A New Method for Quickly Solving Quadratic Assignment Problems


 Job Giles Fleming
 1 years ago
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1 A New Method for Quickly Solving Quadratic Assignment Problems Zhongzhen ZHANG School of Management, Wuhan University of echnology, Wuhan 437, China address: Abstract he quadratic assignment problem is solved by a pivoting based algorithm proposed by the author for quadratic programming where the lower bounds of variables taking or alternatively in the computational process. Experiments are conducted by using instances of Nug, ai, Chr, Had, Rou, Bur, Esc and ho for n = to 4. Most problems including the famous Nug3 are solved to optimum in several seconds to several minutes. About 8 optimal solutions are presented that are never published in the literature. Introduction here are a variety of methods for solving the quadratic assignment problem (QAP) such as the Branch and Bound method, greedy randomizing adaptive search procedure, robust taboo search, simulated annealing approach, and so on. he Branch and Bound method is able to obtain an optimal solution. But it requires too much computational time. It is well known that in Nug3 was solved in a week by computers which began with a very small bound 66 while the minimal value was found to be 64. Most other methods are heuristic and the obtainment of an optimal solution is by chance to some extent. his paper proposes a method for solving QAP in a special way that is completely different from the known methods. he QAP is solved as a quadratic programming by a pivoting based algorithm proposed by the author. It may be called a parameter quadratic programming where the lower bounds of variables taking or alternatively in the computational process. In this method all the n variables x ij are divided into two groups, one group has n variables called nonbasic and another group has n n + variables called basic. Usually all the basic variables are zeros. For a given assignment (the current solution), by setting k basic variables to be, if a feasible solution is formed, it is called a k order basic assignment, k n. For small k, say k = or, the k order basic assignments and their values of the objective function can be calculated in very small amount of computation. In this way a better solution may be found which has a value of the objective function smaller than that of the current one. Another way to search for a better solution is to construct a series of k order basic assignments as the current basic assignments from small k to larger k, k < n. In the meanwhile pivoting operations are carried out in which basic variables with values zero and negative costs are interchanged with nonbasic variables so that a high order basic assignment with small value of the objective function is translated into a low order one and is easily detected by the parameter method. It is called contraction effect of the pivoting operation and is the key of our method. he fundamental method he quadratic assignment problem involves assigning n facilities to n fixed locations so as to minimize the total cost of transferring material between the facilities. Let the n by n matrices D = (d ij ), F = (f ij ) and X = (x ij ), where d ij is the distance from location i to j, f ij is the flow from facility i
2 to j, and x ij = if facility i is assigned to location j, otherwise x ij =. he total cost can be formulated as tr(dxf X ) [] and the problem is as follows: min tr(dxf X ) s.t. j= i= n x ij n x ij =, i =,,,n =, j =,,,n x ij = or, i, j =,,,n. (.) where tr( ) denotes the trace of a square matrix. We will solve it as a quadratic programming problem in which the constraint x ij = or, i, j =,,,n is replaced by x ij l ij, i, j =,,, n where l ij = or, and at the very beginning all the l ij are set to zeros. hat is we shall first solve the problem min x Hx s.t. Ax = b Where H = d( F + F ) df + d F M dnf + d nf x. (.) d d d n F + d ( F + F F M + d F n ) F L L L d nf + d nf d nf + d nf, M d ( + ) nn F F is an n n Hessian matrix, x = (x,, x n, x,, x n,,x n,, x nn ) is an n dimensional column, A is the incidence matrix of the n nodes to n nodes bipartitioned graph and b is a column vector of components and as determined by the first two equality constraints of (.). For brevity, it is supposed that a redundant row of (A b) is deleted. herefore A is an (n ) n matrix. We shall solve (.) by the pivoting algorithm presented in [] which is for solving quadratic programming. his algorithm is divided into three stages: (i)construct an initial table, (ii) preprocessing and (iii) main iterations. he initial table for solving (.) is able. Initial table h a e h e a H A A O b Let A = (A, A ) where A is a nonsingular matrix of dimension (n ) (n ).
