Heuristic Methods for Location-Allocation Problems

Heuristic Methods for Location-Allocation ProblemsAuthor(s): Leon CooperSource: SIAM Review, Vol. 6, No. 1 (Jan., 1964), pp. 37-53Published by: Society for Industrial and Applied MathematicsStable URL: http://www.jstor.org/stable/2027512 .

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SIAM REVIEW

Vol. 6, No. 1, .January, 1964

HEURISTIC METHO1)S FOR LOCATION-ALLOCATIOIN 1110IBLEMS'

LEON COOPER2

I. INTROIUCTION

There are a great mrany combinatorial optimizationi problem-is of the following general type.

Giveni a collectiorn of n elements, for each feasible subset of these elements, there is attached an economic figure of merit. Find the feasible subset which maximizes the figure of merit.

A large variety of production schedulinig problems, sequenlcinig problems, optinmum location problems, problems relating to group organization, etc. have this general character. These problems are characterized further by having in- credibly large numbers of feasible solutionis (frequently of the order of n!) so that solutioni of the problems by complete eniumeration is not feasible even or digital computers.

A fortunate property of many of these problems is the lack of a .sharp maxi- mum, i.e., there are many alternative maxima or near-maximal solutionis. (A maximumii problem is used here for convenienice. Whether oIne is maximizing or minimizing a function is not important.) This is fortunate in. that it makes feasible the use of heuristic algorithms with a reasoiiably high probability that a well constructed heuristic will find one of these near-optimal solutions. If these problems had very sharp optima then the use of heuristics would fre(uently lead to poor solutions, i.e., to solutions which were not near-optimal. How "near", near-optimal needs to be is dictated by the specific problem type, but for many, 2-3 % of optimal or better is achievable by heuristics and is (uite satisfactoiy from an economic point of view.

In this paper we are considering a problem which is most readily visualized in terms of locating factories, warehouses or supply points to serve customers at various destinations. As examples of related problems which are amenable to the same general type of mathematical treatment there is the problem of adding new machines to an existing plant layout (or even the optimum layout of machines in a plant). Another example of this kind of rnathematical approach is in the location of an oil refinery with respect to oil producing wells.

In a pr'evious paper [1] the author called attention to a class of problems, generally called location-allocation problems. The particular one discussed in some detail in that paper is as follows: Given the location of a set of destinations in terms of their co-ordinates and a set of shipping costs (or distances) for the region of initerest, determine the optimum location of a fixed number of sources and the allocation of the destinations to sources. The optimum locations and allocations are those that minimize costs.

R Received bv the editors April 1, 1963 and in revise(d forin October 9, 1963. 2 Washington University, St. Louis, Missouri.

37



38 LEON COOPER

II. PROBLEM FORMULATION

In particular, the location-allocatioil problem can be formulated mathe- matically as follows:

Let the locations of the set of n destiinations be given by: (XDi YD&) i 1 ... , n, their co-ordinates in a Cartesian co-ordinate system. Let the co-ordinates of the m sources that are to be determined be given by (xi, yj) j = 1, 2, ... ,m.

If one assumes that source capacities are not fixed and that shipping costs are constant, i.e., iiidepenident of source output, then no one destination will be supplied by more than one source.

Let aij = 1 or 0 depending upon whether the it"l destination is or is niot served by the jth source. If p is the total cost of supplying n destinations with ni souirces, theen we have

M n

(1) = E aij4(XDi, YDXI XjX yj) j=1 i=1

where 41/(XDi, YDi ) xi, yj) is the cost function for supplying the ilh destination by the jlh source. In particular, if we assume that cost is proportional to distance (which frequently is the case) theii:

(2) 4P(XD.i , yDi, . j, yj) = WAj[(xDi - x)' + (YDi -yj) ]

where wij is a constant weightinig factor relating to multiplicity of supply trips and effectively increases relative distanice. We now have:

m n

(3) S? = E - aiWij[(XDDi - Xi2 + (YDi - Yj)2]1/ j=l i=l

Our problem is to determine xj, yj for all j and the set of aij for all i, j wvliich cause , to be a minimum. This overall problem is a very difficult one. The first approach taken in the previouisly cited paper was to minimize (p for all possible sets of aij.

