The Transportation-Location Problem

The Transportation-Location ProblemAuthor(s): Leon CooperSource: Operations Research, Vol. 20, No. 1 (Jan. - Feb., 1972), pp. 94-108Published by: INFORMSStable URL: http://www.jstor.org/stable/169341 .

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The Transportation-Location Problem

Leon Cooper

Southern Methodist University, Dallas, Texas

(Received January 8, 1971)

This paper defines a problem type, called the transportation-location problem, that can be considered a generalization of the Hitchcock transportation problem in which, in addition to seeking the amounts to be shipped from origins to destinations, it is also necessary to find, at the same time, the optimal locations of these sources with respect to a fixed and known set of destinations. This new problem is characterized mathematically, and exact and approximate methods are presented for its solution.

JN A PREVIOUS paper, [4] I formulated a problem that was a generalization of both the Hitchcock 'transportation problem' (see HADLEY[81) and a problem

that is sometimes referred to as the 'location-allocation' problem with unlimited capacities (see COOPER.12313) Here I will consider both that problem and an even more general formulation in much greater detail.

We are given n fixed locations, to be called destinations, whose positions in Euclidean space are known. Their coordinates will be given by (XDj, YDj), =

1, **, n. We are also given a set of known requirements rj, j= 1, , n, for some commodity or product at each of the n destinations. We wish to locate m sources, where m is a given number, from which the product is to be shipped. These sources are supposed to have certain limitations on their capacity to ship the product. These numbers are known and will be designated ci, i= 1, *>, m. Finally, there may be 'weights' relating to destination requirements, e.g., multiplicity of trips in a time period, or other possible weights. These will be designated jBj, j = 1, , n. We shall designate by (xi, yi), i= 1, * , m, the locations that are to be determined for the sources, and by wij, the amounts to be 'shipped' from source i to destination j, which are also to be determined. Finally, we will define a set of 'cost functions' that depend on the relative locations of the sources with respect to the destinations. These will be designated:

4P(XDj, YDj; xi, Yi) cost of supplying a unit quantity to the jth destination from the ith source.

We assume that the ,6 functions are continuous. Our object is to minimize total cost subject to capacity and requirement constraints. This problem may be formulated as follows:

min z = Ei=i D=1 /3Wij/(XD, YDj; Xi, yi) subject to:

Z=1 wij <ci, i=1, ,m; (1)

Ei=1 wij=rj, j n1, *.,f; wij O, all i,j.

In what follows, let us designate the set W as all w )= {w} satisfying the constraints of (1).

The relation of the problem given in (1) to the transportation problem and the

94



The Transportation-Location Problem 95

location-allocation problem is readily seen. If the t(XDj, YDj; Xi, yi) =aij, a set of constant costs, and if acij/j is designated as yij, then (1) can be written as:

min zi= A-1 'w1 'yiWij, IDEW, (2)

which is a usual form of the transportation problem. Hence, for a fixed set of costs, problem (1) becomes a transportation problem.

Returning to equations (1), we may also note that (a) if the requirements are stated in terms of the Oj and (b) the capacities associated with each source are unlimited, then the set of m+n constraints of (1) are no longer explicitly required and (1) now becomes:

min Z2= X,- i ai-1 aij[3ii(XDj, yDj; Xi, Yi); (3)

Ad i aij=l,1 j=I,- n,n a~j =0, 1.

This is the location-allocation problem discussed in references 2 and 3. The particular form of the functions i that we shall be most interested in

throughout this paper is

t(XDj, YDj; Xi, Y.) = [(XDj-Xi)2+(yDj- yi)2]12, (4)

which is the usual Euclidean distance in a two-dimensional space. However, we shall give some consideration to the more general form given in (1).

CHARACTERISTICS OF THE GENERAL TRANSPORTATION-

LOCATION PROBLEM

WE SHALL REFER to the problem stated in equations (1) as the general transportation-location problem. There are a number of important observations that can be made concerning this problem. THEOREM 1. A necessary and sufficient condition for (1) to have a feasible set of wij is that D-1 r3? zD: 1 ci.