3 Correspondingly, able. is partitioned as able.. able. Initial table (the partitioned form) h I h II a e I e II e III H H A H H A A A O b he preprocessing stage is to solve the inverses A and A which yields able.3 where the columns of basic equality constraints and the corresponding rows are deleted. able.3 Result of the preprocessing e II h I h II A H A e I A A O σ II A b Where H = H H A A A A (H H A A ) is called a reduced Hessian matrix, A A is called a reduced incidence matrix, and σ II = H A b A A H A b. A being nonsingular means that A is corresponding to a spanning tree [3] in the n nodes to n nodes bipartitioned graph. Each branch of the tree is associated with a value of x ij which is the solution of the equality constraint Ax = b and is given by A b here. In the computing programs of the author, n initial feasible solutions are designed for experiments. ake n = 4 for example, they are 343, 3434, 3434, 434 which are called extended assignments meaning that x = x = x 33 = x 44 =, x = x 3 = x 43 =, x = x 3 = x 34 = x 4 =, x = x 33 = x 44 =, x 3 = x 4 = x 3 = x 4 =, x 3 = x 34 = x 4 =, x 4 = x = x 3 = x 43 =, x 4 = x 3 = x 4 = respectively. In the main iterations stage, all the pivots are elements of A A and A A. As soon as a pivoting operation is carried out on a nonzero element ( or ) of A A, another pivoting operation is carried out on the symmetric element in A A. It is called a double pivoting. Another form of the pivoting operation for quadratic programming is called principal pivoting. But it is never used for quadratic assignment problems. 3 he basic assignment and local search Now let us consider a detailed form of able.3 as shown by able 3.. 3
4 able 3. A table in the main iterations stage e e N h N+ h N+M h h N e N+ e N+M w w N w N+, w N+M, w N w NN w N+, N w M+N, N w N+, w N+, N w N+M, w N+M, N σ σ N σ N+ σ N+M Where N = n n +, M = n ; σ N+,,σ N+M are or, i.e., the current solution is feasible; e i is the coefficient vector of x i and h i is the coefficient vector of the complementary inequality of x i. Here we use x i (i =,,, n ) rather than x ij to represent a variable. In able 3., each basic unit vector e j is corresponding to x j. Now let us change the right side terms of k basic inequalities, say x, x,, x k, from to. By (.) of [], the last column becomes k (σ + w,, σ j N + j= k j= w Nj, σ N++ k w N +, j j=,, σ N+M + k w N + M j j=, ). (3.) his operation is denoted by {e, e,, e k } {e +, e +,,e + k } or {e, e,, e k } {e, e,, e k } +. If the last M components of (3.) are or, we have a new feasible solution and say it to be k order basic assignment. Obviously, k can not greater than n. And it is easy to prove there are n! k order basic assignment altogether for k = to n. As a special case, the current feasible solution is called order basic assignment. By (6.) of [], the change of the value of the objective function is w w Δf = (,,,) M w k w w w M k L L L w w w k k M kk M σ σ + (,,, ) M σ k (3.) which is called the cost of the operation {e, e,, e k } {e +, e +,, e + k }. Particularly speaking if we increase the right side term of just one basic inequality, say x j, by, the cost is Δf = wjj + σ j (3.3) he above operation is call + operation and denoted by e j e + j. Since the + operation involves a small amount of computation, it is the most common way to find a better solution. For this reason, we call it to be + search as well. Another way to find a better solution which requires small computation is the + search {e i, e j } {e + i, e + j } whose cost is Δf = (wii + w jj + w ij ) + σ i + σ j. (3.4) Now suppose that we have a table where k basic unit vectors e j s are associated with x j and which determine a feasible solution of the value f of the objective function. We say the k 4