It can be shown [1] that the following equations will yield the xj, yj to minimize so for any fixed set of a.j:

n ai wij(XDi x-) _i [(XI), - Xj)2 + (Yi- =j2i/

0

(4)~- = - j = 1, 2,..*. E4) n

aij Wij(YDi - yj)

X=1 [(XDi - X,)2 + (yDi - =y ) 0]12

Problem 62-11 recently submitted to the SIAM Review [2] also calls attenition to this problem and asks for methods of solution.

In the following ani iteration method [1] is presented for solving for the x,j yj . It has always converged in practice if the starting values are chosen as indicated. This is based on experience witlh several thousand calculations.

Rearrangement of equations (4) will yield the following equations for iteration of xi and y,:



LOCATION-ALLOCATION PROBLEMS 39

E asi, wj) XDi

h+1 i=1 Dk i

whereD~1 = - )2 (Yj -n

vaCij wij

E as; Wij Y,/i xJk+1 i_1 Dkzi

Y v n

tj

Dk 1

wmiere Dijn[(Xli Xj)2 + (sDi _ yj)2jsl2.

The following set of starting values has always yielded, in practica, a conver- Fordent iterative process:

n

E aij wij cl Ii O i=l

L aij

(6) n m) = 1,K2,=K 1 mm E aij Wij Vlvi

0 i_ YJi n

Eaij

Forhe difficulty witn this approach to solving location-allocation problems is that equations (4) must be solved for every possible combination of ail to determine the minimum from this set of solutions. This will be the adsolute teinimum sougl-st and the corresponding source locations and destination allocations will also be known.

For n destinations and mn sources this eintails solvinrg equation.s (4) S(n, 7n) times pvhere S(n, fn) is the Stirliorg iumber of the second kind and is given dby:

(7) 'S(n, m) = ;W!kE(k 1

-)k(M - k)n

Fvor piroblems of industrial interest this caii involve amounts of computation that aire pr-ohibitive even on digital computers. Fvurther, for the general case where costs are non-linear, discontinuous, etc. it may be difficult to even state the cost function and solve the corresponding minimization probrlem in any simple closed form.

In order to treat the problem of very large destination sets (n<500) and non-linear costs, the author has developed some heuristic methods which appear to be fairly good in practice. They have been used to solve many industrial problems witlh apparently good results.

III. CALCULATION OF TIHEORETICAL BOUNDS

Flor small problems it is possible to compare the efficiency of heuristics against the solution obtained by examining all possible allocations and their minimum



40 LEON COOPER

location points. For larger problems this is Inot feasible. In order to have an outside lower bound for the solution, against which to compaie the results obtained for a given problem by the several heuristics devised, some consideration has been given to the calculation of theoretical lower bounds. It is also shown hov to calculate a theoretical upper bound.

THEOREM 1. For an n destination, 1 source location-allocation problem, a lower bound foJr the mninimum value oJ the objective function is

n k-i >I 7dkl

n - 1 k_2 1=1

wherie dkl = [(XrDk - XD1 ) + (YDk - YD2)2].

Proof: Let i = the distatnce froii the minimum point to the ith destiniation. Therefore

n

f= '0i (Pi

For any pair of destiniatiorns it is clear that by the triangle inequality it must be true that:

(Pk + V I - dkl

(the equality would hold if the minimum point and the destinationi poilnts are collinear). Such ine(ualities can be written for each pair of destinatioins:

'Pk + SPO ? dkI (k = 2, * * *, n; I = 1, ... ,k - 1)

There will be n(n - 1)/2 distances dkm and therefore each pi will occur n - 1 times in tlhe summations:

n k-i

(n -1)(ol + (n-1)Vp2 + *s+ (n -1)0 (Ikl k=2 1l1

or 1 k-i

(8) <, 1 E z~~~~~1:dkl. n -1k=2 1=1

Therefore the quantity on the right is a lower bound. As an example of this lower bound consider the following 5 destiination -1

source problem.

