a dynamic transportation model with multiple criteria and multiple constraint levels

PERGAMON Mathematical and Computer Modelling 32 (2000) 1193-1208

MATHEMATICAL

COMPUTER MODELLING

www.elsevier.nl/locate/mcm

A Dynamic Transportation Model with Multiple Criteria

and Multiple Constraint Levels

JUN LI Department of Business Management, Daqing Petroleum Institute

Anda, Heilongjiang 151460, P.R. China

YONG SHI* College of Information Science and Technology

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

yshi@unomaha. edu

(Received September 1999; accepted October 1999)

Abstract-h this paper, a dynamic transportation model with multiple criteria and multiple

constraint levels (DMC2) is formulated by using the framework of MC2 linear programming. An algorithm is developed to solve such DMC2 transportation problems. In this algorithm, dynamic

programming ideology is adopted to find the optimal subpolicies and optimal policy for a given DMC2 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 DMC2 transportation problem. A numerical example is illustrated to demonstrate the applicability of the algorithm in solving the DMC2 transportation problems. @ 2000

Elsevier Science Ltd. All rights reserved.

Keywords-MC2-simplex method, DMC2 transportation problems, Algorithm, Dynamic pro-

gramming.

1. INTRODUCTION

‘IIansportation problems, as an important branch of linear programming, have been extensively

studied by a number of scholars (see, e.g., [l-11]). These studies show the development of

transportation models clearly into three stages:

(1) classical transportation problem (it has only a single criterion (objective) and a single

supply and demand level);

(2) multiple criteria (MC) transportation problem;

(3) multiple criteria and multiple constraint levels (MC2) transportation problem.

Although transportation models have been applied to deal with the real-world problems, the

current studies including recent developments have some limitations in making a sequence of interrelated decisions. Suppose that a transportation problem can be divided into stages, with

a policy decision required at each stage. The solution procedure is designed to find an optimal

*Author to whom all correspondence should be addressed.

0895-7177/00/t - see front matter @ 2000 Elsevier Science Ltd. All rights reserved. Typeset by A+@-@? PII: SO8957177(00)00200-4

1194 J. Ll AND Y. SHI

policy for the overall problem, i.e., a prescription of the optimal subpolicy decision at each stage

for each of the possible states. Obviously, the known techniques fail to handle this case. To

overcome the shortcoming, in this paper we employ dynamic programming ideology to find the

optimal subpolicies and the optimal policy for MC* transportation problems.

The model of MC2 linear programming is supported by both of theory and practice. The

philosophy behind this perspective is that the availability of the resources can fluctuate de-

pending on the decision situation forces, such as the desirability levels believed by the different

decision makers and/or the changeable economic environments 1121. As a particular case, the

MC2 transportation model shares the advantages in solving the real-world problem. To extend

the MC2 transportation model, this paper proposes a dynamic MC2 (DMC2) transportation

model that has a flexibility of dealing with the situation of making a sequence of interrelated

decisions in multiple stages transportation problem.

The notation and definitions of the MC2 linear programming problem and the MC2-simplex

method are briefly sketched as a basis for developing the DMC2 transportation model in Section 2.

The mathematical formulation of DMC2 transportation model will be described in Section 3. In

Section 4, an algorithm is developed to solve such DMC2 transportation problems. In this

algorithm, the traditional Vogel’s rule is used to find an initial basic feasible solution of the

stage-o. 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.

After this the dynamic programming ideology is adopted to extend the process from, a stage to

the next. Repeating the last two steps, we can get the optimal policy for the overall problem.

A numerical example is used to illustrate the algorithm for solving the DMC2 transportation

problems. Section 5 is devoted to the conclusions.

2. MC2 LINEAR PROGRAMMING,

In

from

MC2

AND MC2 TRANSPORTATION MODELS

this section, the solution concepts and the notation of MC2 linear programming adopted

[lO,ll] are sketched to facilitate the discussion on the DMC2 transportation model. An

linear programming problem can be formulated as

minimize XtCX,

s.t. AX = Dy, (2.1)

X 20,

where C E Rqxn, A E Rmxn, and D E Rmxt are matrices of q x n, nx x n, and m x t dimensions,

respectively; X E Rn are decision variables, X E Rq is called the criteria parameter, and y E Rt is

called the constraint level parameter. Both (X, y) are assumed,to be nonnegative and unknown,

REMARK 2.1. It is immediately obvious that the MC2 model is a symmetric extension of the

MC when the constraint level parameter y is known. In fact, for a particular value of y, the

MC2 problem reduces to the MC problem. Then, an efficient solution of (MC,yO) is also an

efficient solution of MC2 at y”.