5 vectors to be catalyzed basic vectors, denoted by e + j s, and say other basic unit vectors to be noncatalyzed basic vectors. When there are catalyzed basic vectors in a table we can also change one of them to noncatalyzed one, say e + j to e j, and the cost is Δf = wjj σ j. (3.5) If the change is feasible and the cost is negative, we get a solution whose value of the objective function is less than f. his operation is denoted by e + j e j and is called search. Similarly there are search and &+ search. he cost of {e + i, e + j } {e i, e j } is and the cost of {e i +, e j } {e i, e j + } is Δf = (wii + w jj + w ij ) σ i σ j (3.6) Δf = (wii + w jj w ij ) σ i + σ j (3.7) We can also do &+, &+ or &+ to search for a better solution. But these operations especially the &+ requires too much amount of computation and seldom used in the computing program of the author. he feasible solutions obtained by operations +, +,,, &+, &+, &+ or &+ are in the vicinity of the current feasible solution or the current basic assignment. herefore the above operations may be called local search methods. 4 he pivoting operation Suppose that we have a table with a feasible solution where some basic unit vectors may be catalyzed and where a part of entries are shown by able 4.. able 4. he current table h s e r able 4. Result of the double pivoting e s h r e r h s w ss * w rs * w rs σ s σ r h r e s w ss / w rs / w rs w rs σ s w ss σ r σ r / w rs In able 4. w rs = or, σ r = or, and e s = e s if x s = or e s = e + s if x s =. Carrying out a double pivoting e r e s and h s h r there will be able 4.. By (6.5) of [] the change of the value of the objective function is Δf = wss σ r w rs σ r σ s (4.) Some special cases called forward (backward) descent pivoting and par pivoting are as follows. (i) Forward descent pivoting It means that (a) e s = e s, and there would be a feasible solution if we performed the + operation e s e + s ; (ii) w rs =, σ r = and wss + σ s <. After the pivoting there will be a feasible solution and the change of the value of the objective function is (ii) Backward descent pivoting wss + σ s. 5
6 It means that (a) e s = e s +, and there would be a feasible solution if we performed the operation e s + e s ; (ii) w rs =, σ r = and wss σ s <. After the pivoting there will be a feasible solution and the change of the value of the objective function is (iii) Par pivoting wss σ s. It means that e s = e s, w rs = and σ r =. he par pivoting is carried out usually when wss + σ s < even if it dose not change the value of the objective function. o see the effect of the par pivoting let us consider able 4.3 where a nonbasic unit vector e 5 is associated with a deviation. able 4.3 A table for par pivoting h h h 3 h 4 e 5 e e e 3 e 4 h 5 w w w 3 w 4 * w w w 3 w 4 w 3 w 3 w 33 w 34 w 4 w 4 w 34 w 44 * σ σ σ 3 σ 4 Carrying out a pivoting on and then on with asterisks, there will be able 4.4. able 4.4 Result of the par pivoting h 5 h h 3 h 4 e e 5 e e 3 e 4 h w w w w 3 +w w 4 w w w +w +w w 3 w +w 3 w w 4 +w 4 w 3 +w w 3 w +w 3 w w 33 +w w 3 w 34 w 4 w 4 w 4 +w 4 w 34 w 4 w 44 σ σ + σ σ 3 σ σ 4 Suppose that performing a + operation {e, e } {e +, e + } in able 4.3, in which some basic unit vectors except for e and e may be catalyzed, there results in a feasible solution. hen the change of the value of the objective function is Δf = (w + w + w ) + σ + σ. On the other hand the same solution and change can be obtained by performing a + operation e e + in able 4.4. If w + σ is negative enough, Δf may be negative and a better solution would be found in able 4.4 which requires a smaller amount of computation than the + operation does in able 4.3. he property of the double pivoting especially the par pivoting that translates some high order basic assignments into lower ones is called contraction effect and is the key to generate a better solution. Of course the double pivoting would change some low order basic assignments to high order ones in the same time. But it is not required by us. he par pivoting is frequently performed in order to generate a low order basic assignment with a small value of the objective function. It takes more 9% of the amount of computation in 6
