4. Example-3
Find dual from primal conversion MIN z = x1 + 2x2 subject to 2x1 + 4x2 <= 160 x1 - x2 = 30 x1 >= 10 and x1,x2 >= 0
Solution: Primal is (Solution steps of Primal by Simplex method)
MIN `z_x` | `=` | `` | `` | `x_1` | ` + ` | `2` | `x_2` |
| subject to | `` | `2` | `x_1` | ` + ` | `4` | `x_2` | ≤ | `160` | `` | `` | `x_1` | ` - ` | `` | `x_2` | = | `30` | `` | `` | `x_1` | | | | ≥ | `10` |
| and `x_1,x_2 >= 0; ` |
Since objective function is minimizing, all `<=` constraints (1) can be converted to `>=` type by multipling both sides by -1
MIN `z_x` | `=` | `` | `` | `x_1` | ` + ` | `2` | `x_2` |
| subject to | ` - ` | `2` | `x_1` | ` - ` | `4` | `x_2` | ≥ | `-160` | `` | `` | `x_1` | ` - ` | `` | `x_2` | = | `30` | `` | `` | `x_1` | | | | ≥ | `10` |
| and `x_1,x_2 >= 0; ` |
In primal, There are `2` variables and `3` constraints, so in dual there must be `2` constraints and `3` variables
In primal, The coefficient of objective function `c_1=1,c_2=2` becomes right hand side constants in dual
In primal, The right hand side constants `b_1=-160,b_2=30,b_3=10` becomes coefficient of objective function in dual
In primal, objective function is minimizing, so in dual objective function must be maximizing
Let `y1,y2,y3` be the dual variables
Since `2^(nd)` constraint in the primal is equality, the corresponding dual variable `y_2` will be unrestricted in sign.
Dual is (Solution steps of Dual by Simplex method)
MAX `z_y` | `=` | ` - ` | `160` | `y_1` | ` + ` | `30` | `y_2` | ` + ` | `10` | `y_3` |
| subject to | ` - ` | `2` | `y_1` | ` + ` | `` | `y_2` | ` + ` | `` | `y_3` | ≤ | `1` | ` - ` | `4` | `y_1` | ` - ` | `` | `y_2` | | | | ≤ | `2` |
| and `y_1,y_3 >= 0; ``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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