As a project manager, Mary is managing 12 different projects. She realized something very interesting related to them. Sometimes they present some unique and particular rules, like the ones for projects #: 3, 6, and 8. She concluded that neither project 6 nor project 8 is selected unless project 3 is selected.

For integer programming problems, which one of the alternatives listed below correctly models this situation?

Question

As a project manager, Mary is managing 12 different projects. She realized something very interesting related to them. Sometimes they present some unique and particular rules, like the ones for projects #: 3, 6, and 8. She concluded that neither project 6 nor project 8 is selected unless project 3 is selected.

For integer programming problems, which one of the alternatives listed below correctly models this situation?

BlackTom AI · Accepted Answer

The implication that "neither project 6 nor project 8 is selected unless project 3 is selected" is modeled by logical implications x6 → x3 and x8 → x3. For binary decision variables (xi ∈ {0,1}), an implication p → q is equivalent to the linear inequality p ≤ q.

1) Relevant knowledge points and concepts
- Binary variables xi ∈ {0,1} represent selection (1 = selected, 0 = not selected).
- Logical implication p → q for binaries is (~p) OR q, which is equivalent to p ≤ q.
- Integer programming uses linear inequalities to encode logical rules.

2) Step-by-step reasoning
- The statement "project 6 is not selected unless project 3 is selected" means if x6 = 1 then x3 must equal 1; symbolically x6 → x3.
- For binary variables, x6 → x3 ⇔ x6 ≤ x3.
- Similarly, "project 8 is not selected unless project 3 is selected" gives x8 → x3 ⇔ x8 ≤ x3.
- Therefore the two constraints x8 ≤ x3 and x6 ≤ x3 correctly encode the rule.

3) Intermediate verification (truth cases)
- If x3 = 0, inequalities give x6 ≤ 0 and x8 ≤ 0 ⇒ x6 = 0, x8 = 0 (neither can be selected).
- If x3 = 1, inequalities give x6 ≤ 1 and x8 ≤ 1 ⇒ x6, x8 may be 0 or 1 (no restriction beyond binary).

4) Why the correct answer is correct
- The pair x8 ≤ x3, x6 ≤ x3 enforces exactly the intended implications and allows all valid combinations: when x3 = 0 both x6 and x8 must be 0; when x3 = 1 they can be chosen freely.

5) Why other options are incorrect
- x3 + x6 + x8 ≤ 1: forbids selecting more than one project; this is far stronger and wrong.
- x3 ≤ x8, x3 ≤ x6: forces x3 = 1 only if both x6 and x8 are 1 (opposite direction).
- x6 ≤ x8, x3 ≤ x8: links everything to x8, wrongly requiring x8 when x3 is chosen and tying x6 to x8.
- None of the above: incorrect because a correct representation is present.

6) Conclusion
The correct formulation is x8 ≤ x3 and x6 ≤ x3, which matches the provided answer.

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