(XDI, yDI) = (31, 11)

(XD2, YD2) = (41, 1 ) (XD3, YD3) = (50, 2)

(XD, YD4) = (42, 12)

(XVD5, YD5) = (48, 13)

The co-ordiniates of the minimuim point are (42.874, 9.200)




5

SP = S S = X 12.010 + 8.413 + 10.130 + 2.933 + 6.381 = 39.867,

n k-i

1 E Edk = (127.389) = 31.847, n - 1 k-2 1=1

and 39.867 > 31.847. Let us now consider the n destination, m source problem. THEOREm 2. For an n destination, ni source location-allocation problem, if there

are at least 3 destinations allocated to each source, then a lower bound for the mini- mnum value of the objective function is one-half the sunm of the n smallest distances in the 11 dkc Jj matrix.

Proof. Let (p,j be the distance from the jth source to the ith destination. If Ij is the set of destinations allocated to the jth source then

'p= S?2 j j=1 iEfIj

In advance we do not know these allocations. Consider some one source, j and the subset of destinations that are allocated to it, Ij . Without loss of generality, let these destinations be denoted D1, D2, - *-, D, with co-ordinates (XDI,

YD1), (XD2, yD2), -

* -

, (XDp , YDP). If the source co-ordinates are (x;, yj), then the distances from the minimum point to this subset of destinations are given by:

=i [(XD, - Xj)2 + (Yi _ y 2]1/2, i = 1, 2, , p.

The distances between any pair of destinations in the subset Ij are given by:

dkz = [(XDk - XDI) + (YDk - YDI)]2 k, 1 = 1, 2, ... 2 P.

For each pair of destinations Dk, DI the following triangle inequality holds:

jkO + ;p >- dkl.

It is clear that each Okj is involved in two inequalities so that: P p-1

(9) 2? (k (=P >= dk,k-1 k=I k-i

There are p distances in the right hand summation. There will be an equality such as (9) for all j = 1, 2, * * n, m, since theie are

171 subsets of destinations and a source for each. Since we do not know in advance the allocation of destinations to sources but we do know that some sum of n distances divided by two is less than or equal to so = j , a lower bound can be obtained by taking one-half of the n least distances in the fl i matrix.

As an example of this lower bound consider the following 7 destination, 2 source problem:

(XDI YDI) = (15, 15)

(XD2, YD2) = (5, 10)



42 LEON COOPER

(XD3, YD3) = (10, 27)

(XD4 X YD4) = (16, 8)

(XD5, YD5) = (25, 14)

(XD6, YD6) = (31, 23)

(XD7, YD7) = (22, 9)

The distance matrix, dkl 11 is:

0 11.18 0 13.00 17.72 0 7.07 11.18 19.92 0

10.05 20.40 19.85 10.82 0 17.89 29.07 21.38 21.21 10.82 0 15.65 25.50 12.17 21.84 15.30 10.82 0

The exact minimum sum of distances, so for this problem is so = 50.45 with destinations 1, 2, 4, 5 allocated to one source and 3, 6, 7 to the other. The lower bound is calculated as one-half of the sum of the seven smallest matrix entries, i.e.,

= 5045 > (7.07 + 10.05 + 10.82 + 10.82 + 10.82 + 11.18 + 11.18)

- 2(71.94) = 35.97

An upper bound is easily calculated as the next result shows. THEOREM 3. For an n destination, m source location-allocation problem, an upper

bound for the minimum value of the objective function is given by: n

[(XDi ) + (YDi - y)2]1/2 i-1

n 1 n

whei e x = -X E XDi , - E YDi nl i n_i_

Proof. The proof is obvious. Either the point (x, y) is the minimum point or it is not. In either case, it is true that

n

_ E [(XDi - X) + (yDi - y)2]1/2 i=l

The point (x, 9) is the starting point for the iterative calculation given by equation (6) for the single source exact calculation of the minimum point.

In summary: For a single source problem,

1 n k-i n

E dkl < (p -< [(XDi 2) + (YDi 1)]12 n-1 k-2 lI i2

and for this problem the bounds are fairly close. For an mn source problem




one-half the sum of (n smallest distances, dkl)

n

- ? -j [(XDi - X) + (YDi

- y)2]1/2

i=

and for this problem the bounds are not very close. However, the lower bound can be used to rank heuristics.