To find all the potential solutions for problem (2.1), we need to identify the corresponding set

of potential bases for (2.1). In the following discussion, we assume that both parameters (X, y) are normalized; i.e., x E Rq with & > 0 and c xk = 1; y E Rt with yk ‘> 0 and c yk = 1.

For a given basis J with its basic variables X(J), we can define the associated basic matrix BJ as the submatrix of A in (2.1) with the columns index of J (i.e., column j of A is in BJ if and

only if j E J), and the associated objective function coefficients Cn as the submatrix of C with

columns index of J. Let X( J*) be the nonbasic variable corresponding to given X(J). Then,

we rearrange the index, if necessary, and decompose A into [BJ, N], where N is the submatrix

of A associated with X( J*); and C into [Cs, CN], where CN is the ‘submatrix of C associated

Dynamic Transportation Model 1195

with X(J*). The initial simplex tableau of (2.1) can be shown in Table 1. Table 1 can be converted by row linear transformation as Table 2, where I, is an m x m identity matrix. By

dropping (A, 7) from Table 2, we can obtain an MC2-simplex tableau with a basis BJ as Table 3.

Table 1.

Table 2.

Table 3.

DEFINITION 2.1. Given a basis J for problem (2.1), define its corresponding

(1) primal parameter set by l?(J) = {y > 0;. B;lDy 2 0); and (2) dual parameter set by A(J) = {A > 0; At[C~ - CBBJ~~V] 2 0).

THEOREM 2.1. Given a basis J for problem (2.1),

(1) the resulting feasible solution X(J, 7) = B;‘Dy 2 0 if and only if y E K’(J); (2) the solution X(J,y) is optimal ifand only ify E r(J) and X1 E A(J).

DEFINITION 2.2. Given a basis J for problem (2.1),

(1) J is said to be a primal potential basis if l?(J) # 0; (2) J is said to be a dual potential basis if A(J) # 0; (3) J is said to be a potential basis if l?(J) x A(J) # 0.

By using the MC2-simplex method discussed above, one can locate a set of potential solutions. Two computer software packages for solving problem (2.1) were developed by Hao and Shi [13] and Chien et al. [14].

Based on the MC2 problem (2.1), Shi [ll] proposed an MC2 transportation problem as follows:

maxz = [XI, X2,. . . ,A,] (2.2)

I 5 xij = [%17%27.. .7Q] [Yl,Y2,. . . ,ytlT

j=l

s.t. Elxij = [bjl,bj2r...,bjt][71,72,...,7~]T

ai, > 0, bjr 2 0, for all T, i, and j,

Xij 2 01 for all i, j,

(2.2)(cont)

1196 J. LI AND Y. SHI

where

xij is the amount of products transported from source i to destination j;

ai, is the sth supply availability level at i, s = 1,. . . , t;

bj, is the sth demand availability level at j, s = 1,. . . , t;

c$) is the “cost” of transporting a unit product from source i to destination j when the

rth criterion is under consideration, r = 1,. . . , q.

Shi [ll] also developed an algorithm for solving the MC2 transportation problem. Since the

algorithm is derived within the framework of MC2-simplex method, the details of the process are

not elaborated here. Hasse and Shi [15] and Huang and Shi [16] implemented this algorithm as

computer software with different approaches for locating the set of potential solutions.

3. THE DMC2 TRANSPORTATION MODEL

In this section, we describe a DMC2 transportation model to coordinate the decisions associated

with different stages. Based on the MC2 problem (2.1), DMC2 transportation problem can be

formulated as

maxt=[Xr,X2,...,X,] ,

c c xijk = [f&l, %2, *. .Y ait1 [x7 729 * . * 7 -YtlT k=l j=l

s.t. k~~~~~xijk=[bjllbj2,...,bj~][yl,T21...rYl]T

air 2 0, bj, 2 0, for all r, i, and j,

where

I xijk 2 0, for all i, j, and k,

xijk is the amount of products transported from source i to destination j at the kth stage;

ai, is the sth supply availability level at i, s = 1,. . . , t;

bj, is the sth demand availability level at j, s = 1,. . . , t;

CL;: is the “cost” of transporting a unit product from source i to destination j at the kth stage

when the rth criterion is under consideration, r = 1,. . . , q.

THEOREM 3.1. Given y” > 0, the DMC2 transportation problem (3.1) has a feasible solution

aty”ifC~n=lai,=C~=n=lbjs,foranys=l,...,t.

THEOREM 3.2. Given a primal potential basis J for problem (3.1), the basic variables X(J)

contain at most m + n - 1 components xijk.

The proofs of Theorems 3.1 and 3.2 can be shown by applying t,he results of Theorems 3.1

and 3.2 of [ll].

