written 7.8 years ago by | modified 2.7 years ago by |
Mumbai University > Mechanical Engineering > Sem 7 > Operations Research
Marks: 5 Marks
Year: May 2016
written 7.8 years ago by | modified 2.7 years ago by |
Mumbai University > Mechanical Engineering > Sem 7 > Operations Research
Marks: 5 Marks
Year: May 2016
written 7.8 years ago by |
For maximization primal all constraints must be ‘≤’ or ‘=’ type, hence multiplying eq. (ii) by -1.
3x + 2y + 2z ≤ 12 .......................... (p)
-2x -2y – z ≤ -8 ............................... (q)
x + 2y + 3z = 15 .............................(r)
Dual of the above primal is
Minimize
S = 12p – 8q + 15r
Subjected to,
x→ 3p – 2q + r ≥ 3
y→ 2p – 2q + 2r ≥ -8
z→ 2p – q + 3z = 4 ....(Since z is unrestricted in primal hence its constraint equation is of ‘=’ type.)
p, q ≥ 0 ........................(R is unrestricted since equation of this variable in primal is of ‘=’ type.)