THREE-DIMENS IONAL TRANSPORTATI ON PROBLEM WITH CAPACITY RESTRICTION *

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Section I deals with the preliminaries. In Section II, a simplex-type technique is developed, along with computational tests for optimality.

1 NZOR volume 9 number 1 January 1981 THREE-DIMENS IONAL TRANSPORTATI ON PROBLEM WITH CAPACITY RESTRICTION * S. MISRA AND C. DAS DEPARTMENT OF MATHEMATICS, REGIONAL ENGINEERING COLLEGE, ROURKELA , INDIA SUMMARY This paper deals with the three dimensional transportation problem with capacity restrictions. Section I deals with the preliminaries. In Section II, a simplex-type technique is developed, along with computational tests for optimality. Section III gives a realistic example of the application of the present model, with a numerical example presented in Section IV. 1. INTRODUCTION The solid (three-dimensional) transportation problem was first introduced and solved by Haley [5,6,7] and various authors [2,3,8,9] have since studied different aspects of the problems. None considered the three-dimensional transportation problem with capacity restrictions. In this paper, an attempt has been made to discuss the three-dimensional capacitated transportation problem. The purpose of this study is to carry out the transformation of an optimal solution when there are changes in the data of the problem (the rim conditions - i.e., the warehouse supplies, market demands, conveyance capacity and the per unit transportation cost). In a subsequent paper we will develop an 'Operator Theory' to study the different aspects of the problem. Such an analysis will prove valuable in incorporating certain aspects of the problem not already captured by its classical multi-dimensional transportation model. 2. FORMULATION AND SOLUTION OF CAPACITATED THREE-DIMEN SIONAL TRANSPORTATION PROBLEM We shall consider the capacitated three-dimensional transportation problem with its classical interpretation as a problem of minimizing the total cost of shipping a homogenous good from m warehouses to n markets by p means of conveyance. Manuscript received June 1980, revised Sept

. = cost of shipping one unit of the good from warehouse i to market j by k means of conveyance; a^ = availability or supply at warehouse i; bj = demand or requirement at market j; ek = the possible

2 48 Define the index sets: (2.1 ) I = {1,2,..,,m}, the set of warehouses (2.2) J = f1,2,...,n}, the set of markets, and (2.3) K = {l,2,...,p}, the set of conveyances. For i e I, j e J and k e K, define x ^ k = amount of the goods shipped from warehouse i to market j by k means of conveyance; c.. = cost of shipping one unit of the good from warehouse i to market j by k means of conveyance; a^ = availability or supply at warehouse i; bj = demand or requirement at market j; ek = the possible amount to be transferred by means of conveyance k. The capacitated three-dimensional transportation problem 'P' for known availabilities, demands, means of conveyance and capacity restrictions can be stated as follows: (2.4) Minimize Z = E c.., x.., >k idk 13k i *j subject to the following constraints: (2.5) E x.., = a., j,k (2.6) E x. = b., i,k ^ for i e I, for j e J, (2.7) for k e K, Z. X ijk = 6k ' 1 /D J (2.8) 0 < x... < U., for i e I, j e J and k e K, ijk ljk where a^, b-; and ek are non-negative numbers and the system (2.5) to (2.8) is assumed to have a feasible solution. Let X = { (xjjk ) (i, j,k) e I x J x K} satisfying the constraints (2.5) to (2.8) be called a feasible solution (or simply solution) to the problem. A basic solution X = {xijk }, corresponding to a basis B, satisfies (2.5), (2.6) and (2.7) and is such that (i,j,k) B implies that either Xijk = 0 or x ijk = Ujifc. If in addition (2.8) is satisfied for (i,j,k) e B, the basic solution is called a primal feasible. We define LB and UB as the set of non-basic variables that are at their lower and upper bounds respectively. We define (B, LB, UB) as a basis structure. It is well known in the ordinary transportation problem [1,4] that given a problem and its basis structure, the associated primal solution is unique.

6), diagonal constraints (2.7) and upper bound constraints (2.8) respectively. Then the dual problem is (2.9) Maximize I a.u.+ Z b.v.+ Z e, w - Z U...t... = F iel 1 1 jej k k i,j,k 13k ^ k subject to the following constraints: (2.

