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.