
AARON W. answered 05/19/25
Experienced Elementary and Middle School Reading and Math Tutor
Hi Sarah,
Here is Proof That an Optimal LP Has an Optimal BFS
Step 1: Representation Theorem Application
By the Representation Theorem, any feasible solution *x* can be written as:
x = Σλᵢvᵢ + Σμⱼdⱼ
where:
- vᵢ are extreme points (BFSs)
- dⱼ are extreme directions
- λᵢ ≥ 0, *Σλᵢ = 1*
- μⱼ ≥ 0
Step 2: Optimality Implies cᵀd = 0
If any cᵀdⱼ > 0, the objective could be made arbitrarily large by taking μⱼ → ∞, contradicting optimality. Thus:
cᵀdⱼ ≤ 0 for all *j*
But if cᵀdⱼ < 0, then *x* wouldn't be optimal (set *μⱼ = 0*). Therefore:
*cᵀdⱼ = 0* for all *j*
Step 3: Compare Objective Values
The objective decomposes as:
cᵀx = Σλᵢ(cᵀvᵢ) + Σμⱼ(cᵀdⱼ) = Σλᵢ(cᵀvᵢ)
Let *v* be the BFS with maximum cᵀvᵢ. Then:
cᵀx = Σλᵢ(cᵀvᵢ) ≤ cᵀv (since *Σλᵢ = 1*)
Step 4: Conclude x = v
By optimality of *x*, we must have cᵀx = cᵀv, which occurs iff:
- *λᵢ = 1* for the maximizing vᵢ
- *λᵢ = 0* for all others
- Thus x = v (a BFS). ∎
Why This Matters for Your Learning:
This proof reveals key insights about LPs:
- BFS Dominance: Only need to check BFSs for optimality
- Degeneracy Implications: Ties in cᵀvᵢ lead to multiple optima
- Algorithm Design: Justifies why simplex works
Common Student Pitfalls I Address:
- Misapplying the Representation Theorem
- Overlooking *cᵀdⱼ = 0* implications
- Confusing BFSs with interior points
Ready to Master LP Proofs?
In our sessions, you'll:
✓ Animate this proof using Python/MATLAB
✓ Solve variants (e.g., minimization, unbounded cases)
✓ Apply to your HW problems with my step-check system
Limited Availability: Book by Friday for a free LP Proof Cheat Sheet
Best Regards,
Aaron