DEFINITION 3.1. The solving procedure involves moving backward stage by stage. The subpro-

cess of problem (3.1) that contains the first s stages is called s-DMC2 problem. AJI optimal

sequence of interrelated decisions associated with the s-DMC2 problem is called an optimal sub- policy ifs < p, and an optimal policy overall problem (3.1) ifs = p.

DEFINITION 3.2. For s-DMC2 problem, cell (i,j) is a nonbasic cell if any kijk (k = 1,. . . , s) is

nonbasic variable; otherwise, cell (i, j) is a basic cell.

Dynamic Transportation Model 1197

DEFINITION 3.3. As boundary c;dy; we assume all the cost parameters at stage-0 are the

same with at stage-l. It means c&, - ciil.

4. AN ALGORITHM FOR THE DMC2 TRANSPORTATION MODEL

Based on the discussion in Sections 2 and 3, this section develops an algorithm for finding an

optimal overall policy for problem (3.1).

DEFINITION 4.1. ai.jle = Atcijr - (ui(X) +Vj(A)) is called the improvement dii$erence of nonbasic

cell (i,j) at the kth stage, where Qj, = min{Xtcijl; 1= 1,. . . , k}.

DEFINITION 4.2. R(T) is called the premise condition set about (X, y) that makes the basis T

exist. Because ai, and bj, are nonnegative for ali r, i, j in DMC2 transportation problem, y E

(0,l) keep the same all the time. Setting Q(T) = (0 < A < 1) when T is an initial basis of

stage-O, otherwise setting R(T) = A(T).

Let s = 0. It means there is only one stage (stage-O) under consideration. A DMC2 trans-

portation problem is reduced to an MC2 transportation problem in this case. Set q” = 0, co = 0.

Find an initial basic solution by using the Vogel’s rule or the other rules that are available.

Q(T) = (0 < X < 1) at this time. Either the t availability levels of supply i are exhausted

simultaneously or demand j is completely satisfied over the t requirement levels if “ijo is a basic

variable (see Example 4.1). The number of the basic variables should be exactly equal to m+n- 1.

Let K be the initial basic solution (basis). By using Definition 2.1, find the primal parameter

set I’(K) and the dual parameter set A(K). Let ui(X) be the simplex multiplier corresponding

to supply i, and vj(X) be the simplex multiplier corresponding to demand j (17-19). Apply (2)

of Definition 2.1; the dual parameter set A(B) of a basis B for problem (3.1) can be identified

by the following theorem.

THEOREM 4.1. A basis B is a dual potential basis for s-DMC2 problem if

(1) Ui(A) + Vj(X) = XtCijw(W 5 S), f or a11 (i, j) such that cell (i, j) is a basic cell and xijw is a

basic variable; and

(2) Ui(X) + ZJj(X) 5 XtCijr(r 5 S), f or all (i, j) such that cell (i, j) is a nonbasic cell and

cijr = min{Xtcijk; k = 1,. . . , s}.

Note that (1) of Theorem 4.1 provides a way to calculate ui(x) and vj(X) corresponding to a

basis B for problem (3.1), while (2) of Theorem 4.1, which is used to find the dual parameter

set A(B), yields a potentiality test for the basis B if the primal parameter set r(B) # 0. The

following lemma is adopted from [ 111.

LEMMA 4.1.

t

c (4.1) s=l

where d, is ai, or bj, for all (i, j) and 7 = [rl, 72, . . . , rtlT is the supply and demand parameter;

(4.2)

for all (i, j) and X = [Xl, X2,. . . , X,IT is the objective parameter.

1198 J. LI ANDY. SHI

LEMMA 4.2. Given a potential solution B (at stage k-l), its nonbasic variable xijr is an ellterillg

variable for the next pivoting (at stage k) if there is a A E A(B) so that ui(x) + ‘uj(X) = Xtcij,,

where cij,. = min{Xtcijl; 1 = 1,. . . , k}.

Calculate ui(A) and vj(X) setting ui(X1) = 0 arbitrarily. Then solve the set of equations (1) of

Theorem 4.1. Set, s = 1 and go to the steps of the following iterations.

STEP 1. From an initial basic solution 2’ that was been built in the last stage, check the basic

cell (i, j) for Xtcijs = Atcijw with CXi = 1 and 0 < Xi < ,l, i = 1,. . . ,q, where cijzu is

described in (1) of Theorem 4.1. There are (m + n - 1) sets can be gotten about X. Take

all the sets satisfying SZ(T) as the interface to divide R(T) and R(T) can be divided at most

into 2m+n-1 subsets. The purpose for doing this step is to make sure that Xtcijs 5 Xtcijw or

Xtcijs > Xtcijlo in any subset of R(T).