IV. HEURISTIC ALGORITHMS

A basic observation regarding location-allocation problems is the following: If, for a set of n destinations and in sources, the location of the sources is known,

the deternmination of the optimal allocation is trivial. It is merely the set of weighted distances (or costs in the general case) which is a minimum. Con- versely, if the allocation is fixed, the determination of the optimum location of the sources is merely the calculation described above with fixed (known) aii . This observation will be used, as well as other considerations to be mentioned later, in the generation of heuristic methods. We will now list and discuss four different heuristic algorithms (as well as combinations of them) that have been employed. Following this, numerical results will be presented comparing results obtained with all of them.

We shall assume that the information available for the solution of these location-allocation problems by any of these methods consists of:

1. (XDi, YDi) i = 1, * , n 2. Values of mn, n. 3. A cost matrix D = dij 1 representing the unit costs of shipping, servicing,

etc. between the ith destination and jth source. In general, dkz # dlk (this is not required if distance matrix is calculated in lieu of costs).

A. DESTINATION SUBSET ALGORITHM

This method makes the assumption that if one considered all possible subsets of m destinations, from the total set of n destinations, as sites at which to locate sources, then one of these should provide a close approximation to the optimum location of sources and, more important, would yield the correct allocations.

Previous work on this method [1] indicated that this was likely. Further, once the "optimum" subset has been found and its allocation noted, one can then solve the exact location problem, with fixed ai,, by means of equations (5) and (6).

An excellent method for generating combinations of r objects taken q at a time within a digital computer is given by Lehmer [3] and was the method programmed for use in this study.

There is no assurance that by considering all possible destinations taken m at a time to be the source locations, we have necessarily included one that has the correct (optimum) allocation to the m sources. (Of course, once the optimum allocation has been found, the optimum source locations can be determined by solving equations (4) and this solution would be the absolute or global optimum



44 LEON COOPER

sought.) However, because of the lack of a sharp minimum it seems likely that this method gives very good answers. There is a great deal of numerical data to substantiate this belief as well as some simple theoretical considerations. (See Cooper [1].)

There are several problems to be discussed concerning this algorithm. The logical thing to do when finding the best set of destinations at which to locate the sources by this heuristic would be to use this allocation and solve equations (4) (hereafter referred to as the "exact location method") for the optimum source locations. This generally results in some improvement in the sum of costs or distances. However, it is by no means clear (and in fact, it is not always true, as will be seen from numerical data) that by taking the next best or even con- ceivably, some far from best set of destinations at which to locate sources (as indicated by the destination subset algorithm) that subsequent improvement by the exact location method might not result in a final answer which is lower than that of the best set of destinations by the destination subset algorithm after it is improved by the exact location method.

Further, it is not even clear that the correct allocation is always included by considering only the destinations as possible source locations.

These doubts will always exist since this method is not exact. However, because of the nature of the minimum, the heuristic appears to be quite adequate [1].

Unfortunately, even this heuristic requires very large amounts of computation time. One has to calculate sums of costs and distances a number of times equal to n!/in!(n - m)! in order to choose the "minimum". For large n, and even more important, for large m, even this simple computation of sums becomes excessive. With very careful hand coding a problem with n = 60, mn = 4 took 3A hours on an IBMI 7072. Problems with larger n and m are certainly of practical importance. Therefore the need exists for less time consuming methods.

B. RANDOM DESTINATION ALGORITHM

In this method one also assumes that the destination set is a very favored set from the point of view of locating sources. The method consists of generating m random numbers normalized to be integers between 1 and n. This set of mn random numbers determines which destinations are to be source location points. The optimum allocation for each destination is determined by choosing the source with the least cost or distance for the particular destination.

This procedure is repeated as many times as desired, the general notion being that a random selection procedure such as this may produce the best or near best set of destinations at which to locate sources with much less computation than the exhaustive examination of all n!/m!(n - m)! cases by the destination subset algorithm.

This general idea of random generation of feasible solutions has been employed by Heller [4] in a study of the job scheduling problem.