Proof. This result is the same as for a transportation problem (see reference 8). In what follows and throughout this paper, it can be shown that the functions

V( XDj, yDJo xi, yi) are such that (1) has a finite minimum. The next theorem is an important result. THEOREM 2. For the problem

minm z= 1 fiw1 0Wij4(xDj, YDj; xi, Yi), ?,5EW,

an optimal solution will occur at an extreme point of the convex set W of feasible solutions to (1).

Proof. Let {x*, y w*} be an optimal solution to (1), where x= (xl, x2, xm), ij (Y1, Y2, , ym), and w= (w11, W12, *, Un, W21, *, W2n, , Wmn). If we fix (x, y) so that (x, g) = (x*, g*), we obtain a resulting problem

min z= D=1 51iw1 0 4Wijk(XDj, YDj; xi*, Yi*) w9EW,

which is a transportation problem. Hence, it has a solution at an extreme point ZC EpW. From this it follows that (x*, y, we) is an optimal solution to the original problem (1).



96 Leon Cooper

THE TRANSPORTATION-LOCATION PROBLEM

LET US NOW consider the problem of equations (1) for a particular form of the function i:

I(XDj, yDj; Xi, Yi) = [(XDj-Xi)2+(YDi-Yi)2]112 (4)

Hence, what we shall now call the transportation-location problem is the following:

mi Z tm a=n Z i WiJ(XDj- (YDj i WW. (7)

We shall now characterize the transportation-location problem. The constraint set is clearly a convex set. What of the objective function? Theorem 3 has some less-than-comforting information. THEOREM 3. The function z= If1 Z}-1 OjWij[(xDj-xi) +(yDj-Yi) ] is neither a convex function nor a concave function of the vector (x, y, z,) for all values of x, y, and w.

Proof. Let us represent the objective function

Z= as-m /3,w4j[( XDj- Xi)2+ (yDj-Yi) ]1"

as z(x, y, w). Consider the special case where all wij=O except w11. Further, let all xi, yi be fixed and known for i> 1 and let Y1=YD1 and XD1>X1. Hence z(x, Y, 7D) = z1( x1, w11) +K, where K is the sum of all the constant terms. If z1 is neither convex nor concave over all of the space defined by (., -, v ), then the same will be true of z. Hence, we examine z1(x1, wil)=01W11(XD1-X1). It is well known[71 that a function such as z1 will be neither convex nor concave if the Hessian matrix associated with z1 is neither positive nor negative semidefinite. The Hessian matrix for z is

za2z1/ax2 a2z1/awnlax1 0 -0( (8)

a2z1/axiaw1l a2z1/awl2 1-/3 0

which is indefinite. Hence z1 is neither convex nor concave in the portion of the space defined by x1 and wil. Therefore, z(G, y, w) is neither convex nor concave over all of the space defined by (x, 7, D).

This proof shows that the objective function z is not a convex or a concave function over all possible values of x, -, -. A more pertinent question is whether or not the objective function is convex or concave over values of x, y, and w limited by the constraint set of the transportation-location problem. The answer to this is given in the following result. THEOREM 4. The function z= I,1 /3wi;(xDJ-xi)2+(yD -yi)2]j is neither a convex nor a concave function of the vector (x, 9, z,) over the constraint set of (7).

Proof. Consider the following problem:

min z = /31wlDl1+/2Wl2Dl2+/3lw2lD21+/2W22D22 subject to (9)

Wii+IW12 = C1, W21 +IW22 = C2, Wi-+IW21 = ri,

W12 + W22 =r2, W11, W12, W21, W22 > 0,

where Dij=[(XDj-X )2+(YDi-yi)2]

In order to simplify the argument but allow sufficient variation over the constraint set, we assume that ( X2, Y2) = ( XD2, YD2) and further that wi =u /ni = v no and




XD1 > XI, XD2 > X1. This simplifies the objective function to:

z( xl, Wi, W12, W 1) = Olwll ( XD1- X1) +I-32W12( XD2- X1) +31KDw21, (10)

where KD = XD1 XD2.

