Pergamon Mathl. Comput. Modelling Vol. 21, No. 4, pp. 13-28, 1995

Copyright@1995 Elsevier Science Ltd Printed in Great Britain. All rights reserved

0895-7177(95)00002-X 0895-7177195 $9.50 + 0.00

A Transportation Model with Multiple Criteria and

Multiple Constraint Levels*

Y. SHI Department of Information Systems and Quantitative Analysis

College of Business Administration University of Nebraska at Omaha, Omaha, NE 68182, U.S.A.

(Received and accepted October 1994)

Abstract-In this paper, a transportation model with multiple criteria and multiple constraint levels (MC2) is formulated by using the framework of MC2 linear programming. An algorithm is developed to solve such MC2 transportation problems. In this algorithm, the traditional northwest corner rule is adopted to find an initial basic feasible solution for a given MC2 transportation problem. Then the MC2-simplex method is applied to locate the set of all potential solutions over possible changes of the objective coefficient parameter and the supply and demand parameter for the MC2 transportation problem. A numerical example is illustrated to demonstrate the applicability of the algorithm in solving the MC2 transportation problems.

Keywords-MC2 linear programming, MC?-simplex method, MC2 transportation problems,

Algorithm.

1. INTRODUCTION

The traditional transportation problem was initiated by Hitchcock [l] and Koopmans [2].

Dantzig [3] applied his simplex method to solve the transportation problems. Charnes and

Cooper [4] developed the stepping stone method and Reinfeld and Vogel [5] proposed the modi-

fied distribution method as an alternative way to find an optimal solution. The northwest corner

rule is often used to find an initial basic feasible solution in the process of solving the transporta-

tion problems. Houthakker [6], Vogel [5], and Russell [7] also developed methods of finding an

initial basic feasible solution. The traditional transportation model, however, may have limita-

tions in dealing with the real world problems because it has only a single criterion (objective)

and a single supply and demand level. For example, a transportation problem may involve two

objectives: the minimization of total energy cost consumed in transportation and the minimiza-

tion of total product deterioration during transportation; and two supply and demand levels:

the supply and demand quantities when the economy is in recession and the supply and demand

quantities when the economy is booming. The traditional transportation model certainly fails to

handle this case.

A transportation problem with two criteria was proposed by Aneja and Nair [8]. By using the

techniques in multiple criteria (MC) linear programming [g-12], they developed an algorithm to

*This research has been partially supported by a University Research Fellowship Award from University of Ne- braska at Omaha. The author is grateful for the constructive comments of P.-L. Yu of the University of Kansas, Y.-H. Liu and J. D. Stolen of the University of Nebraska at Omaha on an earlier version of this paper.

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13

14 Y. SHI

identify the set of nondominated extreme points in the criteria space. Other formulations and

solution methods for MC transportation problems were studied by Lee and Moore [13], Diaz [14],

Isermann 1151, Current and Min [16], Ringuest and Rinks [17] and the citations therein. But, all of

those approaches have assumed there is only a fixed supply and demand level in the transportation

problem. Thus, they cannot be used to deal with a transportation problem involving fluctuations

of supply and demand.

Seiford and Yu [18] formulated a mathematical model of multiple criteria and multiple con-

straint level (MC2) linear programming. In their paper, a parametric approach was used to extend

the traditional simplex method to an MC2-simplex method for solving the MC2 problems. The

main aim of this paper is to propose a mathematical formulation of an MC2 transportation prob-

lem within the framework of MC2 linear programming and to develop an algorithm to solve such

transportation problems.

The model of MC2 linear programming is supported at least by the following two facts. First,

from the linear system structure’s point of view, the criteria and constraints may be “interchange-

able” [ll]. Thus, like multiple criteria, multiple constraint (resource availability) levels can be

considered. Second, from the application’s point of view, it is more realistic to consider multiple

resource availability levels (discrete right hand sides) than a single resource availability level in

isolation. The philosophy behind this perspective is that the availability of the resources can

fluctuate depending on the decision situation forces, such as the desirability levels believed by

the different decision makers and/or the changeable economic environments (see [19]). As a par-

ticular case, the MC2 transportation model shares the above advantages in solving the real-world

problems. This paper proceeds as follows.

In the next section, the notation and definitions of the MC2 linear programming problem and

the MC2-simplex method of [11,18] are briefly sketched as a basis for developing the MC2 trans-

portation model. This model is formulated in Section 3. In Section 4, an algorithm is developed

to solve such MC2 transportation problems. In this algorithm, the traditional northwest corner

rule is adopted to find an initial basic feasible solution for a given MC2 transportation problem.

Then the MC2-simplex method is applied to locate the set of all potential solutions over possible

changes of the criteria coefficient parameter and the supply and demand parameter for the MC2

transportation problem. A numerical example is used to determine the algorithm for solving

the MC2 transportation problems. Finally, the conclusions and some further research problems

are given in Section 5.

2. MC2 LINEAR PROGRAMMING

In this section, the solution concepts of MC2 linear programming adopted from [11,18] are

sketched to facilitate the discussion on the MC” transportation model.