It is clear that the random destination algorithm can be no more successful than the destination subset algorithm. However, since all but one (or perhaps




few) of the allocations examined by the destination subset algorithm are of no interest, it is certainly possible that a random generation method may find an acceptable (if not optimal) solution fairly rapidly.

The question of how many trials are required to achieve a reasonable degree of confidence in the answer with the random destination method is of some importance. As will be indicated later (see Figure I), the distribution of solutions in the sample space of the random destination method is not a normal distribution. One can assume a normal distribution (this is not too bad since we are only interested in the left hand side of the distribution) and use various sequential estimation techniques [5], [6] for estimation of mean and variance to decide when to terminate the generation of random trials. For this problem, the gain in knowledge obtained by this procedure was outweighed by the additional computer time and effort required. In a practical problem one would compute as large a sample of the total number of C,,' = n!/m!(n - m)! solutions as one could afford. A practical albeit crude rule, based on empirical observation, is as follows: The minimum solution was found in all cases tested to lie between 12 and 2 standard deviations from the mean so this simple statistic was calculated at intervals in the computer program and the computation was terminated either by a preset number of trials (if the problem was a small one) or when a solution was found within a certain specified number of standard deviations from the mean.

To use the example of the problem whose distribution of solutions is shown in Figure I:

(?.in = 145.657

Standard deviation (a) = 38.817

illean (rp) = 220.602

ss- a = 181.785 o- 2a = 142.968

After choosing the lowest sum of distances by this method and the corresponding allocation, we can employ the exact location method with the allocation found. This always results in a smaller sum of distances. Interestingly enough, this combination of methods has been superior, in a few cases, to the best three solutions of the destination subset algorithm after subsequent improvement by the exact location method. This will be examined later in more detail in the numerical results.

C. SUCCESSIVE APPROXIMATIONS ALGORITHM

The basic idea here is again to find an approximate algorithm which will be less time consuming than the destination subset algorithm. We do this by first applying the destination subset algorithm for a two source problem, i.e., we generate all possible combinations of the destinations taken two at a time and

allocate optimally. This is very rapid since (2) is not an unreasonably large

number of combinations to calculate for ranges of n of interest.



46 LEON COOPER

We now consider that we will add a source. We do this by placing the third source at a destination and then determiining if some of the destinations previously allocated to the first two sources might not be better allocated to the new source. We repeat this by considering the placing of the third source at each of the destinations and finally choose the best of all of these destinations for the placing of the third source. We also have the new allocation of destinations to the three sources. The entire process is now repeated for a fourth source, etc. until the number of sources equals m.

This is a finite and fairly rapid calculation with reasonable but not unusually good results as the numerical comparisons will show.

D. ALTERNATE LOCATION AND ALLOCATION ALGORITHM

The concept of this algorithm is extremely simple. The idea is to divide the destination set into m subsets (approximately equal). For each of these in subsets we determine the optimum single source location using the exact location method. Now we examine each destination to determine if it is possibly closer to one of the other sources than the one to which it was initially allocated. If there are any changes in allocation we again locate the sources using the exact location method.

This process is continued, alternately re-allocating destinations and then optimally locating sources, until no further changes are possible. This is a monotone-decreasing convergent process but it is not necessarily true that it will converge to the optimum solution.

It is fairly obvious that the destination subset algorithm, the random destination algorithm, and the successive approximations algorithm are related to each other, whereas the alternate location and allocation algorithm is distinctly different from the other three.

The destination subset algorithm examines as possible sites for sources all possible subsets of m members of the total destination set and chooses the best. This is, both from practical and theoretical considerations, an excellent approximation but costly in terms of computation time. If D is the set of all possible subsets containing m members of the total set of n destinations, then it is clear that the random destination method merely generates, in a random fashion, members of D. If continued for a sufficient period of time all solutions (members) in D would be genlerated. Hence, if R is the set of solutions generated by the random destination method, R c D.

The successive approximations algorithm also considers the destination set as the basic set from which source locations will be chosen. However, it does this in a rather complex fashion. There is no simple relationship between the "building-up" of the solution by the successive approximations algorithm and the destination subset algorithm, as there was between the random destination and destination subset algorithms. However, the final solution is a single member of D, i.e., if s is the solution obtained by the successive approximations algorithm then, s E D.