Let us consider the Hessian matrix of z(x1, W11, W12, W21):

(2 (2Z (2 ( 2Z

ax12 4aXw1 ax14w12 ax14w21 0 -01 -02 0

aX.aWl aW1 w - 1 0 0 0

(2Z (2 ( 2Z (2 (1 l dz daz c~z a~z [ -B

ax~awi2 3w11W1 (W2 -0au /2 0 0 0 (X1(W12 fW12 12 1W121W21

axfiw2l aWfllW2l aw2tw21 aw21 0 0

Since the Hessian matrix is indefinite for the problem given by (9), we see that, in general, the objective function is neither convex nor concave over the constraint set of the transportation-location problem.

A consequence of Theorem 4 is that the transportation-location problem, which is given by equations (7), is a nonconvex nonlinear programming problem in which we are minimizing a nonconvex objective function over a convex set. However, because of Theorem 2, we can state the equivalent result for the transportation- location problem. THEOREM 5. For the problem

Mi n z= i 1 'aEWj[(XDj- (yDj Y] W (7)

some optimal solution will occur at an extreme point of the convex set of feasible solutions to (7).

Proof. The problem above is a special case of the general formulation proven in Theorem 2.

The importance of this result is that it makes it unnecessary to consider any but basic feasible solutions to the constraints of (7). This will be important when we consider computational approaches in the next section of this paper.

AN EXACT ALGORITHM FOR THE TRANSPORTATION-LOCATION PROBLEM

THE FIRST algorithm we shall consider is exact and relatively simple in concept. However, its use will be limited, as will be evident, to relatively small problems.

We may note, according to Theorems 2 and 5, that an optimal solution to the transportation-location problem will occur at an extreme point of the convex set of solutions, JV = {I ijlAw 7b, w O }. We know that the extreme points of W1 cor- respond to basic feasible solutions of Aw < b, w 0. If the number of basic feasible solutions is designated as NBFS, for an m-source, n-destination problem, NBFS-<

( mn 1 This follows from the basic theory of the transportation problem.18'



98 Leon Cooper

Actually, (m2 + ) is the number of basic solutions, most of which are infeasible.

Hence, it is usually the case that NBF<< (m+n- For example, in the simple

example we shall present, with m=2, n= 4 = 8!/5!3=

However, there are only nine different nondegenerate basic feasible solutions. According to results of DEMUTH 5] AND DOIG,16] the minimum number of basic

feasible solutions, for n< n, to a transportation problem of order in Xn is i!/(n- rn+l)! For m=2, n=4, we have 4!/3!=4. Hence, the number of bases we have had to examine, 9, is much closer to 4, the minimum number, than to the unrealistic upper bound, 56. This is somewhat encouraging.

In any case, NBFS is a finite number. Suppose we generate all basic feasible solutions. Let us designate any such solution w = {I ijI }. (We consider, subse- quently, how to do this.) For each such solution, we can then solve the problem:

m n= z =1 Eli}1 ij [(XDj-X) +(YDj-Y)2]1/2 (12)

by considering this problem as a set of problems of the form:

min = - i Vij [(XDj-Xi)2 +(YDj-Y) 2]12* (i=1 *, *m) (13)

We can do this, since the {I '6j } are simply known nonnegative constants or weights. An iterative technique for solving problems of the form given by equations (13)

was given by the author in references 2 and 3, and is repeated here for convenience. The initial estimates of (xi, ye) are given by

Xt?= ( E =n

' fijD (E2n1W j ) 0i (Z 1 WijXDj)/( ji -1 (i=1 . *m) (14)

yi1 (Z 1in' f~ V

and the general iteration equations are:

Xk.+1 = ( Ei~ k

1W X tj )l/( E i1 t n

D kj) xi +1 (E 1 W *jDj/Dj )/(E=1 Wij/Dj), (i=1, ** ,m) (15)

Y1 ( Z=1 w ijyDj/D

J)/( Z ibijI/Dj),

where the superscript on the xi and yi is the iteration parameter and

Di j= [( XDj-X)+ (yDj-y )]/ ( 16)

If we now designate the minimum value of z for the lth basic feasible solution as zi then it is obvious that the optimal value of the objective function z? for the transportation-location problem will be:

z =min V, zi*. (17)

If the minimum is taken on at 1=8, then the optimal values of the variables are (Xi., ys), i= 1, ,m, and ?bj, i = 1 *, m, j= 1, ,n, where the designation (xi,, yis) indicates (xi, ye) for the sth basic feasible solution and ?ibj, are the set of ?rij for this solution.