STEP 2. Check the basic cell (i,j) in a subset of Q(T). Move the allocations to the stage-s

for the basic cells Xtcijs 2 Xtcijw and keep the allocations in the stage-w for the basic cells

XtCijs > XtCijw. The goal for doing Step 2 is to obtain an initial basic solution of the s-DMC2

transportation problem (see Definition 3.1). One subset of Q(T) corresponds to an initial basic

solution of the s-DMC 2 problem. Choose an initial basic solution of th&’ s-DMC 2 problem; the

course goes to the next step. .

STEP 3. By using (3) of Definition 2.2, if I< is not a potential solution, go to Step 4, otherwise

set K = Bj and r$ = 77j-l U {Bj}. For every given (X,7) in the subset of fit(T), if there is a B

of $ such that (X,7) E l?(B) x A(B), then $ = 17 and go to Step 9; otherwise go to Step 5,

where (X, y) is the parameter space associated with the initial basic solution. . .

STEP 4. Determine the entering basic variable: select the nonbasic variable zcijr associated with

nonbasic cell (i,j) having the largest upper bound of ui(x) +vj(X) > Atcijr by (2) of Lemma 4.1,

where Cijr = IIliIl{XtCijl; 1 = 1,. . . , k} and if T is not unique, we select the largest one. Go to

Step 6.

STEP 5. Select the nonbasic variable xijr that satisfies Lemma 4.2 as the entering basic variable.

If xijr is not unique, choose one arbitrarily.

STEP 6. Determine the leaving variable: identify the loop formed by the nonbasic cell (i,j)

corresponding to the entering basic variable xijr, and the basic cells in the s-DMC 2 transportation

simplex tableau. Number the cells in this loop consecutively: starting with cell (i, j), set ‘I+” 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 (1) of Lemma 4.1.

STEP 7. For an adjacent basis Q of I<, determine the new basic variables: add the values of the

leaving basic variable to the entering basic variable xijr and all the other basic variables allocated

in the cell with “+” in the loop, respectively, for the t supply or demand levels. Subtract these

values from the leaving basic variable and all the other basic variables allocated in the cell

with “-” in the loop, respectively, for the t supply or demand levels. The result will be a new

simplex tableau for the basis Q.

STEP 8. For the basis Q, first find the primal parameter set I’(Q) # 0 by .using (1) of Defini-

tion 2.1. Second, calculate ui(X) and uj(X) by setting ui(X) = 0. Then, solve the equality set (1)

of Theorem 4.1. If the inequality set (2) of Theorem 4.1 is tenable, then Q is a dual potential

solution, that is, A(Q) # 0, according to (3) of Definition 2.2. Otherwise A(Q) # 0. Go to Step 3.

STEP 9. Set ci = ei-‘Uq. If there is a B of Ei such that (X,7) E I’(B) xR(B) for every 0 < A < 1

and O < y < 1, then go to Step 10. Otherwise, if the initial basis that in the course has been

finished, there is a B of ci such that (X,7) E I’(B) x A(B) for every given (X,7) in Q(T), go

to Step 1; if the initial basis has not been finished, go to Step 2 from another subset of o(T)

entering the next iteration.

STEP 10. If s < p, let s = s + 1 and go to Step 1, if s = p, the whole procedure terminates.

Dynamic Transportation Model 1199

The following simple example is used to show how the above algorithm can be employed to solve the DMC 2 transportation problems.

EXAMPLE 4.1. Consider a transportation problem involving three suppliers, denoted by PI, Pz,

P3, and three demanders, denoted by D1, Dz, and Ds. A particular product is transported from the ith supplier to the j th demander, i = 1,2,3 and j = 1,2,3. Both the supply level and demand level change with the economic situations. Suppose that if the economy is in recession, the supply capacity level of PI is seven units, Pz is five units, and P3 is eight units; while the demand requirement level of D1 is six units, DZ is ten units, and Ds is four units. If the economy is booming, the supply capacity level of PI is eight units, Ps is ten units, and P3 is 12 units; while the demand requirement level of D1 is ten units, D-J is 15 units, and 0s is five units. Assume there are two criteria under consideration:

(1) the minimization of total transportation expenses (C(r)) consumed in transportation, and (2) the minimization of total product deterioration (Cc2)) during transportation.

Assume the transportation can take place in every month of the quarter. The transportation expense and the deterioration for transportation one unit products at every month are given in Table 4, respectively. The goal is to locate the set of all potential solutions for the above

three-stage transportation problem.

Table 4.

Note that in the above problem, the total amount of products transported is equal to the total amount received in the both levels. 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 + 72 = 1 and yr,yz > 0. Let X be the criteria parameter over the two criteria, such that Ar + Xz = 1 and X1, A2 > 0. Let q be the set of all potential solutions overall possible changes of (X, 7) for the three stages transportation problem. In the following, the algorithm discussing above is used to locate q. Set 7’ = 0.

