Jsun Yui Wong
The computer program listed below seeks to solve the following nonlinear programming problem from page 447 of Lu [14]:
Minimize 2700 * X(5) * X(1) + 7200 * X(6) * X(2) + 240 * X(7) * X(3) + 900 * X(8) * X(4)
subject to
X(5) = 5.555*( X(9)- 395),
X(6) = 3.125*( X(10)- 398),
X(7) = 4.545*( X(11)- 365),
X(7) = 5.555*( 575- X(9) ),
X(8) = 3.571*( X(12)- 358),
X(8) = 3.125*( 718- X(10) ),
X(5) + X(6)=1000,
55.55 <=X(5)<= 978.125,
21.875 <= X(6) <= 944.45,
21.775 <=X(7) <= 944.35,
55.55 <=X(8)<= 978.125,
.007407407 <=X(1) <= .02,
.004807692 <=X(2)<= .019417476,
.004862759 <=X(3) <= .04736842,
.002912615 <=X(4) <= .01502718.
0 DEFDBL A-Z
2 DEFINT K
3 DIM B(199), N(199), A(199), H(199), L(199), U(199), X(1111), D(111), P(111), PS(133)
12 FOR JJJJ = -32000 TO 32000 STEP .01
14 RANDOMIZE JJJJ
16 M = -1D+37
91 A(1) = .007407407 + RND * (.02 – .007407407)
92 A(2) = .004807692 + RND * (.019417476 – .004807692)
93 A(3) = .004862759 + RND * (.04736842 – .004862759)
94 A(4) = .002912615 + RND * (.01502718 – .002912615)
95 A(5) = 55.55 + RND * (978.125 – 55.55)
96 A(6) = 21.875 + RND * (944.45 – 21.875)
97 A(7) = 21.775 + RND * (944.35 – 21.775)
98 A(8) = 55.55 + RND * (978.125 – 55.55)
128 FOR I = 1 TO 200000
129 FOR KKQQ = 1 TO 12
130 X(KKQQ) = A(KKQQ)
131 NEXT KKQQ
132 REM GOTO 163
133 FOR IPP = 1 TO (1 + FIX(RND * 12))
151 J = 1 + FIX(RND * 12)
153 r = (1 – RND * 2) * A(J)
157 X(J) = A(J) + (RND ^ (RND * 10)) * r
161 NEXT IPP
197 IF X(5) < 55.55 THEN GOTO 1670
199 IF X(5) > 978.125 THEN 1670
201 IF X(6) < 21.875 THEN 1670
203 IF X(6) > 944.45 THEN 1670
205 IF X(7) < 21.775 THEN 1670
207 IF X(7) > 944.35 THEN 1670
209 IF X(8) < 55.55 THEN 1670
211 IF X(8) > 978.125 THEN 1670
227 IF X(1) < .007407407 THEN 1670
229 IF X(1) > .02 THEN 1670
231 IF X(2) < .004807692 THEN 1670
233 IF X(2) > .019417476 THEN 1670
235 IF X(3) < .004862759 THEN 1670
237 IF X(3) > .04736842 THEN 1670
239 IF X(4) < .002912615 THEN 1670
241 IF X(4) > .01502718 THEN 1670
261 X(9) = (X(5) / 5.555) + 395
264 X(7) = 5.555 * (575 – X(9))
267 X(6) = 1000 – X(5)
269 X(10) = (X(6) / 3.125) + 398
272 X(8) = 3.125 * (718 – X(10))
275 X(12) = (X(8) / 3.571) + 358
278 X(11) = (X(7) / 4.545) + 365
279 GOTO 365
280 REM X(10) = (X(6) / 3.125) + 398
281 FOR J99 = 4 TO 5
283 IF X(J99) < 0 THEN X(J99) = X(J99) ELSE X(J99) = 0
285 NEXT J99
311 FOR j44 = 1 TO 12
312 REM IF X(j44) < .00001 THEN 1670
314 NEXT j44
365 POBA = -2700 * X(5) * X(1) – 7200 * X(6) * X(2) – 240 * X(7) * X(3) – 900 * X(8) * X(4)
466 P = POBA
1111 IF P <= M THEN 1670
1452 M = P
1454 FOR KLX = 1 TO 12
1459 A(KLX) = X(KLX)
1460 NEXT KLX
1557 REM GOTO 128
1670 NEXT I
1889 REM IF M < -30000 THEN 1999
1900 PRINT A(1), A(2), A(3), A(4), A(5), A(6)
1903 PRINT A(7), A(8), A(9), A(10), A(11), A(12)
1911 PRINT M, JJJJ
1999 NEXT JJJJ
This BASIC computer program was run with qb64v1000-win [27]. The output through JJJJ = -31999.98 is summarized below:
.
.
.
-22909.1358249864 -32000
.
.
.
-22909.13345539065 -31999.99
7.407406589624948D-03 4.807692020734168D-03 4.862758593874687D-03
2.912614845634454D-03 978.1250274698625 21.87497253013748
21.77497253013745 978.1250274698625 571.080112955871
404.999991209644 369.7909730539357 631.907876636758
-22909.13300472901 -31999.98
Above there is no rounding by hand; it is just straight copying by hand from the monitor screen. The computational results above can be compared with the results in Table 4 of Lu [14, p. 448].
On a personal computer with a Pentium Dual-Core CPU E5200 @2.50GHz, 2.50 GHz, 960 MB of RAM and qb64v1000-win [27], the wall-clock time for obtaining the output through JJJJ=-31999.98 was 11 seconds, total–most of these seconds were for creating the .EXE file.
Acknowledgment
I would like to acknowledge the encouragement of Roberta Clark and Tom Clark.
References
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