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Chapter 3 • Theory & Derivations

Duality Theory, Complementary Slackness & Post-Optimality Sensitivity Analysis

Symmetric and asymmetric Primal-Dual linear programming formulations and economic shadow price interpretation; Weak Duality Lemma, Strong Duality Theorem, and Farkas' Lemma of linear inequalities; Complementary Slackness Theorem and primal-dual optimality verification; The Dual Simplex Algorithm for restoring primal feasibility while maintaining dual optimality; Post-optimality sensitivity analysis: perturbations in objective coefficients, right-hand-side resource vectors, and structural column additions.

§3.1 Symmetric and Asymmetric Primal-Dual Formulations and the Involution Theorem

Every Linear Programming Problem, which we designate as the **Primal Problem (P)**, is intrinsically paired with a companion linear optimization problem constructed from the exact same data arrays $(\mathbf{A}, \mathbf{b}, \mathbf{c})$, termed the **Dual Problem (D)**. Originating in John von Neumann's 1947 notes linking linear programming to zero-sum matrix games and formalized by David Gale, Harold Kuhn, and Albert Tucker (1951), Duality Theory reveals that optimizing a primal system in $\mathbb{R}^n$ subject to $m$ constraints is mathematically equivalent to optimizing a dual system in $\mathbb{R}^m$ subject to $n$ constraints. ### Symmetric Primal-Dual Pairings Consider the **Symmetric Canonical Primal Problem (P)** in $n$ non-negative decision variables $\mathbf{x} = (x_1, \dots, x_n)^T \in \mathbb{R}_+^n$ and $m$ resource inequalities defined by $\mathbf{A} = [a_{ij}] \in \mathbb{R}^{m \times n}$, $\mathbf{b} \in \mathbb{R}^m$, and $\mathbf{c} \in \mathbb{R}^n$: $$\text{(P)} \quad \text{Maximize } Z = \mathbf{c}^T \mathbf{x} = \sum_{j=1}^n c_j x_j \quad \text{subject to} \quad \mathbf{A}\mathbf{x} \le \mathbf{b} \;\; \left(\sum_{j=1}^n a_{ij} x_j \le b_i, \; i=1,\dots,m\right), \quad \mathbf{x} \ge \mathbf{0}$$ Associated with each of the $m$ primal constraints $i \in \{1, \dots, m\}$, we introduce a scalar **dual variable** $y_i$ to form the dual column vector $\mathbf{y} = (y_1, y_2, \dots, y_m)^T \in \mathbb{R}^m$. The **Symmetric Dual Problem (D)** reverses the sense of optimization from maximization to minimization, interchanges the roles of the right-hand-side capacity vector $\mathbf{b}$ and the objective coefficient vector $\mathbf{c}$, and transposes the technological matrix $\mathbf{A}$ to $\mathbf{A}^T \in \mathbb{R}^{n \times m}$: $$\text{(D)} \quad \text{Minimize } W = \mathbf{b}^T \mathbf{y} = \sum_{i=1}^m b_i y_i \quad \text{subject to} \quad \mathbf{A}^T \mathbf{y} \ge \mathbf{c} \;\; \left(\sum_{i=1}^m a_{ij} y_i \ge c_j, \; j=1,\dots,n\right), \quad \mathbf{y} \ge \mathbf{0}$$ ### Asymmetric Formulations and the General