If you’ve ever seen terms like linear programming, integer optimisation, or Branch and Bound and felt lost — this post is for you.
We’ll build a complete MILP solver in pure Rust (zero external crates) that solves a real factory planning problem. Every term is explained from scratch. The code compiles and runs today.
By the end you’ll understand:
- What MILP is and why it’s hard
- The full mathematical formulation
- What Branch and Bound actually does (with the real tree output)
- Why this matters for quantum optimisation
The Problem: A Furniture Factory
A furniture factory makes two products:
| Product | Profit | Machine hrs needed | Wood units needed |
|---|---|---|---|
| Table (x₁) | £50 each | 3 hrs | 4 units |
| Chair (x₂) | £30 each | 2 hrs | 3 units |
Available resources:
- Machine hours: 120 total
- Wood: 160 units total
- Market demand caps: max 30 tables, max 40 chairs
Question: How many tables and chairs to make to maximise profit?
You can’t make 16.7 tables. They must be whole numbers. That’s the key challenge — and it’s what makes this MILP.
What Is MILP? (Every Word Explained)
Linear → All relationships are proportional. Profit scales linearly with units. profit = 50 × tables + 30 × chairs — no powers, no curves, just straight lines.
Programming → Old term for “optimisation”. Nothing to do with coding.
Integer → Some variables must be whole numbers. You can’t produce 2.5 tables or hire 0.7 people.
Mixed → Some variables can be continuous (e.g. temperature, weight, price) while others must be integer (units, people, machines). Our problem has all integer variables, but MILP handles both.
Why is it hard?
Pure LP (without integers) is solvable in polynomial time — milliseconds even for massive problems. The moment you add integer constraints, the problem becomes NP-hard. You can’t just solve equations — you need to search through combinations. A factory with 100 products has more valid integer combinations than atoms in the universe.
The Mathematics
Decision Variables
x₁ = number of Tables to produce (integer, ≥ 0)
x₂ = number of Chairs to produce (integer, ≥ 0)
These are what the solver decides. Everything else is fixed input.
Objective Function
Maximise: Z = 50·x₁ + 30·x₂
Every table adds £50 to profit. Every chair adds £30. We want Z as large as possible.
Constraints
3·x₁ + 2·x₂ ≤ 120 ← machine hours: can't exceed 120 total
4·x₁ + 3·x₂ ≤ 160 ← wood: can't exceed 160 units
x₁ ≤ 30 ← market can only absorb 30 tables
x₂ ≤ 40 ← market can only absorb 40 chairs
x₁, x₂ ≥ 0 ← can't produce negative quantities
x₁, x₂ ∈ ℤ⁺ ← must be whole numbers (the "Integer" in MILP)
Matrix Form
Solvers work with matrices. Written as A·x ≤ b:
A x b
[ 3 2 ] [ x₁ ] [ 120 ]
[ 4 3 ] · [ x₂ ] ≤ [ 160 ]
Objective vector c = [50, 30]
Maximise cᵀ·x subject to A·x ≤ b, lb ≤ x ≤ ub, x ∈ ℤ
What Does the Feasible Region Look Like?
Plot x₁ on the x-axis, x₂ on the y-axis. Each constraint is a line. The feasible region is the polygon where all constraints are satisfied simultaneously. The optimal solution always sits at a corner (vertex) of this polygon.
x₂ (Chairs)
40 |---.
| \ ← demand cap x₂ ≤ 40
| \
| ◆ ← OPTIMAL: (30, 13), Profit = £1890
| .\
| . \ ← Wood: 4x₁ + 3x₂ = 160
| . \
| . \__ Machine: 3x₁ + 2x₂ = 120
|___________ x₁ (Tables)
0 30
The Algorithm: Branch and Bound
This is how every serious MILP solver works at its core.
Step 1: LP Relaxation
First, forget the integer constraint. Solve as pure LP (allowing fractions). This gives the upper bound — the best profit achievable if fractions were allowed.
For our problem, LP relaxation gives: x₁ = 30, x₂ = 13.33, profit = £1,900.
But x₂ = 13.33 is fractional. We can’t make 13.33 chairs.
