Part I · trunk A · linear programming · spine

Duality — the load-bearing unit

Every LP has a shadow LP that bounds it, prices its constraints, certifies its optimum, and re-solves it cheaply after a change.

needs 02~2½ h slides + 2½ h lablab: uv run co lab 03

The curriculum calls this unit load-bearing, and asks you to spend longer on it than feels justified. The reason is on the second-to-last slide: cutting planes, column generation, Lagrangian relaxation, Benders, the Hungarian algorithm, primal–dual approximation and KKT are each duality in a different setting. If this unit is solid, those units are applications. If it is shaky, they are seven separate things to memorize.

There are four views of the dual in this unit, and each gets a section:

a bound (weak and strong duality);
a certificate (complementary slackness);
a set of prices (sensitivity);
an algorithm (the dual simplex).

How to read this deck

One LP runs through the whole unit: $\max 3x + 2y$ subject to $x + y \le 4$, $2x + y \le 6$, $x, y \ge 0$, optimal at $(2, 2)$ with value 10 and dual $(1, 1)$. Each face of the dual is shown on it with numbers. Proofs are on the slides only where they are one line; the longer ones are in the notes under Full proof.

03 · overview

Four faces of one idea

1 · a bound

Nonnegative combinations of constraints bound the objective. The best such bound is an LP, and it is exact.

2 · a certificate

A pair $(x, y)$ with equal objectives proves both optimal. Complementary slackness is the pointwise version.

3 · a set of prices

$y_i$ is what one more unit of $b_i$ is worth, exactly, within a computable range: a shadow price.

4 · an algorithm

The dual simplex method keeps optimality and restores feasibility. It is the method used after adding a cut or a bound.

getting around

Words with a dotted underline are glossary terms: click one for its definition. G lists every term in the unit, N shows the notes, where the longer proofs live. One LP, $\max 3x + 2y$, runs through every section: a solid underline like that one shows it on hover. Optional extra: interactive figures, linked from the slides they support by a small ▸ chip; skip them freely.

section 1

The dual as a bound

Adding up constraints, done as well as possible.

03 · bound

Where the dual comes from

$\max\ 3x + 2y\;$ s.t. $\;x + y \le 4,\ \ 2x + y \le 6,\ \ x, y \ge 0$.

Call this the primal. Multiply the constraints by multipliers $u, v \ge 0$ and add:

$$(u + 2v)\,x + (u + v)\,y \;\le\; 4u + 6v$$

If $u + 2v \ge 3$ and $u + v \ge 2$, the left side is at least $3x + 2y$ for every $x, y \ge 0$. So $4u + 6v$ is an upper bound on the optimum.

try some multipliers

$u = 3, v = 0$: $\ 3x + 3y \le 12$, bound 12.
$u = 2, v = 1$: $\ 4x + 3y \le 14$, bound 14.
$u = 1, v = 1$: $\ 3x + 2y \le 10$, bound 10. Can any do better?

The best such bound is itself an LP:

the dual $$\begin{aligned}\min\ & 4u + 6v\\ \text{s.t. } & u + 2v \ge 3\\ & u + v \ge 2\\ & u, v \ge 0\end{aligned}$$

$u = v = 1$ gives the bound $10$, and $x = y = 2$ attains $10$. Both are optimal, and neither needed a solver to check.

The example box. Each choice must pass both checks. $u = 3, v = 0$: $u + 2v = 3 \ge 3$ and $u + v = 3 \ge 2$. $u = 2, v = 1$: $4 \ge 3$ and $3 \ge 2$. $u = 1, v = 1$: $3 \ge 3$ and $2 \ge 2$, both tight, and the bound 10 is attained at $(2, 2)$, so no choice can do better. A choice like $u = 0, v = 1$ fails the second check ($u + v = 1 < 2$): constraint 2, $2x + y \le 6$, alone does not bound $3x + 2y$, because $y$ gets coefficient 1 where the objective needs 2.