7 solving a QAP. o see the effect of the par pivoting in details, let us consider the reduced incidence matrix and the associated deviation for n = 4 which is shown by able 4.5 where e ij is the coefficient vector of x ij, i, j =,, 3, 4. able 4.5 A reduced incidence matrix of n = 4 e e 3 e 4 e 3 e 4 e 3 e 34 e 4 e 4 e e e 33 e 44 e e 3 e 43 * From able 4.5 we see that the current solution is 34 and other 3 basic assignments are as follows. (i) order basic assignments (6 ones): e + 34, e , e , e , e , e (ii) order basic assignments (6 ones): {e, e 34 } + 43, {e 3, e 3 } + 34, {e 4, e 3 } + 43, {e 4, e 4 } + 43, {e 4, e 4 } + 43, {e 4, e 4 } (iii) 3 order basic assignments (8 ones): {e, e 3, e 3 } + 34, {e, e 4, e 3 } + 43, {e, e 4, e 4 } + 43, {e 3, e 4, e 4 } + 34, {e 3, e 34, e 4 } + 34, {e 3, e 34, e 4 } + 34, {e 4, e 3, e 4 } + 43, {e 3, e 34, e 4 } (iv) 4 order basic assignments (3 ones): {e, e 3, e 34, e 4 } + 34, {e 3, e 4, e 3, e 4 } + 34, {e 4, e 3, e 3, e 4 } Let e enter and e 3 leave the basis there will be able 4.6. able 4.6 Result of the pivoting e e 3 e 4 e 3 e 4 e e 34 e 4 e 4 e e e 33 e 44 e 3 e 3 e 43 From able 4.6 we can get 3 basic assignments as follows. (i) order basic assignments (5 ones): e , e , e , e , e (ii) order basic assignments (7 ones): 7
8 {e, e 3 } + 34, {e, e 4 } + 43, {e, e } + 34, {e 3, e } + 34, {e 4, e } + 43, {e 4, e 4 } + 43, {e 4, e 4 } (iii) 3 order basic assignments (9 ones): {e, e 4, e 4 } + 43, {e, e, e 34 } + 43, {e 3, e 4, e 4 } + 34, {e 3, e 4, e 4 } + 34, {e 3, e 34, e 4 } + 34, {e 4, e 3, e 4 } + 43, {e 4, e 3, e 4 } + 43, {e 4, e, e 4 } + 43, {e 3, e 34, e 4 } (iv) 4 order basic assignments ( ones): {e, e 3, e 34, e 4 } + 34, {e 3, e, e 34, e 4 } We see that 34 and 43 are order basic assignments in able 4.5 and are order ones in able 4.6; 34 and 43 are 3 order basic assignments in able 4.5 and are order ones in able 4.6; 34 and 43 are 4 order basic assignments in able 4.5 and are 3 order ones in able 4.6 due to e 3 leaving the basis. Example 4. Consider a QAP of n = 4 where the distance matrix and flow matrix are D = 4 4 and F = he table associated with the extended assignment 343 is as following. able 4.7 he initial table in the main iterations stage h h 3 h 4 h 3 h 4 h 3 h 34 h 4 h 4 e e e 33 e 44 e e 3 e 43 e e 3 e 4 e 3 e 4 e 3 e 34 e 4 e 4 h h h 33 h 44 h h 3 h 43 σ i * * Where e ij is the coefficient vector of x ij and h ij is the coefficient vector of the associated inequality, i, j =,, 3, 4. he last column is the deviation of the nonbasic vector. he value of the objective function of the current solution is 58. From able 4.7 we see that the cost of e e + is 3 / 8 =. herefore the resulting order basic assignment 34 has a value of 58 = 46. Performing a forward descent 8
9 pivoting e e and h h to yield able 4.8 where the current solution is 34 with a value f = 46. able 4.8 Result of the forward descent pivoting e e 3 e 4 e 3 e 4 e 3 e 34 e 4 e 4 h h h 33 h 44 h h 3 h 43 σ i h h 3 h 4 h 3 h 4 h 3 h 34 h 4 h 4 e e e 33 e 44 e e 3 e * * Performing a + search in able 4.8 there are e + 34, e , e , e with costs, 6, 6, 8 respectively. herefore we can not perform a forward descent pivoting anymore. he cost of e 3 e + 3 is 6 / = 4 <. But it results in an infeasible solution. Let us perform a par pivoting e 3 e 3 and h 3 h 3 to yield able 4.9 where the current solution has no change. able 4.9 Result of the par pivoting e e 3 e 4 e 3 e 4 e 3 e 34 e 4 e 4 h h h 33 h 44 h h 3 h 43 σ i h h 3 h 4 h 3 h 4 h 3 h 34 h 4 h 4 e e e 33 e 44 e e 3 e