First, let us just consider stage-O of the DMC 2 transportation problem. Using the Vogel’s rule according to Definition 3.3, we can obtain the initial basis TO of stage-0 (see Table 5), such that

R(Te) = (0 < Xl < 1) ( see Definition 4.2). In the allocation cell (i, j) of Table 5, the figure in the left upper corner is the related allocations when the economy is in recession and the figure in

Table 5.

D1 DZ D3

6(O) l(O)

Pl (L-2)

10(O) -‘w)

5w 9 (533) (497)

10(O)

5(O) 3(O)

p3 (-W)

5(O) 7(O)

1200 J. LI AND -f. SHI

the right lower corner is the related allocations when the economy is in booming. For example,

the allocations in cell (1,l) of Table 5 are 6yr + 1Oyz units and the zeroes that follow 6 and 10

in brackets express the allocations take place in stage-o. The figures in middle brackets are the

improvement differences of nonbasic cells.

For nonbasic cell (i,j), the improvement differences at stage-0 is (see Definition 4.i)

cijo = Xtcijo - (w(X) -I- Wj(X)) .

Since the basic variables Zijo of TO satisfy (1) of Theorem 4.1, the following equalities are

obtained:

for 1~110: 3X1 + 6X2 = Q(X) + vi(x);

for 21313: 2x1 + 4x2 = Q(x) + ‘us(x);

for ~220: 4x1 + 8X2 = uz(X) + 212(X);

for X320: 12x1 + 11X2 = U3(X) + U2(X);

for ~330: lOXi f 8A2 = 113(A) + v3(X).

Set z&(x) = 0, then VI(~) = 3X1+6X2, v3(x) = 2x1 $4ii2, ‘u3(x) = 8X1+ 4x2, 212(x) = 4x1 + 7x2,

and Q(X) = X2. For example, the number (5,3) in cell (2,l) of Table 5 represents ozlo = 5X1+3X2.

Set s = 1 and go to the steps of the first iteration.

Steps of the Iteration in Stage-l

STEP 1. From the initial basic solution TO that was been built in stage-O, check the basic cells for

Xtcijr = Xtcijs with Xz + X2 = 1 and 0 < Xi < 1, i = 1,2. We can get the same set (0 < Xr < 1)

as the boundary points to divide s2(To) because cijr = cijs (Definition 3.3). Thus, R(To) keep

the same.

STEP 2. Check the basic cell (i,j) in s2(Ta). Because Xtcijr = Xtcijo for every (i,j), move the

allocations to stage-l for all the basic cells. Now we get an initial basic solution in stage-l (see

Table 6). For nonbasic cells a~1 = Xtcij,. - (ui(X) + uj (A)), w h ere r = 0 or 1, there is (~ijr = aija

for all nonbasic cells because cijo = cijr for all i, j. Let Ki be the resulting initial basic solution,

then

x(Kl) = (~111,~131,~221,~321,~331)~

Set X(Kr) 2 0, obviously the primal parameter set l’(Kr) = {2/3 < yr < 1). Set gijr > 0 for

all i,j, then the dual parameter set h(Kr) = 8. Go to the next step.

STEP 3. By (3) of Definition 2.2, the basis Kr is not a potential solution since R(Kr) = 8. Go

to Step 4.

Table 6.

D1 D2 D3

A 60) l(l)

-_ (L-2) +

10(l) -2(l)

5(l)

p2 (593) (497)

10(l)

5(l) 30)

p3 (-w%+

- -

5(l) 7(l)

Dynamic Transportation Model 1201

STEP 4. Determine the entering basic variable corresponding to Kr . By (2) of Lemma 4.1, the

upper bounds for crijr < 0 can be obtained as ~11 = -2Ar + -AZ 5 ,/2((-2)2 + 02) M 2.83.

Therefore, select 5311 as the entering variable since its upper bound 2.83 is the largest one (in

fact, there is only 0311 satisfying flijr < 0). Go to Step 6.

STEP 6. Determine the leaving variable corresponding to Kr. Select a loop formed by the cells:

(3,1), (3,3), (193)Y and (l,l), corresponding to the variables 2311, 1~331, 2131, 2111, respectively

(see Table 6). Then cell (3,1),(1,3) are marked with “-I-” and (3,3),(1,1) with “-“. From the

cells with “--‘I, the upper bounds of value of the basic variables are obtained by using (1) of

Lemma 4.1 (see Table 6)

~111 = 671 + 1072 5 d@%i@ M 16.49,

2331 = 371 + 772 < JqKq M 10.77.

