a pivoting on wrs, let I2 := I2\{r}∪{s} and I1 := I1\{s}∪{r}, return to (i).

Some remarks are given as follows.

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inequality with the most negative deviation to enter the basic system, i.e., if σr = min{σi : σi <0, i∈I2}, then ar enters the basis.

Rule 2. (The largest distance rule) Among nonbasic inequalities with negative deviations, select a nonbasic inequality farthest from the current basic solution to enter the basic system, i.e., if σr / || ar ||2 = min{σi / || ai ||2 : σi < 0, i∈I2}, then ar enters the basis.

Rule 3. (The smallest index rule) Among nonbasic inequalities with negative deviations, select a nonbasic inequality with the smallest index to enter the basic system; among basic inequalities taking the minimal ratio, select a basic inequality with the smallest index to leave the basic system, i.e., if r = min{i∈I2 : σi < 0}, then ar enters the basis; if s = min{k : w0k / wrk = θ, k∈I1} where θ = min{ w0j / wr : wrj > 0, j∈I1}, then as leaves the basis.

Like Bland’s anti-cycling rule (Chvatal, 1983), it can be proved (Zhang, 2004) that when the smallest index rule is applied, cycling will be avoided.

A Numerical Example Let us show how to solve a linear program by Algorithm 1.

Example 1. Solve the linear program min z = 7x1 + 7x2 − 3x3 + x4 s.t. x1 − 2x2 – x3 + x4 = 7, 2x1 + x2 − 3x3 − x4 = 12, −x1 − 4x2 − 2x3 − x4 ≥ 4, 2x1 + x2 + x3 + 2x4 ≥ 2, x1 ≥ 0, x2 ≥ −1, 1 ≥ x3 ≥ −5, x4 free.

Solution. Let a1 = (1, −2, −1, 1), b1 = 7, a2 = (2, 1, −3, −1), b2 = 12, a3 = (−1, −4, −2, −1), b3 = 4, a4 = (2, 1, 1, 2), b4 = 2, e1 = (1, 0, 0, 0), l1 = 0, e2 = (0, 1, 0, 0), l2 = −1, e3 = (0, 0, 1, 0), l3 = −5, u3 = 1.

Since x4 is free and the coefficient at x4 of the objective function is positive, we introduce an artificial inequality x4 ≥ −M into the problem and let e4 = (0, 0, 0, 1), l4 = −M. In order to represent the coefficient vector of the objective function to be a nonnegative linear combination of basic vectors, we let e1, e2, −e3, e4 be initial basic vectors. Correspondingly, the initial basic

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system is x1 ≥ 0, x2 ≥ −1, −x3 ≥ −1, x4 ≥ −M and the initial basic solution is x(0) = (0, −1, 1, −M)T with a value −M −10 of the objective function. Vectors a1, a2, a3, a4 and e3 are nonbasic with deviations σ1 = a1 x(0) − b1 = −M − 6, σ2 = M − 16, σ3 = M − 2, σ4 = −2M − 2 and 1 − (−5) = 6 respectively. Table 3 is the initial table where the inequality constraint x3 ≥ −5 is temporarily ignored since − x3 ≥ −1 is basic.

TABLE 3. INITIAL TABLE

e1

e2

−e3

e4

σi

c

7

7

3

1

−M−10

a1

1

−2

1

1*

−M−6

a2

2

1

3

−1

M−16

a3

−1

−4

2

−1

M−2

a4

2

1

−1

2

−2M−2

Since a1 is the coefficient vector of an equality constraint, a1 enters the basis first. Since the deviation of a1 is σ1 = −M − 6 < 0 and min {w0j / w1j : w1j > 0, j∈I1} = min {7/1, 3/1, 1/1} = 1, e4 leaves the basis and the entry 1 with an asterisk is the pivot. Table 4 gives the result of pivoting where the column of a1 is eliminated.