Let us now return to a point we glossed over, earlier, viz., generating all the basic feasible solutions to the constraints of the transportation-location problem, i.e., the constraints:

D_1 wij3cO, i=1, *,m =1, wjrj, n; (8) wijO" , i=l,- .,Ml j=l, -n.




As has already been mentioned, these are simply the constraints of the standard transportation problem. We can make use of the transportation problem tableau (see reference 8) to advantage in order to generate only basic feasible solutions. The alternative would be to find all basic solutions and discard the infeasible solu-

TABLE I TABLEAUX FOR THE EXAMPLE

1 2 3 cG 1 2 3 Ci

1 70 10 80 1 80 80

2 80 40 120 2 70 10 40 120

ri 70 90 40 ri 70 90 40

Tableau 1 Tableau 2

1 2 3 ci 1 2 3 C}

1 70 10 80 1 40 40 80

2 90 30 120 2 70 50 120

ri 70 90 40 ri 70 90 40

Tableau 3 Tableau 4

1 2 3 Ci

1 40 40 80

2 30 90 120

ri 70 90 40

Tableau 5

tions. This may require orders of magnitude more work. The method described below is more efficient.

As an example, consider a problem with m=2, n=3. There will be at most 2+3-1=4 nonzero values of wij in a basic feasible solution. Suppose cl =80, C2= 120, r1 =70, r2=90, r3 =40. An initial tableau is shown in Tableau 1 of Table I. The blank squares have zero values of wqj, i.e., w13 =0, w21 =0. It is a simple matter to find all basic feasible solutions



100 Leon Cooper

from this initial Tableau 1. We can use the standard 'loop' method for allowing a zero variable to become positive and still remain feasible (see reference 8). This is merely the application of the simplex method to the special case of the transportation problem. From Tableau 1, we can generate two new tableaux. These are given in Tableaux 2 and 3. From Tableau 2 we can generate one new tableau given in Tableau 4. From Tableau 3 we can generate Tableau 5. From Tableau 4 we can generate Tableau 5, and from Tableau 5 back to Tableau 4.

We have now generated all the basic feasible solutions. The relations between them for this example can be represented as the directed graph shown in Fig. 1.

It can be seen that there are five basic feasible solutions. If all basic solutions had been

found, we would have had to solve four simultaneous equations in four variables, (6)=15 times. The above procedure is simpler and very much less work.

Figure 1.

We now state the algorithm we have been discussing for solving the transportation-location problem and then present a numerical example.

Enumeration Algorithm for the Transportation-Location Problem

1. Using the transportation-problem tableau, starting with any basic feasible solution, generate the connected graph of all basic feasible solutions. Designate each such solution,

iWijI}, 1 =1, *.., T, where there are T basic feasible solutions.

2. For each such solution, solve the set of location problems:




min zii= 51- Wiil[(XDj-Xil)2+ (YDj-Yi1)]"I, i=1, M,

and set zl*= Zt=n Z'i

3. The optimal solution is found by z* =min,=, T Zl*, with Wall and (xa, Y*l) being the corresponding values of the variables.

A sample problem. Let (XD1, YD1) = (0, 0); (XD2, YD2) = (0, 1); (XD3, YD3) = (1, 1); (XD4,

YD4) ==(1, 0); m=2, n=4; cl=50, c2=100; ri =20, r2=40, r3=60, r4=30. A basic feasible solution is given in Tableau 1 of Table II. From Tableau 1 we can generate Tableaux 2, 3 and 4. From Tableau 2 we can generate Tableaux 5 and 6. From 3 we generate 7 and 4. From 4 we generate 8 and 4. From 5 we generate 6 and 7. From 6 we generate 5 and 8.