Select 2331 as the leaving variable since its upper bound 10.77 is the smallest. Continue to Step 7.

STEP 7. For an adjacent basis K2, determine the new basic variables: add 3yr + 7~2, the value of

the leaving basic variable 2331, to the allocations of (3,l) and (1,3) in the loop, and then subtract

3yr + 772 from the allocations of (3,3) and (1,l) in the loop. This result is shown in Table 7.

Table 7.

Dl D2 D3

30) 40)

Pl (-L--2)

30) 5(l)

5(l)

p2 (793) (6,7)

10(l)

30) 5(l)

p3 (290)

STEP 8. The primal parameter set I’(Kz) # 0 and the dual parameter set A(K2) = 0 can be

identified by (1) of Definition 2.1 (see Table 7). Go to Step 3.

Through the second iteration, we can identify the adjacent base Ks. Here Ks is a potential

solution because I?(&) # S,A(K3) # 0 can be identified by using (1) and (2) of Definition 2.1.

Same as ahead, obviously l?(Ks) = (0 < 71 < 1) and A(&) = {2/3 5 X1 < 1) (see Table 8).

Set Ks = B1 and q1 = nOU{Br} = {Bl}. Th ere is only Br in $ and for R(Ts) = (0 < Xr < l},

there exists some (XI, yr) 4 II’ x A(Bl), so go to Step 5.

Table 8.

Dl D2 03

30) 4(l)

4 (1,2)

3(l) 5(l)

5(l)

9 (7>3) (575)

100)

60) 2(l)

p3 (L-2)

100) 2(l)

1202 J. LI AND Y. SHI

Table 9.

Dl D2 D3

50) 2(l)

9 cm

5(l) 3(l)

5(l)

p2 (VI (5,5)

10(l)

60) 2(l)

p3 (-w

1W) 2(l)

STEP 5. From the simplex tableau (Table 8) of Ks, check the nonbasic variable z.,~~ for

l&(X) + Vj(X) - PCijl = 0.

Obviously, only 2331 satisfy the above condition, so choose 2331 as entering variable. Go to the

next step. Through Steps 6-8, we can obtain Table 9.

We can identify the basis K4 as a potential solution with l?(K4) = (0 < yr < l} and h(K4) =

(0 < Xi < 2/3} ( see Table 9). Set K4 = B-J and v2 = q1 U {Bz} = {Bi,&}. For every (Xr,-yi)

in R(Te) = (0 < Xi < l}, there is a Bj of v2 such that (Xi,yi) E l?(Bj) 5 A(Bj). Set q2 = ~1

and go to Step 9.

STEP 9. Set I’ = co U r]i = qi. For every 0 < Xi < 1 and Oyi < 1, there is a Bj of E1 such that

(Xi,yi E l?(Bj) x A(Bj), go to Step 10.

STEP 10. Because s = 1 < 3, set s = s + 1 = 2 and go to the steps of the iteration in stage-2.

For the sake of space, the details of the process in stage-2 are not elaborated here. The results

of stage-2 are shown from Tables 10-15.

Steps of the Iteration in Stage-3

STEP 1. Let T = L1 in Table 10, so Q(T) = R(L1) = (0 < X1 5 l/2}. Check basic cell (i,j) in

Table 10 for Pcija = Xtcijw with Xi +X2 = 1 and Xi, X2 E (0, l), where ti is 1 or 2. For example,

it is cizw = (c$$, cK$, but cgiw = ( &, &‘,)Y

for 2122: 6x1 + 8X2 = 4x1 + 3x2; then xi = g,

for 2212: 6x1 + 8x2 = 4x1 + 6X2; then no solution for Xi,

for 2311: 12x1 + 10X2 = 9x1 + 10X2; then Xr = 0,

for ~322: 8X1+9X2 = lOXi + 8X2; then Xi = i,

for 2332: 4x1 i- 7x2 = 6x1 i- 8x2; then xi = -1.

We can only obtain Xi = l/3 (cell (3,2)) satisfy Q(T) = (0 < A1 5 l/2}, so we divide 0(T)

into R(T*) and R(T**), where

STEP 2. Check the basic cell (i,j) of Table 10 in R(T*). Through the step, we can obtain

Table 16. In Table 16, the allocations that are followed by (‘(l)“, “(2)“) ‘or ,‘(3)” express the

transportation take place in stage-l, stage-2, or stage-3. Table 16 gives a potential solution Ki

and it can be expressed as: transport 7yi + 872 units from PI to Ds at stage-2, 5yi + 10y2 units

Dynamic Transportation Model 1203

Table 10.