This is unit 00's certificate idea in its original setting. There, a matching was a lower bound on vertex cover because we combined constraints (one per matched edge). Here the combination has real-valued weights, and the best weights form an LP.

It is also unit 01's Farkas argument turned into optimization. Farkas says a nonnegative combination of constraints can prove $0 \le -1$. Duality says a nonnegative combination can prove "objective $\le$ bound", and the best provable bound is the optimum.

Keep this LP. It is SMALL in the lab and the tableau example in unit 02, and its dual $(1, 1)$ is the first thing lab step 2 checks.

03 · bound

The rules, once

The dualization rules, for a max primal. In general the multipliers are $y = (y_1, \dots, y_m)$; on the textbook LP they are $(u, v)$.

primal (max $c^\top x$)	dual (min $b^\top y$)
constraint $i$: $a_i x \le b_i$	$y_i \ge 0$
constraint $i$: $a_i x \ge b_i$	$y_i \le 0$
constraint $i$: $a_i x = b_i$	$y_i$ free
$x_j \ge 0$	dual constraint $j$: $(A^\top y)_j \ge c_j$
$x_j \le 0$	dual constraint $j$: $(A^\top y)_j \le c_j$
$x_j$ free	dual constraint $j$: $(A^\top y)_j = c_j$

how to rebuild it

Multipliers must keep inequalities pointing the right way: $\le$ constraints get $y \ge 0$. The combination must dominate the objective on the variable's allowed sign: $x_j \ge 0$ needs coefficient $\ge c_j$.

involution

For a min primal, flip every sign. Then $\text{dual}(\text{dual}(P)) = P$ exactly, and lab step 1 tests this on 40 random LPs.

03 · bound · example

Dualizing, line by line

primal $$\begin{aligned}\max\ & x_1 + 2x_2\\ \text{s.t. } & x_1 + x_2 \le 3 \\ & x_1 - x_2 \ge -1 \\ & x_1 \ge 0,\ \ x_2 \text{ free}\end{aligned}$$

dual $$\begin{aligned}\min\ & 3y_1 - y_2\\ \text{s.t. } & y_1 + y_2 \ge 1 \\ & y_1 - y_2 = 2 \\ & y_1 \ge 0,\ \ y_2 \le 0\end{aligned}$$

primal piece	rule	dual piece
constraint 1 is $\le$	$\le$ constraint	$y_1 \ge 0$
constraint 2 is $\ge$	$\ge$ constraint	$y_2 \le 0$
$x_1 \ge 0$, column $(1, 1)$, cost 1	$x_j \ge 0$	$y_1 + y_2 \ge 1$
$x_2$ free, column $(1, -1)$, cost 2	$x_j$ free	$y_1 - y_2 = 2$
right-hand sides $(3, -1)$	→ costs	$3y_1 - y_2$

both solved

Primal: $x = (1, 2)$, value $1 + 4 = \mathbf 5$.
Dual: $y = (\tfrac32, -\tfrac12)$, value $\tfrac92 + \tfrac12 = \mathbf 5$.

Check the dual: $\tfrac32 - \tfrac12 = 1 \ge 1$ ✓, $\ \tfrac32 + \tfrac12 = 2$ ✓, signs ✓. Equal values, so both are optimal.

Why $y_2 \le 0$ makes sense here. Constraint 2 is $x_1 - x_2 \ge -1$. Multiplied by $y_2 = -\tfrac12$ it flips to $-\tfrac12 x_1 + \tfrac12 x_2 \le \tfrac12$, a $\le$ constraint that can be added to $\tfrac32 \times$ constraint 1, $\tfrac32 x_1 + \tfrac32 x_2 \le \tfrac92$. The sum is $x_1 + 2x_2 \le 5$: exactly the objective, bounded by 5.

Why the free variable gives an equation. The sum's coefficient on $x_2$ must equal 2, not just be at least 2. If it were 3, the sum $x_1 + 3x_2 \le$ bound would not bound $x_1 + 2x_2$ when $x_2$ is negative.