TABLE 4. RESULT OF THE FIRST PIVOTING

c e4 a2 a3 a4

e1

e2

−e3

σi

6 −1 3 0 0

9 2 −1 −6 5

2 −1 4* 3 −3

−4 M+6 −22 −8 10

In Table 4 let a2 enter and −e3 leave the basis to yield Table 5 in which the coefficient vector −e3 of −x3 ≥ −1 has become a nonbasic vector. It is the time to consider x3 ≥ −5. The deviation of x3 ≥ −5 is equal to u3 − l3 − 11/2 = 6 − 11/2 = 1/2 which is listed in the last column of Table 5.

TABLE 5. RESULT OF THE SECOND PIVOTING

e1

e2

σi

σi

c

9/2

19/2

7

a1

−1/4

7/4

M+1/2

-

−e3

−3/4

1/4

11/2

1/2

a3

−9/4

−21/4

17/2

-

a4

9/4*

17/4

−13/2

-

The unique negative deviation in Table 5 is −13/2 taken by a4 therefore a4 enters the basis. Since min {

9/2 9/4

,

19 / 2 17 / 4

}=2

taken by e1, e1 leaves the basis. A pivoting on 9/4 gives Table 6.

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TABLE 6. RESULT OF THE THIRD PIVOTING

c e4 −e3 a3 e1

a4

e2

σi

σi

2 −1/9 −1/3 −1 4/9

1 20/9 5/3 −1 −17/9

20 M−2/9 10/3 2 26/9

8/3 -

Up till now all the deviations of nonbasic vectors are nonnegative, the iteration process is finished.

Since e2 is basic, x2 = l2 = −1. Since e1, −e3 and e4 are nonbasic, x1 = 26/9 + l1 = 26/9 + 0 = 26/9, x3 = u3 − 10/3 =1− 10/3 = − 7/3, and x4 = M −2/9 + l4 = M −2/9 + (−M) = −2/9.

Therefore the optimal solution is x = (26/9, −1, − 7/3, −2/9)T with a value 20 of the objective function.

Relationship between Pivoting Algorithm and Simplex Algorithm The Simplex Algorithm Simplex method was proposed by G. B. Dantzig in 1947. In the past decades it has been the most frequently employed method for solving linear programming.

Before linear programs are solved by the simplex algorithm, preliminary procedures have to be done as follows. (i) Whenever a problem contains general inequality constraints, these constraints are transformed into equalities by introducing slack or surplus variables. (ii) Whenever a problem contains free variables, each of which is be replaced with two nonnegative variables. (iii) Whenever the lower bound of a variable is not zero, this variable is replaced by a variable with zero as its lower bound.

In this way a standard form is constructed as following max z = cx s.t. Ax = b, x ≥ 0.

(7)

where c is an n-dimensional row vector, x is an unknown n-dimensional column vector, A is an m × n matrix, b is an m-dimensional column vector and m < n as in usual. The fundamental concepts in simplex method are defined upon the standard form as follows where the rank of A is supposed to be m.

A set of m linearly independent columns of A is called

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a basis of (7). Columns of A in the basis are called basic, otherwise called nonbasic. The full rank matrix formed by basic vectors is called a basis matrix. Variables associated with basic vectors are called basic, otherwise called nonbasic. The solution to Ax = b by letting nonbasic variables be zero is called a basic solution. If every component of the basic solution is nonnegative, it is called a basic feasible solution and the basis is called a feasible basis.

In order to initiate the formal calculation, the standard form (7) is needed to be further transformed into the following form called canonical form. max z = z0 + σDxD s.t. xB + WxD = w0, xB ≥ 0, xD ≥ 0

(8)

where xB is a column vector formed by m components of x, xD is formed by other components of x, σD is an (n − m)-dimensional row vector, W is an m × (n − m) matrix and w0 is an m-dimensional column vector. The canonical form explicitly determines a basic solution xB = w0 and xD = 0 called current basic solution.

In the computational process of (8) by simplex algorithm, all the components of the right hand side term of equality constraint keep nonnegative, but some coefficients of nonbasic variables of the objective function are positive. As soon as all the coefficients of the objective function are nonpositive, the current basic solution is optimal.

