The following computer program seeks to solve Instance P17 on page 18 of Hungerlaender and Anjos [21]; the data are from Anjos [6]. This is a corridor allocation problem [4].
Because of line 111, the first row can have 9, 10, 11, or 12 facilities.
0 REM DEFDBL A-Z
1 DEFINT I,J,K,X,A
2 DIM B(99),N(99),A(2002),H(99),L(99),U(99),X(2002),D(111),P(111),PS(33),J44(2002),J(99),AA(99),HR(32),HHR(32),Y(33),C(33),CC(33)
3 DIM HS(49,49)
4 DIM PE(49,49)
5 DIM SD(49,49)
27 HR(1)=20:HR(2)=3:HR(3)=9:HR(4)=3:HR(5)=7
28 HR(6)=3:HR(7)=7:HR(8)=5:HR(9)=9:HR(10)=6
29 HR(11)=5:HR(12)=3:HR(13)=9:HR(14)=3:HR(15)=7
30 HR(16)=3:HR(17)=7
31 FOR IL=1 TO 17
32 FOR JL=1 TO 17
33 READ HS(IL,JL)
34 NEXT JL
35 NEXT IL
61 DATA 999,0,5,0,5,2,10,3,1,5,5,5,0,0,5,4,4
62 DATA 0,999,3,10,5,1,5,1,2,4,2,5,0,10,10,3,0
63 DATA 5,3,999,2,0,5,2,4,4,5,0,0,0,5,1,0,0
64 DATA 0,10,2,999,1,0,5,2,1,0,10,2,2,0,2,1,5
66 DATA 5,5,0,1,999,5,6,5,2,5,2,0,5,1,1,1,5
67 DATA 2,1,5,0,5,999,5,2,1,6,0,0,10,0,2,0,1
68 DATA 10,5,2,5,6,5,999,0,0,0,5,10,2,2,5,1,2
69 DATA 3,1,4,2,5,2,0,999,1,1,10,10,2,0,10,2,5
70 DATA 1,2,4,1,2,1,0,1,999,2,0,3,5,5,0,5,0
71 DATA 5,4,5,0,5,6,0,1,2,999,5,5,0,5,1,0,0
72 DATA 5,2,0,10,2,0,5,10,0,5,999,5,2,5,1,10,0
73 DATA 5,5,0,2,0,0,10,10,3,5,5,999,2,10,5,0,1
74 DATA 0,0,0,2,5,10,2,2,5,0,2,2,999,2,2,1,0
75 DATA 0,10,5,0,1,0,2,0,5,5,5,10,2,999,5,5,1
76 DATA 5,10,1,2,1,2,5,10,0,1,1,5,2,5,999,3,0
77 DATA 4,3,0,1,1,0,1,2,5,0,10,0,1,5,3,999,0
78 DATA 4,0,0,5,5,1,2,5,0,0,0,1,0,1,0,0,999
88 FOR JJJJ=-32000 TO 32000
89 RANDOMIZE JJJJ
90 M=-1D+37
91 FOR J44=1 TO 17
93 A(J44)=J44
94 NEXT J44
111 IF RND<1/4 THEN IMAX=9 ELSE IF RND<1/3 THEN IMAX=10 ELSE IF RND<1/2 THEN IMAX=11 ELSE IMAX=12
126 REM IMAR=10+FIX(RND*1000)
128 FOR I=1 TO 500
129 FOR KKQQ=1 TO 17
130 X(KKQQ)=A(KKQQ)
131 NEXT KKQQ
133 III=1+FIX(RND*17)
134 JJJ=1+FIX(RND*17)
221 X(III)=A(JJJ)
223 X(JJJ)=A(III)
231 FOR J44=1 TO 17
233 FOR J45=1 TO 17
234 IF X(J44)=J45 THEN HHR(J44)=HR(J44) ELSE GOTO 238
237 Y(J45)=J44
238 NEXT J45
253 FOR ISE20 =1 TO IMAX
254 C(ISE20)=.5*HHR(Y(ISE20))
258 FOR ISE2000 =ISE20+1 TO IMAX
259 C(ISE20)=C(ISE20)+HHR(Y(ISE2000) )
263 NEXT ISE2000
269 NEXT ISE20
386 FOR ISE20N =IMAX+1 TO 17
387 C(ISE20N)=.5*HHR(Y(ISE20N))
388 FOR ISE2000N =ISE20N+1 TO 17
389 C(ISE20N)=C(ISE20N)+HHR(Y(ISE2000N) )
390 NEXT ISE2000N
391 NEXT ISE20N
395 NEXT J44
411 PROD=0
412 FOR J44=1 TO 17
413 FOR J45=J44+1 TO 17
414 PROD=PROD -HS(Y(J44),Y(J45))*ABS( C(J44)-C(J45) )
415 NEXT J45
416 NEXT J44
422 P=PROD
999 REM
1111 IF P<=M THEN 1670
1452 M=P
1453 FOR KLX=1 TO 17
1454 CC(KLX)=C(KLX)
1455 A(KLX)=X(KLX)
1456 NEXT KLX
1559 IIMAX=IMAX
1657 GOTO 128
1670 NEXT I
1889 IF M<-4700 THEN 1999
1901 PRINT A(1),A(2),A(3),A(4),A(5)
1902 PRINT A(6),A(7),A(8),A(9),A(10)
1903 PRINT A(11),A(12),A(13),A(14),A(15)
1904 PRINT A(16),A(17)
1905 PRINT CC(1),CC(2),CC(3),CC(4),CC(5)
1906 PRINT CC(6),CC(7),CC(8),CC(9),CC(10)
1907 PRINT CC(11),CC(12),CC(13),CC(14),CC(15)
1908 PRINT CC(16),CC(17)
1919 PRINT M,JJJJ,IIMAX
1999 NEXT JJJJ
This BASIC computer program was run with Microsoft's GW BASIC 3.11 interpreter. Selected candidate solutions through JJJJ=-31829 are shown below. What follows is a hand copy from the computer-monitor screen; immediately below there is no rounding by hand.
10 14 16 3 2
8 11 13 9 7
4 5 17 15 6
12 1
46.5 39.5 34.5 30.5 26.5
21.5 15 10.5 4.5 49
35.5 30.5 26.5 22.5 19.5
13.5 4.5
-4672 -31972 9
11 6 16 12 2
9 3 5 10 8
13 14 17 7 15
4 1
49.5 42.5 35.5 30.5 26.5
22.5 19.5 15 10.5 4.5
46 34.5 30.5 26.5 21.5
13.5 4.5
-4673 -31970 10
.
.
.
10 6 8 5 2
16 11 12 9 15
13 4 17 14 3
7 1
47.5 40.5 33.5 28.5 25.5
22.5 19.5 13.5 4.5 48
34.5 28.5 23.5 19.5 15
10.5 4.5
-4666 -31829 9
On a personal computer with a Pentium Dual-Core CPU E5200 @2.50GHz, 2.50 GHz, 960 MB of RAM, and the IBM basica/D interpreter, version GW BASIC 3.11, the wall-clock time through JJJJ= -31829 was eleven minutes.
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