Solving the dual by hand. From the equation, $y_1 = 2 + y_2$. The objective is $3(2 + y_2) - y_2 = 6 + 2y_2$, so make $y_2$ as small as allowed. The first dual constraint gives $2 + 2y_2 \ge 1$, so $y_2 \ge -\tfrac12$. Hence $y_2 = -\tfrac12$, $y_1 = \tfrac32$, value 5.

Solving the primal by hand. $x_2$ is capped by $3 - x_1$ and by $x_1 + 1$. The two caps meet at $x_1 = 1$, $x_2 = 2$. Left of that the objective is $x_1 + 2(x_1 + 1)$, rising; right of it, $x_1 + 2(3 - x_1)$, falling. So the optimum is $(1, 2)$ with value 5.

In the lab this is GeneralLP("max", (1, 2), ((1, 1), (1, -1)), ("<=", ">="), (3, -1), (">=0", "free")). The reference dual returns exactly the LP on the slide, and HiGHS confirms both values.

03 · bound

Weak and strong duality

weak duality

For $\max\{c^\top x : Ax \le b,\, x \ge 0\}$ and its dual: if $x$ and $y$ are feasible, $c^\top x \le b^\top y$.

Why: $c^\top x \le (A^\top y)^\top x = y^\top (Ax) \le y^\top b$: the first step uses $x \ge 0$, the second $y \ge 0$. It is "add up the constraints", written with matrices.

strong duality

If the primal has an optimal solution, so does the dual, and the optimal values are equal.

Why: if every dual bound stayed above $z^* + \varepsilon$, the constraints plus "objective $\ge z^* + \varepsilon$" would have no solution. Farkas (unit 01) hands over weights proving that, and those weights are a dual solution with a bound below $z^* + \varepsilon$. Contradiction.

full proof in the notes (N)

on the running example

primal $(x, y)$	$3x + 2y$	dual $(u, v)$	$4u + 6v$
$(1, 1)$	5	$(2, 1)$	14
$(3, 0)$	9	$(3, 0)$	12
$(2, 2)$	10	$(1, 1)$	10

Every primal value is below every dual value: weak duality, nine comparisons at once. The last row has no gap: strong duality, and both are optimal.

The difference $b^\top y - c^\top x$ is the duality gap of a pair. For $(1,1)$ against $(2,1)$ it is 9; it closes to 0 only at the optimum.

The example table. Feasibility of each entry: $(1,1)$ has $2 \le 4$ and $3 \le 6$; $(3, 0)$ has $3 \le 4$ and $6 \le 6$; the three duals were checked on "Where the dual comes from". Each of the primal values 5, 9, 10 is at most each of the dual values 14, 12, 10, as weak duality promises, without solving anything.

Full proof of strong duality

For $\varepsilon > 0$, the system $Ax \le b,\ -c^\top x \le -(z^* + \varepsilon),\ x \ge 0$ has no solution. Farkas for $\{x \ge 0 : Ax \le b,\ -c^\top x \le -(z^*+\varepsilon)\}$ gives $(y, t) \ge 0$ with $y^\top A - t c^\top \ge 0$ and $y^\top b - t (z^* + \varepsilon) < 0$ (the form with $x \ge 0$; see the notes of unit 01's Farkas slide). If $t = 0$, then $y^\top A \ge 0$ and $y^\top b < 0$, which is a certificate that $Ax \le b$, $x \ge 0$ is infeasible, a contradiction. So $t > 0$. Divide by $t$: $A^\top y \ge c$, $y \ge 0$, and $b^\top y < z^* + \varepsilon$. By weak duality $b^\top y \ge z^*$. The dual optimum therefore lies in $[z^*, z^* + \varepsilon)$ for every $\varepsilon$, and since an LP attains its optimum (unit 01, FTLP) it equals $z^*$.