Let I1 and I2 be index sets for basic and nonbasic variables respectively. The component of σD associated with the nonbasic variable xi is denoted by σi and the column of W associated with xi is denoted by wi. The components of w0 and wi associated with a basic xj are denoted by w0j and wij respectively. The simplex algorithm for the canonical form (8) is carried out as follows where w0 ≥ 0.

Step 1. If σD ≤ 0, the current basic solution is optimal, stop. Otherwise select a nonbasic variable xr (r∈I2) associated with the largest component of σD as the entering variable.

Step 2. If the column wr ≤ 0 that is associated with xr, (8) is unbounded from above, stop. Otherwise if w0s / wrs = min{w0j / wrj : wrj > 0, j∈I1}, then xs is the leaving variable.

Step 3. Carry out a pivoting on wrs, that is to carry out elementary row transformations to convert the current canonical form into a new canonical form where the

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index sets for basic and nonbasic variables are {r}∪I1\{s} and {s}∪I2\{r} respectively. Then return to step 1.

hand, y is feasible for (9) and is hence optimal.

A variation of the simplex algorithm is dual simplex algorithm. Applying this method to (8), all the coefficients of nonbasic variables of the objective function keep non-positive and some components of the left hand side term of the equality constraint are negative. Once every components of the left hand side term of the equality constraint become nonnegative, i.e., the basic solution becomes feasible, it is the optimal solution.

The coefficient at a nonbasic xi in the objective function of (10) is

Correspondence between Two Algorithms We shall show that the solving of the standard form by simplex algorithm is corresponding to the solving of its dual by pivoting algorithm.

Let A = (p1,…,pn) in the standard form (7), then the dual of (7) is min f = yb s.t. ypi ≥ ci, i = 1,…,n

(9)

where y = (y1,…,ym).

Without loss of generality, suppose the matrix B = (p1,…,pm) formed by the first m columns of A is a basis in line with simplex method. The matrix formed by nonbasic vectors is denoted by D. Correspondingly, partition c into two vectors cB and cD and partition x into two vectors xB and xD. Then (7) is written as max z = cB xB + cD xD s.t. BxB + DxD = b xB ≥ 0, xD ≥ 0.

−1

max z = cB B b + (cD – cB B D) xD s.t. xB + B−1 D xD = B−1 b xB ≥ 0, xD ≥ 0.

(10)

Denote B−1b = w0 = (w01,…,w0m)T, then b = B (B−1 b) = B w0 = w01 p1 +…+ w0m pm.

If B−1 b = w0 ≥ 0, B is a feasible basis of (7), and xB = B−1 b and xD = 0 is a basic feasible solution. On the other hand, the system of ypi ≥ ci, i = 1,…,m is a positive basic cone of (9) whose vertex y = cB B−1 is a solution to the system yB = cB, i.e., the system of ypi = ci, i = 1,…,m.

Therefore a basic feasible solution of (7) corresponds to the vertex of a positive basic cone of (9).

If there is cD – cB B−1 D = cD – y D ≤ 0 also, then xB = B−1 b and xD = 0 is the optimal solution of (7). On the other

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σi = ci – cB B−1 pi = ci − y pi, i = m + 1,…, n.

Let B−1pi = wi = (wi1,…,wim)T, i = m + 1,…,n, then the ith nonbasic vector of (9) can be represented as pi = B (B−1 pi) = B wi = wi1 p1 +…+ wim pm, i = m + 1,…,n.

On the other hand, the coefficient at nonbasic xD in the equality constraint of (10) is B−1D = (wm+1,…, wn).

Suppose B−1b = (w01,…, w0m)T ≥ 0. For an r∈{m + 1,…, n}, if the coefficient σr = cr – y pr > 0 of a nonbasci variable xr, then xr is a possible entering variable for (10); and the deviation −σr = y pr − cr < 0 of ypr ≥ cr with respect to y , hence ypr ≥ cr is a possible entering inequality for (9). If there is wr = (wr1,…, wrm)T ≤ 0 also, then (10) has an unbounded solution, while (9) has no feasible solution.

