Find dual from primal conversion
MAX z = x1 - 2x2 + 3x3
subject to
-2x1 + x2 + 3x3 = 2
2x1 + 3x2 + 4x3 = 1
and x1,x2,x3 >= 0
Solution:
Primal is (Solution steps of Primal by Simplex method)
MAX `z_x` | `=` | `` | `` | `x_1` | ` - ` | `2` | `x_2` | ` + ` | `3` | `x_3` |
|
subject to |
` - ` | `2` | `x_1` | ` + ` | `` | `x_2` | ` + ` | `3` | `x_3` | = | `2` | `` | `2` | `x_1` | ` + ` | `3` | `x_2` | ` + ` | `4` | `x_3` | = | `1` |
|
and `x_1,x_2,x_3 >= 0; ` |
In primal, There are `3` variables and `2` constraints, so in dual there must be `3` constraints and `2` variables
In primal, The coefficient of objective function `c_1=1,c_2=-2,c_3=3` becomes right hand side constants in dual
In primal, The right hand side constants `b_1=2,b_2=1` becomes coefficient of objective function in dual
In primal, objective function is maximizing, so in dual objective function must be minimizing
Let `y1,y2` be the dual variables
Since `1^(st),2^(nd)` constraints in the primal are equalities, the corresponding dual variables `y_1,y_2` will be unrestricted in sign.
Dual is (Solution steps of Dual by Simplex method)
MIN `z_y` | `=` | `` | `2` | `y_1` | ` + ` | `` | `y_2` |
|
subject to |
` - ` | `2` | `y_1` | ` + ` | `2` | `y_2` | ≥ | `1` | `` | `` | `y_1` | ` + ` | `3` | `y_2` | ≥ | `-2` | `` | `3` | `y_1` | ` + ` | `4` | `y_2` | ≥ | `3` |
|
and ```y_1,y_2` unrestricted in sign |
This material is intended as a summary. Use your textbook for detail explanation.
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