3 49 We now define the dual problem to 'P'. Let u^ for i e I, Vj for j e J, Wk for k e K and tijk for i e I, j e J and k e K be dual variables (simplex multipliers) associated with the row constraints (2.5), column constraints (2.6), diagonal constraints (2.7) and upper bound constraints (2.8) respectively. Then the dual problem is (2.9) Maximize I a.u.+ Z b.v.+ Z e, w - Z U...t... = F iel 1 1 jej k k i,j,k 13k ^ k subject to the following constraints: (2.10) u. + v. + w. - t... < c., for iel, jej and kek i 3 k 13k - 13k ' J (2.11) t... > 0, for iel, jej and kek. 1 3k J Given a basis B, one can determine a one parameter family of solutions Ui, Vj and w^ to the equations (2.12) + Vj + wk = c i _k for (i,j,k) e B such that ui + Vj + Wk = dijk is unique for all iel, jej and kek [1,4]. The above reference also provides an algorithm for determining these variables. We refer to D = {dijk} as the three-index dual matrix, since the dijk's are uniquely It is well known [1,4] that a primal basic feasible solution is optimal if its dual solutions satisfy (2.13) u. + v. + w, < c. for (i,j,k) e L B 1 3 k 13k (2.14) u. + v. + w, > c... for (i,j,k) e U B 1 3 k 13k A basis B, or more accurately, a basis structure (B, LB, UB) is said to be dual feasible if the ui, vj and wk determined from (2.12) satisfy (2.13) and (2.14). If we define (2.15) t. = Max(0, u. + v. + w, - c. ) for iel, j e J and k ek 13k 1 3 k 13k it may be verified that (2.12) to (2.15) imply (2.10) and (2.11). By the duality theorem of linear programming, [1,4] a basic solution is optimal if it is both primal and dual feasible. Furthermore, for such a solution: (2.16) Z = Z c. x. = Ea.u. + Eb.v. + Ee w, - U t = F... 13k ink I D, k k ink ink i,],k J 1 D k i,j,k J J For simplification the optimality criteria can be summarized as follows: (2.17) 0 < x... < U... =* u. + v. + w, = c... 13k 13k 1 3 k 13k

50 =* U. + V. + w. >c 1 J k - ijk =* U. + V. + w. < c i j k ijk With the help of these criteria, we can reduce the solution of our problem to a modification of the steppingstone method.

The additional row, column and diagonal are required to collect the quantities of produce which could not be placed on to the cost elements Cij^ in the first step.

4 50 =* U. + V. + w. >c 1 J k - ijk =* U. + V. + w. < c i j k ijk With the help of these criteria, we can reduce the solution of our problem to a modification of the steppingstone method. We start the calculation with a diagram as Figure 1, but we add one more row, one more column and one more diagonal to the three-index cost matrix. The additional row, column and diagonal are required to collect the quantities of produce which could not be placed on to the cost elements Cij^ in the first step. An initial solution can be obtained by using any one of the known methods from the transportation problem, e.g., the north-west corner rule or the penalty method. It is a known fact in the transportation problem that each non-degenerate basic solution will contain (m + n + p - 2 ) positive components. However, the upper bound restrictions (2.8) imply that the corresponding cost elements must not receive an amount more than Uij^, i.e., x^jj^ = Uijk must always be taken into account. The occupied elements for which xijk = u ijk are basic elements proper only if, during the construction of the initial solution, a source or a destination or a conveyance is cancelled when Xij^ is assigned to cij^. The elements with their upper bounds are called saturated elements. The problem is soluble only if the additional row, column and diagonal can be ultimately dispensed with. 3. AN APPLICATION A commonly occurring problem in the three-dimensional distribution system [1 0] is the optimal allocation of n different products from m different factories to p different consumer zones, so as to minimize the total transportation cost. Consider a soap manufacturer who has m factories in various parts of a country. Each of m plants can manufacture n different types of soap. The soap is to be distributed from the factories to p different consumer zones. Define ai as the number of units of n different types of soap available in ith factory; b^ as the number of units of jth soap available in m different plants; e^ as the demand at the kth consumer zone; xijk as the amount of jth type made in ith plant shipped to kth consumer zone; c^j^ as the cost of shipping one unit of the jth type from ith plant to kth consumer zone. This problem has the structure of the model discussed in Section A NUMERICAL EXAMPLE Consider the problem with a^, = 80, a2 =90, a3 =55, b^ = 70, b9 = 60, b 3 = 35, b4 = 60, e-^ = 100, e2 = 55, and

is imposed on the new cells rising from the inclusion of one more row, column and diagonal.