10 From able 4.9 we see that the cost of e 3 e + 3 is 6 / = < and which results in a feasible solution 34 with f = 46 = 44. he above example is very small. Generally speaking, after a table is set up we shall not only do the + search but also do the + search to find a better solution. In the case that there are catalyzed basic vectors in the table we may do the searches of,, &+, &+ and &+ besides the + and + searches. Meanwhile the descent pivoting and the par pivoting are carried out if the cost of the leaving unit vector is negative. 5 Results of experiments Several computing programs are made in Delphi 6.. hey are conducted in this way: First run the program from the first extended assignment until the optimal solution is obtained or a certain number of operation processes are carried out. If the optimal solution is not obtained, run the program from the second extended assignment, and so on. An entire operation process which doesn t find out a better solution is as follows. For a given feasible solution, (i) Do + search and + search combining with pivoting operations where no basic unit vectors are catalyzed; (ii) Construct a order basic assignment from which the number of catalyzed basic vectors is increased by the operation &+. Meanwhile local searches and pivoting operations are performed. (iii) Update the table by transferring the last k order basic assignment into a order basic assignment or construct another feasible solution by using previous feasible solutions, then return to (i). he operation &+ of (ii) is imposed no matter whether the cost is negative or not so as to increase the number of catalyzed basic vectors by every time until a certain number, say n \. Other operations are done only if their costs are negative. If a better solution is found in the midway of the above operation process, the operation process ends and then another one follows. he difficulty is how many operation processes should be set if each of which can not find out a better solution. If the number of operation processes is small, a better solution may be missed. And if the number is large, there would run much time without finding a better solution. able 5. gives the results of the program DelQAPpvt run on a personal computer of a GHz CPU and a G memory. Since the minimal values of the objective functions are known from QAPLIB Home Page, the experiment is designed as that as soon as the minimal value occurs or operation processes are carried out without finding a better solution the experiment ends for each initial extended assignment. he Max No. of processes is the maximal number of operation processes between two better solutions. Usually the value of the objective function of the initial extended assignment is big. Many better solutions (values of the objective functions become smaller and smaller) occur just in one or two operation processes. As the value becomes very small, the number of operation processes increases rapidly. From table 5. we see that most instances are solved to optimum in several seconds to several minutes.
11 able 5. Computing results of DelQAPpvt ime Initial Max No. of ime Initial Max No. of (seconds) solution processes (seconds) solution Processes Nug.9 3 Rou.4 3 Nug Rou Nug5.4 3 Rou Nug6a.47 4 Chra.8 9 Nug6b.7 Chrb Nug Chrc Nug Chr5a 7.9 nd 39 Nug Chr5b.9 69 Nug Chr5c Nug Chr8a Nug Chr8b.69 4 Nug Chra rd Nug Chrb Nug Chrc Nug Chra aia.7 3 Chrb 5.4 4th 55 aib.37 4 Chr5a.7 nd 7 ai5a Bur6a ai5b.7 4 Bur6b ai7a.3 96 Bur6c aia Bur6d nd 45 aib.36 Bur6e ai5a Bur6f ai5b ai3a rd Bur6g Bur6h st 4 8 ai3b Esc3a ai35a nd 67 Esc3b ai4b nd 4 Esc3c.3 Had Had st 43 8 Esc3d Esc3e.5.56 Had6 Had8 Had st Esc3g Esc3h ho st 5 9 ho Some new solutions are given in table 5. that are different from those published on QAPLIB Home Page. Each of Esc3ah has many optimal solutions. Just one of them is listed.
