Jsun Yui Wong
The complete computer program listed below seeks to solve a five-department space allocation problem (Armour and Buffa, 1963). The departmment sizes for department 1 through department 5 are 20x50, 30x40, 10x20, 60x70, and 40x50, respectively; the interdepartmental flows are shown in line 1624, line 1625, and line 1626 of the following program.
0 DEFSNG A-Z
3 DEFINT I,J,K
4 DIM X(466),A(466),L(466),K(466),P(466),B(466),S(466),J(466),HS(99)
6 DIM T(11,11,5),TZ(11,11),TL(11)
65 FOR JJJJ=-32000 TO 32000
74 RANDOMIZE JJJJ
76 M=-1D+17
85 FOR I=1 TO 10
88 A(I)=FIX(RND*240)
89 NEXT I
126 IMAR=10+FIX(RND*20000)
128 FOR I=1 TO IMAR
129 FOR KK=1 TO 10
131 X(KK)=A(KK)
132 NEXT KK
223 IJL=1+FIX(RND*10)
234 X(IJL)=FIX(RND*240)
401 HS(1)=ABS(X(1)-X(2))-25
402 HS(2)=ABS(X(1)-X(3))-15
403 HS(3)=ABS(X(1)-X(4))-40
404 HS(4)=ABS(X(1)-X(5))-30
405 HS(5)=ABS(X(2)-X(3))-20
406 HS(6)=ABS(X(2)-X(4))-45
407 HS(7)=ABS(X(2)-X(5))-35
409 HS(9)=ABS(X(3)-X(4))-35
410 HS(10)=ABS(X(3)-X(5))-25
411 HS(11)=ABS(X(4)-X(5))-50
413 HS(13)=ABS(X(6)-X(7))-45
414 HS(14)=ABS(X(6)-X(8))-35
415 HS(15)=ABS(X(6)-X(9))-60
416 HS(16)=ABS(X(6)-X(10))-50
417 HS(17)=ABS(X(7)-X(8))-30
418 HS(18)=ABS(X(7)-X(9))-55
419 HS(19)=ABS(X(7)-X(10))-45
420 HS(20)=ABS(X(8)-X(9))-45
421 HS(21)=ABS(X(8)-X(10))-35
422 HS(22)=ABS(X(9)-X(10))-60
551 IF HS(1)<-.0001 AND HS(13)<-.0001 THEN TL(1)=999999! ELSE TL(1)=0
553 IF HS(2)<-.0001 AND HS(14)<-.0001 THEN TL(2)=999999! ELSE TL(2)=0
554 IF HS(3)<-.0001 AND HS(15)<-.0001 THEN TL(3)=999999! ELSE TL(3)=0
555 IF HS(4)<-.0001 AND HS(16)<-.0001 THEN TL(4)=999999! ELSE TL(4)=0
557 IF HS(5)<-.0001 AND HS(17)<-.0001 THEN TL(5)=999999! ELSE TL(5)=0
559 IF HS(6)<-.0001 AND HS(18)<-.0001 THEN TL(6)=999999! ELSE TL(6)=0
560 IF HS(7)<-.0001 AND HS(19)<-.0001 THEN TL(7)=999999! ELSE TL(7)=0
565 IF HS(9)<-.0001 AND HS(20)<-.0001 THEN TL(8)=999999! ELSE TL(8)=0
568 IF HS(10)<-.0001 AND HS(21)<-.0001 THEN TL(9)=999999! ELSE TL(9)=0
569 IF HS(11)<-.0001 AND HS(22)<-.0001 THEN TL(10)=999999! ELSE TL(10)=0
1611 SUMTL=0
1613 FOR IR=1 TO 10
1615 SUMTL=SUMTL+TL(IR)
1618 NEXT IR
1624 PR1=-111*ABS(X(1)-X(2))-222*ABS(X(1)-X(3))-110*ABS(X(1)-X(4))-340*ABS(X(1)-X(5))-88*ABS(X(2)-X(3))-260*ABS(X(2)-X(4))-70*ABS(X(2)-X(5))
1625 PR2=-250*ABS(X(3)-X(4))-90*ABS(X(3)-X(5))-280*ABS(X(4)-X(5))-111*ABS(X(6)-X(7))-222*ABS(X(6)-X(8))-110*ABS(X(6)-X(9))-340*ABS(X(6)-X(10))
1626 PR3=-88*ABS(X(7)-X(8))-260*ABS(X(7)-X(9))-70*ABS(X(7)-X(10))-250*ABS(X(8)-X(9))-90*ABS(X(8)-X(10))-280*ABS(X(9)-X(10))
1650 P=PR1+PR2+PR3-999*SUMTL
1651 IF P<=M THEN 1670
1657 FOR KEW=1 TO 10
1658 A(KEW)=X(KEW)
1659 NEXT KEW
1661 M=P
1662 MM=PR1+PR2+PR3
1666 GOTO 128
1670 NEXT I
1890 IF M>-95555! THEN 1912 ELSE 1999
1912 PRINT A(1),A(2),A(3),A(4),A(5)
1914 PRINT A(6),A(7),A(8),A(9),A(10)
1915 PRINT MM,M,JJJJ
1999 NEXT JJJJ
This BASIC computer program was run with Microsoft's GW BASIC 3.11 interpreter, and the output produced during the first 25 minutes of running is presented below. What immediately follows is a manual copy from the computer screen.
113 78 98 33 143
54 54 54 54 54
-95325 -95325 -31788
73 108 88 153 43
85 85 85 85 85
-95325 -95325 -31737
105 140 120 185 75
186 186 186 186 186
-95325 -95325 -31597
158 63 143 108 188
137 137 137 137 137
-92265 -92265 -31584
146 141 131 96 176
90 45 90 90 90
-93620 -93620 -31427
143 238 158 193 113
167 167 167 167 167
-92265 -92265 -31402
Interpreted in accordance with line 1912 through line 1915, the output through JJJJ=-31402 was produced during the first 25 minutes of running on a personal computer with an Intel 2.66 GHz. chip and the IBM basica/D interpreter, which is not a compiler.
Reference
G. C. Armour and E. S. Buffa, "A Heuristic Algorithm and Simulation Approach to Relative Location of Facilities," Management Science 9, 294-309 (1963).