An MC2 linear programming problem can be formulated as

Minimize X%X

subject to Ax = Dy (1)

x 2 0,

where C E R’Jxn, A E RmX”, and D E RmxP, are matrices of q x n, m x n, and m x p dimensions,

respectively; x E Rn are decision variables; X E R’J is called the criteria parameter and y E R*

is called the constraint level parameter. Both (7, X) are assumed unknown.

REMARK 2.1. Problem (1) has q criteria (objectives) and p constraint levels. If the constraint

level parameter y is known, then (1) is reduced to an MC linear programming problem [20];

in addition, if the criteria parameter X is known, then (1) is reduced to a linear programming

problem [21,22]. The details for the theoretical connection of MC2 linear programming and MC

linear programming can be referred to Gyetvan and Shi [23].

A Transportation Model 15

An MC2-simplex method for solving (1) can be described as follows. Given basic variables

{Xj,****, xj, ) for (l), we denote the index set of the basic variables by J = {jr,. . . , jm}. Note that the basic variables may contain some slack variables. Without confusion, J is also called a basis for (1). Because a potential basis J depends on parameters (y, X), we have the following definitions.

DEFINITION 2.1. A basis J is a @ma1 potential basis for problem (I) if and only if there exist some y > 0 and X = 0 such that J is an optj~ai basis for (1).

DEFINITION 2.2. A basis J is a dual potential basis for problem (1) if and only if there exist

some X > 0 and y = 0 such that J is an optimal basis for (1).

DEFINITION 2.3. A basis J is a potential basis for problem (1) if and only if there exist a y > 0 and X > 0 such that J is an optimal basis for (1).

REMARK 2.2. From Definitions 2.1-2.3, we say that a basis J is a potential basis if and only if J is both a primal and a dual potential basis.

Let x(J) be the basic variables corresponding to a potential basis J. We call Z(J) the potential solution of problem (1). To find all the potential solutions for problem (l), we need to identify the corresponding set of potential bases for (1). In the following discussion, we assume that both parameters (y, A) are normalized; i.e.,

yERPwith rk>Oand eyk=landXER’withXk>Oand g&=1. k=l k=l

Given a basis J with its basic variables x(J), we define the associated basis m&ix BJ as the submatrix of A with column index of J (i.e., column j of A is in BJ if and only if j E J), and the associated objective ~~ct~o~ coefficient CJ as the submatrix of C with column index of J. Let Z( J’) be the nonbasic variables corresponding to given x(J). Then, we rearrange the index, if necessary, and decompose A into [BJ, R], where R is the submatrix of A associated with X( J’), and C into [CJ, CJ,], where CJ, is the submatrix of C associated with z(J’). The initial simplex

tableau of (1) can be

44 4-J’) RHS

pgtq

(2)

(3)

Tableau (2)-(3) can be rewritten as

44 4J’) RHS

~1

(4

(5)

where iT, is an m x m identity matrix. Note that row block (4) is the result of premultiplying (2) by Bil. Row block (5) is obtained

by first premultiplying (4) by xtC~ and then adding the result to row block (3). Tableau (4)-(5) implies that z(J, y) = B;lDy is a basic solution associated with (J, 7) and its objective value is given by V(J) = XtC~BflL)y when (y, X) are specified.

By dropping (7, X) from Tableau (4),(5), we obtain an MC2-simplex tableau with a basis BJ as follows.

x(J) 4J’) RHS

Ln BJIR B;‘D (6)

0 CJB;‘R - Cy C.&D (71

16 Y. SHI

DEFINITION 2.4. Given a bask J for problem (l), define its corresponding

(i) primal parameter set by

r(J) = (7 > 0 1 BJ’Dy 2 0); and

(ii) dual parameter set by

A(J) = {A > 0 1 At [CJBJ’R - CJ,] 5 0).

The following theorem is adopted from [ll, Chapter 81.

THEOREM 2.1. Comparing (4),(5) with (6),(7), we see that given J:

(i) The resulting feasible solution z(J, y) = BylDy 2 0 if and only if y E I’(J);

(ii) The solution X( J, y) is optimal if and only if y E I’(J) and X E A(J);

(iii) J is a primal potential basis if and only if r( J) # 8;

(iv) J is a dual potential basis if and only if A(J) # 8; and

(v) J is a potential basis if and only if I’(J) x A(J) # 0.

By using the MC2-simplex method discussed above, one can locate a set of potential solutions,

denoted by J = { J1, . . . , Jr}. For every (y, A), if there is a J of 3 such that (y, A) E l?(J)XA(J),

then 2 is called the set of all potential solutions over all possible changes of parameters (y, A)

for a given MC2 problem (1). The separation, adjacency, and connectedness of the potential

solutions can be referred to [ll, Section 8.4.41.

3. AN MC2 TRANSPORTATION MODEL

Baaed on the MC2 problem (l), an MC2 transportation problem is proposed as follows.

m n

Minimize C C Ctj Xij

i=l j=I

2 f: Cij Xij

i=l j=l

Subject to 2 Xij = (ail,. . . , sip), i = l,...,m, j=l

m

c Xij = (bjl,...,bjp), j=l,...,n,

i=l

Xij > 0 for all (i,j),

where C$ = the “cost” of transporting a unit product from source i to destination j when the

kth criterion is under consideration, k = 1,. . . ,q.

ai, = the sth supply availability level at i, s = 1, . . . ,p, which is assumed to be positive,

bj, = the sth demand availability level at j, which is assumed to be positive,

xij = the amount of the products transported from i to j.

