Solving a Nonlinear Programming Problem with Continuous Signomial Terms

Jsun Yui Wong

The computer program listed below seeks to solve the following nonlinear programming problem:

Minimize   .6224 * X(1) * X(3) * X(4) + 1.7781 * X(2) * X(3) ^ 2 + 3.1661 * X(1) ^ 2 * X(4)+ 19.84 * X(1) ^ 2 * X(3)

subject to

– X(1) + .0193 * X(3)<=0,

– 3.141592654 * X(3) ^ 2 * X(4) – (4 / 3) * 3.141592654 * X(3) ^ 3 + 1296000 <= 0,

– X(2) + .00954 * X(3)<=0,

0 <= X(1) <= 2.919,

0 <= X(2) <= 2.919,

0 <= X(3) <= 500,

10<= X(4) <= 240,

where all four variables are continuous.
.
The problem above is based on Li et al.’s pressure vessel optimization problem, [16, pp. 931-932], which is

minimize   .6224 * X(1) * X(3) * X(4) + 1.7781 * X(2) * X(3) ^ 2 + 3.1661 * X(1) ^ 2 * X(4)+ 19.84 * X(1) ^ 2 * X(3)

subject to

– X(1) + .0193 * X(3)<=0,

– 3.141592654 * X(3) ^ 2 * X(4) – (4 / 3) * 3.141592654 * X(3) ^ 3 + 1296000 <= 0,

– X(2) + .00954 * X(3)<=0,

0 <= X(1) <= 2.919,

0 <= X(2) <= 2.919,

0 <= X(3) <= 500,

10<= X(4) <= 240,

where X(4) is continuous and the other three are discrete.

X(5) through X(7) 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 REM

67 A(1) = RND * 2.919
69 A(2) = RND * 2.919
70 A(3) = RND * 500
77 A(4) = 10 + RND * 230

128 FOR I = 1 TO 200000

129 FOR KKQQ = 1 TO 4

130 X(KKQQ) = A(KKQQ)
131 NEXT KKQQ
133 FOR IPP = 1 TO (1 + FIX(RND * 4))

151 J = 1 + FIX(RND * 4)

 

183 r = (1 – RND * 2) * A(J)
187 X(J) = A(J) + (RND ^ (RND * 10)) * r

191 NEXT IPP
195 REM

197 REM

201 IF X(1) < 0 THEN 1670
203 IF X(1) > 2.919 THEN 1670
211 IF X(2) < 0 THEN 1670
213 IF X(2) > 2.919 THEN 1670

221 IF X(3) < 0 THEN 1670
223 IF X(3) > 500 THEN 1670

 

231 IF X(4) < 10 THEN 1670
233 IF X(4) > 240 THEN 1670

 

305 X(5) = X(1) – .0193 * X(3)

306 X(6) = -1296000 + 3.141592654 * X(3) ^ 2 * X(4) + (4 / 3) * 3.141592654 * X(3) ^ 3

 

307 X(7) = X(2) – .00954 * X(3)

 

325 FOR J99 = 5 TO 7

 

330 IF X(J99) < 0 THEN X(J99) = X(J99) ELSE X(J99) = 0

331 NEXT J99

 

359 POBA = -.6224 * X(1) * X(3) * X(4) – 1.7781 * X(2) * X(3) ^ 2 – 3.1661 * X(1) ^ 2 * X(4) – 19.84 * X(1) ^ 2 * X(3) + 1000000 * (X(5) + X(6) + X(7))

 

466 P = POBA

1111 IF P <= M THEN 1670

 

1452 M = P
1454 FOR KLX = 1 TO 7

 

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

1670 NEXT I

1889 IF M < -5815 THEN 1999

1900 PRINT A(1), A(2), A(3), A(4), A(5)
1903 PRINT A(6), A(7), M, JJJJ

1999 NEXT JJJJ

This BASIC computer program was run with qb64v1000-win [33]. The complete output through JJJJ = -31998.08000000031 is shown below:

.7341974261537864          .3629141681609905          38.0413174172946
234.3433308095876          0
0          0          -5814.338558493735          -31998.8000000002

.7296919488290422          .3606871083849268          37.80787299632342
238.1857223382523          0
0          0          -5807.524953049913          -31998.08000000031

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 [33], the wall-clock time for obtaining the output through
JJJJ= -31998.08000000031 was 90 seconds, total, including the time for creating the .EXE file. One can cautiously compare the computational results here with those in Table 5 of Li et al. [16, p. 932] because here the four variables are continuous whereas there three are discrete.

Acknowledgment

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

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