5 51 e 3 = 70, and the tri-dimensional matrix of costs, with upper bounds on the variables, denoted by ' in Figure 1. Figure 2 exhibits an extended tri-dimensional matrix where one more row, one more column and one more diagonal are added with very large cost M, except for the cell (4,5,4), where C 454 = 0* No bound is imposed on the new cells rising from the inclusion of one more row, column and diagonal. An initial feasible solution, obtained by using the penalty method, is exhibited in Figure 3, with '*' distinguishing the saturated cells. We circle the costs of the basic cells. The quantities above the circle represent the allocation on the basic cells. We determine the dual ables; ui / i e I, Vj,- j e J and Wfc,r k e K so that U1 + V1 + W1 C lll = 3 U 3 + V2 + W2 C 322 = 5 U1 + V 3 + W 3 C 132 = 8 U 3 + V2 + U2 + V 5 + W2 C252 = M U 4 + V2 + U 3 + V1 + W1 C311 = 1 U4 + V 5 + U 3 + V1 + W2 C 312 = 4 S u> I I u2 + V2 + W2 C222 = 4 U 3 + V4 + C 323 = 6 c 341 = 5 C424 = M rhc451 = M The solution of this system is: u^ = 0, U 2 = -3, U 3 = -2, U4=0, = 0, V2 = 1* V3 = 2, V4=4, V5 = M - 3, wi = 3, W 2 = 6, W 3 = 7, W 4 = M 41* 1. These dual variables are shown in Figure 3. The intermediate solutions are shown in Figure 4 and Figure 5, and the optimal solution in Figure 6, where the additional row, column and diagonal are ultimately dispensed with. 1 I REFERENCES 1 c h a r n e s, A. and COOPER, w.w., Management Models and Industrial Applications of Linear Programming, Vols. I-ll John Wiley and Sons. (1961). 2 CORBAN, A., 'A Multidimensional Transportation Problem', Res. Roum. Math. Pures. and Appl. 9 (8), (1964). 3 CORBAN, A., 'A Multidimensional Transportation Model II', St. Cere. Math. 27, (1965). 4 HADLEY, G., Linear Programming, Addison-Wesley (1962). 5 HALEY, K.B., 'The Solid Transportation Problem', Opns. Res. 10, (1962). 6 HALEY, K.B., 'The Multi-Index Problem', Opns. Res. 11, (1963). 7 HALEY, K.B., 'The Existence of a Solution to the Multi-Index Problem', Opns. Res. Quart. 16, (1965).

6 52 FIG.I

7 53 FIG. 2

8 54 FIG. 3 (M-3)

9 FIG. 4

10 FIG.5

11 57 FIG.6

9 MORAVEK, J. and VALCH, M., 'On the Necessary Conditions for the Existence of a Solution of the Multi-Index Problem', Opns. Res.

12 58 8 MORAVEK, J. and VALCH, M., 'Determination of the Feasible Solution of the Three-Index Transportation Problem', Eknomicko Matematicky Obzor 1 (2) (1965). 9 MORAVEK, J. and VALCH, M., 'On the Necessary Conditions for the Existence of a Solution of the Multi-Index Problem', Opns. Res. 15, (1967). 10 SCHELL, E., 'Distribution of a Product by Several Properties', Directorate of Management Analysis : Proceedings of the Second Symposium in Linear Programming, H. Antosiewiez (ed), Vols I and II, (1955).

TIME MINIMIZATION IN TRANSPORTATION PROBLEMS

75 M O P vozmme 6 number J Jan^arz/ TIME MINIMIZATION IN TRANSPORTATION PROBLEMS J,K. SHARMA D.N. POST-GRADUATE COLLEGE, MEERUT. KANT I SWARUP INDIAN INSTITUTE OF PUBLIC ADMINISTRATION, NEW DELHI. Summary