Jsun Yui Wong

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

Maximize (4 * X(1) + 2 * X(2) + 10) / (X(1) + 2 * X(2) + 5)

subject to

X(1) + 3 * X(2)<=30,

- X(1) + 2 * X(2)<=5,

X(1), X(2) >=0,

X(1) and X(2) are integer variables.

The problem above is based on problem f1 in Raouf and Hezam [26], which is

Maximize (4 * X(1) + 2 * X(2) + 10) / (X(1) + 2 * X(2) + 5)

subject to

X(1) + 3 * X(2)<=30,

- X(1) + 2 * X(2)<=5,

X(1), X(2) >=0.

X(3) and X(4) 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

71 FOR J40 = 1 TO 2

74 A(J40) = (RND * 50)

77 NEXT J40

128 FOR I = 1 TO 100

129 FOR KKQQ = 1 TO 2

130 X(KKQQ) = A(KKQQ)

131 NEXT KKQQ

133 FOR IPP = 1 TO (1 + FIX(RND * 1))

181 J = 1 + FIX(RND * 2)

183 R = (1 - RND * 2) * A(J)

187 X(J) = A(J) + (RND ^ (RND * 10)) * R

222 NEXT IPP

223 FOR J41 = 1 TO 2

225 X(J41) = INT(X(J41))

235 NEXT J41

256 FOR J47 = 1 TO 2

257 IF X(J47) < 0 THEN 1670

258 REM IF X(J47) > 60 THEN 1670

259 NEXT J47

311 X(3) = 30 - X(1) - 3 * X(2)

313 X(4) = 5 + X(1) - 2 * X(2)

315 REM X(5) = 1 - X(1) + X(2)

322 FOR J44 = 3 TO 4

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

327 NEXT J44

333 REM

337 REM

339 POBA = (4 * X(1) + 2 * X(2) + 10) / (X(1) + 2 * X(2) + 5) + 1000000 * (X(3) + X(4))

466 P = POBA

1111 IF P <= M THEN 1670

1452 M = P

1454 FOR KLX = 1 TO 4

1455 A(KLX) = X(KLX)

1456 NEXT KLX

1557 REM GOTO 128

1670 NEXT I

1889 REM IF M < -.000000000003 THEN 1999

1900 PRINT A(1), A(2), A(3), A(4)

1902 PRINT M, JJJJ

1999 NEXT JJJJ

This BASIC computer program was run with qb64v1000-win [35]. The complete output through JJJJ =-31999.96000000001 is shown below:

26 0 0 0

3.67741935483871 -32000

29 0 0 0

3.705882362941177 -31999.99

0 0 0 0

2 -31999.98

30 0 0 0

3.714285714285714 -31999.97000000001

30 0 0 0

3.714285714285714 -31999.96000000001

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 [35], the wall-clock time for obtaining the output through JJJJ= -31999.96000000001 was 2 seconds, not including creating .EXE file time. One can compare the computational results here with those in Table 1 of Raouf and Hezam [26, Page 5 of 10].

**Acknowledgment**

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

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