Primal-Dual Correspondence Table What if the Primal Problem contains a mixture of $\le$ inequalities, $\ge$ inequalities, and strict equalities ($=$), alongside non-negative, non-positive, and unrestricted variables? By converting any general Primal LP into symmetric canonical form (replacing $\mathbf{a}_i^T\mathbf{x} \ge b_i$ by $-\mathbf{a}_i^T\mathbf{x} \le -b_i$, replacing $\mathbf{a}_i^T\mathbf{x} = b_i$ by the pair $\mathbf{a}_i^T\mathbf{x} \le b_i$ and $-\mathbf{a}_i^T\mathbf{x} \le -b_i$ with dual variables $y_i^+, y_i^- \ge 0$ so $y_i = y_i^+ - y_i^- \in \mathbb{R}$ is unrestricted, and splitting unrestricted primal variables $x_j = x_j^+ - x_j^-$), we obtain an exact one-to-one structural correspondence between the $m$ primal constraints and $m$ dual variables, and between the $n$ primal variables and $n$ dual constraints:
Maximization Primal (P)Minimization Dual (D)Structural Intuition
Objective $\max Z = \mathbf{c}^T\mathbf{x}$Objective $\min W = \mathbf{b}^T\mathbf{y}$Primal RHS $\mathbf{b}$ becomes Dual cost; Primal cost $\mathbf{c}$ becomes Dual RHS
Constraint $i$ is $\le$ (Sensible)Dual variable $y_i \ge 0$ (Sensible)Upper resource bound commands a non-negative shadow price
Constraint $i$ is $=$ (Odd)Dual variable $y_i \in \mathbb{R}$ unrestricted (Odd)Exact equality constraint can impede objective in either direction
Constraint $i$ is $\ge$ (Bizarre)Dual variable $y_i \le 0$ (Bizarre)Reversed primal inequality reverses dual variable sign
Variable $x_j \ge 0$ (Sensible)Dual constraint $j$ is $\ge$ (Sensible)Non-negative primal activity requires imputed cost $\ge$ unit return
Variable $x_j \in \mathbb{R}$ free (Odd)Dual constraint $j$ is $=$ (Odd)Unrestricted primal activity forces exact zero net arbitrage ($\mathbf{a}_j^T\mathbf{y} = c_j$)
Variable $x_j \le 0$ (Bizarre)Dual constraint $j$ is $\le$ (Bizarre)Non-positive primal activity reverses dual inequality
#### Theorem 3.1 (Involutory Property: The Dual of the Dual is the Primal): For any Linear Programming Problem (P), constructing the dual of its dual problem (D) recovers the exact original Primal Problem (P). #### Proof: Write the symmetric Dual (D) as a maximization problem by negating its objective and constraints: $\max (-W) = (-\mathbf{b})^T \mathbf{y}$ subject to $(-\mathbf{A}^T)\mathbf{y} \le -\mathbf{c}, \mathbf{y} \ge \mathbf{0}$. Applying the canonical dual transformation to this maximization problem yields a minimization problem in vector $\mathbf{x} \ge \mathbf{0}$: $\min (-\mathbf{c})^T \mathbf{x}$ subject to $(-\mathbf{A}^T)^T \mathbf{x} \ge -\mathbf{b}, \mathbf{x} \ge \mathbf{0}$. Negating the objective back to maximization and multiplying the constraints by $-1$ produces $\max Z = \mathbf{c}^T\mathbf{x}$ subject to $\mathbf{A}\mathbf{x} \le \mathbf{b}, \mathbf{x} \ge \mathbf{0}$, which is identically (P).

