Jsun Yui Wong
The computer program listed below seeks to solve the following integer problem:
Maximize ( n1x / d1x )+ (n2x / d2x )
where
n1x = X(1) ^ 2 – 4 * X(1) + 2 * X(2) ^ 2 – 8 * X(2) + 3 * X(3) ^ 2 – 12 * X(3) – 56,
n2x = 2 * X(1) ^ 2 – 16 * X(1) + X(2) ^ 2 – 8 * X(2) – 2,
d1x = X(1) ^ 2 – 2 * X(1) + X(2) ^ 2 – 2 * X(2) + X(3) + 20,
d2x = 2 * X(1) + 4 * X(2) + 6 * X(3),
subject to
X(1) + X(2) + X(3)<=10,
-X(1) – X(2) + X(3)<=4,
X(j)>=1, j=1, 2, 3,
X(1) through X(3) are integer variables.
One notes the preceding sentence, which is X(1) through X(3) are integer variables. The integer problem above is based on Example 3 in Shen, Duan, and Pei [23, p. 155].
X(4) and X(5) below are slack variables.
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
75 A(1) = 1 + RND * 7
77 A(2) = 1 + RND * 7
1
78 A(3) = 1 + RND * 7
128 FOR I = 1 TO 2000
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 * 3)
183 r = (1 – RND * 2) * A(J)
187 X(J) = A(J) + (RND ^ (RND * 10)) * r
191 NEXT IPP
193 REM GOTO 209
196 FOR J99 = 1 TO 3
199 X(J99) = INT(X(J99))
204 NEXT J99
209 IF X(1) < 1 THEN 1670
212 IF X(1) > 8 THEN 1670
214 IF X(2) < 1 THEN 1670
216 IF X(2) > 8 THEN 1670
218 IF X(3) < 1 THEN 1670
222 IF X(3) > 8 THEN 1670
306 X(4) = 10 – X(1) – X(2) – X(3)
307 X(5) = 4 + X(1) + X(2) – X(3)
325 REM FOR J99 = 4 TO 7
326 XX(4) = X(4)
327 XX(5) = X(5)
329 IF X(4) < 0 THEN X(4) = X(4) ELSE X(4) = 0
330 IF X(5) < 0 THEN X(5) = X(5) ELSE X(5) = 0
331 REM NEXT J99
340 n1x = X(1) ^ 2 – 4 * X(1) + 2 * X(2) ^ 2 – 8 * X(2) + 3 * X(3) ^ 2 – 12 * X(3) – 56
342 n2x = 2 * X(1) ^ 2 – 16 * X(1) + X(2) ^ 2 – 8 * X(2) – 2
349 d1x = X(1) ^ 2 – 2 * X(1) + X(2) ^ 2 – 2 * X(2) + X(3) + 20
352 d2x = 2 * X(1) + 4 * X(2) + 6 * X(3)
357 POBA = n1x / d1x + n2x / d2x + 1000000 * (X(4) + X(5))
466 P = POBA
1111 IF P <= M THEN 1670
1452 M = P
1454 FOR KLX = 1 TO 5
1456 XXX(KLX) = XX(KLX)
1459 A(KLX) = X(KLX)
1460 NEXT KLX
1557 REM GOTO 128
1670 NEXT I
1889 IF M < -99999 THEN 1999
1900 PRINT A(1), A(2), A(3), A(4), A(5), XXX(4), XXX(5), M, JJJJ
1999 NEXT JJJJ
This BASIC computer program was run with qb64v1000-win [29]. The complete output through JJJJ = -31999.90000000002 is shown below:
1 2 7 0 0
0 0 -.6923076923076923 -32000
1 1 6 0 0
2 0 -1.755952380952381 -31999.99
8 1 1 0 0
0 12 -.919683257918552 -31999.98
1 1 6 0 0
2 0 -1.755952380952381 -31999.97000000001
1 8 1 0 0
0 12 -.4588235294117647 -31999.96000000001
1 1 6 0 0
2 0 -1.755952380952381 -31999.95000000001
1 1 6 0 0
2 0 -1.755952380952381 -31999.94000000001
1 8 1 0 0
0 12 -.4588235294117647 -31999.93000000001
1 1 6 0 0
2 0 -1.755952380952381 -31999.92000000001
8 1 1 0 0
0 12 -.919683257918552 -31999.91000000001
1 2 7 0 0
0 0 -.6923076923076923 -31999.90000000002
Above there is no rounding by hand; it is just straight copying by hand from the monitor screen. The candidate solution above at
JJJJ=-31999.9600000001, for example, is optimal, Shen, Duan, and Pei [23, p. 155].
On a personal computer with a Pentium Dual-Core CPU E5200 @2.50GHz, 2.50 GHz, 960 MB of RAM and qb64v1000-win [29], the wall-clock time for obtaining the output through
JJJJ= -31999.90000000002 was 2 seconds, not including the time for creating the .EXE file.
Acknowledgment
I would like to acknowledge the encouragement of Roberta Clark and Tom Clark.
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