The alternate location and allocation algorithm is basically different from




the other three algorithms in that, from the outset one employs a technique (the exact location calculation) to leave the destination set in locating sources. Here one obtains a single solution rather than examining a vast array of possible candidates. One could regard the monotone-decreasing sequence of solutions obtained on the path to the final converged solution a set of solutions but they are a set of solutions never considered by the other algorithms, i.e., if A is this sequence (including the final solution), then in general:

DnA = 0, RnA = 0, s E A.

V. DISTRIBUTION OF SOLUTIONS

Before showing comparisons of numerical results for the various algorithms it is of some interest to consider the point previously alluded to, namely, the existence of many near-optimal solutions, by examining a particular problem.

Let us consider a problem with 15 destinations and 3 sources and with the following destination set:

Destination Co-ordinates (XDi YDi)

1 (5, 9) 2 (5, 25) 3 (5, 48) 4 (13, 4) 5 (12, 19) 6 (13, 39) 7 (28, 37) 8 (21, 45) 9 (25, 50)

10 (31, 9) 11 (39, 2) 12 (39, 16) 13 (45, 22) 14 (41, 30) 15 (49, 31)

The optimum solution to this problem is a minimum sum of distances = 143.209 with sources located at:

(xI, Y2) = (8.888, 14.466)

(x2, Y2) = (20.997, 44.998)

(X3, Y3) = (40.361, 17.968)

with allocations as follows:

n 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

mrl 1 1 2 1 1 2 2 2 2 3 3 3 3 3 3

The best solution found by the destination subset algorithm was 145.057 and this solution (with the correct allocation) after subsequent improvement by the exact location method yielded 143.209.



48 LEON COOPER

There are two ways we can indicate the lack of a sharp minimum. The first is by indicating the distribution of the sum of distances obtained by the destination subset algorithm. There are a total of C36 solutions calculated and C16 = 455 solutions. The solutions (sum of distances) obtained between the minimum (145.057) and 160.00 (about 10% higher) are:

145.057 146.238 147.035 148.217 155.198 155.537 156.289 156.379 157.092 16 solutions. 157.177 157.868 158.268 158.358 159.049 159.330 159.847

E 16

+ ? ? t > c o ? ? jg r

0 a

? D

0 a

a P= MnIMUMn Sum c>; )isA01"ce5

FIiGURE I



LOCATION-ALLOCATION PROBLEAIS 49

Figure I shows a frequency distribution of the entire set of solutions agaiiist the sum of distances. The worst solution found was 358.879.

It can be seen that, from an economic point of view, even without subsequent improvement by the exact location method, any one of the first 4-5 % of the solutions would be acceptable. In other words, 1 % or even more for some ap- plications, of the total number of solutions would be acceptable.

Another way of graphically portraying the lack of a sharp minimum can be seen in Figure II. The fifteen destinations are shown and the 3 destinations at which the sources were to be located corresponding to the minimum sum of distances are circled. The arrows indicate the allocations. If this allocation is used and the equations solved by the exact location method, we find that the optimal source locations are given by the crosses and that the minimum sum of distances is 143.209. It can be seen, then, that a relatively small change in the objective function (sum of distances) corresponds to a relatively large perturbation in the location of the sources. It is this relative insensitivity to source location with correct or near-correct allocations which makes the use of heuristics feasible in this problem. This same situation frequently prevails in other combinatorial optimization problems of economic importance.

3404/

xo~~ I0 _ . X 7

O & .~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~.

?10 L.o 30 40

FIGURE II



50 LEON COOPER

TABLE I

Case Destination Random* Successive Alternate Loc. DS Subset Destination Approximation and Alloc. + Exact RD. + Exact

1 364.354 375.276 379.238 389.893 360.018 363.371 2 314.709 328.879 326.965 311.668 311.670 311.670 3 327.398 329.085 351.421 313.941 323.666 313.944 4 204.476 214.415 226.194 204.093 204.090 204.090 5 359.258 369.439 359.258 355.399 354.341 366.331 6 348.344 355.512 369.963 344.010 343.744 353.496 7 233.147 237.203 247.734 262.797 231.872 231.872 8 318.466 328.572 333.120 317.820 313.541 323.033 9 211.907 218.650 211.910 231.878 211.089 211.089

10 201.579 204.516 227.116 198.707 198.708 201.428

* 400 trials were generated.