A Transportation Model 17

As mentioned before, the “cost” of transporting a unit product in problem (8) could be the

energy cost consumed in transportation, the transportation time, or the product deterioration

during transportation. The supply and demand levels could be the desirability levels believed by

the different decision makers or the economic situations. The interpretation of the criteria and

supply and demand levels in the MC2 transportation problem (8) should depend on the features

of the real world problems.

Mathematically speaking, problem (8) is a special case of the MC2 linear programming prob-

lem (1). Thus, all of the results presented in Section 2 can be applied to problem (8).

By putting the criteria parameter X and the constraint level parameter 7 on problem (8), it

becomes:

Minimize Pl, x2,. . . , A,)

Subject to 2 Xij = (ail, j=l

m

C xij = (bjl, i=l

n

Ii2 1 Cij Xij

i=l j=l m n

C C Cfj Xij i=l j=l

m 71’

C C Cf.j Xij

i=l j=l

i = 1,

j = 1,

. . .

. . .

(9)

, n,

Xij > 0 for all (i,j).

The MC2 transportation problem (9) is a parametric linear programming problem. If the values

of (7, X) were known, it could be solved by a traditional transportation algorithm [3-71. If y was

known, problem (9) could be solved by some MC transportation algorithm [8,13-171. However,

when parameters (y, X) cannot be known ahead of decision time, these approaches would not

be effective because there are infinitely many possible combinations of (y, X). In addition, the

changes of y may make the original choice infeasible, while that of X can render the choice

not optimal. With unknown (y, X), this paper applies the MC2-simplex method to develop an

algorithm for solving problem (9). In order to do so, some theoretical properties of problem (9)

are studied as follows.

LEMMA 3.1. For any y” > 0, Cz, X:=1 ai, 7: = Cj”=, C&, bjs 7: if and only if cz, ai, =

C,“=, bjs, for any s = 1,. . . ,P.

PROOF. The proof of sufficiency is trivial. We only show the necessity: Suppose that CL”=, ais #

C,“=, bj,,, for some SO. Without loss of generality, we assume that cz1 aiso > CT=“=, bjsO. We choose yS > 0, such that ^(s approaches zero for all s # SO. Then, we see that ysO = 1 - &S0 yS

approaches 1 as CszsO ys approaches zero. This implies that we can properly choose 7 so that

Cz”=, X:=1 ais?, > C,Z, Cf=, bjsYs* I

THEOREM 3.1. Given y” > 0, the MC2 transportation problem (9) has a feasible solution at y”

if Cy=“=, ai, = cy=, bjs, for any s = 1,. . . ,p.

18 Y. SHI

PROOF. Given a y” > 0, we suppose Cz”=, ai, = cy=, bjS, for any s = 1, . . . ,p. By Lemma 3.1,

we have CT=“=, EL, ai, $’ = Cy=, Ck, bj, 7,” = I-I, for any 7,” > 0. This implies the existence

of a feasible solution, namely xij = (~~=, ai, $!)(~~=i bj, $)/HI for all (i,j) (see page 321

of [25]). I

The condition cz”=, ai, = cy=, bjs, for any s = 1, . . . , p, of Theorem 3.1 is imposed in the

following discussion to guarantee that problem (9) has a feasible solution for a y. This condition

means that when the .sth supply and demand level is under consideration, the total amount of

products transported is equal to the total amount received.

With the above condition, it is observed that there are at most m + n - 1 linearly independent

constraints in problem (9). This leads to the following property.

THEOREM 3.2. Given a primal potential basis J for problem (9), the basic variables x(J) contain

at most m + n - 1 components Xij.

The proof of Theorem 3.2 can be readily shown by integrating the results in Section 2 and the

corresponding theoretical result in the traditional transportation problem [24-261. If the number

of the basic variables with positive in x(T) is exactly equal to m + n - 1, then the solution T

is called a nondegenerate solution for all y E y(T). If the number of the basic variables with

positive in z(T) is less than m + n - 1, then the solution T is called a degenerate solution for

all y E y(T). In this paper, the MC2 transportation problems having nondegenerate solution are

examined. It should be noted that the degenerate case can be similarly studied.

4. AN ALGORITHM FOR THE MC2 TRANSPORTATION MODEL

Based on the discussion in Sections 2,3, this section develops an algorithm for solving the MC2

transportation problem (9) (or (8)). S ince the algorithm is derived within the framework of the

MC?-simplex method of [18], it is called an MC2 algorithm for problem (9).

Let Cij = (Cij,... ,cfj) be the vector of q criteria coefficients when xij is transported from

supply source i to demand destination j. Let ui(x) be the simplex multiplier (implicit price)

corresponding to supply i, and vj (X) be the simplex multiplier corresponding to demand j [3,27].

Both ui(X) and Vj(X) are functions of X in the situation. Apply (ii) of Definition 2.4, the dual

parameter set A(T) of a basis T for problem (9) can be identified by the following theorem.

