max24−A−B−C
subject to,
A+B+C≤24B+C≥8A≥0,B≥0,C≥0
max24−A−B−C+0S1+0S2−Ma1
subject to,
A+B+C+S1=24B+C−S2+a1=8A,B,C,S1,S2,a1≥0
Iteration-1
a1 is goin out from the basis and B is come into the basis.
Iteration-2
since all the Z−Cj≥0 optimum solution is obtain.
optimum solution is,
A=0B=8C=0Zmax=24−8Zmax=16
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