TABLE II

n = 60 m= 4

Destination Subset ............................................. 620.566 * Random Destination .* ....... .....638.494 Successive Approximation .................... ................. 652.799 Alternate Location and Allocation .............................. 636.769 Destination Subset + Exact .................................... 619.656 Random Destination + Exact .................................. 627.710

* 4000 trials were generated.

In the course of these investigations some 400 problems have been solved. In addition, some of these techniques have been applied to the solution of industrial problems in a certain company. All the problems examined exhibited the kind of minimum property shown in the example and in Figure II.

VI. NUMERICAL RESULTS

Table I shows some typical results for 10 problems for which n = 30, m = 3. The column headings indicate the algorithm used. Since the exact location method could be used to refine the best answer from either the destination subset algorithm or the random destination algorithm, this was done and the results are also shown in the table. Table II shows a comparison of results on a single problem with n = 60 and m = 4.

Earlier studies [1] indicated that the destination subset algorithm after improvement by the exact location method gave the optimum or a near optimum result. The optimum is achieved if the allocation (the division of the destinations into mn subsets) is the correct one. The results in Table I (with the excep- tion of the interesting case 3) bear this out. The lowest sum of distances obtained (except for case 3) was that of the destination subset algorithm followed by subsequent exact location given under the heading "D.S. + Exact". How- ever, it should be borne in mind that the destination subset algorithm is pro-




hibitive in terms of computation time for large problems. Hence for such problems one of the other methods is mandatory. A comparison of computation times is instructive. For the problem given in Table II where n = 60 and m = 4 the computation times are as follows:

Destination Subset 3 hrs. 30 min (18 hours) Random Destination 8 min. 26 sec. Successive Approximation 48 sec. Alternate Location & Allocation 2 min. 15 sec. D.S. + Exact Approximately equal to D.S. R.D. + Exact Approximately equal to R.D.

All computer programs except the destination subset method were coded in FORTRAN and computations run on the IBM 7072. The destination subset algorithm was hand coded in Autocoder. We estimate it would have run 18 hours if written in FORTRAN. Even with the carefully hand coded version this problem took 32 hours. It is clear that although, in general, it gives extremely good results it is too time consuming. The next best method appears to be the random destination algorithm and subsequent improvement by the exact location method. It was conjectured earlier [1] that this might indeed prove to be the case. It is indeed quite striking that in the case given in Table II with only 4000 trials (less than 1 % of the possible allocations) an answer was obtained in 8 minutes that was only 1 % higher than a 32 hour calculation and which was 1.8 standard deviations from the mean solution of the 4000 trials. The same general behavior is evident in Table I.

The alternate location and allocation algorithm is also satisfactory from a practical point of view. The successive approximation algorithm is somewhat less satisfactory but extremely rapid. A larger number of statistical data to substantiate these points will be presented later in this discussion.

It is difficult to compare, in any simple way, the relative accuracy of eaclh heuristic for a given amount of computer time. The difficulty arises in that while, for example, the successive approximation algorithm achieves accuracy/ minute at a rate which is 10 times that of the random destination algorithm it can not continue to improve its value beyond a certain point (the converged value) while the random destination algorithm can continue to improve and in fact to approach the results of the destination subset algorithm. The limitations on accuracy are evident in Table III.

Case 3 in Table I is an extremely interesting one since this was the first example the author obtained, in which the destination subset algorithm did not lead to the optimal allocation although this had been conjectured. Previously [1] no such cases had been found. It can be seen here that the combined use of the random destination algorithm and the exact location method has resulted in a lower sum of distances than the combined use of the destination subset algorithm and the exact location method. Therefore, the destination subset algorithm result is not the optimal allocation.

In order to obtain better numerical comparisons between the methods, a



52 LEON COOPER

computer program was written to randomly generate 100 destination sets with n = 40 and to solve these 100 problems with nt = 3 with all the methods previously discussed. In addition, not only was the best allocation from the destination subset algorithm subsequently improved by the exact location method but so were the next two best allocations.