§3.2 Economic Interpretation of Duality: Imputed Resource Valuations and Shadow Prices

Duality Theory is far more than a formal algebraic transposition; as discovered by Leonid Kantorovich (1939) and Tjalling Koopmans (1951), the Dual Linear Program embodies the decentralized price mechanism of competitive economic equilibrium and statistical resource valuation.

Suppose the Primal Problem $\max Z = \sum_{j=1}^n c_j x_j$ subject to $\sum_{j=1}^n a_{ij} x_j \le b_i$ ($i=1,\dots,m$), $x_j \ge 0$, models a manufacturing enterprise or statistical research institute that uses $m$ scarce resources (each available in total quantity $b_i$ units) to produce $n$ distinct outputs (in quantities $x_j$, each generating net revenue $c_j$ dollars per unit and consuming $a_{ij}$ units of resource $i$ per unit of output $j$). Dimensional analysis immediately illuminates the meaning of the dual variables $y_i$:

  • Because $b_i$ is measured in [units of resource i] and the dual objective $W = \sum_{i=1}^m b_i y_i$ is measured in [dollars], each dual variable $y_i$ must have physical dimensions of [dollars per unit of resource i].

Thus, $y_i$ represents the imputed unit valuation (or accounting price / shadow price) assigned internally to one unit of scarce resource $i$. Now consider an outside entrepreneur or resource broker who offers to rent or purchase the entire endowment $(b_1, \dots, b_m)$ of the enterprise's resources at unit prices $\mathbf{y} = (y_1, \dots, y_m)^T \ge \mathbf{0}$:

1. The Dual Constraint ($\sum_{i=1}^m a_{ij} y_i \ge c_j$): To produce one unit of activity $j$, the enterprise consumes bundle $(a_{1j}, a_{2j}, \dots, a_{mj})$ of the $m$ resources. At unit resource prices $y_i$, the total imputed opportunity cost of the resources tied up in producing one unit of activity $j$ is $\sum_{i=1}^m a_{ij} y_i$. For the enterprise to be willing to cede its resources rather than operate activity $j$ internally, the imputed value of the resources required to make one unit of product $j$ must be at least as large as the direct profit $c_j$ that product $j$ would generate! If $\sum_{i=1}^m a_{ij} y_i < c_j$, an arbitrage profit opportunity exists: purchasing the required resources at prices $y_i$ and running activity $j$ yields a strictly positive net economic rent $c_j - \sum_{i=1}^m a_{ij} y_i > 0$.

2. The Dual Objective ($\min W = \sum_{i=1}^m b_i y_i$): Subject to offering competitive resource valuations that leave zero excess arbitrage profit across all $n$ production activities ($\mathbf{A}^T\mathbf{y} \ge \mathbf{c}$), the buyer seeks to minimize the total expenditure $W = \sum_{i=1}^m b_i y_i$ paid for the resource bundle $\mathbf{b}$.

At the joint primal-dual optimum $(\mathbf{x}^*, \mathbf{y}^*)$, Strong Duality ($\mathbf{c}^T\mathbf{x}^* = \mathbf{b}^T\mathbf{y}^*$) states that the maximum total profit achievable by physical production equals the minimum total imputed value of the scarce resources! Moreover, if the optimal primal basis $\mathbf{B}$ is non-degenerate, the optimal dual vector $\mathbf{y}^* = (\mathbf{B}^{-1})^T \mathbf{c}_B$ equals the exact marginal partial derivative (shadow price) of the optimal objective value function $Z^*(\mathbf{b}) = \mathbf{c}_B^T \mathbf{B}^{-1}\mathbf{b}$ with respect to each resource availability $b_i$:

$$y_i^* = \frac{\partial Z^*(\mathbf{b})}{\partial b_i}, \qquad i = 1, 2, \dots, m$$

That is, $y_i^*$ quantifies the exact rate at which maximum achievable profit $Z^*$ increases per additional unit of resource $i$ made available (within the feasibility range of the current optimal basis $\mathbf{B}$).