Step 2: Branch
Pick the fractional variable (x₂ = 13.33). Split into two sub-problems:
Branch A: x₂ ≤ 13 (round down)
Branch B: x₂ ≥ 14 (round up)
Push both onto the search stack. Now solve each one’s LP relaxation.
Step 3: Bound
For each sub-problem, the LP relaxation gives an upper bound. If that upper bound is worse than the best integer solution found so far — prune the branch (don’t explore further). This is what makes B&B efficient.
Step 4: Repeat
Keep branching on fractional variables, pruning hopeless branches, until every branch is either pruned or yields an integer solution.
Our Actual Search Tree (from the code output)
[Node 1] depth=0 LP=1900.0 → BRANCH x₂=13.33 (≤13 | ≥14)
[Node 2] depth=1 LP=1895.0 → BRANCH x₁=29.5 (≤29 | ≥30)
[Node 3] depth=2 → INFEASIBLE, pruned
[Node 4] depth=2 LP=1890.0 → BRANCH x₂=14.67 (≤14 | ≥15)
[Node 5] depth=3 LP=1887.5 → BRANCH x₁=28.75 (≤28 | ≥29)
[Node 6] depth=4 → INFEASIBLE, pruned
[Node 7] depth=4 LP=1880.0 → ✓ INTEGER (x1=28, x2=16) ★ £1880
[Node 8] depth=3 LP=1870.0 ≤ best=1880 → PRUNED
[Node 9] depth=1 LP=1890.0 → ✓ INTEGER (x1=30, x2=13) ★ £1890
Only 9 nodes to find the global optimum. Without B&B, brute force would check 30×40 = 1,200 combinations. For larger problems (1,000 variables), B&B reduces billions of checks to thousands.
The Rust Code
Pure Rust. Zero external dependencies. The full solver from scratch.
Data Structures
/// The LP problem in standard form: Maximise c·x subject to A·x ≤ b
#[derive(Clone, Debug)]
struct LpProblem {
c: Vec<f64>, // objective coefficients
a: Vec<Vec<f64>>, // constraint matrix
b: Vec<f64>, // rhs of constraints
lb: Vec<f64>, // lower bounds per variable
ub: Vec<f64>, // upper bounds per variable
n_vars: usize,
}
/// One node in the Branch and Bound search tree.
/// Each node represents a sub-problem with tightened bounds.
#[derive(Clone, Debug)]
struct BbNode {
lb: Vec<f64>, // current lower bounds (tightened by branching)
ub: Vec<f64>, // current upper bounds (tightened by branching)
depth: usize, // depth in tree (for logging)
}
LP Relaxation Solver (Coordinate Ascent)
For each node we solve the LP relaxation — the same problem but without integer constraints.
We use coordinate ascent: move each variable as far as possible in its profit-improving direction while staying feasible. For small bounded problems like ours, this converges to the exact LP optimum.
fn solve_lp_relaxation(prob: &LpProblem, lb: &[f64], ub: &[f64]) -> LpResult {
let n = prob.n_vars;
// Start at lower bounds (always feasible when b ≥ 0)
let mut x: Vec<f64> = lb.to_vec();
for _iter in 0..10_000 {
let mut improved = false;
for i in 0..n {
let dir = prob.c[i]; // gradient: improving direction for x[i]
if dir.abs() < 1e-10 { continue; }
// Find how far we can move x[i] before hitting a constraint
let step = if dir > 0.0 {
max_step_up(prob, &x, i, ub) // move up
} else {
max_step_down(prob, &x, i, lb) // move down
};
if step.abs() > 1e-9 {
x[i] += step;
if dot(&prob.c, &x) > dot(&prob.c, &x) + 1e-10 {
improved = true;
} else {
x[i] -= step; // revert if no improvement
}
}
}
if !improved { break; } // converged
}
LpResult::Optimal(dot(&prob.c, &x), x)
}
max_step_up computes the maximum safe step by checking each constraint:
fn max_step_up(prob: &LpProblem, x: &[f64], i: usize, ub: &[f64]) -> f64 {