For these 100 problems, the mean per cent error from the best (lowest) sum of distances found was calculated in an effort to compare the methods. The results are given in Table III.

The mean per cent errors given in Table III indicate the relative accuracy of each of the heuristics. The errors shown confirm the evaluations given earlier. A comparison of the results obtained by lowest, second lowest and third lowest sum of distances found with the destination subset algorithm and subsequent improvement by the exact location method indicates that despite the occasional occurrence of an incorrect allocation with the lowest sum of distances, on the average this is the best policy to follow if one is employing the destination subset algorithm. The relatively small per cent differences between the lowest, second lowest and third lowest ansswers found after subsequent improvement by the exact location method is another indication of the lack of a sharp minimum and why heuristic methods can be employed with some confidence. The relatively small difference in the mean per cent errors of the combined use of random destination algorithm and exact location method compared to the combined destination subset algorithm and exact location method is of extreme importance when one contrasts the difference in calculation time for each (e.g., for the

TABLE III MEAN PER CENT ERRORS (100 PROBLEMS)

Destination Subset ............................................. 0.948 Random Destination ........................................... 2.518 Successive Approximation ...................................... 7.086 Alternate Location and Allocationi .............................. 2.582 R. D. + Exact ................................................. 0.578 D. S. + Exact (First) .......................................... 0.212 D. S. + Exact (Second) ........................................ 0.292 D. S. + Exact (Third) . ........................................ 0.355

TABLE IV MEAN PER CENT DEVIATION FROM LOWER BOUND (100 PROBLEMS)

Destination subset .............................................. 360.1 Random Destination ........................................... 367.2 Successive Approximation ...................................... 387.9 Alternate Location and Allocation .............................. 367.7 R. D. + Exact ................................................ 35 R. I D. S. + Exact (First) .......................................... 356.8 D. S. + Exact (Second) ........................................ 357.1 D. S. + Exact (Third) ..................| 357.4



LOCATION-ALLOCATION PROBLEMNS 53

n = 60, mn = 4 problem, 81 minutes for the random destination compared to 31 hours for the destination subset) it is apparent that the best practical method of solving large location-allocation problems is with the use of the random destination algorithm and subsequent improvement by a single calculation of the exact location method.

In order to check the evaluation of the various heuristics given above which is based on the mean per cent errors from the lowest sum of distances found (Table III), the lower bound given in Theorem II was calculated for all 100 problems and the mean deviation from the lower bounds calculated. This is shown in Table IV.

It can be seen from Table IV that the ranking of the heuristics is exactly the same as that given in Table III. Further, the relative differences between heuristics is approximately the same. This result provides some additional assurance that the previous conclusions are valid.

SUMMARY

A number of heuristic algorithms have been devised for solving large location- allocation problems. The algorithms have been extensively tested and several are in use in at least one industrial corporation. It has been found that one of the algorithms, the random destination algorithm, appears to provide satis- factorily close approximations to the optimal solution to location-allocation problems in reasonable amounts of computation time.

ACKNOWLEDGMENT

The author gratefully acknowledges the support of the Washington Univer- sity Computer Center through NSF grant G-22296.

REFERENCES

1. L. COOPER, Location-Allocation Problems, Operations Research 11 No. 3 (1963). 2. SIAM REVIEW, Vol. 4, No. 4, October 1962, p. 394 Problem 62-11, submitted by K. Eise-

mann. 3. D. H. LEHMER, Teaching Combinatorial Tricks to a Comtputer, Proceedings of Symposia

in Applied Mathematics, Vol. X, Combinatorial Analysis, American Mathemati- cal Society.

4. J. HELLER, Some Numerical Experiments For An MXJ Flow Shop and Its Decision- Theoretical Aspects, Operations Research 8 No. 2 (1960).

5. J. MOSHMAN, The Application of Sequential Estimation To Computer Simtulation and Monte Carlo Procedures, J. Association for Computing Machinery 5 No. 4 (1958).

6. A. WALD, Sequential Analysis, John Wiley and Sons, New York (1952).



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Heuristic Methods for Location-Allocation Problems