A second proof is constructive: the simplex method terminates (with Bland's rule) at a basis with all reduced costs $\le 0$, and $y^\top = c_B^\top A_B^{-1}$ is then dual feasible with $b^\top y = c^\top x$. That is section 2's reading of the tableau.

03 · bound

Four possible outcomes, not nine

The outcome table: which primal status can pair with which dual status.

primal \ dual	optimal	unbounded	infeasible
optimal	✓ equal values	✗	✗
unbounded	✗	✗	✓
infeasible	✗	✓	✓ possible

Unbounded primal ⇒ infeasible dual: any dual solution would bound it.

Both infeasible happens, for example with $\max\ x_1$ subject to $x_1 - x_2 \le -1,\ -x_1 + x_2 \le -1$.

It is why presolve sometimes can only say "infeasible or unbounded". In the lab's random general LPs, HiGHS found 246 optimal pairs and 154 unbounded/infeasible pairs out of 400.

The table follows from weak duality (the ✗ in "optimal / unbounded" and "unbounded / unbounded") and strong duality (the ✗ in "optimal / infeasible").

Check the doubly infeasible example: adding the two primal constraints gives $0 \le -2$, so the primal is infeasible. The dual is $\min -y_1 - y_2$ subject to $y_1 - y_2 \ge 1$, $-y_1 + y_2 \ge 0$, $y \ge 0$; adding its constraints gives $0 \ge 1$, so it is infeasible too. (The second dual constraint is $\ge 0$ because $x_2$ has cost 0.)

This is why solvers report "infeasible or unbounded" from presolve. Sometimes presolve can see that the dual is infeasible, which leaves both possibilities open for the primal. Unit 02's then met this: HiGHS labelled some unbounded LPs infeasible until presolve was turned off.

section 2

The dual as a certificate

Where the multipliers must be zero, and where they can be read off.

03 · certificate

Complementary slackness

theorem

Feasible $x$ and $y$ are both optimal if and only if

$$y_i\,(b_i - a_i x) = 0 \ \text{ for every constraint } i, \qquad x_j\,\big((A^\top y)_j - c_j\big) = 0 \ \text{ for every variable } j.$$

why: the gap splits into pieces

$b^\top y - c^\top x = \underbrace{y^\top (b - Ax)}_{\ge 0} + \underbrace{x^\top (A^\top y - c)}_{\ge 0}$

Every piece is (nonnegative) × (nonnegative). The gap is 0 exactly when every piece is 0, and a zero gap means both are optimal.

In words: a constraint with slack has price 0. A variable that is used has zero reduced cost. Only a binding constraint can carry a price.

Checking takes two matrix–vector products: an optimality certificate, lab step 3. The next slide runs the checks on the example.

Each product in the sum is a nonnegative number times a nonnegative number, which is why a zero total forces each one to zero. That identity is the proof of weak duality written as an equation instead of an inequality.

Complementary slackness is often the fastest way to find a dual solution when you already have a primal optimum. The constraints with slack force some $y_i = 0$, the positive variables force some dual constraints to hold with equality, and what remains is a square linear system. Lab step 3's rejection tests include a pair that is primal and dual feasible and fails only this condition: $x = (0,0)$, $(u, v) = (1,1)$ on the textbook LP.

It generalizes directly to KKT conditions (unit 29). There, complementarity says an inactive constraint has zero multiplier, which is the same statement with gradients in place of rows.

03 · certificate · example

Complementary slackness, on the example

the optimal pair

$x = (2, 2)$ and the dual's $(u, v) = (1, 1)$:

piece	slack	paired with
constraint 1	$4 - 4 = 0$	$u = 1$
constraint 2	$6 - 6 = 0$	$v = 1$
variable $x$	$3 - 3 = 0$	$x = 2$
variable $y$	$2 - 2 = 0$	$y = 2$

All four products are 0, so the gap $10 - 10$ is 0.