let mut max_step = ub[i] - x[i]; // can't exceed upper bound
for (k, &bk) in prob.b.iter().enumerate() {
let a_ki = prob.a[k][i];
if a_ki > 1e-10 {
let lhs: f64 = dot(&prob.a[k], x);
let step = (bk - lhs) / a_ki; // slack / coefficient = max safe step
if step < max_step { max_step = step; }
}
}
max_step.max(0.0)
}
Branch and Bound
fn branch_and_bound(prob: &LpProblem, integer_vars: &[bool]) -> Option<MilpSolution> {
let mut best_obj = f64::NEG_INFINITY;
let mut best_x: Option<Vec<f64>> = None;
let mut stack: Vec<BbNode> = vec![BbNode { lb: prob.lb.clone(),
ub: prob.ub.clone(), depth: 0 }];
while let Some(node) = stack.pop() {
// 1. Solve LP relaxation for this node
let (lp_obj, lp_x) = match solve_lp_relaxation(prob, &node.lb, &node.ub) {
LpResult::Infeasible => continue, // prune: infeasible
LpResult::Optimal(o, x) => (o, x),
};
// 2. Prune if LP bound can't beat current best
if lp_obj <= best_obj + 1e-6 { continue; }
// 3. Find most fractional integer variable (closest fraction to 0.5)
let branch_var = integer_vars.iter().enumerate()
.filter(|&(i, &is_int)| {
if !is_int { return false; }
let frac = lp_x[i].fract();
frac > 1e-6 && frac < 1.0 - 1e-6
})
.min_by(|&(i, _), &(j, _)| {
let fi = (lp_x[i].fract() - 0.5).abs();
let fj = (lp_x[j].fract() - 0.5).abs();
fi.partial_cmp(&fj).unwrap()
})
.map(|(i, _)| i);
match branch_var {
None => {
// All integer vars are integral → valid integer solution!
if lp_obj > best_obj {
best_obj = lp_obj;
best_x = Some(lp_x);
}
}
Some(vi) => {
let floor_v = lp_x[vi].floor();
let ceil_v = lp_x[vi].ceil();
// Child 1: x[vi] ≤ floor (round down branch)
let mut ub2 = node.ub.clone();
ub2[vi] = ub2[vi].min(floor_v);
stack.push(BbNode { lb: node.lb.clone(), ub: ub2, depth: node.depth+1 });
// Child 2: x[vi] ≥ ceil (round up branch)
let mut lb2 = node.lb.clone();
lb2[vi] = lb2[vi].max(ceil_v);
stack.push(BbNode { lb: lb2, ub: node.ub.clone(), depth: node.depth+1 });
}
}
}
best_x.map(|x| MilpSolution { objective: best_obj, variables: x,
nodes_explored: 0 })
}
Defining the Problem
fn main() {
let prob = LpProblem::new(
vec![50.0, 30.0], // objective: max 50x₁ + 30x₂
vec![
vec![3.0, 2.0], // 3x₁ + 2x₂ ≤ 120 (machine hours)
vec![4.0, 3.0], // 4x₁ + 3x₂ ≤ 160 (wood supply)
],
vec![120.0, 160.0], // rhs
vec![0.0, 0.0 ], // lower bounds: x₁, x₂ ≥ 0
vec![30.0, 40.0 ], // upper bounds: x₁ ≤ 30, x₂ ≤ 40
);
let integer_vars = vec![true, true]; // both must be integers
let solution = branch_and_bound(&prob, &integer_vars);
}
Output
╔══════════════════════════════════════════════════════════╗
║ MILP Solver — Pure Rust, Zero Dependencies ║
║ Factory Production Planning (Branch and Bound) ║
╚══════════════════════════════════════════════════════════╝
BRANCH AND BOUND TREE:
──────────────────────────────────────────────────────────
[Node 1] depth=0 obj=1900.0 → BRANCH x2=13.333 (≤13 | ≥14)
[Node 2] depth=1 obj=1895.0 → BRANCH x1=29.500 (≤29 | ≥30)
[Node 3] depth=2 → INFEASIBLE, pruned
[Node 4] depth=2 obj=1890.0 → BRANCH x2=14.667 (≤14 | ≥15)
[Node 5] depth=3 obj=1887.5 → BRANCH x1=28.750 (≤28 | ≥29)
[Node 6] depth=4 → INFEASIBLE, pruned
[Node 7] depth=4 obj=1880.0 → ✓ INTEGER (x1=28, x2=16) ★ £1880
[Node 8] depth=3 LP=1870.0 ≤ best=1880.0 → PRUNED
[Node 9] depth=1 obj=1890.0 → ✓ INTEGER (x1=30, x2=13) ★ £1890
╔══════════════════════════════════════════════════════════╗
║ ✅ OPTIMAL SOLUTION ║