12 able 5. New solutions for some instances of QAP value Solution Nug Nug5 Nug6b Nug8 Nug Nug Nug Nug4 Nug5 Nug (5, 6,,, 4, 8,,,, 7, 9, 3) (9, 8, 3,,,, 7, 4, 3, 4,, 5, 6, 5, ) (, 5, 6, 5,, 4, 3, 4, 7,,,, 3,8, 9) (, 5,6, 5,,, 7, 4, 3, 4, 9, 8, 3,, ) (5,, 6, 4, 3, 7,, 5,, 9,, 4, 8, 3,, 6) (5, 3,, 8,, 7, 9, 3, 6,,,, 4, 5, 4, 6) (9, 3,, 6, 3, 4,,, 7, 5, 8, 5, 8,, 7, 6, 4, ) (6,, 7, 5, 7, 3, 8,, 5, 9, 6,,,, 4, 9, 3,, 4, 8) (9, 3,, 4, 8, 6,,,, 4, 3, 8,, 5, 9, 6,, 7, 5, 7) (7, 5, 7,, 6, 9, 5,, 8, 3, 4,,,, 6, 8, 4,, 3, 9) (5,, 3, 9, 3,, 4, 5, 6,, 6,, 8, 4, 7,,, 7, 8, 9, ) (7,,, 7, 8, 9,, 5, 6,, 6,, 8, 4, 5,, 3, 9, 3,, 4) (, 9, 8, 7,,, 7, 4, 8,, 6,, 6, 5, 4,, 3, 9, 3,, 5) (5, 3, 6,, 6,,, 8, 4, 4, 5,,, 9,, 7, 3,, 9, 8,, 7) (5, 4, 4, 8,,, 6,, 6, 3, 5, 7,, 8, 9,, 3, 7,, 9,, ) (7,, 8, 9,, 3, 7,, 9,,, 5, 4, 4, 8,,, 6,, 6, 3, 5) (, 5, 3, 6,, 4,, 6, 9, 7,,, 4, 3, 8,, 9, 5,, 4, 3,, 8, 7) (7, 4,, 9, 8,, 5,,, 8, 3, 4, 5, 6,, 7, 3,,, 3,, 9, 6, 4) (, 4, 3,, 8, 7, 4, 3, 8,, 9, 5,, 6, 9, 7,,,, 5, 3, 6,, 4) (4,, 6, 3, 5,,,, 7, 9, 6,, 5, 9,, 8, 3, 4, 7, 8,, 3, 4, ) (5,, 8, 4, 7,, 5, 6,,,, 8, 3, 4, 4, 5, 9, 6, 7, 3,,, 9,, 3) (3,, 9,,, 3, 7, 6, 9, 5, 4, 4, 3, 8,,,, 6, 5,, 7, 4, 8,, 5) (3, 3, 4,, 7,, 7, 4,, 4, 9, 6, 3, 6, 8,, 9, 8, 5,,, 5,,, 5) (7, 4, 8,, 5,,, 6, 5,, 4, 4, 3, 8,, 3, 7, 6, 9, 5, 3,, 9,, ) (,, 9,, 3, 5, 9, 6, 7, 3,, 8, 3, 4, 4,, 5, 6,,, 5,, 8, 4, 7) (5,, 5, 7,,, 3, 6,, 3, 4,,, 9, 8, 6, 4, 7, 4, 8, 9, 5, 6, 3,,, 7) (5, 8,, 9, 9, 7, 3,, 5, 5,, 8, 6,,,, 6, 3, 4,, 7, 3, 4, 7, 4, 6,) (3,, 5, 7,,,, 4,,, 5,,, 9, 8, 6, 6, 6, 3, 8, 9, 4, 7, 4, 5, 3, 7)
13 able 5. (continue) value Solution Nug8 Nug3 Chr8b Chra Chrb Chr5a Had4 Had Bur6a Bur6b (, 8,, 8,, 9, 8, 6, 6, 7, 9,, 5, 7, 4, 7, 4, 3, 5, 6,,, 5, 3, 4,,, 3) (, 4, 6,, 7, 3, 8, 4,, 9, 7,,,, 8, 3, 4, 9, 5,, 8,, 3, 5, 6, 6, 5, 7) (8, 5, 3, 7,, 6,, 5, 9,,, 7, 8, 8,, 6, 5,, 4, 7,, 9, 6, 4, 3, 3, 4, ) (5, 8, 7, 3, 4,, 3,,, 6, 3, 4, 7,, 8, 7, 9, 5, 6, 4,, 9, 9, 8, 5,, 6, 3,, ) (, 4, 3, 7, 8, 5, 4, 3, 6,,, 3, 5, 9, 7, 8,, 7, 8, 9, 9,, 4, 6,,, 3, 6,, 5) (,, 3, 6,, 5, 8, 9, 9,, 4, 6, 5, 9, 7, 8,, 7, 4, 3, 6,,, 3,, 4, 3, 7, 8, 5) (,, 3, 4, 