Jsun Yui Wong
The computer program listed below seeks to solve the following multi-objective nonlinear integer programming problem from Drici and Moulai [21, p. 7]:
Maximize ( X(1) + 2 * X(2) + 1) * (-X(2) + 10)
maximize (3 * X(1) + 1 * X(2) + 2) * (2 * X(1) + 1 * X(2) + 2)
maximize ( X(1) + 3 * X(2) + 2) * (2 * X(1) + 3 * X(2) + 2)
subject to
3 * X(1) + 4 * X(2) <= 24
2 * X(1) + X(2) <= 10
X(1) >= 0, X(2) >= 0 and integers.
This paper tries to alleviate the following issue: "One issue with the approach is that it is necessary to preselect which objective to minimize and the epsilon values," Pike-Burke [57, p. 2] on the epsilon-constraint method. In the following computer program, (1 * X(1) + 2 * X(2) + 1) * (-X(2) + 10)
is used as the principal objective, Sharma, Dahiya, and Verma [62, p. 1914], but another objective can also be so used. 89 epsi2 = -25 + RND * 200 and 90 epsi3 = -25 + RND * 200 try different epsilon values.
One notes lines 92, 93, and 128.
0 DEFDBL A-Z
1 DEFINT K
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), LHS(44), PLHS(44), LB(22), UB(22), PX(22), CC(20), RR(20), WW(20), AL(50), SW(50), SV(50), C2(22), C3(22), C4(22), C5(22)
81 FOR JJJJ = -32000 TO 32000
85 RANDOMIZE JJJJ
86 M = -3E+50
88 REM epsi1 = -25 + RND * 200
89 epsi2 = -25 + RND * 200
90 epsi3 = -25 + RND * 200
92 A(1) = FIX(RND * 15)
93 A(2) = FIX(RND * 15)
128 FOR I = 1 TO 100
129 FOR KKQQ = 1 TO 2
130 X(KKQQ) = A(KKQQ)
131 NEXT KKQQ
151 FOR IPP = 1 TO FIX(1 + RND * 0)
153 J = 1 + FIX(RND * 2)
155 GOTO 162
156 r = (1 - RND * 2) * A(J)
159 X(J) = A(J) + (RND ^ (RND * 15)) * r
161 GOTO 169
162 IF RND < .5 THEN X(J) = A(J) - INT(RND * 2) ELSE X(J) = A(J) + INT(RND * 2)
163 GOTO 169
164 IF A(J) = 0 THEN X(J) = 1 ELSE X(J) = 0
169 NEXT IPP
170 FOR J44 = 1 TO 2
172 X(J44) = INT(X(J44))
177 NEXT J44
215 FOR J44 = 1 TO 2
216 IF X(J44) < 0 THEN 1670
219 NEXT J44
222 IF 3 * X(1) + 4 * X(2) > 24 THEN 1670
225 IF 2 * X(1) + X(2) > 10 THEN 1670
321 REM IF (1 * X(1) + 2 * X(2) + 1) * (-X(2) + 10) < epsi1 THEN 1670
324 IF (3 * X(1) + 1 * X(2) + 2) * (2 * X(1) + 1 * X(2) + 2) < epsi2 THEN 1670
325 IF (1 * X(1) + 3 * X(2) + 2) * (2 * X(1) + 3 * X(2) + 2) < epsi3 THEN 1670
497 PDU = (1 * X(1) + 2 * X(2) + 1) * (-X(2) + 10)
666 P = PDU
1111 IF P <= M THEN 1670
1450 M = P
1454 FOR KLX = 1 TO 2
1455 A(KLX) = X(KLX)
1456 NEXT KLX
1527 g1star = (1 * X(1) + 2 * X(2) + 1) * (-X(2) + 10)
1529 g2star = (3 * X(1) + 1 * X(2) + 2) * (2 * X(1) + 1 * X(2) + 2)
1533 g3star = (1 * X(1) + 3 * X(2) + 2) * (2 * X(1) + 3 * X(2) + 2)
1557 GOTO 128
1670 NEXT I
1889 IF M < -900000 THEN 1999
1933 PRINT g1star, g2star, g3star, JJJJ
1956 PRINT A(1), A(2), M, JJJJ
1999 NEXT JJJJ
This BASIC computer program was run with QB64v1000-win [70]. Selected candidate solutions of a single run through JJJJ= -31865 are shown below:
72 192 192 -31999
4 2 72 -31999
60 90 342 -31961
1 5 60 -31961
70 154 238 -31954
3 3 70 -31954
60 204 84 -31943
5 0 60 -31943
52 64 400 -31925
0 6 52 -31925
66 120 288 -31865
2 4 66 -31865
Above there is no rounding by hand; it is just straight copying by hand from the monitor screen. On a personal computer with a Pentium Dual-Core CPU E5200 @2.50GHz, 2.50 GHz, 960 MB of RAM and QB64v1000-win [70], the wall-clock time (not CPU time) for obtaining the output through JJJJ = -31865 was 1 or 2 seconds, not including the time for “Creating .EXE file" (6 seconds, total, including the time for “Creating .EXE file"). One can compare the computational results above with those in Drici and Moulai [21, pp. 7-13].
Acknowledgment
I would like to acknowledge the encouragement of Roberta Clark and Tom Clark.
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