Monday, April 2, 2018

Solving a Free-Sign Pure Discrete Signomial Programming Problem

Jsun Yui Wong

The computer program listed below seeks to solve the last instance in Table 5 of Lu, [32, p. 114]: 

Minimize        X(1) ^ 3 * X(2) * X(3) ^ 3 * X(4) + X(1) ^ 3 * X(2) * X(3) * X(4) ^ 2

subject to

         X(1) ^ 3 * X(2) * X(3) ^ 2 + X(3) * X(4)<=-500

        - X(1) ^ 3 * X(2) * X(3) + X(3) ^ 2 * X(4)<=500

         -6<=X(1) <= 6.75

        -6<= X(2) <= 6.75

        -1<= X(3) <= 9.2

        -9<= X(4) <= 6.3

where X(1) through X(4) are free-sign discrete variables; here the number of discrete values is 512. 

X(5) and X(6) 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 32111 STEP .01

    14 RANDOMIZE JJJJ

    16 M = -1D+37


    61 A(1) = -6 + FIX(RND * 512) * 12.75 / 511

    63 A(2) = -6 + FIX(RND * 512) * 12.75 / 511

    65 A(3) = -1 + FIX(RND * 512) * 10.2 / 511


    67 A(4) = -9 + FIX(RND * 512) * 15.3 / 511

    128 FOR I = 1 TO 2000


        129 FOR KKQQ = 1 TO 4


            130 X(KKQQ) = A(KKQQ)
        131 NEXT KKQQ
        133 FOR IPP = 1 TO (1 + FIX(RND * 2))


            181 j = 1 + FIX(RND * 4)


            183 r = (1 - RND * 2) * A(j)
            187 X(j) = A(j) + (RND ^ (RND * 10)) * r
        222 NEXT IPP


        268 IF X(1) < -6 THEN 1670
        269 IF X(1) > 6.75 THEN 1670


        272 IF X(2) < -6 THEN 1670
        273 IF X(2) > 6.75 THEN 1670


        274 IF X(3) < -1 THEN 1670
        275 IF X(3) > 9.2 THEN 1670
        284 IF X(4) < -9 THEN 1670
        285 IF X(4) > 6.3 THEN 1670


        308 X(5) = -500 - X(1) ^ 3 * X(2) * X(3) ^ 2 - X(3) * X(4)


        309 X(6) = 500 + X(1) ^ 3 * X(2) * X(3) - X(3) ^ 2 * X(4)


        401 FOR J47 = 5 TO 6

            402 IF X(J47) < 0 THEN X(J47) = X(J47) ELSE X(J47) = 0


        403 NEXT J47

        443 POBA = -X(1) ^ 3 * X(2) * X(3) ^ 3 * X(4) - X(1) ^ 3 * X(2) * X(3) * X(4) ^ 2 + 1000000 * (X(5) + X(6))


        466 P = POBA

        1111 IF P <= M THEN 1670


        1450 M = P
        1454 FOR KLX = 1 TO 6

            1455 A(KLX) = X(KLX)
        1456 NEXT KLX
        1557 GOTO 128

    1670 NEXT I
    1889 IF M < 72810 THEN 1999


    1907 PRINT A(1), A(2), A(3), A(4), A(5), A(6), M, JJJJ

1999 NEXT JJJJ


This BASIC computer program was run with QB64v1000-win [49]. The complete output through JJJJ = -31995.0000000008 is shown below:

3.229896900873914        -1.315536597736727       6.060244169953957
6.299724378997351       0      0                 72813.92169807518
-31998.09000000031

3.309100297995073        -1.23243415832176       6.043621618088537
6.299938047635811       0      0                 72816.111093616
-31997.68000000037

-2.231428778900335        3.97735652428387       6.066947644653092
6.299999972210325       0      0                 72813.31551915896
-31997.22000000045

3.441393159311623        -1.092310317177698       6.050514577356587
6.299999660072656       0      0                 72816.09250444798
-31995.0000000008

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 [49], the wall-clock time for obtaining the output through
JJJJ = -31995.0000000008 was 32 seconds, total, including the time for “Creating .EXE file.”  One can compared the computational results above with those on page 114 of Lu [32, Table 5].           
                                               
Acknowledgment

I would like to acknowledge the encouragement of Roberta Clark and Tom Clark.

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