Sunday, November 19, 2017

Solving a Speed Reducer Design Problem

Jsun Yui Wong

The computer program listed below seeks to solve the following speed-reducer weight minimization problem in Li et al. [16, p. 932], Kashan [13, p. 68], and Kashan [12, p. 1790].

Minimize     

.7854 * X(1) * X(2) ^ 2 * (3.3333 * X(3) ^ 2 + 14.9334 * X(3) - 43.0934) - 1.508 * X(1) * (X(6) ^ 2 + X(7) ^ 2) + 7.4777 * (X(6) ^ 3 + X(7) ^ 3) + .7854 * (X(4) * X(6) ^ 2 + X(5) * X(7) ^ 2)

subject to

         27 / (X(1) * X(2) ^ 2 * X(3))  <=1,

         397.5 / (X(1) * X(2) ^ 2 * X(3) ^ 2)  <=1,

         1.93 * X(4) ^ 3 / (X(2) * X(3) * X(6) ^ 4) <=1,

        1.93 * X(5) ^ 3 / (X(2) * X(3) * X(7) ^ 4) <=1,


         (1 / (110 * X(6) ^ 3)) * (((((745 * X(4)) / (X(2) * X(3))) ^ 2 + 16900000))) ^ .5  <=1,


         (1 / (85 * X(7) ^ 3)) * (((((745 * X(5)) / (X(2) * X(3))) ^ 2 + 157500000))) ^ .5  <=1,


        X(2) * X(3) / 40  <=1,

         5 * X(2) / X(1) <=1,

        X(1) / (12 * X(2)) <=1,

         (1.5 * X(6) + 1.9) / X(4) <=1,

         (1.1 * X(7) + 1.9) / X(5)  <=1,

       2.6 <=  X(1) <= 3.6,
         .7 <= X(2) <= .8 ,

         17 <= X(3) <= 28,
         7.3 <= X(4) <= 8.3,
         7.3 <= X(5) <= 8.3,

         2.9 <= X(6) <= 3.9,

        5<=  X(7) <= 5.5,

where X(3), the number of teeth, is an integer, and the other six variables are continuous.

X(8) through X(18) below are slack variables.

One notes line 193, which is 193 X(3) = INT(X(3)), where X(3) is the number of teeth, Kashan [13, p. 68].


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

    87 A(1) = 2.6 + FIX(RND * 1000) * .001

    89 A(2) = .7 + FIX(RND * 1000) * .0001


    91 A(3) = 17 + FIX(RND * 1000) * .011

    94 A(4) = 7.3 + FIX(RND * 1000) * .001

    96 A(5) = 7.3 + FIX(RND * 1000) * .001



    98 A(6) = 2.9 + FIX(RND * 1000) * .001

    99 A(7) = 5 + FIX(RND * 1000) * .0005



    128 FOR I = 1 TO 100000


        129 FOR KKQQ = 1 TO 7

            130 X(KKQQ) = A(KKQQ)
        131 NEXT KKQQ
        133 FOR IPP = 1 TO (1 + FIX(RND * 7))


            151 J = 1 + FIX(RND * 7)
            155 IF J = 1 GOTO 167 ELSE IF J = 2 THEN GOTO 169 ELSE IF J = 3 THEN GOTO 172 ELSE IF J = 4 THEN GOTO 174 ELSE IF J = 5 THEN GOTO 176 ELSE IF J = 6 THEN GOTO 178 ELSE IF J = 7 THEN GOTO 180

            167 X(1) = 2.6 + FIX(RND * 1000) * .001
            168 IF RND < .5 THEN 170

            169 X(2) = .7 + FIX(RND * 1000) * .0001

            170 IF RND < .5 THEN 173
            172 X(3) = 17 + FIX(RND * 1000) * .011
            173 IF RND < .5 THEN 175
            174 X(4) = 7.3 + FIX(RND * 1000) * .001
            175 IF RND < .5 THEN 177
            176 X(5) = 7.3 + FIX(RND * 1000) * .001

            177 IF RND < .5 THEN 179
            178 X(6) = 2.9 + FIX(RND * 1000) * .001
            179 IF RND < .5 THEN 181
            180 X(7) = 5 + FIX(RND * 1000) * .0005
            181 REM