THEOREW 4.1. A basis T is a dual potential basis for problem (9) if

(i) ?&(A) + Vj(X) - XtCij = 0, f or every (i, j) such that xij is a basic variable; and

(ii) Ui(X) + Vj(X)- XtCij < 0, f or every (i, j) such that xij is a nonbasic variable.

(i) of Theorem 4.1 provides a way to calculate Us and 1-‘j (X) corresponding to a basis T for

problem (9), while (ii) of Theorem 4.1, which is used to find the dual parameter set A(T), yields

a potentiality test for the basis T if the primal parameter set y(T) # 0.

There are important tasks for developing such an MC2 algorithm:

(i) find an initial basic feasible solution;

(ii) select the “entering basic variable” and the ‘{leaving basic variable” at an iteration where

the solution is not a potential solution; and

(iii) select the “entering basic variable” and the “leaving basic variable” at an iteration where

the solution is a potential solution.

The third task is not needed for the traditional transportation problem because the traditional

algorithm usually terminates if an optimal solution is found. However, for a given MC2 trans-

portation problem (9), this task is definitely needed to pivot from current potential solution to

other adjacent potential solutions (see [ll, Chapter 81).

In this algorithm, for convenience, the traditional northwest corner rule [4] is adopted to find

an initial basic feasible solution for problem (9). Other alternative methods, such as Vogel’s

A Transportation Model 19

approximation method and Russell’s approximation method could also be used to find a better initial basic feasible solution for problem (9). With the initial basic feasible solution, the MC2- simplex method then can be applied to locate the set of all potential solutions over possible changes of (y, X) for problem (9). According to Section 2, a variable xij for problem (9) is a function of y and the objective coefficient of the variable is a function of X. Thus, finding a way for determining the entering basic variable and the leaving basic variable at each iteration requires an estimate of the “upper bound” of the value of the chosen variable or the value of the objective coefficient (If the MC2 transportation problem has a maximizing criterion, the “lower bounds” of the values of the objective coefficients should be considered). The following lemma can be used for this purpose.

LEMMA 4.1.

(i)

where d, IS ai, or bjs, for all (i, j) and y = (71,. . . , Y~)~ is the parameter,

(ii)

supply and demand

for all (i,j) where X = (Al,. . . , X,)t is the objective parameter.

PROOF. (i) By Chauchy inequality,

Taking the square root on both sides of

(i) is derived. Similarly, the proof of (ii) can be shown.

Letir={Tl,..., ??k} be the set of all potential solutions over all possible changes of (y, X) for problem (9). The following lemma allows one to choose an entering variable for pivoting from a known potential solution Ti to an adjacent feasible (primal potential) or potential solution Tj.

LEMMA 4.2. Given a potential solution T, its nonbasic variable xij is an entering variable for

the next pivoting if there is a X E A(T) so that ui(X) + Vj(X) - Xtcij = 0.

Lemma 4.2 is a special case of the definition of the “effective constraint” in [11,18] for the MC2 linear programming problem. By integrating the above discussion, an MC2 algorithm for solving the MC2 transportation problems is proposed as follows.

20 Y. SHI

ALGORITHM 4.1

STEP 0. (Initialization Step) Set 7 o = 0. Find an initial basic feasible (primal potential)

solution by using the northwest corner rule (41. This means that select ~11, which corresponds

to the northwest comer (allocation cell) (1,l) of the MC2 transportation simplex tableau (see

Example 4.1), as the first basic variable. Either the p av~iability levels of supply 1 are exhausted

simultaneously or demand 1 is completely satisfied over the p requirement levels. Note that the sth

availability level of supply 1 is provided to only the sth requirement level of demand 1, s = 1, . . . ,p.

If the p availability levels of supply 1 are exhausted, then 521 will be processed; otherwise, if the

p requirement levels of demand 1 are satisfied, then 212 will be chosen. Therefore, if z:ij is the

last selected variable, next select zi+i,j if the p availability levels of supply i are exhausted.

Otherwise, next select si,j+l. In such a way, an initial basic feasible solution can be found. In

the case of the nondegenerate solution, the number of the basic variables with positive values

is exactly equal to m + n - 1. Let K be the initial basic feasible solution (basis). By using

Definition 2.4, find the primal parameter set I’(K) and the dual parameter set A(K). Calculate

ui(x) and uj(X) by setting its ur(X) = 0. Then, solve the set of equations ui(X) + vjfx) = Xtcij,

for every (i: j) such that xij is the basic variable (see (if of Theorem 4.1). Go to the steps of the

following iterations.

STEPS OF THE ith ITERATION:

STEP 1. By using (v) of Theorem 2.1, if a basis K is not a potential solution, then go to Step 2.

If a basis K is a potential solution, then set K = ‘T’, 7j = Ij-’ U {Tj}. For every (7, X), if there

is a 2’ of 7j such that (y, X) E I’(T) x A(T), then 7j = 7 and go to Step 7; otherwise, go to

Step 3.