feasible, but not optimal

$x = (0, 0)$ with $(u, v) = (1, 1)$. Constraint 1 has slack 4 but price 1: product 4. Constraint 2: slack 6, price 1: product 6. The gap is $10 - 0 = 4 + 6$, exactly the sum. The certificate check rejects it on constraint 1.

using it to find the dual

Given only $x = (2, 2)$: both variables are positive, so both dual constraints must be tight:

$u + 2v = 3,\quad u + v = 2 \;\Rightarrow\; v = 1,\ u = 1.$

Two equations, no optimization. Then check $u, v \ge 0$: yes, so $(2, 2)$ is proved optimal.

a constraint with slack gets price 0

Add the constraint $x \le 3$. At $(2, 2)$ it has slack 1, so its price must be 0. Solving confirms duals $(1, 1, 0)$, optimum still 10. Loosening a constraint that isn't pressing changes nothing.

The slack column for variables is the dual constraint's surplus, $(A^\top y)_j - c_j$ with $y = (u, v)$. For variable $x$ the column of $A$ is $(1, 2)$, so $u + 2v = 3$ against cost 3: the table's $3 - 3$. For variable $y$ the column is $(1, 1)$: $u + v = 2$ against cost 2. Both are tight, so it does no harm that $x$ and $y$ are positive.

The failing pair $x = (0,0)$, $(u, v) = (1,1)$: the variable pieces are $0 \times 0$ each, and the constraint pieces are $1 \times 4$ and $1 \times 6$. The identity on the previous slide says the gap $(4u + 6v) - (3x + 2y) = 10 - 0$ equals $4 + 6 + 0 + 0$, and it does. The reference certify_optimal returns (False, "complementary slackness fails on primal constraint 0").

Why "find the dual" works in general: at a nondegenerate optimum, the positive variables give exactly as many equations as there are prices left unknown on the tight constraints. Degeneracy can leave it underdetermined, and then several duals are optimal. The extra-constraint box was solved with unit 02's reference two_phase, which returned duals $(1, 1, 0)$.

03 · certificate

Reading the dual off the final tableau

At an optimal basis, $y^\top = c_B^\top A_B^{-1}$, and the reduced cost of every column is $\bar c_j = c_j - y^\top a_j$.

A slack $s_i$ has cost 0 and column $e_i$, so $\bar c_{s_i} = -y_i$. The objective row, which stores $-\bar c$, holds $y_i$ in slack column $i$.

The optimal dual solution is the slack columns of the objective row (reading duals off the tableau). Every simplex solve returns a primal solution and a dual certificate at once.

Unit 02's final tableau:

y	0	1	2	−1	2
x	1	0	−1	1	2
z	0	0	1	1	10

$(u, v) = (1, 1)$, and $4u + 6v = 10 = 3x + 2y$.

section 3

The dual as prices

What a unit of each resource is worth, and for how long.

03 · prices

Shadow prices

Optimum of the textbook LP as the right-hand side of $x + y \le b_1$ varies. Shaded: the range where the current basis stays optimal.

theorem

If the optimal basis is nondegenerate, $\;\dfrac{\partial z^*}{\partial b_i} = y_i$.

More generally the value function $z^*(b)$ is concave and piecewise linear, and $y$ is a supergradient.

Here $y_1 = u = 1$ on $b_1 \in [3, 6]$. Below 3, constraint 2 stops binding and the price jumps to 3. Above 6, constraint 1 itself stops binding and the price drops to 0.

explore interactively

The curve was computed with the exact simplex at 17 values of $b_1$: slopes 3, 1, 0, with breakpoints at $b_1 = 3$ (value 9) and $b_1 = 6$ (value 12).

Why concave: $z^*(b)$ is the minimum over dual feasible $y$ of $b^\top y$. The dual feasible set does not depend on $b$, so $z^*$ is a minimum of linear functions of $b$, which is concave. The pieces correspond to the dual vertices, one per optimal basis. This picture comes back in unit 11: the Lagrangian dual function is also a minimum of linear functions, and subgradient methods climb it.

