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problem with the objective and the subjective functions. solve this using linear programming in excel?

Xij=(1,2),(1,6),(1,10),(1,9),(2,4),(2,5),(2,3),(3,5).etcâ€¦
Min
Z=4X1,2+X1,6+5X1,10+3X1,4+3X2,3+3X2,3+2X2,4+4X3,8+2X3,5+2X4,7+3X4,6+3X5,8+5X5,12+4X6,7+
4X6,11+4X6,10+3X7,12+2X8,13+2X8,10+2X9,14+2X10,11+4X10,15+2X11,16+5X11,15+3X12,17+3X1
3,17+4X14,18+4X15,20+7X15,19+3X16,21+6X18,20+3X10,22+3X17,21+4X18,10+20X18,27+23X19,2
7+9X20,19+18X20,27+12X20,24+20X21,24+14X22,24+3X22,23+5X23,28+8X23,25+15X24,26+12X24,
25+4X25,28+6X25,29+8X26,24+24X26,34+23X27,34+6X28,30+4X28,31+4X29,31+25X29,34+3X30,31
+2X30,32+5X31,33+8X31,32+31X31,37+3X32,33+432,35X+5X38,40+2X39,43+5X34,37+9X34,39+2X3
5,37+28X35,43+5X36,38+2X37,39+4X38,40+24X39,43+22X39,41+26X39,42+12X40,44+3X41,47+9X4
2,48+8X42,50+6X43,40+18X44,45+9X45,46+15X46,3+17X46,2+3X47,50+4X47,48+5X48,51+5X49,52
+5X50,52+3X50,53+2X51,53+4X51,54+X52,53+9X52,2+7X53,2+10X53,1+7X54,1
S.t.
X53,1+X54,1=X1,2+X1,6+X1,10+X1,9
X1,2=X2,3+X2,5+X2,4
X2,3+X46,3=X3,5+X3,8
X2,4=X4,6+X4,7
X2,5+X3,5=X5,8+X5,12
X1,6+X4,6=X6,7+X6,11+X6,10
X4,7+X6,7=X7,12
X3,8+X5,8=X8,13
X1,9=X9,10+X9,14
X1,10+X9,10+X6,10=X10,19+X10,15+X10,11
X6,11+X10,11=X11,15+X11,16
X7,12+X5,12=X12,17
X14,18=X18,19+X18,27
X20,19+X15,19+X10,19+X18,19=X19,27
X15,20+X16,20=X20,27+X20,24
X10,21++X17,21=X21,24
X17,22=X22,24+X22,23
X22,23=X23,25+X23,28
X22,24+X21,24+X20,24=X24,26+X24,25
X24,25+X23,25=X25,28+X25,29
X24,26=X26,29+X26,34
X20,27+X19,27+X18,27=X27,34
X25,28+X23,28=X28,31+X28,30
X25,29+X26,29=X29,31+X29,34
X28,30=X30,31+X30,33
X30,31+X28,31+X29,31=X31,32+X31,33
X31,32=X32,33+X32,35
X30,33+X31,33+X32,33=X33,36
X26,34+X29,34=X34,37+X34,39
X32,35=X35,37+X35,43
X33,36=X36,38
X34,37+X31,33+X35,37=X37,39
X36,38=X38,40
X37,39+X34,29=X39,43+X39,41+X39,42
X38,40=X40,44
X39,41=X41,47
X39,42=X39,42
X39,43+X35,43=X43,49
X40,44=X44,45
X44,45=X45,46
X45,46=X46,2+46,3
X41,47=X47,49+X47,50
X42,48=X48,51
X43,49+X47,49=X49,52
X47,50+X42,50=X50,52+X50,53
X48,51=X51,53+X51,54
X50,52+X49,52=X52,53+X52,2
X50,53+X52,53+X51,53=X53,2+X53,1
X51,54=X54,1
X1,54+X1,53+X1,2+X1,6+X1,9=2
Xij>=0
Xij=Binary