Otherwise we calculate the minimum ratio min{w0j / wrj : wrj > 0, j∈{1,…,m}} to determine a leaving variable for (10) or a leaving inequality for (9). Thus the solving of (7) by simplex is corresponding to the solving of (9) by pivoting algorithm one by one.

Dual Solution of Linear Programming For some problem it is more convenient to obtain the optimal solution by solving its dual. Let us describe such a method.

Consider linear programming

By using B−1, it is further written as −1

Therefore the criteria for optimal solutions of (7) and (9) are consistent.

max z = cx s.t. ai x = bi, i = 1,…,l, ai x ≤ bi, i = l + 1,…,m, xj ≥ 0, j∈J.

(11)

Where c = (c1,…,cn), x = (x1,…,xn)T, ai = (ai1,…,ain), bi is a real number, i = 1,…,m, and J⊆{1,…,n}.

The dual of (11) is defined by min w = yb s.t. y pj = cj, j∈{1,…,n}\J, y pj ≥ cj, j∈J, yl +1,…,ym ≥ 0.

(12)

Where y = (y1,…,ym), b = (b1,…,bm)T and pj = (a1j,…,amj)T, j = 1,…,n.

Theorem 1. Suppose that x* = (x1*,…,xn*)T is the optimal solution of the primal problem (11) and

International Journal of Advances in Management Science (IJ-AMS) Volume 3 Issue 3, August 2014

c = ∑ w0i ai + ∑ w0i ai + ∑ w0 j (−e j ) i∈E

i∈I

j∈Z

(13)

where w0i (i∈I) and w0j (j∈J) are nonnegative, and E, I, Z are active sets with respect to x*, i.e., E = {i ∈{1,…,l} : ai x* = bi}, I = {i ∈{l+1,…,m} : ai x* = bi}, Z = {j ∈{1,…,n} : −ejx* = 0},

− ej x* = 0, j∈Z, multiply (13) through by x* on the right to yield cx* = ∑ w0i bi + ∑ w0i bi + ∑ w0 j ⋅ 0 i∈E l

= ∑

i= 1

where ej is the j row of the identity matrix of order n.

Then (14)

j∈I '

i∈Z '

xj* = w0j, j∈E’∪I’, x = 0, j∈{1,…,n}\E’\I’

l

n

i= 1

i = l +1

TABLE 7. INITIAL TABLE

cj = ∑ w0i aij + ∑ w0i aij = ∑ yi* aij + ∑ yi* aij =∑

i =1

i∈I

yi* aij

= y*pj, j∈{1,…,n}\Z

or cj = ∑ w0i aij + ∑ w0i aij − w0j i∈E

i∈I

l

n

= ∑

i= 1

yi* aij

min w = 8y1 + y2 + 2y3 s.t. y1 + 2y2 − y3 = 5, 2y1 + y2 + 4y3 ≥ 4, y1 − 3y2 + y3 ≥ −6, y1 − y2 ≥ −1, y1, y2, y3 ≥ 0.

Denote the coefficient vectors of the four general constraints by a1, a2, a3, a4. Denote the rows of the identity matrix of order 3 by e1, e2, e3. The initial table is given by table 7.

Proof. The jth component of (13) is

m

= y*b.

Solution. The dual of this problem is

(16)

is the optimal solution of the primal problem (11).

i∈E

i = l +1

max z = 5 x 1 + 4x2 − 6x3 – x4 s.t. x1 + 2x2 + x3 + x4 ≤ 8 2x1 + x2 − 3x3 – x4 ≤ 1, −x1 + 4x2 + x3 ≤ 2, x2, x3, x4 ≥ 0.

(15)

where w0j (j∈I’) and w0i (i∈Z’) are nonnegtive and E’, I’, Z’ are active sets with respect to y* which are associated with equality constraints, inequality constraints and nonnegative restrictions respectively.