The economic reading: if constraint $i$ is a resource (machine hours, budget), $y_i$ is the most you should pay for one more unit, and a constraint with slack is worth nothing at the margin. This is also where the name "shadow price" comes from.

03 · prices

Ranging: how long the price holds

Ranging: change $b_i$ to $b_i + \delta$. Reduced costs don't involve $b$, so the basis stays optimal for as long as it stays feasible:

$$x_B(\delta) = A_B^{-1}(b + \delta e_i) = \bar b + \delta\, d$$

where $d$ is slack column $i$ of the final tableau.

Feasible while $\bar b_r + \delta d_r \ge 0$ for all rows:

$$\max_{d_r > 0} \frac{-\bar b_r}{d_r} \;\le\; \delta \;\le\; \min_{d_r < 0} \frac{-\bar b_r}{d_r}$$

example

Slack column 1 is $d = (2, -1)$ with $\bar b = (2, 2)$: $\delta \in [-1, 2]$, so $b_1 \in [3, 6]$. Inside, $z^* = 10 + 1\cdot\delta$.

HiGHS's ranging report gives $[3, 6]$ with objective 9 and 12 at the ends. The lab's then checks 103 binding constraints, and all agree.

Outside the range the basis changes and the slope changes. A linear prediction there is wrong, which is why lab step 5 must refuse it.

explore interactively

Degeneracy caveat: at a degenerate optimum, several optimal bases exist and each has its own range, and the left and right derivatives of $z^*$ differ. Solvers report the range for the basis they stopped at. Treat ranging output as "this basis's range", not "the" range.

Cost ranging, the same analysis for the objective coefficients: changing $c_j$ keeps the basis feasible and changes reduced costs, so the basis stays optimal while the objective row stays nonnegative. Solvers report both kinds (HiGHS's getRanging has col_cost_up/col_cost_dn next to the row bounds used in the lab).

The non-binding constraints the lab's then skips: when slack $s_i$ is basic with value $v$, $d = e_r$, so the range is $\delta \ge -v$ with no upper end. You can loosen a non-binding constraint forever, and tighten it until it just binds. HiGHS's report uses a different convention for those constraints, so the comparison is restricted to binding ones.

03 · prices

Sign conventions: an afternoon, as promised

The same LP, $\max 3x + 2y$, $x + y \le 4$, $2x + y \le 6$:

source	$u$	$v$
your tableau	1	1
HiGHS, `sense=max`	1	1
HiGHS, written as $\min -c^\top x$	−1	−1
SciPy `linprog` marginals	−1	−1

one sign convention covers all four

A reported dual is $\partial(\text{objective})/\partial(\text{rhs})$ in the solver's own objective sense.

Maximizing, loosening a $\le$ constraint helps, so the dual is $\ge 0$. Minimizing $-c^\top x$, the same loosening lowers the objective, so it is $\le 0$.

Every solver documents its convention somewhere, and every modeller has at some point shipped a sign error from forgetting it. Before trusting duals from a solver, solve a two-variable LP whose duals you know, like this one, and look. That is what the curriculum means by "which will take you an afternoon, and should".

Two more places where conventions differ: duals of $\ge$ constraints (some solvers report them with the opposite sign to $\le$ constraints, some don't), and reduced costs of variables at their upper bound in bounded-variable simplex (which have the opposite sign to those at their lower bound). HiGHS, Gurobi and CPLEX each document these; they don't all agree.

Gurobi, for reference, reports Pi for constraint duals and RC for reduced costs. It isn't installed here, and no lab depends on it.

section 4

The dual as an algorithm

Keep optimality, restore feasibility.

03 · dual simplex

The dual simplex method

State: a basis whose objective row is nonnegative (dual feasible) but with some negative right-hand sides (no primal feasibility).

