6. Example-5
Find dual from primal conversion MAX z = 3x1 + x2 + 2x3 - x4 subject to 2x1 - x2 + 3x3 + x4 = 1 x1 + x2 - x3 + x4 = 3 and x1,x2 >= 0 and x3,x4 unrestricted in sign
Solution: Primal is (Solution steps of Primal by Simplex method)
MAX `z_x` | `=` | `` | `3` | `x_1` | ` + ` | `` | `x_2` | ` + ` | `2` | `x_3` | ` - ` | `` | `x_4` |
| subject to | `` | `2` | `x_1` | ` - ` | `` | `x_2` | ` + ` | `3` | `x_3` | ` + ` | `` | `x_4` | = | `1` | `` | `` | `x_1` | ` + ` | `` | `x_2` | ` - ` | `` | `x_3` | ` + ` | `` | `x_4` | = | `3` |
| and `x_1,x_2 >= 0; ``x_3,x_4` unrestricted in sign |
In primal, There are `4` variables and `2` constraints, so in dual there must be `4` constraints and `2` variables
In primal, The coefficient of objective function `c_1=3,c_2=1,c_3=2,c_4=-1` becomes right hand side constants in dual
In primal, The right hand side constants `b_1=1,b_2=3` becomes coefficient of objective function in dual
In primal, objective function is maximizing, so in dual objective function must be minimizing
The `x_3,x_4` variables in the primal are unrestricted in sign, therefore the `3^(rd),4^(th)` constraints in the dual shall be equalities.
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` | `=` | `` | `` | `y_1` | ` + ` | `3` | `y_2` |
| subject to | `` | `2` | `y_1` | ` + ` | `` | `y_2` | ≥ | `3` | ` - ` | `` | `y_1` | ` + ` | `` | `y_2` | ≥ | `1` | `` | `3` | `y_1` | ` - ` | `` | `y_2` | = | `2` | `` | `` | `y_1` | ` + ` | `` | `y_2` | = | `-1` |
| and ```y_1,y_2` unrestricted in sign |
This material is intended as a summary. Use your textbook for detail explanation. Any bug, improvement, feedback then
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