Cutting planes and branch-and-cut

The cutting-plane method:

1. Solve the LP: $x^*$.
2. If $x^*$ is integral, it is optimal.
3. Separate. If a cut is found, add it and re-solve with the dual simplex (unit 03). Otherwise branch (unit 07).

On a two-variable example, three slides on, two rounds of this loop take the LP from $(1, 1.5)$ to the integer optimum $(1, 1)$.

A family whose separation problem is hard is useless in practice, however strong its members. Separation is the central question for every family.

For $Ax \le b$, $x \ge 0$ integer, and any multipliers $u \ge 0$:

$$\begin{aligned} u^\top A\, x &\le u^\top b \\ \Longrightarrow\quad \lfloor u^\top A \rfloor x &\le u^\top b \quad (x \ge 0) \\ \Longrightarrow\quad \lfloor u^\top A \rfloor x &\le \lfloor u^\top b \rfloor \quad (x \text{ integer}) \end{aligned}$$

Step 1 uses $x \ge 0$ (rounding coefficients down only lowers the left side). Step 2 uses integrality: the left side is an integer.

The number of rounds needed is the CG rank. It can be large: the matching polytope's odd-set inequalities have rank 1, but some inequalities for TSP have rank that grows with $n$.

A tableau row $\;x_i + \sum_{j \in N} \bar a_{ij} z_j = \bar b_i\;$ with $\bar b_i$ fractional. With the fractional part $f(v) = v - \lfloor v \rfloor$, the Gomory fractional cut is

$$\sum_{j \in N} f(\bar a_{ij})\, z_j \;\ge\; f(\bar b_i)$$

It is valid for all integer solutions, and violated by the current vertex, where every nonbasic $z_j = 0$ and so the left side is $0 < f(\bar b_i)$.

It is the CG cut with multipliers taken from row $i$ of $A_B^{-1}$, rearranged.

Worked: $\max x_2$, $3x_1 + 2x_2 \le 6$, $-3x_1 + 2x_2 \le 0$. LP optimum $(1, \tfrac32)$. Final tableau row for $x_2$:

$$x_2 + \tfrac14 s_1 + \tfrac14 s_2 = \tfrac32$$

Cut: $\tfrac14 s_1 + \tfrac14 s_2 \ge \tfrac12$. Substitute $s_1 = 6 - 3x_1 - 2x_2$ and $s_2 = 3x_1 - 2x_2$:

$$\tfrac14(6 - 4x_2) \ge \tfrac12 \iff x_2 \le 1$$

One more round adds $x_2 \le x_1$, and the LP reaches the integer optimum $(1, 1)$. Lab step 1's test is this example; the next slide draws it.

One knapsack constraint $\sum_j w_j x_j \le c$, with $x \in \{0,1\}^n$ and $w \ge 0$.

Solvers usually separate heuristically, sorting by $(1-x_j^*)/w_j$, and exactly only on small constraints.

Constraint: $5x_0 + 5x_1 + 5x_2 + 5x_3 + 8x_4 \le 12$. LP point $x^* = (1,\ 0.7,\ 0.7,\ 0,\ 0)$, weight $5 + 3.5 + 3.5 = 12$: feasible for the LP.

cover $C$	weight	$\sum_{C}(1 - x^*_j)$	violated?
$\{0, 1, 2\}$	15	$0 + 0.3 + 0.3 = \mathbf{0.6}$	yes
$\{0, 4\}$	13	$0 + 1 = 1$	no (tight)
$\{1, 4\}$	13	$0.3 + 1 = 1.3$	no
$\{0, 1, 3\}$	15	$0 + 0.3 + 1 = 1.3$	no

Cost below 1 means violated. The cheapest cover is $\{0,1,2\}$, and indeed $x^*_0 + x^*_1 + x^*_2 = 2.4 > 2$.

The reference separate_cover (a DP over weights, lab step 3) returns $(0, 1, 2)$. The next slide strengthens the resulting cut by lifting.

$5x_0 + 5x_1 + 5x_2 + 5x_3 + 8x_4 \le 12$, cover $C = \{0,1,2\}$: $\;x_0 + x_1 + x_2 \le 2$.

Lifting. Lift $x_3$ (weight 5): with $x_3 = 1$, capacity 7 remains, and at most one item of $C$ fits. So $\alpha_3 = 2 - 1 = 1$.

Lift $x_4$ (weight 8): with $x_4 = 1$, capacity 4 remains, and nothing fits. So $\alpha_4 = 2 - 0 = 2$.

$$x_0 + x_1 + x_2 + x_3 + 2x_4 \le 2$$

At the root (the root cut loop of branch-and-cut): several rounds of solve, separate every family, select, add, and re-solve, until the bound stops moving enough.

Selection scores candidates by:

• efficacy: the Euclidean distance the cut moves $x^*$, $(a^\top x^* - \beta)/\|a\|$. Gomory's cut 1: 0.5, cut 2: 0.24;
• parallelism: skip cuts nearly parallel to one already chosen;
• numerics: skip huge coefficient ranges.

Reading

Wolsey, ch. 8–9.
Conforti, Cornuéjols & Zambelli, ch. 5–7.
Cornuéjols (2008), Math. Programming B 112, the survey (optional).

Next

09 separation ⇔ optimization, for exponentially many cuts · 10 column generation, the same idea with columns · 27 modelling for a real solver.