TABLE II TABLEAUX FOR THE ENUMERATION-ALGORITHM PROBLEM

1 2 3 4 Ci 1 2 3 4 Ci

1 20 30 50 1 10 40 50

2 10 60 30 100 2 10 60 30 100

ri 20 40 60 30 ri 20 40 60 30

Tableau 1 Tableau 2

1 2 3 4 Ci 1 2 3 4 Ci

1 20 30 50 1 20 0 30 50

2 40 30 30 100 2 40 60 100

rj 20 40 60 30 ri 20 40 60 30

Tableau 3 Tableau 4

1 2 3 4 Ci 1 2 3 4 C;

1 40 10 50 1 40 10 50

2 20 50 30 100 2 20 60 20 100

ri 20 40 60 30 ri 20 40 60 30

Tableau 5 Tableau 6



102 Leon Cooper

TABLE II-Cont.

1 2 3 4 cs 1 2 3 4 ci

1 50 50 1 20 30 50

2 20 40 10 30 100 2 20 20 60 100

ri 20 40 60 30 ri 20 40 60 30

Tableau 7 Tableau 8

1 2 3 4 ci

1 20 30 50

2 20 40 40 100

ri 20 40 60 30

Tableau 9

From 7 we generate 9 and 5. From 8 we generate 6 and 9. From 9 we generate 4 and 8. Hence we are done. All of these tableaux are shown in Table II. Figure 2 gives the graph of the nine basic feasible solutions.

For each of these nine basic feasible solutions to the constraints of the transportation- location problem, two location problems were solved with the results shown in Table III. It can be seen that the minimum occurs for solution 2. Hence the solution to our problem is: Wil = 10, w12 =40, w13 =0, w14 =0; w21 =10, w22 =0, w23 =60, w24 =30; (xi, Yi) = (0, 1), (x2, Y2)=

(1, 1)2; zG 54.143.

Fi.2.Gap f outostote apl 'rolm




We have made use, in the above algorithm, of the usual transportation-problem method. A worthwhile point to investigate is whether more efficient methods than this 'loop' method might not be found to generate all possible basic feasible solutions. MOTZKIN, RAIFFA, AND THRALL1101 have presented a method for finding all the vertices of a convex polyhedral set, as has BALWNSKI.11] WITZGALL AND

WETS111" have also examined this problem. One of the most comprehensive stud- ies, and perhaps the most promising, is due to IKOHLER;1"' it considers a more general problem, but can be used to find all extreme-point solutions. It is quite possible, and this is under investigation, that, if either of these methods is applied to the specific polyhedral convex set of the transportation problem, a simplified

TABLE III RESULTS FOR THE ENUMERATION-ALGORITHM PROBLEM

Basic feasible (xi, yi) ZlZ solution

1 (0, 1) 20 60 (1, 1) 40

2 (0, 1) 10 5.4 (1, 1) 44.143 54.143

3 (1, 1) 28.286 93_77 (0.641, 0.792) 65.490 93.776

4 (1,0) 20 6 (1, 1) 40 60

5 (0, 1) 10 6.8 (1, 1) 58.284 68.284

6 (0, 1) 14.142 62.428 (1, 1) 48.286

7 (1,1) 0 6.2 (0.287, 0.596) 69.126 69.126

8 (1, 10) 28.286 76.572 (1, 1) 48.286 7.7

9 (1,0) 20 7.7 (0.173, 0.924) 59.579

method for finding all the basic feasible solutions will result that is more efficient than the usual 'loop' method. This would greatly improve the algorithm presented here.

HEURISTIC ALGORITHM NO. 1 FOR THE TRANSPORTATION-

LOCATION PROBLEM

THE HEURISTIC algorithm described in this section was a first attempt to devise a rapid suboptimal method for solving transportation-location problems. It was



104 Leon Cooper

suggested by a heuristic method previously developed for the pure location-allocation problem and described in reference 3. This method, called the 'alternate location-allocation method,' is as follows:

1. Select some subset consisting of m of the n destinations that are given and consider these as source locations.

2. Allocate each of the remaining n - m destinations to the closest of the m sources given in Step 1.

3. Within each of the m sets of destinations determined in Step 2, use the iteration method given in references 2 and 3 [also used in equations (14) and (15) of this paper] to find the exact location of the optimal source location.