D1 DZ D3

7(2)

4 (091) c&l) W’)

5w 4 (091) (3,l)

10(Z)

l(l) 3(2) 40) p3

00) 7(2) 5(2)

Here r(Ll) = (0 < ~1 < 1) and A(Lq) = (0 < Xl 5 l/2).

Table 11.

Dl D2

7(2) A (2,-l)

8(2)

5(2) 9 (--WI

100)

l(2) 30)

p3

w4 70)

Here r(Lz) = (0 < y1 < I} and A(&) = {l/2 5 X1 5 3/5}.

D3

(-1,2)

(lv3)

4(2)

5(2)

from PZ to DI at stage-2, y1 units from PS to DI at stage-l, 3yl+7y2 units from P3 to D2 at stage-

2, 4yl+5yz units from P3 to 03 at stage-3. The other tables give the rest potential solutions have

the same meaning, where II’ = (0 < y1 < 1) and h(K1) = (0 < X1 5 l/3}. The improvement

difference aijs must be paid more attention, where cijr = min{Xtcijl; 1 = 1,2,3} and ui(X), Vj(A)

must be connected with the stage that basic variable takes place (see Definition 4.1). For instance,

(where A(K1) = (0 < XI 2 l/3}), c22r = min{(4,8), (5,5), (8,lO)) = (5,5) = ~222,

for X122: AtC122 = 4x1 + 3x2 = Ui(x) + Q(x),

for 1212: Xtc212 = 4x1 + 6x2 = uz(X) + VI(X),

for 2311: Xtc311 = 9x1 +10X2 = u3(X)+z~(X),

for X322: XtC322 = 10X1 + 8A2 = 213(X)+ V2(X),

for 2333: XtC333 = 4x1 +7x2 = U3(x)+2)3(x).

Set u,(x) = 0, then ‘up(x) = 4x1 + 3x2, 213(x) = 6X1 + 5x2, VI(~) = 3A1 + 5x2, ‘Q(/\) = x1 + x2,

and v3(X) = -2X1 + 2X2, SO 0223 = (5,5) - [(l, 1) + (4,3)] = (0,l).

STEP 3. Set KI = B1 and $ = q”U {Bl} = {I$}. F or every given (X, y) in Q(T*) = (0 < X1 _<

l/3}, then is a B1 of q1 such that (Xl,y~) E l?(Bl) x A(B1). Then set q1 = 71, and go to Step 9.

STEP 9. Set c1 = to u ~1 = {BI}. There is only B1 in c1 and there exists some (Al,~l) $!

I’(&) x A(&) for 0 < X1 < 1 and 0 > y1 < 1. For a given (X, y) in M(L1) = (0 < X1 < l/2},

there exists some (Xl, yl) #, I’(B1) x A(B1) either. Thus, the initial T = L1 in Table 10 has not

been finished, so go to Step 2.

STEP 2. Check the basic cell (i,j) of Table 10 in R(T**). Through the step we can obtain

Table 17. Table 17 gives a potential solution Kz. I’(K2) = (0 < y1 < l}, A(K2) = {l/3 I X1 5

l/2)*

1204 J. h AND Y. SHI

Table 12.

Dl D2

7(2)

Pl (092)

B(2)

2(2) 3(2)

9

3(2) 7(2)

4(2)

p3 (2,-3)

7(2)

Here r(L3) = (0 < yl < 1) and h(h) = {3/5 5 X1 5 2/3}.

D3

(0>4)

(L3)

4(2)

5(2)

Table 13.

DI D2

7(2)

Pl uY4

g(2)

2(2) 3(2)

p2

3(2) 7(2)

4(2)

p3 (2,-3)

7(2)

Here r(L4) = (0 < y1 < 1) and A(L4) = {2/3 < X1 < 3/4}.

03

(0>4)

(L3)

4(2)

5(2)

Table 14.

DI D2

7(2)

4 (-G)

g(2)

2(2) 3(l)

p2

3(2) 7(l)

4(2)

p3 (3,--C)

7(2)

Here I = (0 < 71 < 1) and A(L5) = {3/4 < XI I 5/6}.

D3

(-L7)

(L3)

4(2)

5(2)

Table 15.

DI D2 03

2(l) 5(2)

PI (092)

30) 5(2)

5(l)

9 (L-5) (2,-Z)

10(l)

4(2) 4(2)

p3 (Z-1)

7(2) 5(2)

Here r(h) = (0 < YI < 1) and A(h) = {S/6 5 Xl < 1).

Dynamic Transportation Model

Table 16.

1205

9

4

p3

Dl

(04

5(2)

10(Z)

10)

O(l)

D2 I 03

7(2)

(4,2) S(2)

W) (5,2)

3(2) 4(3)

7(2) 5(3)

Table 17.