6, 5, 9, 8,, 7, 5,, 8, 3, 7, 4, 6, ) (, 3,, 6, 4, 5, 7, 8,, 9, 3,, 6,, 7, 4, 8, 5) (6, 3, 9,,,, 5, 4, 8, 5, 7, 8, 4, 7, 3,, 6, ) (7, 4,,, 3,, 6, 3, 7, 6, 4, 9,,, 8, 5, 5, 8) (9,, 6,, 5,, 8, 3, 7, 6, 4, 7,, 4,, 5, 3, 8) (4, 5,,, 8, 9, 7, 6,, 5, 3, 4, 6,, 7,, 8, 3) (3,, 7, 8, 9,, 9, 4,,,, 6, 5, 8, 5,, 4, 6, 7, 3) (, 9, 3,,,, 6, 4, 7, 8, 7, 3,, 5,,, 9, 5,, 4, 8, 6) (5,, 5, 3, 8, 4, 6, 8,,, 4, 6, 3, 5, 4, 9, 3,,,, 7,,, 7, 9) ( 8, 3,,,, 5,, 4, 3, 6, 7,, 9, 4) (8, 5,, 6, 4, 9, 7,, 6,,, 7,,, 5, 3, 4, 9, 8, 3) (8, 5,, 4, 6, 9, 7,, 6,,, 7,,, 5, 3, 4, 9, 8, 3) (8, 5, 6, 6, 4, 9, 7,,,,, 5, 3,,, 7, 4, 9, 8, 3) (8, 5, 6, 6, 4, 9, 7,,,,, 7,,, 5, 3, 4, 9, 8, 3) (8, 5, 6, 4, 6, 9, 7,,,,, 5, 3,,, 7, 4, 9, 8, 3) (, 5, 6, 7, 4,, 3, 6,, 8, 9, 5,,, 8, 4, 3, 9,,, 7, 5, 6, 4,, 3) (5,, 6, 7, 4,, 3, 6,, 8, 5, 9,,, 8, 4, 3, 9,, 7,, 5, 4, 6, 3, ) (6,, 5, 7, 4,, 3, 6,, 8,, 9, 5,, 8, 4, 3, 9,, 5,, 7, 6, 4, 3, ) (, 5, 5, 7, 4, 6, 4,, 3, 8, 9, 5,,, 8,, 3, 9,,, 7, 6, 6, 4, 3, ) (7,, 5, 7, 4, 3,,, 3, 8, 5, 9,,, 8,, 3, 9,, 5,, 6, 6, 4, 4, 6) (5, 5,, 7, 4,, 3,, 3, 8,, 5, 9,, 8,, 3, 9,, 6, 7,, 4, 6, 4, 6) 3
14 able 5. (continue) Value Solution Bur6c Bur6d Bur6e Bur6f Bur6g Bur6h Esc3a Esc3b Esc3c (6, 7,, 7, 4,, 3,, 3, 8,, 5, 9,, 8,, 3, 9,, 5, 5,, 4, 6, 6, 4) (,, 3, 3, 6,, 5,, 5,9, 8, 8, 9,, 4,,, 4, 5, 6, 4,, 3, 7, 7, 6) (, 3,, 3, 6, 5,, 5,, 9, 8, 9, 8,, 4,,, 5, 4, 4,, 6, 3, 7, 7, 6) (3, 3,,, 6, 8, 6,, 5, 9, 9,, 8,, 3, 5, 4,, 5, 6,, 4, 7, 4,, 7) (6, 4, 3,, 6,, 7, 5,, 9, 9, 8,,, 3, 5, 4, 5,, 3,,, 4, 7, 8, 6) (,, 6,, 6,, 7, 5,, 9, 8,, 9,, 3, 5, 4, 5,, 4, 3, 3, 7, 4, 6, 8) (4, 3,,, 6,, 7,, 5, 9, 8, 9,,, 3, 5, 4,, 5,, 6, 3, 7, 4, 6, 8) (4, 4, 3, 7, 6, 6, 5,, 7, 5,,, 8, 9, 3, 8,, 5, 9, 6, 4,,,, 3, ) (4, 4, 3, 7, 6, 6, 5, 7,, 5,,, 8, 9, 3, 8,, 9, 5, 6, 4,,,,, 3) (3, 4, 4, 7, 6, 6, 5, 7,, 5, 8,,, 9, 3, 8,, 9, 5,, 4, 6,,,, 3) (3, 3, 6, 7, 6, 6, 3,,, 5, 8, 9,,, 4, 5,, 5, 9, 4, 7,,, 4,, 8) (4,, 7, 7, 6,, 8,,, 5, 8,, 9,, 4, 5,, 5, 9, 3, 6, 3,, 4, 6, 3) (4, 6, 7, 7, 6, 6, 3,,, 5, 8,, 9,, 4, 5,, 5, 9, 3,, 3,, 4, 8, ) (,,, 3, 3, 5, 4, 8,,, 4, 7,, 8,, 5, 9, 9, 5, 6, 6, 6, 4, 3,, 7) (3,,,, 6,, 5,, 8,, 7,, 4, 8, 4, 