4. Determine for each destination whether or not it is closer to another of the sources located in Step 3 than the one to which it is allocated. This defines a new grouping of m subsets of destinations.

5. Repeat Steps 3 and 4 until no further changes are possible.

It is shown in reference 3 that this algorithm is a moderately successful one, but by no means the best of the several heuristics tested for the pure location-allocation problem. A modification of this method for the transportation-location problem can be made as follows.

Alternating Transportation-Location Heuristic

1. Select arbitrarily m of the (XDj, YDj) and let these be the m initial source locations. This then yields a set of distances between each of the destinations and the assumed sources.

2. Using these distances as cost coefficients yij [as in equations (2)], we can solve an ordinary transportation problem to find a set of {wij}.

3. Using the {wijj from Step 2, we can solve a location problem using equations (14) and (15) and find a new set of source locations.

4. We now iterate Steps 2 and 3 until no further changes, to within some tolerance, are obtained in two successive cycles.

It can be seen that the general notion behind this approximate method is to locate sources alternately given a pattern of allocations and to determine an allocation given a set of source locations. The location-allocation problem method and the usual transportation-problem method are alternately applied to perform the calculations. It can readily be seen that this iteration method yields a convergent monotone nonincreasing sequence of values for z. However, there is no guarantee that it will converge to the global maximum we seek. Still, what experience we have with this and similar algorithms indicates that the result, when not optimal, lies within -1O percent and usually within 2-3 percent of the optimal solution.

Table IV indicates the results with this heuristic method for the first seven problems given in reference 2. These are location-allocation problems with m=2, n=7. They were solved as transportation-location problems by using a set of capacities and requirements such that each destination was supplied by only one source. The starting points in each of these problems were completely random. For problems 3 and 4 the optimal solutions were obtained. In the others, results of varying degrees of closeness to optimality are obtained.




TABLE IV ALTERNATING TRANSPORTATION-LOCATION HEURISTIC RESULTS

Problem Allocations Optimal Z Optimal no. obtained allocations obtained Z

1 (1, 2, 3, 4, 5) (1, 2, 4, 5) 52.118 50.450 (6, 7) (3, 6, 7)

2 (1, 3,4, 6, 7) (1, 3,4, 6) 81.764 72.000 (2, 5) (2, 5, 7)

3 (1, 2, 3, 4) (1, 2, 3, 4) 38.323 38.323 (5, 6, 7) (5, 6, 7)

4 (1, 2, 3, 7) (1, 2, 3, 7) 48.850 48.850 (4, 5, 6) (4, 5, 6)

5 (1, 2, 3, 4, 5) (1, 2, 3, 5) 38.560 38.033 (6, 7) (4, 6, 7)

6 (1, 2, 3) (1, 2, 3, 4) 44.564 36.175 (4, 5, 6, 7) (5, 6, 7)

7 (1, 5, 6, 7) (1, 3, 4, 5, 6, 7) 61.935 59.716 (2, 3, 4) (2)

In order to study how the results obtained by the heuristic are related to the initial choice of the sources, 17 different (randomly selected) starting values were chosen and the heuristic then applied for Problem 1 of Table IV. Table V indicates the results obtained. It will be noted, from Table IV, that, for problem num-

TABLE V SUMMARY OF LOCAL MINIMA FOUND FOR PROBLEM 1

Allocation z No. of times (out of 17) obtained obtained this solution occurs

(1, 2, 3, 4, 5) 52.118 3 (6, 7)

(1, 2, 4, 5) 50.450 3 (optimal solution) (3, 6, 7)

(1, 2, 3, 4) 52.001 4 (5, 6, 7)

(1, 3, 4, 5, 6, 7) 59.705 3 (2)

(1, 2, 3, 7) 60.128 2 (4, 5, 6)

(1, 2, 4, 5, 6) 57.672 2 (3, 7)



106 Leon Cooper

ber 1, the optimal allocations are (1, 2, 4, 5), (3, 6, 7) and the optimal value of z is 50.450.