Dl D2 D3

7(2)

fS (-2,2) (2,3)

8(2)

5(2)

p2 (LO) (5,2)

lO(2)

l(l) 3(3) 4(3)

s

VI 7(3) 5(3)

Table 18.

Dl D2 D3

7(2) Pl W) (2,3)

8(2)

5(2)

9 052) (3-4)

lO(2)

l(2) 3(3) 4(3) 4

WI 7(3) 5(3) _

STEP 3. Set Kz = B2 and q2 = q1 U {Bz} = {BI, Bz}. F or every given (A, -y) in R(T** = {l/3 <

XI < l/2}, there is a basis BZ of v2 such that (X1,71) E I'(Bz) x R(B2). Set q2 = 772 and go to

Step 9.

STEP 9. Set E2 = E1 U~Z = 71 UQZ = {BI, Bz}, for 0 < A1 < 1 and 0 < 71 < 1, there exists some

(X1,71) $ V(Bl)u W32)) x {WV uW2)). F or all (X,7) in R(L1) = (0 < X1 5 l/2}, there

is a basis Bj of q2 such that (X1,71) E I'(Bj) x h(Bj), SO go to Step 1.

Let T = L2 in Table 11, we can obtain Table 18. It gives a potential solution KS, where

r(K3) = (0 < y1 < 1) and A(K3) = {l/2 I X1 < 3/5}.

From the initial solution in Tables 12-14, we can obtain the potential solutions the same as the

one in Table 18. Set c6 = E5 U qs = { BI, . . . , Bs} = (0 < Xl I 5/6}. Table 18 gives a potential solution when {l/2 5 X1 5 5/6}.

Let T = Lc~ in Table 15, we can obtain Table 19. It gives a potential solution K4. Here

l?(Kd) = (0 < 71 < l}, and A(K1) = {5/6 I X1 < 1). Set 1(4 = B, and q7 = q6 U {B7} =

{Bl,.. . , BT}. For every given (A, 7) in R(T) = {5/6 5 X1 < l}, there exists B7 of q7 such that

(X1,71) E I'(B7) x A(B7). Set v7 = r]7 and go to Step 9.

1206 J. h AND Y. SHI

Table 19.

DI D2 D3

2(l) 50)

Pl (2,3)

3(l) 5w

5(l)

p2 (h-5) (4,-l)

10(l)

4v.1 4(3)

p3 uw)

7w 5(3)

ocn, <l

04.,12/3

Table 9

I

21354 <l

Table 8

0-q 1112 1/21A, 1213 2/3d, 1314 3/4s& cl

Table 10

I-

I Table 11 I

l.--zzlmII 1 III I

Figure 1.

STEP 9. Set E7 = [“U?p = {BI,... ,&}, for 0 < Xr < 1 and 0 > yl < 1, there is a basis

of t7 such that (Xr,yi) E I’(Bj) x h(Bj). Go to Step 10.

STEP 10. The whole procedure terminates because s = 3 = p.

The potential solutions with respect to X in every stage can

5. CONCLUSIONS

be shown in Figure 1.

A transportation model with multiple stages, multiple criteria, and multiple constraint lev-

els (DMC2) has been formulated within the framework of MC2 linear programming. An al-

gorithm has been developed to solve such DMC2 transportation problems. In this algorithm,

Dynamic Transportation Model 1207

Vogel’s rule is adopted to find an initial basic solution for initial stage of the DMC2 problem.

Then, the MC2-simplex method is used to locate the set of all potential solutions overall possible

changes of the parameters in every stage. The significance of this algorithm is that the solving

process is a sequence of iteration. Thus it is possible to develop a software package that can solve

large-size DMC 2 transportation problems, based on [15,16]. At this moment, our computational

experience with 20 problems in different stage,order reveals that the algorithm is effective. One

more advantage for the algorithm is that when obtaining the optimal policy overall problem, one

will get all the optimal subpolicies over any of the subprocedure at the same time.

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

tential solution of the DMC2 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 A 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 DMC2 transportation problem need to be constructed

to recover the feasibility and maintain the optimality. All theoretical result of [21-231 can be

readily constructed to various optimal contingency plans. If the supply and demand level is an

unknown variable, then the de nova programming approaches (24,251 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 studying and the results will be reported in the future.

Transportation problem as an important branch of linear programming is of a broadly practical

field. MC transportation model was proposed to deal with the problem in multiple criteria

spaces. MC2 transportation model was formulated to handle the problem involving fluctuations

of supply and demand in multiple criteria spaces. The practice tells us every step of development

in transportation problem has advanced its usage to deal with the world problems. Since this

DMC2 model is developed based on MC2 model and dynamic programming ideology, it will

certainly be used the real-world problem solving.

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