Leaving row: any row $r$ with $\bar b_r < 0$. The most negative is the usual choice.
Entering column: among $j$ with $T_{rj} < 0$, minimize $\;\dfrac{T_{mj}}{-T_{rj}}$, the dual ratio test.
Pivot. If no $T_{rj} < 0$, row $r$ reads "nonnegative combination = negative": the LP is infeasible.

why it works

It is exactly the primal simplex method applied to the dual LP, run on the same tableau.

The dual ratio test keeps every objective-row entry nonnegative after the pivot, and the objective value decreases monotonically toward the optimum from above.

Row $r$ in the infeasible case, an infeasible row, is a Farkas certificate: unit 01 again.

The next slide runs one pivot of it on the example.

explore interactively

Why the ratio test has that form. Pivoting on $T_{rs}$ subtracts $\frac{T_{ms}}{T_{rs}} \times$ row $r$ from the objective row. For column $j$ the new entry is $T_{mj} - T_{ms}\frac{T_{rj}}{T_{rs}}$. With $T_{rs} < 0$ and $T_{rj} < 0$, this stays $\ge 0$ exactly when $\frac{T_{mj}}{-T_{rj}} \ge \frac{T_{ms}}{-T_{rs}}$, which is what choosing the minimum ratio guarantees. Columns with $T_{rj} \ge 0$ only get larger.

Infeasibility: row $r$ says $\sum_j T_{rj} x_j = \bar b_r < 0$ with every coefficient $\ge 0$ and every variable $\ge 0$. The left side cannot be negative. The row is a combination of the original rows, so its multipliers are a Farkas certificate. Lab step 4's third test builds this case on purpose, with the cut $x + y \ge 5$ against $x + y \le 4$.

In modern solvers the dual simplex is the default LP algorithm for most problems, not just for re-solves. Dual steepest-edge pricing is cheap to maintain, and on many real LPs the dual is easier to get feasible.

03 · dual simplex · example

One dual simplex pivot, on the example

From the optimum $(2, 2)$, add the cut $x \le 1.5$ with slack $s_3$. Written in the current basis, its row is $s_1 - s_2 + s_3 = -\tfrac12$:

	x	y	s₁	s₂	s₃	rhs
y	0	1	2	−1	0	2
x	1	0	−1	1	0	2
s₃	0	0	1	−1	1	−½
z	0	0	1	1	0	10

1. Objective row still $\ge 0$: dual feasible. Only $s_3 = -\tfrac12$ is wrong, so row $s_3$ leaves.
2. Negative entries in that row: only $s_2$, ratio $1 / 1$. $s_2$ enters.
3. Pivot. New point $(1.5, 2.5)$, $z = 9.5$, every rhs $\ge 0$: optimal after one pivot.

A cold re-solve from the origin takes 2 pivots. New duals $(2, 0, 1)$: constraint 2 now has slack $\tfrac12$, and price 0.

Shaded: what the cut removes. Ring: the old optimum, now infeasible. The pivot slides along $x + y = 4$ to the new corner.

explore interactively

Writing the cut in the basis. The cut is $x + s_3 = 1.5$. The tableau's $x$ row says $x - s_1 + s_2 = 2$. Subtract it: $s_1 - s_2 + s_3 = 1.5 - 2 = -\tfrac12$. That is the new row. The objective row is untouched, which is why the basis stays dual feasible.

The dual ratio test. Candidates are columns with a negative entry in the leaving row: only $s_2$, entry $-1$. Its objective-row entry is 1, so the ratio is $1 / 1$. With one candidate there is nothing to compare, but the test is what keeps the objective row nonnegative: after the pivot it reads $(0, 0, 2, 0, 1 \mid \tfrac{19}{2})$.

After the pivot. Row $s_2$ becomes $-s_1 + s_2 - s_3 = \tfrac12$. Adding it to the $y$ row gives $y = 2 + \tfrac12 = \tfrac52$; subtracting it from the $x$ row gives $x = 2 - \tfrac12 = \tfrac32$. The objective is $3 \cdot 1.5 + 2 \cdot 2.5 = 9.5$, down by $\tfrac12$ from 10.