§3.3 The Weak Duality Lemma, Farkas' Lemma, and the Strong Duality Theorem

We now establish the rigorous mathematical theorems governing the relationship between the Primal Problem (P): $\max Z = \mathbf{c}^T\mathbf{x}$ subject to $\mathbf{A}\mathbf{x} \le \mathbf{b}, \mathbf{x} \ge \mathbf{0}$, and the Dual Problem (D): $\min W = \mathbf{b}^T\mathbf{y}$ subject to $\mathbf{A}^T\mathbf{y} \ge \mathbf{c}, \mathbf{y} \ge \mathbf{0}$. ### Lemma 3.1 (The Weak Duality Theorem): Let $\mathbf{x} \in \mathbb{R}_+^n$ be any feasible solution of the Primal Problem (P) ($\mathbf{A}\mathbf{x} \le \mathbf{b}, \mathbf{x} \ge \mathbf{0}$), and let $\mathbf{y} \in \mathbb{R}_+^m$ be any feasible solution of the Dual Problem (D) ($\mathbf{A}^T\mathbf{y} \ge \mathbf{c}, \mathbf{y} \ge \mathbf{0}$). Then the primal objective value never exceeds the dual objective value: $$Z(\mathbf{x}) = \mathbf{c}^T \mathbf{x} \le \mathbf{b}^T \mathbf{y} = W(\mathbf{y})$$ #### Proof: Because $\mathbf{y}$ is dual feasible ($\mathbf{A}^T\mathbf{y} \ge \mathbf{c}$, or $\mathbf{c}^T \le \mathbf{y}^T\mathbf{A}$) and $\mathbf{x} \ge \mathbf{0}$ is non-negative, taking the inner product with $\mathbf{x}$ preserves the direction of inequality: $$\mathbf{c}^T \mathbf{x} \le (\mathbf{y}^T \mathbf{A}) \mathbf{x} = \mathbf{y}^T (\mathbf{A}\mathbf{x})$$ Similarly, because $\mathbf{x}$ is primal feasible ($\mathbf{A}\mathbf{x} \le \mathbf{b}$) and $\mathbf{y} \ge \mathbf{0}$ is non-negative, taking the inner product with $\mathbf{y}^T$ preserves the inequality: $$\mathbf{y}^T (\mathbf{A}\mathbf{x}) \le \mathbf{y}^T \mathbf{b} = \mathbf{b}^T \mathbf{y}$$ Chaining these two inequalities together yields $\mathbf{c}^T\mathbf{x} \le \mathbf{y}^T\mathbf{A}\mathbf{x} \le \mathbf{b}^T\mathbf{y}$, completing the proof. The non-negative difference $W(\mathbf{y}) - Z(\mathbf{x}) = \mathbf{b}^T\mathbf{y} - \mathbf{c}^T\mathbf{x} \ge 0$ is called the **duality gap** at the feasible pair $(\mathbf{x}, \mathbf{y})$. #### Immediate Corollaries of Weak Duality: 1. **Sufficient Certificate of Optimality**: If $\mathbf{x}^*$ is primal feasible, $\mathbf{y}^*$ is dual feasible, and their objectives coincide ($\mathbf{c}^T\mathbf{x}^* = \mathbf{b}^T\mathbf{y}^*$), then $\mathbf{x}^*$ is globally optimal for (P) and $\mathbf{y}^*$ is globally optimal for (D), because for every primal feasible $\mathbf{x}$ and dual feasible $\mathbf{y}$, $\mathbf{c}^T\mathbf{x} \le \mathbf{b}^T\mathbf{y}^* = \mathbf{c}^T\mathbf{x}^* \le \mathbf{b}^T\mathbf{y}$. 2. **Primal Unboundedness Implies Dual Infeasibility**: If (P) is unbounded above ($\sup \mathbf{c}^T\mathbf{x} = +\infty$), then (D) must be strictly infeasible ($\mathcal{F}_D = \emptyset$), since any dual feasible vector $\mathbf{y}$ would provide a finite upper bound $\mathbf{b}^T\mathbf{y} < \infty$ on $\mathbf{c}^T\mathbf{x}$. 3. **Dual Unboundedness Implies Primal Infeasibility**: Symmetrically, if (D) is unbounded below ($\inf \mathbf{b}^T\mathbf{y} = -\infty$), then (P) must be strictly infeasible ($\mathcal{F}_P = \emptyset$). (Note: It is possible for **both** (P) and (D) to be simultaneously infeasible!) ### Farkas' Lemma and the Strong Duality Theorem The bridge from Weak Duality to zero duality gap at optimality is anchored by **Farkas' Lemma** (Gyula Farkas, 1902), a foundational theorem of the alternative deduced from separating hyperplanes: for any matrix $\mathbf{A} \in \mathbb{R}^{m \times n}$ and vector $\mathbf{b} \in \mathbb{R}^m$, **exactly one** of the following two linear systems possesses a solution: - **System I**: $\mathbf{A}\mathbf{x} = \mathbf{b}$ and $\mathbf{x} \ge \mathbf{0}$ (meaning $\mathbf{b}$ lies inside