STEP 2. When a basis K is not a potential solution, determine the entering basic variable: Select

the nonbasic variable zrij having the largest upper bo?md of ~i(x)+~j(x) --At qj > 0 with llix = 1,

for 0 < Xk =c 1, k: = 1,. . . , q, 1: = (1,. . . , 1) E Rp, by (ii} of Lemma 4.1.

STEP 3. When a basis K is a potential solution, select the nonbasic variable zaj which satisfies

Lemma 4.2 as the entering basic variable. If there are several Xij satisfying Lemma 4.2, choose

one arbitrarily.

STEP 4. Given a basis K, determine the leaving variable: Identify the loop formed by the

nonbasic cell (i, j) corresponding to the entering basic variable fcij, and the basic cells in the

MC2 transportation simplex tableau. Number the cells in this loop consecutively: starting with

the (i, j), set “+” on the odd-numbered cells, and “-” on the even-numbered cells. Among the

cells with “-“, select the basic variable having the smallest upper bound of the value by using (i)

of Lemma 4.1.

STEP 5. For an adjacent basis Q of K, determine the new basic variables: Add the values of the

leaving basic variable to the allocation cell with “-t-” in the loop, respectively, for the p supply or

demand levels. Subtract these values from the allocation cell with “-” in the loop, respectively,

for the p supply or demand levels. The result will be a new MC2 transportation simplex tableau

for the basis Q.

STEP 6. For the basis Q, first fmd the primal parameter set l?(Q) # 0 by using (i) of De&i-

tion 2.4. Second, calculate Us and v~(X) by selecting the row having the largest number of

allocations and setting its u{(X) = 0. Then, solve the set of equations ui(X) + vj(X) = Xtcij, for

all (i,j) such that xij is a basic variable (see (i) of Theorem 4.1). If ui(X) + vj(X) - Xtcij < 0,

for every (i, j) such that xij is a nonbasic variable (see (ii) of Theorem 4,1), then Q is a dual potential solution; that is, A(Q) # 0, according to (iv) of Theorem 2.1. Go back to Step 1 of the i + lth iteration.

STEP 7. The algorithm terminates since 7j is the set of all potential solutions over a11 possible

changes of (y, X) for problem (9).

A Transportation Model 21

The following example, adopted from Reinfeld and Vogel [5] is used to show how the above

algorithm can be employed to solve the MC2 transportation problems.

EXAMPLE 4.1. Consider a transportation problem involving three factories, denoted by Fi, F2,

and F3, and four warehouses, denoted by WI, Wz, Ws, and W,. A particular product is trans-

ported from the ith factory to the jth warehouse, i = 1,2,3, and j = 1,2,3,4. Both the supply

level and demand level vary with the economic situations. Suppose that if the economy is in

recession, the supply capacity level of FI is 30 tons, Fz is 40 tons, and F3 is 53 tons; while the

demand requirement level of WI is 22 tons, Wz is 35 tons, Ws is 25 tons, and Wd is 41 tons. If

the economy is booming, the supply capacity level of FI is 20 tons, F2 is 55 tons, and F3 is 60

tons; while the demand requirement level of WI is 30 tons, Wz is 30 tons, Ws is 22 tons, and W4 is 53 tons. Assume there are two criteria under consideration:

(i) the minimization of total energy cost consumed in transportation, and

(ii) the minimization of total product deterioration during transportation.

The energy cost for transporting one ton products and the deterioration percentage of one ton

products during transportation are given in Table 1 and Table 2, respectively. The goal is to

locate the set of all potential solutions for the above transportation problem.

Table 1. The energy cost consumed for transporting one ton products ($10 per unit cost).

Table 2. The deterioration percentage of one ton products during transportation.

Wl w2 w3 w4

Fl 1 23 1 27 1 16 1 18

F2 12 17 20 51

F3 22 28 12 32

Wl w2 w3 w4

FI 4 2 3 1

F2 2 3 4 2

F3 1 5 2 3

Note that in the above problem when the sth supply and demand level is under consideration,

s = 1,2, the total amount of products transported is equal to the total amount received. That

is, when the economy is in recession, 30 + 40 + 53 = 123 = 22 + 35 + 25 + 41; and when the

economy is booming, 20 + 55 + 60 = 135 = 30 + 30 + 22 + 53. By Theorem 3.1, the problem

has a feasible solution. Let y be the supply and demand parameter over the economic situations

such that yi + yz = 1 and yi > 0, i = 1,2, and X be the criteria parameter over two criteria

such that Xi + Xz = 1 and Xi > 0, i = 1,2. Let 7 be the set of all potential solutions over all

possible changes of (y, X) for the transportation problem. In the following, Algorithm 4.1 is used

to locate 7.