╠══════════════════════════════════════════════════════════╣
║ Tables (x₁) : 30 units ║
║ Chairs (x₂) : 13 units ║
║ Max Profit : £1890 ║
╠══════════════════════════════════════════════════════════╣
║ Machine hrs : 116 / 120 (96.7% utilised) ║
║ Wood used : 159 / 160 (99.4% utilised) ║
║ B&B Nodes : 9 explored ║
╠══════════════════════════════════════════════════════════╣
║ CONSTRAINT CHECK: ║
║ 3(30)+2(13) = 116 ≤ 120 ✓ ║
║ 4(30)+3(13) = 159 ≤ 160 ✓ ║
║ x₁=30 ≤ 30 ✓ ║
║ x₂=13 ≤ 40 ✓ ║
╚══════════════════════════════════════════════════════════╝
WHY NOT A SIMPLER GUESS?
──────────────────────────────────────────────────────────
All tables (30, 0) profit=£1500 ✓
All chairs (0, 40) profit=£1200 ✓
Equal split (20, 20) profit=£1600 ✓
Optimal (30, 13) profit=£1890 ✓ ← solver wins
The solver found £1,890 — significantly better than every naive guess, and it verified all constraints automatically.
Running It
cargo new milp_solver
# paste main.rs (link below)
cargo run --release
No Cargo.toml dependencies needed. Pure Rust standard library only.
Key Terms — Quick Reference
| Term | Meaning |
|---|---|
| Decision variable | What the solver decides (x₁, x₂) |
| Objective function | What you’re maximising or minimising (Z = 50x₁ + 30x₂) |
| Constraint | A rule the solution must satisfy (3x₁ + 2x₂ ≤ 120) |
| Feasible region | All points satisfying every constraint |
| LP relaxation | Same problem without integer constraint — gives upper bound |
| Branch | Split a fractional variable into two sub-problems |
| Bound | Prune branches whose LP bound can’t beat the current best |
| Prune | Discard a branch — guaranteed to not contain the optimum |
| Integer solution | All integer variables have whole-number values |
Why This Matters for Quantum Computing
Classical MILP solvers (HiGHS, Gurobi, OR-Tools) handle problems up to roughly 10,000–100,000 variables well. Beyond that, Branch and Bound’s exponential tree becomes intractable.
Quantum optimisation attacks this differently. Instead of sequentially searching the B&B tree, quantum algorithms explore combinations simultaneously via superposition. But to feed problems to a quantum computer, you need to convert them to QUBO (Quadratic Unconstrained Binary Optimisation):
MILP → convert integer vars to binary → add constraint penalties → QUBO
Once in QUBO form, quantum algorithms like QAOA or quantum-inspired algorithms like Simulated Bifurcation can solve it — and they scale better on large, dense combinatorial problems where classical B&B degrades.
This Rust implementation is the foundation. In the next post, we’ll convert this exact MILP into a QUBO matrix and run it through a Simulated Bifurcation solver — still in pure Rust.
What’s Next
- Part 2: QUBO conversion — turning this MILP into a matrix a quantum solver can eat
- Part 3: Simulated Bifurcation in Rust — quantum-inspired solver from scratch, no hardware needed
- Part 4: Benchmarking against HiGHS on large-scale problems
I’m Mayuresh, Founder & CTO at AmbiCube. Building quantum-classical optimisation infrastructure in Rust. My vision is to build world’s best schedule solution company, if anyone is interested in learning something real computer science then get in touch and help me. Follow for posts on Rust, quantum algorithms, and applied AI systems
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