Then j*

+ ∑

yi*bi

Example 2. Solve the optimal solution for the linear program

Conversely, if the dual problem (12) has an optimal solution y* and j∈E '

m

Similarly, (16) can be proved to be the optimal solution of problem (11).

is the optimal solution of the dual problem (12).

b = ∑ w0 j p j + ∑ w0 j p j + ∑ w0i eiT

j∈Z

i∈I

yi*bi

Therefore (14) is the optimal solution of problem (12).

th

yi* = w0i, i∈E∪I, yi* = 0, i∈{1,…,m}\E\I

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+ ∑ yi* aij − w0j = y*pj – w0j, j∈Z. i = l +1

c a1 a2 a3 a4

e1 8 1 2 1 1

e2 1 2* 1 −3 −1

e3 2 −1 4 1 0

0 −5 −4 6 1

Since w0j ≥ 0, j∈Z, the above expression can be written as

Since a1 is the coefficient vector of the equality constraint, a1 enters the basis first. The deviation of a1 is − 5 < 0 and

cj ≤ y*pj, j∈Z.

min {w0j / w1j : w0j > 0, j∈I1} = min {8/1, 1/2} = 1/2, e2 leaves the basis. A pivoting on 2* gives Table 8.

Besides

TABLE 8. RESULT OF THE FIRST PIVOTING

yi* = w0i ≥ 0, i∈I, y = 0, i∈{l + 1,…,m}\I. i*

Therefore (14) is a feasible solution of problem (12).

Since ai x* = bi, i∈E∪I,

c e2 a2 a3 a4

e1 15/2 −1/2 3/2 5/2 3/2*

a1 1/2 1/2 1/2 −3/2 −1/2

e3 5/2 1/2 9/2 −1/2 −1/2

5/2 5/2 −3/2 −3/2 −3/2

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www.ij-ams.org

International Journal of Advances in Management Science (IJ-AMS) Volume 3 Issue 3, August 2014

In Table 8, deviations of nonbasic a2, a3, a4 are negative.

Let us apply the largest distance rule to choose a vector to enter the basis. Since min {σ2 / || a2 ||2, σ3 / || a3 ||2, σ4 / || a4 ||2} = min {−1.5 / 21 , −1.5 / 11 , −1.5 / 2 } = −1.5 / 2 , a4 enters the basis. In Table 8 the unique positive entry in the row of a4 is 3/2, e1 leaves the basis and the constraint y1 ≥ 0 associated with e1 is redundant. A pivoting on 3/2 gives Table 9 where the row of e1 is eliminated.

TABLE 9. RESULT OF THE SECOND PIVOTING

c e2 a2 a3

a4 5 −1/3 1 5/3

a1 3 1/3 1 −2/3

e3 5 1/3 5 1/3

10 2 0 1

Up till now all the deviations of nonbasic vectors are nonnegative.

Since c = 5a4 +3a1 +5e3, the optimal solution for the original problem is x1 = 3, x4 = 5, x2 = x4 = 0 and the value of the objective function is 10.

Conclusions This paper presented a pivoting-based method for solving linear programming. This method is novel in that it never requires additional variables to change the form of the original problem and is very easy to use.

The advantage of such a method is threefold. One is that it can easily find out redundant constraints and

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eliminate them in the iteration process. The second one is that it saves the computation of pivoting operation in tableau form, even for integer programming where branch variables or cutting inequalities are frequently added to the problem in the computational process. The third one is that it can be easily converted into graphic algorithms (Zhang, 2004) for many network optimization problems such as the shortest path problem and minimum cost flow problem. Because the pivoting row and column can be directly obtained from the underling graph with a forest or spanning tree so that a pivoting is conducted at very small amount of computation.

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Robert E. Bixby. Solving Real-World Linear Programs: A Decade and More of Progress. Operations Research, Volume 50 Issue 1, January-February 2002.

Zhang

Zhongzhen.

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Convex Portfolio

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and

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