the finitely generated convex cone $\text{cone}(\mathbf{a}_1, \dots, \mathbf{a}_n)$). - **System II**: $\mathbf{A}^T\mathbf{y} \le \mathbf{0}$ and $\mathbf{b}^T\mathbf{y} > 0$ (meaning there exists a hyperplane with normal $\mathbf{y}$ separating $\mathbf{b}$ from $\text{cone}(\mathbf{a}_1, \dots, \mathbf{a}_n)$). ### Theorem 3.2 (The Strong Duality Theorem of Linear Programming): If either the Primal Problem (P) or the Dual Problem (D) possesses a finite optimal solution, then **so does the other**, and their optimal objective values are **strictly equal** (zero duality gap): $$Z^* = \max_{\mathbf{x} \in \mathcal{F}_P} \mathbf{c}^T \mathbf{x} = \min_{\mathbf{y} \in \mathcal{F}_D} \mathbf{b}^T \mathbf{y} = W^*$$ Moreover, if $\mathbf{B}$ is an optimal basis matrix for the standard-form Primal Problem ($\max \mathbf{c}^T\mathbf{x} + \mathbf{0}^T\mathbf{s}$ subject to $\mathbf{A}\mathbf{x} + \mathbf{I}_m\mathbf{s} = \mathbf{b}, \mathbf{x},\mathbf{s} \ge \mathbf{0}$), then the **Simplex multiplier vector** $\mathbf{y}^* = (\mathbf{B}^{-1})^T \mathbf{c}_B$ is an **optimal Basic Feasible Solution of the Dual Problem (D)**! #### Constructive Simplex Proof of Strong Duality: Suppose (P) has a finite optimal solution. Adding slack variables $\mathbf{s} \ge \mathbf{0}$ with cost $\mathbf{c}_s = \mathbf{0}$, the Simplex Algorithm (with Bland's anti-cycling rule) terminates at an optimal basis $\mathbf{B}$ with primal BFS $\mathbf{x}_B^* = \mathbf{B}^{-1}\mathbf{b} \ge \mathbf{0}$ and optimal objective $Z^* = \mathbf{c}_B^T \mathbf{B}^{-1}\mathbf{b}$. At termination, the Simplex optimality condition holds across all $n$ structural columns $\mathbf{a}_j$ of $\mathbf{A}$ and all $m$ slack columns $\mathbf{e}_i$ of $\mathbf{I}_m$. Define $\mathbf{y}^* = (\mathbf{B}^{-1})^T \mathbf{c}_B$, so $(\mathbf{y}^*)^T = \mathbf{c}_B^T \mathbf{B}^{-1}$. 1. For the $n$ structural columns $j = 1, \dots, n$, the optimal net evaluation satisfies: $$z_j - c_j = \mathbf{c}_B^T \mathbf{B}^{-1} \mathbf{a}_j - c_j = (\mathbf{y}^*)^T \mathbf{a}_j - c_j \ge 0 \implies \mathbf{A}^T \mathbf{y}^* \ge \mathbf{c}$$ 2. For the $m$ slack columns $s_i$ with unit column $\mathbf{e}_i$ and cost $c_{s_i} = 0$, the optimal net evaluation satisfies: $$z_{s_i} - 0 = \mathbf{c}_B^T \mathbf{B}^{-1} \mathbf{e}_i - 0 = (\mathbf{y}^*)^T \mathbf{e}_i = y_i^* \ge 0 \implies \mathbf{y}^* \ge \mathbf{0}$$ Thus $\mathbf{y}^* = (\mathbf{B}^{-1})^T \mathbf{c}_B$ satisfies both $\mathbf{A}^T\mathbf{y}^* \ge \mathbf{c}$ and $\mathbf{y}^* \ge \mathbf{0}$, proving that **$\mathbf{y}^*$ is Dual Feasible**! Notice further that $y_i^* = z_{s_i} - c_{s_i}$, meaning **the optimal dual variables $y_1^*, \dots, y_m^*$ appear directly in the final Simplex $z_j - c_j$ row under the $m$ slack columns**! Finally, evaluating the dual objective at $\mathbf{y}^*$ gives: $$W(\mathbf{y}^*) = \mathbf{b}^T \mathbf{y}^* = (\mathbf{y}^*)^T \mathbf{b} = \mathbf{c}_B^T \mathbf{B}^{-1} \mathbf{b} = \mathbf{c}_B^T \mathbf{x}_B^* = Z^*$$ By Corollary 1 of Weak Duality, equality $W(\mathbf{y}^*) = Z^*$ proves that $\mathbf{y}^*$ is globally optimal for (D) and $Z^* = W^*$.
Primal (P) / Dual (D) StatusDual Optimal (Finite)Dual UnboundedDual Infeasible
Primal Optimal (Finite)Possible (Strong Duality $Z^* = W^*$)Impossible (violates Weak Duality)Impossible (violates Strong Duality)
Primal UnboundedImpossible (violates Weak Duality)Impossible (violates Weak Duality)Possible (Primal ray $\implies \mathcal{F}_D = \emptyset$)
Primal InfeasibleImpossible (violates Strong Duality)Possible (Dual ray $\implies \mathcal{F}_P = \emptyset$)Possible (Both $\mathcal{F}_P = \emptyset$ and $\mathcal{F}_D = \emptyset$)

