Jsun Yui Wong
The computer program listed below seeks to solve the following problem from p. 116-118 of the 2017 article by Lu [12]:
Minimize .25 * 3.141592654 ^ 2 * X(1) ^ 2 * X(2) * X(3) + .5 * 3.141592654 ^ 2 * X(1) ^ 2 * X(2)
subject to
(3.141592654) ^ -1 * 1000 * (8 * X(1) ^ -3 * X(2) ^ 2 + 2.92 * X(1) ^ -2 * X(2) - 4.92 * X(1) ^ -1) - 189000 * (X(2) - X(1))<=0,
8 * (11.5D+06) ^ -1 * 1000 * X(1) ^ -4 * X(2) ^ 3 * X(3) + 1.05 * X(1) * X(3) + 2.1 * X(1) - 14<=0,
.009 - X(1)<=0,
X(2) - 4<=0,
3 * X(1) - X(2)<=0,
8 * (11.5D+06) ^ -1 * 300 * X(1) ^ -4 * X(2) ^ 3 * X(3) - 6<=0,
1.25 - 8 * (11.5D+06) ^ -1 * 700 * X(1) ^ -4 * X(2) ^ 3 * X(3)<=0,
.009<=X(1) <=.5,
.6<= X(2) <=4,
1<= X(3) <=120,
where X(1) and X(2) are positive discrete variables and X(3) ls a positive integer variable.
X(4) through X(10) below are slack variables, which are added.
Like Lu [11, p. 23], the following computer program uses discreteness of .002 for X(1) and discreteness of .02 for X(2). That gives line 67 and line 69, which are
67 A(1) = .009 + INT(RND * 250) * .002 and 69 A(2) = .6 + INT(RND * 170) * .02, respectively.
0 DEFDBL A-Z
2 DEFINT K
3 DIM B(99), N(99), A(99), H(99), L(99), U(99), X(1111), D(111), P(111), PS(33)
12 FOR JJJJ = -32000 TO 32000 STEP .01
14 RANDOMIZE JJJJ
16 M = -1D+37
22 REM
67 A(1) = .009 + INT(RND * 250) * .002
69 A(2) = .6 + INT(RND * 170) * .02
77 A(3) = 1 + FIX(RND * 120)
79 REM
128 FOR I = 1 TO 10000
129 FOR KKQQ = 1 TO 3
130 X(KKQQ) = A(KKQQ)
131 NEXT KKQQ
133 FOR IPP = 1 TO (1 + FIX(RND * 2))
181 J = 1 + FIX(RND * 2)
183 r = (1 - RND * 2) * A(J)
187 X(J) = A(J) + (RND ^ (RND * 10)) * r
191 NEXT IPP
196 REM
201 IF X(1) < .009 THEN 1670
203 IF X(1) > .5 THEN 1670
255 REM
258 IF X(2) < .6 THEN 1670
266 IF X(2) > 4 THEN 1670
277 IF X(3) < 1 THEN 1670
288 IF X(3) > 120 THEN 1670
304 X(4) = -(3.141592654) ^ -1 * 1000 * (8 * X(1) ^ -3 * X(2) ^ 2 + 2.92 * X(1) ^ -2 * X(2) - 4.92 * X(1) ^ -1) + 189000 * (X(2) - X(1))
306 X(5) = -8 * (11.5D+06) ^ -1 * 1000 * X(1) ^ -4 * X(2) ^ 3 * X(3) - 1.05 * X(1) * X(3) - 2.1 * X(1) + 14
307 X(6) = -.009 + X(1)
317 X(7) = -X(2) + 4
320 X(8) = -3 * X(1) + X(2)
321 X(9) = -8 * (11.5D+06) ^ -1 * 300 * X(1) ^ -4 * X(2) ^ 3 * X(3) + 6
323 X(10) = -1.25 + 8 * (11.5D+06) ^ -1 * 700 * X(1) ^ -4 * X(2) ^ 3 * X(3)
325 FOR J99 = 4 TO 10
330 IF X(J99) < 0 THEN X(J99) = X(J99) ELSE X(J99) = 0
331 NEXT J99
359 POBA = -.25 * 3.141592654 ^ 2 * X(1) ^ 2 * X(2) * X(3) - .5 * 3.141592654 ^ 2 * X(1) ^ 2 * X(2) + 1000000 * (X(9) + X(10) + X(4) + X(5) + X(6) + X(7) + X(8))
466 P = POBA
1111 IF P <= M THEN 1670
1452 M = P
1454 FOR KLX = 1 TO 10
1459 A(KLX) = X(KLX)
1460 NEXT KLX
1557 REM GOTO 128
1670 NEXT I
1889 IF M < -2.62 THEN 1999
1900 PRINT A(1), A(2), A(3), A(4), A(5)
1903 PRINT A(6), A(7), A(8), A(9), A(10)
1907 PRINT M, JJJJ
1999 NEXT JJJJ
This BASIC computer program was run with qb64v1000-win [24]. The complete output through JJJJ =-31998.28000000028 is shown below:
.2916175494333378 1.384178625681288 7
0 0
0 0 0 0 0
-2.613976714086823 -31998.49000000024
.2916175518899017 1.384178658661224 7
0 0
0 0 0 0 0
-2.613976820408248 -31998.28000000028
Above there is no rounding by hand; it is just straight copying by hand from the monitor screen. One can compare the two solutions above to the solutions given in Lu [12, Table 7, p. 120] and in Lu [11, Table 8, p. 24].
On a personal computer with a Pentium Dual-Core CPU E5200 @2.50GHz, 2.50 GHz, 960 MB of RAM and qb64v1000-win [24], the wall-clock time for obtaining the output through JJJJ= -31998.28000000028 was 20 seconds, including the seconds for creating the .EXE file.
Acknowledgment
I would like to acknowledge the encouragement of Roberta Clark and Tom Clark.
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