Complementary slackness again. The new duals are the slack columns, $(2, 0, 1)$. Constraint 2, $2x + y \le 6$, has $3 + 2.5 = 5.5$, slack $\tfrac12$, so its price dropped to 0; constraint 1 is tight and now worth 2; the cut is worth 1.

Every number is from the reference add_constraint and dual_simplex, and from a cold two_phase on the LP with the cut added, which took 2 pivots.

03 · dual simplex

Why branch-and-cut lives on it

After an optimal solve, add a cut $a^\top x \le \beta$ (unit 08) or tighten a bound $x_j \le \lfloor x_j^* \rfloor$ (unit 07):

the new row, written in the current basis, leaves the objective row unchanged, so the basis is still dual feasible;
only the new row's rhs can be negative, so the tableau is still almost primal feasible.

That is the dual simplex's starting state, and it is typically a few pivots away from optimal: a warm start, not a cold start.

Lab then, 163 random LPs with one added cut:

	warm	cold
your dual simplex (pivots)	0.98	3.34
HiGHS (iterations)	0.96	3.47

On 4×5 LPs. The ratio widens with size: a cold solve's pivots grow with the LP, while a warm re-solve's pivots grow with the size of the change.

03 · the load

This unit, wearing seven hats

08 cutting planes	cuts are new constraints; the dual simplex re-optimizes after each round; Gomory cuts are read off the tableau.
10 column generation	the master's duals are the pricing problem's objective; a column with positive reduced cost enters.
11 Lagrangian relaxation	price the hard constraints, and the best prices give a bound: a partial dual, climbed by subgradients.
12 Benders	subproblem duals become cuts on the master: duals as rows instead of columns.
15 matching	the Hungarian algorithm keeps dual potentials feasible and grows the matching on tight edges: primal–dual, and complementary slackness.
24 primal–dual approximation	raise duals until constraints go tight, buy what goes tight, and bound the cost by weak duality.
29 KKT	the conditions of this unit for curved objectives and constraints: feasibility, dual feasibility, complementary slackness.

When one of those units feels hard, come back to the slide here that it generalizes: the seven hats are one idea.

03 · lab

Lab 03 — what you build

about 2½ hours · uv run co lab 03 · on unit 02's tableaux

The dual transformer for every constraint sense and variable sign, as an exact involution.
Duals from the tableau: one slice.
An optimality certificate checked exactly, with the reason when it fails.
Warm start: add a cut, then the dual simplex. Agrees with a cold re-solve.
Sensitivity: ranges, and predictions that must be exact inside and refused outside.

uv run co test 03
uv run co then 03     # sign conventions · ranging vs HiGHS · warm vs cold

03 · checkpoint

Done when

checkpoint

All five lab steps are ✓.
Given a solved LP, you can read off the shadow price of each constraint, predict the objective change from a small right-hand-side perturbation before re-solving, and be right.
You can say why the dual simplex is the method used after adding a cut.

Reading

Bertsimas & Tsitsiklis, ch. 4 (§4.1–4.5) and §5.1–5.2.
Chvátal, ch. 5 and 10.
Gärtner & Matoušek, §6.5–6.6 (optional).

05 formulations (IP begins) · 04 ellipsoid and interior point (branch) · 11 Lagrangian relaxation, which needs 16 as well · 29 convexity, if you want the continuous view now.

Duality — the load-bearing unit

Four faces of one idea

The dual as a bound

Where the dual comes from

The rules, once

Dualizing, line by line

Weak and strong duality

Four possible outcomes, not nine

The dual as a certificate

Complementary slackness

Complementary slackness, on the example

Reading the dual off the final tableau

The dual as prices

Shadow prices

Ranging: how long the price holds

Sign conventions: an afternoon, as promised

The dual as an algorithm

The dual simplex method

One dual simplex pivot, on the example

Why branch-and-cut lives on it

This unit, wearing seven hats

Lab 03 — what you build

Done when

Reading

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