Friday, October 27, 2017

Solving a Geometric Programming Problem with the Method of the Present Blog

Jsun Yui Wong

The computer program listed below seeks to solve the illustrative example in Wu [27]:   

Minimize           X(1) * X(2) ^ .5 * X(3) ^ 1.2+ 2 * X(1)

subject to

         2 * X(1) + X(2) + X(3)>=8,

         X(2) + 2 * X(3)>=10.5,

        X(1) + 2 * X(3)<=10,

1<= X(1), X(2), X(3)<=5.

X(4), 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 32000 STEP .01


    14 RANDOMIZE JJJJ
    16 M = -1D+37

    22 FOR J55 = 1 TO 3


        67 A(J55) = 1 + FIX(RND * 405) * .01


    71 NEXT J55

    128 FOR I = 1 TO 30000


        129 FOR KKQQ = 1 TO 3

            130 X(KKQQ) = A(KKQQ)
        131 NEXT KKQQ
        133 FOR IPP = 1 TO (1 + FIX(RND * 3))

            181 J = 1 + FIX(RND * 3)

            183 REM r = (1 - RND * 2) * A(J)
            187 REM  X(J) = A(J) + (RND ^ (RND * 10)) * r
            189 X(J) = 1 + FIX(RND * 405) * .01


        191 NEXT IPP
        200 FOR J44 = 1 TO 3

            201 IF X(J44) < 1 THEN 1670
            203 IF X(J44) > 5 THEN 1670
        255 NEXT J44

        301 X(4) = -8 + 2 * X(1) + X(2) + X(3)

        305 X(5) = -10.5 + X(2) + 2 * X(3)
        308 X(6) = 10 - X(1) - 2 * X(3)

        333 FOR J44 = 4 TO 6

            335 IF X(J44) < 0 THEN X(J44) = X(J44) ELSE X(J44) = 0

        339 NEXT J44


        359 POBA = -X(1) * X(2) ^ .5 * X(3) ^ 1.2 - 2 * X(1) + 1000000 * (X(4) + X(5) + X(6))


        466 P = POBA

        1111 IF P <= M THEN 1670


        1452 M = P
        1454 FOR KLX = 1 TO 6


            1459 A(KLX) = X(KLX)
        1460 NEXT KLX
        1557 REM GOTO 128

    1670 NEXT I

    1889 IF M < -9.5 THEN 1999
    1900 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 [25]. The complete output through JJJJ = -31999.90000000002 is shown below:

1       1.52        4.49         0        0
0      -9.475106820436846       -31999.99

1       1.5        4.5         0        0
0      -9.445616203848971       -31999.94000000001

1       1.5        4.5         0        0
0      -9.445616203848971       -31999.91000000001

1       1.5        4.5         0        0
0      -9.445616203848971       -31999.90000000002

Above there is no rounding by hand; it is just straight copying by hand from the monitor screen.  One can compare the computational results above to the results in Wu [27].

On a personal computer with a Pentium Dual-Core CPU E5200 @2.50GHz, 2.50 GHz, 960 MB of RAM and qb64v1000-win [25], the wall-clock time for obtaining the output through JJJJ=   -31999.90000000002 was 15 seconds; 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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