        191 NEXT IPP
        193 X(3) = INT(X(3))

        201 IF X(1) < 2.6 THEN 1670
        203 IF X(1) > 3.6 THEN 1670
        211 IF X(2) < .7 THEN 1670
        213 IF X(2) > .8 THEN 1670


        231 IF X(3) < 17 THEN 1670
        233 IF X(3) > 28 THEN 1670
        235 IF X(4) < 7.3 THEN 1670
        237 IF X(4) > 8.3 THEN 1670
        239 IF X(5) < 7.3 THEN 1670
        241 IF X(5) > 8.3 THEN 1670

        247 IF X(6) < 2.9 THEN 1670
        249 IF X(6) > 3.9 THEN 1670

        251 IF X(7) < 5 THEN 1670
        253 IF X(7) > 5.5 THEN 1670



        305 X(8) = 1 - 27 / (X(1) * X(2) ^ 2 * X(3))
        306 X(9) = 1 - 397.5 / (X(1) * X(2) ^ 2 * X(3) ^ 2)


        307 X(10) = 1 - 1.93 * X(4) ^ 3 / (X(2) * X(3) * X(6) ^ 4)

        309 X(11) = 1 - 1.93 * X(5) ^ 3 / (X(2) * X(3) * X(7) ^ 4)


        320 X(12) = 1 - (1 / (110 * X(6) ^ 3)) * (((((745 * X(4)) / (X(2) * X(3))) ^ 2 + 16900000))) ^ .5



        322 X(13) = 1 - (1 / (85 * X(7) ^ 3)) * (((((745 * X(5)) / (X(2) * X(3))) ^ 2 + 157500000))) ^ .5



        323 X(14) = 1 - X(2) * X(3) / 40

        325 X(15) = 1 - 5 * X(2) / X(1)

        327 X(16) = 1 - X(1) / (12 * X(2))

        329 X(17) = 1 - (1.5 * X(6) + 1.9) / X(4)

        330 X(18) = 1 - (1.1 * X(7) + 1.9) / X(5)

        335 FOR J99 = 8 TO 18



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

        341 NEXT J99



        359 POBA = -.7854 * X(1) * X(2) ^ 2 * (3.3333 * X(3) ^ 2 + 14.9334 * X(3) - 43.0934) + 1.508 * X(1) * (X(6) ^ 2 + X(7) ^ 2) - 7.4777 * (X(6) ^ 3 + X(7) ^ 3) - .7854 * (X(4) * X(6) ^ 2 + X(5) * X(7) ^ 2) + 1000000 * (X(8) + X(9) + X(10) + X(11) + X(12) + X(13) + X(14) + X(15) + X(16) + X(17) + X(18))



        466 P = POBA

        1111 IF P <= M THEN 1670



        1452 M = P
        1454 FOR KLX = 1 TO 18



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

    1670 NEXT I

    1889 IF M < -2996 THEN 1999

    1900 PRINT A(1), A(2), A(3), A(4), A(5)
    1903 PRINT A(6), A(7), A(8), A(9), A(10)

    1950 PRINT A(11), A(12), A(13), A(14), A(15), A(16), A(17), A(18), M, JJJJ

1999 NEXT JJJJ


This BASIC computer program was run with qb64v1000-win [33]. The complete output through JJJJ =  -31999.74000000004 is shown below:   

3.5      .7      17      7.301       7.718
3.351     5.287    0    0     0
0      0    0    0    0
0    0     0     -2994.958416306808
-31999.99

3.5      .7      17      7.301       7.716
3.351     5.287    0    0     0
0      0    0    0    0
0    0     0     -2994.914508725582
-31999.98

3.5      .7      17      7.301       7.716
3.351     5.287    0    0     0
0      0    0    0    0
0    0     0     -2994.914508725582
-31999.78000000004

3.5      .7      17      7.301       7.718
3.352     5.287    0    0     0
0      0    0    0    0
0    0     0     -2995.213455221353
-31999.74000000004

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=-31999.74000000004 was 15 seconds, not including the time for creating the .EXE file.   One can compare the computational results here with those in Table 6 of Li et al. [16, p. 933].     

Acknowledgment

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

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