STEP 0. (Initialization Step) Set 7’ = 8. The initial MC2 transportation simplex tableau of the

problem is presented by Table 3. In the allocation cell (i,j) of Table 3, the figure in the left upper

small square is the coefficient of the first criterion, and the figure in the right upper small square

is the coefficient of the second criterion for the variable ~ij. The circled figures of Table 3 are

related allocations. For the northwest corner (allocation cell) of Table 3, (1, l), when the economy

is in recession, 22 tons are transported to WI from Fl; and when the economy is booming, 30 tons

are transported to WI from Fl. Since WI is satisfied, the allocation cell (1,2) proceeds. In (1,2), when the economy is in recession, 8 tons are transported to Wz from F,; and when the economy

is booming, -10 tons are transported to Wz from Fl, which means that Fl owes 10 tons to Wz

because in the cell (1,1) 10 more tons are transported from Fl to WI. Since Fl is exhausted, the

allocation cell (2,2) proceeds. In (2,2), when the economy is in recession, 27 tons are transported

to W2 from Fz; and when the economy is booming, 40 tons are transported to Wz from Fz. This

leads to proceed the allocation cell (2,3), because Wz is satisfied. In (2,3), when the economy is in

recession, 13 tons are transported to Ws from Fz; and when the economy is booming, 15 tons are

transported to Ws from Fz. Because FZ is exhausted, the allocation cell (3,3) proceeds. In (3,3),

22 Y. SHI

when the economy is in recession, 12 tons are transported to Ws from Fs; and when the economy

is booming, 7 tons are transported to Ws from Fs. Now, the allocation cell (3,4) proceeds,

because Ws is satisfied. In (3,4), when the economy is in recession, 41 tons are transported to

W, from Fs; and when the economy is booming, 53 tons are transported to Wd from F3. Let

Ki be the resulting initial feasible solution (basis). Then, z(Ki) = (xii,xi2, x22,223,233,234),

where the number of basic variable is 6 (= 3 + 4 - 1). The values of xij, in x(Ki), the primal

parameter set l?(K r and the dual parameter set h(Ki) are given in Table 4. Since the basic )

variables xij of Kr satisfy (i) of Theorem 4.1, the following are obtained:

For 211: Xtcii = 23x1 + 4x2 = ui(X) + q(x).

For x12: Xtci2 = 27x1 + 2X2 = ui(X) + ‘uz(X).

For 222: At&z = 17x1 -t 3x2 = Up + r&(x).

For x23: Xtc23 = 20x1 -l-4x2 = ZL~(X)+V~(X).

For 233: Xtcs3 = 12X1 + 2x2 = us(X) + ‘us(X).

For x34: Xt~34 = 32X1 + 3x2 = u3(x) + ?J~(X).

Set ui(X) = 0, then vi(X) = 23x1 + 4&,212(x) = 27x1 + 2X2,212(X) = -lOXi + X2,v3(X) =

3Oii1+3x2,~3(x)= -18X1 -x2, and ?~4(x)= 50X1+4X2.

Go to the steps of the first iteration.

STEPS OF THE FIRST ITERATION:

STEP 1. By Theorem 2.1, the basis Ki is not a potential solution since h(Ki) = 8. Go to Step 2.

STEP 2. Determine the entering basic variable corresponding to Ki based on results of Step 0:

By (ii) of Lemma 4.1, the corresponding nonbasic variable xij are checked and the upper

bounds of Us + uj(X) - XtCij 2 0 with Xi + X2 = 1 and 0 < Xi < 1, i = 1,2, are obtained as

follows:

For x13: ui(X) + us(X) - Xtcis = 14x1 + 2x2 I 20.08;

For 214: ui(X) +2)4(X) - Xtc 14 = 32x1 + 3x2 5 45.64; and

For x21: ?Q(X) f vi(x) - XtC21 = x1 -I- 3x2 5 4.49

(Us and Vj(X) are referred to Step 0).

Therefore, select xi4 as the entering variable since its upper bound 45.64 is the largest one.

Go to Step 4.

STEP 4. Determine the leaving variable corresponding to Ki based on Table 3.

Select a loop formed by the cells: (1,4), (1,2), (2,2), (2,3), (3,3), and (3,4), corresponding

to the variables x14,x12,222,x23,x33, and x34, respectively (see Table 3). Note that xi4 is the

entering variable and 212,x22, ~23, x33, and x34 are the basic variables in Table 3. Then, number

(1,4) as “+“, (1,2) as “-“, (2,2) as “+“, (2,3) as “-‘I, (3,3) as ‘I+“, and (3,4) as “-“. From

the cells with “-“, the upper bounds of the value of the basic variables are obtained by using

Lemma 4.1 (see Table 4).

x12 = 8yi - 10y2 5 18.19;

x23 = 1371 + 1572 5 28.19; and

x34 = 41 yi + 53y2 5 95.15.

Select 212 as the leaving variable since its upper bound 18.19 is the smallest one. Go to Step 5.

STEP 5. For an adjacent basis K2 of Kl, determine the new basic variables: Add 8yi - 1072,

the value of the leaving basic variable 512, to the allocations of (1,4)+, (2,2)+, (3,3)+ in the

A Transportation Model

Table 3. The initial MC2 transportation simplex tableau of K1.

b12 30 30 22 53 135

Table 4. The initial feasible (primal potential) solution of K1.

23

xc(K1) r(Kd A(K5) Value of Zij (Kl)

(111,212,x22,r23rr33,234) Q<XIl 0 xl1 = 2271 + 3O-y2

x12 = 871 - 1072

X22 = 2771 + 4O-y~

123 = 1371 + 1572

233 = 1271 + 772

X34 = 4171 + 5372

loop, respectively. Then, subtract 871 - 1072 from the allocations of (1,2)-, (2,3)-, and (3,4)-

in the loop, respectively. This results in the MC2 transportation simplex tableau for basis Ks.

