Solving a Multiple Objective Programming Problem Using a Problem Obtained from the Original Problem

 

Jsun Yui Wong

The following formulation is from Jain and Lachhwani [15, pp. 46-47]:

Maximize {Z1(X), Z2(X)}

where Z1(X)= 2 * X(1) + (2 * X(1)+6*X(2)) / (X(1)+ X(2) + 1),

Z2(X)= 2 * X(2) + ( X(1)+X(2)) / (X(1)- X(2) + 1),

subject to

X(1) +X(2) + X(3)=4,

3*X(1) + X(2) – X(4)=6,

X(1) -X(2+0* X(3))+0* X(4)=0,

where X(1) through X(4) >=0.

The X(1) above, for example, is not the same as the X(1) below– see Jain and Lachhwani [15, pp. 46-47].
The X(4) beow can be better denoted by lambda–see Jain and Lachhwani [15, p. 47].

In [15] Jain and Lachhwani show their procedure for transforming the problem above to the following problem [15, p. 47]:

Maximize X(4)

subject to

X(1)+ 2 * X(3)=4,

– X(2) + 4 * X(3)=6,

6 * X(4)<= 2 * X(3) + (8 * X(3)) / (2 * X(3) + 1),

6 * X(4)<= 4 * X(3),

where X(1) through X(4) >=0.

The following computer program uses the second formulation shown above.

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 31999 STEP .01

 

14 RANDOMIZE JJJJ
16 M = -1D+37

 

71 FOR J40 = 1 TO 4

75 A(J40) = RND

 

77 NEXT J40
128 FOR I = 1 TO 10000

 

129 FOR KKQQ = 1 TO 4

 

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

 

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

 

223 FOR j41 = 1 TO 4

 

225 IF X(j41) < 0 THEN 1670

 

235 NEXT j41

261 X(1) = 4 – 2 * X(3)
264 X(2) = -6 + 4 * X(3)

 

171 IF X(1) < 0 THEN GOTO 1670

173 IF X(2) < 0 THEN GOTO 1670

 

318 X(5) = 0 + 2 * X(3) + (8 * X(3)) / (2 * X(3) + 1) – 6 * X(4)
319 X(6) = 0 + 4 * X(3) – 6 * X(4)

337 FOR J44 = 5 TO 6

 

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

 

339 NEXT J44

 

433 POBA = X(4) + 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 < 1.2 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 [43]. The complete output through JJJJ = -31999.78000000004 is shown below:

1.213658951115804D-08          1.999999975726821         1.999999993931705
1.199999997653591          0          0          1.199999997653591
-31999.96000000001

1.952032668839365D-08          1.999999960959347          1.999999990239837
1.19999999622607          0          0          1.19999999622607
-31999.93000000001

2.275460575518196D-08          1.99999995449079          1.999999988622697
1.199999995600776          0          0          1.199999995600776
-31999.80000000003

7.748838015686488D-09          1.999999984502324          1.999999996125581
1.19999999850189          0          0          1.19999999850189
-31999.78000000004

This solution (7.748838015686488D-09      1.999999984502324      1.999999996125581) at JJJJ= -31999.78000000004 is translated to (2 2 0 2) for the original problem–see Jain and Lachhwani [15, p. 48].

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 [43], the wall-clock time for obtaining the output through
JJJJ= -31999.78000000004 was 15 seconds, total, including the time for “Creating .EXE file.” One can compare the computational results above to those on page 48 of Jain and Lachhwani [15].

Acknowledgment

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

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