From the results of Table V we can see that in three of the 17 trials we obtained the optimal solution. In roughly 60 percent of the trials we obtained a solution no worse than about 3 percent of optimal and usually better than this. The maximum error was about 15 percent. However, it can be seen that repeated use of the heuristic method, with varying starting values, is apt to give a reasonably good approximation to the optimal solution, if not the optimal solution itself.

HEURISTIC ALGORITHM NO. 2 FOR THE TRANSPORTATION-

LOCATION PROBLEM

THIS ALGORITHM is based on earlier work reported in reference 4. It seems to be extremely efficient, as will be seen. The heuristic method that has been developed is as follows.

Transportation-Location Heuristic Method

1. Use one of the heuristic methods reported in reference 3 (the "alternate location- allocation" method referred to in the previous section is one of these) to find a solution with unspecified destination requirements and unlimited source capacities. Let the matrix A = llaijll be the allocation matrix for the solution obtained, i.e., axij = 1 if source i supplies destination j and axij =0 otherwise. A is a matrix of zeros and ones such that each column contains exactly one '1.' There are no restrictions on the rows, since one source may supply more than one destination. The allocation matrix A is merely a convenient way of indicating the subdivision of n destinations into m subsets, i.e., m subsets served by each of the m sources.

2. Replace the nonzero elements of A by their respective rj, thus forming a new matrix W1 = 1Iw' ill where

I rito , ati 1

0O, aij=O. 3. Using the original matrix A of allocations, we now derive two new matrices, analogous

to W1 of Step 2 above, as follows. Find the pair of points being served by the same source such that the distance between them is a maximum. Let the sources be s., u = 1, . .., m, and let Qu be the subsets of destinations. Then we wish to find a pair of points p,, pt as follows:

DU=maxk, Uf QU [(XDk-XDl )2+ (YDk-YD)2]1/2,

[(XD.-XDt)2+ (YD8-yDt )2]1/2= max DU, u=1, ...,m.

Let the subset in which this occurs be Qh with source Sh* We then eliminate source Sh and replace it with points p8, Pt. We now have a set of m+1 sources (su, u==1, * *, m; u h), p8, Pt. Therefore, we have one more source than is desired. Let R = { (su, u = 1, ** , m; u-h), p8, pt}. The source coordinates are (xu, yu) for uih, (XDS, YDS), (XDt, YDt). For notational simplicity, we rename the sources as s,. Therefore, R =I{sIV =1, --, m+1 }, where the first m-i sources are su(u #h) and where sm =ps, sm+ =pt. We now wish to find the pair of sources sa and Sb that are closest together. Let us designate the set of indices v, corresponding to s8eR, as V, i.e., V = {vlsveR }. The pair of sources Sa and sb is now de-




termined by [(Xa-Xb)2+ (ya-yb)2]'2=miink, lV;k3d I [(Xk- XI)2+ (Yk-yl)2I'12-

Having found the pair of sources sa, 8b that are closest together, first we eliminate se and apply the 'alternate' method of successive location and allocation until convergence, as referred to under step 1 above. This will determine a second solution. Next we restore source Sa and eliminate 8b, apply the 'alternate' procedure and obtain a third solution. We calculate two new matrices W2 and W3 corresponding to these solutions, as we did in step 2.

4. Sum the requirements for each subset of destinations served by one source, i.e., calculate Ai rj for each subset Si, i =1, *, m. We now calculate the differences, c-Ejsirj, i= 1, . ., m. A negative difference implies a capacity deficit and a positive difference indicates a capacity surplus.

5. The iterative calculation begins with this step. Let I_ be any set with a capacity deficit and let 1+ be any set with a capacity surplus. Choose a destination point Pk in the set I_ such that the difference in the distances from Pk to the source of I_ and from Pk to the source of I+ is a minimum. Symbolically, if Pk has coordinates (Xvk, Ypk) and the source location of I_ is (x_, y-) and that of I+ is (x+, y+), then we choose Pk as the index j from I_ such that

= Min,,,_ { (XPj-X_) 2+ (YDj-Y-)2]12- [(XDi-X+)2+ (YDj- y )2]1/2}.