STEP 6. For basis K2, 2(Ks) = (~11, z14,z22, x23, x33,234), the primal parameter set I’(K2) # 8,

and the dual parameter set A(K2) = 8 can be identified (see [28]). Since the basic variables qj

of Ks satisfy (i) of Theorem 4.1, the following are obtained:

For ~11: Aterr = 23x1 +4X2 = ur(x) + WI(~).

For 514: Xtcr4 = 18x1 + X2 = ur(X) +214(X).

For 222: Xtc22 = 17 X1 + 3x2 = uz(X) +212(X).

For ~23: Xtc23 = 20 x1 -I- 4x2 = ‘@(A) i- V3(x).

For x33: Xtc33 = 12 Xr + 2 AZ = 213(X) + 213(X).

For x34: Atcad = 32 XI + 3 X2 = us(X) + 2)4(X).

Setur(X) = O,thenvr(X) = 23Xr+4&,~4(X) = 18Xr+Xg,~a(X) = 14Xr+2X:!,‘us(X) = -2X1,

us(x) = 22x1 + 4x2, and 212(x) = -5x1 - x2.

Go to the steps of the second iteration. Through the second and third iterations, we can

identify that two adjacent bases {KS, K4). Here, Ks is a primal potential solution; while K4 is

a potential solution with x(K4) = (x~~,z~~,x~~,z~~,x~~,x~~), I’(K4) = (71 > 0 1 0 < yr 5 1,

24 Y. SHI

?'I+% = I), A(K4) = {AI > 0 1 Z/31 5 XI 51,X1 +X2 = 1). The objective payoff ofK4 is:

130 20

I Yl ( > 1 -Yl

5 25

X(K4)x(K4) = (Xi,1 -A,) > 35 30

17 5

25 22

\ll 33

which is a function of (71, Xl) E I’(K4) x A(K4). The Q(X) and q(X) are obtained:

For 214: Xtcr4 = 18x1 + X2 = ur(X) + Q(X).

For 221: Atc2r = 12 X1 + 2X2 = 9u2(X) + Q(X).

For 222: Xtc22 = 17x1 + 3x2 = 24x) + ~(4.

For 231: Xtcsr = 22 Xr + X2 = us(X) + Q(X).

For 233: Xtcss = 12 X1 + 2x2 = us(X) +213(X).

For x34: Xtcs4 = 32Ar + 3x2 = us(x) + v4(x).

Set u3(X) = 0 (corresponding to the row having the largest number of allocations in the simplex

tableau of K4), then 214(X) = 32x1 +3&,213(X) = 12x1 + 2Xz,wi(X) = 22x1 + Xs,ui(X) =

-14X1 - 2Xs,us(X) = -10X1 +X2, and W(X) = 27X1+2&.

For the sake of space, the details of the process are not elaborated here, but can be found

in [28]. In the following, we show the fourth and further iterations.

STEPS OF THE FOURTH ITERATION:

STEP 1. Since K4 is a potential solution, set K4 = Tl and 7’ = 7’ U {Ti} = {Tr}. Note that

there exists some (71, Xr) $ II’ x R(Tr). Go to Step 3.

STEP 3. From the simplex tableau of K4, check the nonbasic variable xij for ~i(X)+~j(X)-X~cij =

OwithXi+X~=landO<A~<l,i=1,2:

For xii: u,(X) + q(X) - Xtc ii = 15 xi + 5 x2 = 0; then xi = -a.

For 212: ui(X) + ,us(X) - Xtc 12 = 14x1 + 2x2 = 0; then Xi = -f.

For 213: ui(X) + us(X) - Xtc 13=18Xr+3Xs=O; thenAl=-;.

For ~23: u~(X)+U~(X) - Xtc 23 = 18x1 +X2 = 0; then Xr = -A.

For ~24: u~(X)+TJ~(X)- Xtc 2

24=29X1-2X2=0; thenXr=-. 31

For 232: t~3(~)f?.9(~)-~~C32 =x1 +3x2 = 0; then x1 = 5.

Thus, both ~24 and ~32 satisfy Lemma 4.2. Choose ~24 arbitrarily as entering variable. Go to Step 4.

STEP 4. Determine the leaving variable corresponding to K4 based on its simplex tableau: Select a loop formed by the cells: (2,4)+, (2,1)-, (3,1)+, and (3,4)-, corresponding to the

nonbasic variables ~24, and the basic variables ~21, 231, and x34, respectively, in the simplex

tableau of K4. From the cells with LL-“, the upper bounds of the value of the basic variables are obtained by using Lemma 4.1 [28]:

221 = 5 yi + 25 y2 I 36.20 and x34 = 11 y1 + 33 y2 I 49.39.

Select 221 as the leaving variable since its upper bound 36.20 is the smallest one. Go to Step 5.

A Transportation Model

Table 5. The MC2 Transportation Simplex Tableau of KS.