5, 9, 5, 9, 3,, 6, 3, 6, 4, 7) (3,,,, 6,, 5,, 8,, 7, 4,, 8, 4, 5, 9, 5, 9, 3, 6,, 6, 3, 4, 7) (, 6, 3,, 6, 5,, 8,,,, 4, 7, 8, 4, 5, 9, 9, 5,,, 3, 3, 6, 4, 7) (7, 9,, 8, 7, 5, 8, 6,, 3, 9,,, 3, 4, 7, 3,,, 3, 6, 9, 8, 5, 3, 5, 4, 6,, 4,, 3) (5, 7, 6, 8, 5, 6,, 3, 7, 3, 9, 3,, 4, 8, 3,, 4,,, 7,, 8,, 5,4, 3, 9, 6,, 3, 9) (7, 5, 9, 8, 4, 6, 5, 8,,, 9,,, 3, 4, 5, 6,, 3, 7,,,, 3, 4, 6, 7, 8, 9, 3, 3, 3) 4
15 able 5. (continue) Value Solution Esc3d Esc3e Esc3g Esc3h ai3a ai35a ho3 ho (,, 3, 8, 9, 7, 6, 5, 8,, 4,, 6,, 5, 6, 3, 4, 9,,,, 7, 3, 4, 5, 7, 8, 9, 3, 3, 3) (8,, 3, 4, 5, 6,, 7,,,,, 9, 3, 4, 6, 5, 7, 8, 9,,, 3, 4, 5, 6, 7, 8, 9, 3, 3, 3) (5,,, 3, 4, 6, 4, 7, 8, 9,,,, 3, 5, 6, 7, 8, 9,,,, 3, 4, 5, 6, 7, 8, 9, 3, 3, 3) (, 9, 9,,, 4, 3, 5, 9, 7, 7,,, 6, 5, 3, 4, 3,, 8, 8, 3, 3, 6, 7,, 3, 5, 4, 8,, 6) (9, 8, 4, 4, 3, 5, 5, 7,,, 8,,, 3, 9, 6, 8,, 7,,, 9,, 5, 3, 4, 6, 7, 3, 6) (9, 9, 8,, 7, 33, 3, 6, 5,, 3, 6, 4, 7,, 5, 3, 3, 9,, 6, 5,, 3, 34,, 8, 4,,, 4, 8, 3, 35, 7) (, 7, 4, 6, 3,,, 5, 8,, 3, 5, 4, 4, 7, 8, 6, 3, 6,, 9, 9,, 7,, 8, 3, 5,, 9) (3, 35, 7, 6, 3, 5, 37, 38, 5, 9, 8,, 3,, 4, 9, 36,, 6, 3, 6, 33, 8, 8, 7,,,, 5,, 4, 39, 3, 34, 7, 4, 3, 9,, 4) ai4a is difficult to solve. It is just found a solution of the value It is greater than by the RoS method. he contraction effect of pivoting operations occurs to us the formation of substance that is under certain conditions or control while the formation of a good solution of QAP is depending on the appropriate control of pivoting operations. It is hopeful to discover the deep mechanism of QAP and develop more efficient computing method. References [] G Finke, R E Burkard, F Rendl. Quadratic assignment problems. Annals of Discrete Mathematics, 987(3): 68 [] Zhongzhen Zhang. An efficient method for solving the local minimum of indefinite quadratic programming, on QUADRAIC PROGRAMMING PAGER, N.I.M. Gould and Ph.L.oint, [3] Zhongzhen Zhang. Convex Programming: Pivoting Algorithms for Portfolio Selection and Network Optimization. Wuhan University Press, 4(in Chinese) 5
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