In: Statistics and Probability
Use implicit enumeration to solve the following zero-one IP
Objective function: max z = 0.2A + 0.3B + 0.5C + 0.1D
s.t. 0.5A + 1B + 1.5C + 0.1D ≤ 3.1
0.3A + 0.8B + 1.5C + 0.4D ≤ 2.5
0.2A + 0.2B + 0.3C + 0.1D ≤ 0.4
A, B, C, D = 0, 1
| 
 Solution  | 
 A  | 
 B  | 
 C  | 
 D  | 
 Feasibility  | 
 Z value  | 
| 
 1  | 
 0  | 
 0  | 
 0  | 
 0  | 
 Feasible  | 
 0  | 
| 
 2  | 
 1  | 
 0  | 
 0  | 
 0  | 
 Feasible  | 
 0.2  | 
| 
 3  | 
 0  | 
 1  | 
 0  | 
 0  | 
 Feasible  | 
 0.3  | 
| 
 4  | 
 0  | 
 0  | 
 1  | 
 0  | 
 Feasible  | 
 0.5  | 
| 
 5  | 
 0  | 
 0  | 
 0  | 
 1  | 
 Feasible  | 
 0.1  | 
| 
 6  | 
 1  | 
 1  | 
 0  | 
 0  | 
 Feasible  | 
 0.5  | 
| 
 7  | 
 1  | 
 0  | 
 1  | 
 0  | 
 Infeasible  | 
 0.7  | 
| 
 8  | 
 1  | 
 0  | 
 0  | 
 1  | 
 Feasible  | 
 0.3  | 
| 
 9  | 
 0  | 
 1  | 
 1  | 
 0  | 
 Infeasible  | 
 0.8  | 
| 
 10  | 
 0  | 
 1  | 
 0  | 
 1  | 
 Feasible  | 
 0.4  | 
| 
 11  | 
 0  | 
 0  | 
 1  | 
 1  | 
 Feasible  | 
 0.6  | 
| 
 12  | 
 1  | 
 1  | 
 1  | 
 0  | 
 Infeasible  | 
 1  | 
| 
 13  | 
 1  | 
 0  | 
 1  | 
 1  | 
 Infeasible  | 
 0.8  | 
| 
 14  | 
 1  | 
 1  | 
 0  | 
 1  | 
 Infeasible  | 
 0.6  | 
| 
 15  | 
 0  | 
 1  | 
 1  | 
 1  | 
 Infeasible  | 
 0.9  | 
| 
 16  | 
 1  | 
 1  | 
 1  | 
 1  | 
 Infeasible  | 
 1.1  | 
The complete enumeration (i.e., the list of all possible solution sets) for this model is as follows.
The solution 7, 9, 13 & 14 can be eliminated because they violate the third constraint. The solution 12, 15 & 16 can also be eliminated because they violate 2nd and 3rd constraints. Among the remaing solution sets 1 can be eliminated as it chooses none of the recreational values. After evaluating the objective function value of these remaining solutions, we find the best solution to be 11, with A=0, B=0, C=1, D=1, as it has the maximum Z value=0.6.