25

b12 I 30 1 30

20

55

60

135

Table 6. The Potential Solution of K5

W5) UK51 WK5)

O<YlIl 2

o<x1<- 31

Value of Xij (KS)

x14 = 3071 + 2072

222 = 35’y1 + 3o’j’z

524 = 5’71 + 2572

x31 = 2271 + 3072

x33 = 2571 +22-y~

x34 =6y1 +8-y2

STEP 5. For an adjacent basis KS of TI, determine the new basic variables: Add 5 y1 + 25 y2,

the value of the leaving basic variable 521, to the allocations of (2,4)+ and (3,1)+ in the loop,

respectively. Then, subtract 5 yr + 25 yz from the allocations of (2,1)- and (3,4)- in the loop,

respectively. The result is the MC2 transportation simplex tableau for the basis KS given in

Table 5.

STEP 6. For the basis KS, ~(K.F,) = ( 214, 522, ~24, 531, 233, x34), the primal parameter set

y(K5) # 0, and the dual parameter set h(K5) # 8 are identified as in Table 6. By (v) of

Theorem 2.1, KS is a potential solution since I’(K5) x A(K5) # 8. From Table 5, the objective

payoff of KS is

At C(K.5) +G) = (XI, 1 - XI) 18 17 51 22 12 32

1 3 2 1 2 3

i

30 20

35 30

5 25

22 30

25 22

6 8

Yl ( > l -Y1

26 Y. SHI

The pi and vj(X) are obtained:

For 214: Xtci4 = 18X1 + A:! = ul(A) +214(X).

For 222: Xtc22 = 17 Xi + 3 X2 = q?(X) + vz(X).

For $24: Xtc24 = 51 X1 + 2 X2 = 212(X) + 214(X).

For 231: Xtcsi = 22 Xi + X2 = u3(X) + q(X).

For 133: Xtcs3 = 12 Xi + 2 X2 = u3(X) + us(X).

For 234: Xtc34 = 32 Xi + 3 Xz = u3(X) + 214(X).

Set ug(X) = 0 ( corresponding to the row having the largest number of allocations in the

simplex tableau of K4), then vi(X) = 22x1 + 2Xz,va(X) = 12X1 +2X2, Q(X) = 32x1 + 3x2,

uz(x) = 19x1 -AZ, vz(X) = -2x1 +4x2, and ui(x) = -14x1 - 2x2.

Go to the steps of the fifth iteration.

STEPS OF THE FIFTH ITERATION:

STEP 1. Since Ks is a potential solution, set Ks = T2 and ‘T2 = 7’ U {Tz} = {Tl, T2}. For

every (n,xi), there is a Tj of ‘T2 such that (n,Xi) E l?(Tj) x h(Tj). Go to Step 7.

STEP 7. The process terminates since 72(= 7) is the set of all potential solutions over all

possible changes of (~1, Xi) for the given transportation problem.

5. CONCLUSIONS

A transportation model with multiple criteria and multiple constraint levels (MC2) has been

formulated within the framework of MC2 linear programming. An algorithm has also been de-

veloped to solve such MC2 transportation problems. In this algorithm, the traditional northwest

corner rule is adopted to find an initial basic feasible solution for a given MC2 transportation

problem. Then, the MC2-simplex method is applied to locate the set of all potential solutions

over possible changes of the criteria (objective) parameter and supply and demand parameter

for the MC2 transportation problem. There are a number of research problems that remain

unexplored.

From a computational point of view, it is of particular interest to develop a computer soft-

ware that can solve large-size MC2 transportation problems. Based on the idea of a computer

program for standard MC2 problems developed by Chien, Shi and Yu [29], Haase and Shi [30]

have developed a computer software called “MC2 Trasp” that can solve a median-size MC2 trans-

portation problem. The further development and improvement of this program towards solving

large-size MC2 transportation problems is an ongoing project.

The framework of the MC2 transportation problem described in this paper can be immediately

applied to many research areas, such as aggregate production planning in production/operations

management [31]. The aggregate production planning problem usually involves multiple conflict-

ing criteria, such as inventory cost versus shortage cost [32], and multiple resource availability

levels, such as fluctuations of capacity and demand levels. With the potential solutions of the MC2

aggregate production planning problem, one can study trade-offs between multiple criteria and

capacity and demand levels. Optimal polices that can cope with capacity and demand uncertainty

can be developed in the basis of these trade-offs.

Another interesting research problem deals with constructing contingency plans for each po-

tential solution of the MC2 transportation problem. Recall from Section 2 that, for a given

potential solution, if there is some y which does not belong to the primal parameter set for the

solution, then the solution becomes infeasible; and if there is some X which does not belong to the

dual parameter set for the solution, then the solution is not optimal. In either case, contingency

plans for the potential solution of the MC2 transportation problem need to be constructed to

A Transportation Model 27

recover the feasibility and maintain the optimality. All theoretical results of [33-391 can be read-

ily applied to construct various optimal contingency plans. If the supply and demand level is an

unknown variable, then the de no~o programming approaches [40,41] can be applied to calculate

an optimal supply and demand level from infinitely many possible choices. As a result, we can

design an optimal transportation problem with multiple criteria. These problems are currently

under investigation and the significant results will be reported in the near future.

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