6. We now reallocate part or all (if possible) of the requirement at Pk from the source in I_ to the source in I+. The amount reallocated depends on the size of the deficit at I_ and the surplus at I+. The four cases are

(a) If the (requirement at Pk) > (deficit for I-) and the (surplus for I+) ? (deficit for I_), then we reallocate the deficit from I_ by supplying this amount to pk from I+ instead of I_.

(b) If the (requirement at Pk) _ (deficit for I-) and the (surplus for I+) < (deficit for 1_), then we reallocate the amount of the surplus at I+ by supplying this amount to pk from I+ instead of I_.

(c) If the (requirement at Pk) < (deficit for I-) and the (surplus at I+) _ (requirement at Pk), then we reallocate the entire requirement at Pk from I_ to I+.

(d) If the (requirement at Pk) < (deficit for I-) and the (surplus at I+) < (requirement at Pk), then we reallocate the amount of the surplus at I+ by supplying this amount to Pk from I+ instead of I_.

7. With the new allocations, new source locations are computed for each subset of destinations Si by the use of equations (14) and (15), using the allocations as weights.

8. Each subset of destination points Si is now examined for the points that might be closer to a different source with an excess capacity. If possible, we reallocate part or all of its requirement, depending on the amount of the surplus at this source.

9. We now repeat steps 7 and 8 (exact source location and subsequent reallocation to satisfy requirements), until no further change in the allocation matrix occurs.

10. The entire reallocation process (steps 5-9) is now repeated until all capacity deficits are removed. The value of z for this allocation matrix WI is computed.

11. The procedure of steps 4-10 is repeated for allocation matrices W2 and W3.

12. The minimum of the three values of z obtained is chosen as the solution, together with its source locations and destination allocations.

This heuristic method was tested on the eight two-source, seven-destination problems listed in reference 2. The requirements were generated randomly. The capacities were chosen all equal, with their sum 5 percent higher than the sum of the



108 Leon Cooper

requirements. In order to have exact solutions to compare the heuristic against, the exact extremal equations (14) and (15) were used to solve for all possible allocations that, for these test problems, included the optimal solutions. For these eight problems, the heuristic method produced the optimal solution.

In reference 4 this basic method was also applied to 100 randomly generated problems with apparently good results, although the correct solutions were not known in advance.

REFERENCES

1. M. BALINSKI, "An Algorithm for Finding All Vertices of Convex Polyhedral Sets," J. Soc. Ind. & Appl. Math 9, 72-88 (1961).

2. L. COOPER, "Location-Allocation Problems," Opns. Res. 11, 331-343 (1963). 3. --, "Heuristic Methods for Location-Allocation Problems," SIAM Review 6, 37-52

(1964). 4. - , "Solutions of Generalized Locational Equilibrium Models," J. of Regional Science

7, 1-18 (1967). 5. 0. DEMUTH, "A Remark on the Transportation Problem," Casopis pest. mat. 86, 103-1 10

(1961). 6. A. DOIG, "The Minimum Number of Basic Feasible Solutions to a Transport Problem,"

Opnal. Res. Quart. 14, 387-391 (1963). 7. W. H. FLEMING, Functions of Several Variables, Addison-Wesley, Reading, Mass., 1965. 8. G. HADLEY, Linear Programming, Addison-Wesley, Reading, Mass., 1962. 9. D. A. KOHLER, "Projections of Convex Polyhedral Sets," ORC 67-29, University of

California, Berkeley, 1967. 10. T. S. MOTZKIN, H. RAIFFA, AND R. AM. THRALL, "The Double Description Method,"

in H. W. KUHN AND A. W. TUCKER (eds.), Contributions to the Theory of Games, Vol. II, Annals of Mathematics Study No. 28, Princeton U. Press, Princeton, 1953.

11. C. WITZGALL, AND R. J. WETS, Journal of Research of the National Bureau of Standards 71B, 1-7 (1967).



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The Transportation-Location Problem