§3.4 The Complementary Slackness Theorem and Primal-Dual Optimality Verification

The algebraic chain of inequalities in the proof of the Weak Duality Lemma ($\mathbf{c}^T\mathbf{x} \le \mathbf{y}^T\mathbf{A}\mathbf{x} \le \mathbf{b}^T\mathbf{y}$) yields a powerful necessary and sufficient condition linking the positive variables of one problem to the binding constraints of the other, known as the Complementary Slackness Theorem.

Let $\mathbf{x} \in \mathbb{R}_+{}^n$ be primal feasible for (P) with primal slack vector $\mathbf{s} = \mathbf{b} - \mathbf{A}\mathbf{x} \ge \mathbf{0}$ ($s_i = b_i - \sum_{j=1}^n a_{ij}x_j \ge 0$ for $i=1,\dots,m$), and let $\mathbf{y} \in \mathbb{R}_+{}^m$ be dual feasible for (D) with dual surplus vector $\mathbf{e} = \mathbf{A}^T\mathbf{y} - \mathbf{c} \ge \mathbf{0}$ ($e_j = \sum_{i=1}^m a_{ij}y_i - c_j \ge 0$ for $j=1,\dots,n$). Expressing the duality gap $W(\mathbf{y}) - Z(\mathbf{x})$ in terms of $\mathbf{s}$ and $\mathbf{e}$, we substitute $\mathbf{b} = \mathbf{A}\mathbf{x} + \mathbf{s}$ and $\mathbf{c} = \mathbf{A}^T\mathbf{y} - \mathbf{e}$:

$$W(\mathbf{y}) - Z(\mathbf{x}) = \mathbf{b}^T\mathbf{y} - \mathbf{c}^T\mathbf{x} = (\mathbf{A}\mathbf{x} + \mathbf{s})^T\mathbf{y} - (\mathbf{A}^T\mathbf{y} - \mathbf{e})^T\mathbf{x} = \mathbf{x}^T\mathbf{A}^T\mathbf{y} + \mathbf{s}^T\mathbf{y} - \mathbf{y}^T\mathbf{A}\mathbf{x} + \mathbf{e}^T\mathbf{x} = \mathbf{s}^T\mathbf{y} + \mathbf{e}^T\mathbf{x}$$

Writing out the two inner products componentwise:

$$W(\mathbf{y}) - Z(\mathbf{x}) = \sum_{i=1}^m s_i y_i + \sum_{j=1}^n e_j x_j = \sum_{i=1}^m \left( b_i - \sum_{j=1}^n a_{ij} x_j \right) y_i + \sum_{j=1}^n \left( \sum_{i=1}^m a_{ij} y_i - c_j \right) x_j \ge 0$$

Theorem 3.3 (The Complementary Slackness Theorem):

Let $\mathbf{x}$ be feasible for the Primal Problem (P) and $\mathbf{y}$ be feasible for the Dual Problem (D). Then $\mathbf{x}$ and $\mathbf{y}$ are simultaneously optimal for (P) and (D), respectively, if and only if they satisfy the Complementary Slackness Conditions:

$$\begin{aligned} \text{1. Primal Complementary Slackness:} \quad & s_i y_i = \left( b_i - \sum_{j=1}^n a_{ij} x_j \right) y_i = 0 \qquad \forall i = 1, 2, \dots, m \\ \text{2. Dual Complementary Slackness:} \quad & e_j x_j = \left( \sum_{i=1}^m a_{ij} y_i - c_j \right) x_j = 0 \qquad \forall j = 1, 2, \dots, n \end{aligned}$$

Equivalently, in logical implication form at any primal-dual optimal pair $(\mathbf{x}^*, \mathbf{y}^*)$:

  • If a primal constraint is non-binding (has positive slack $s_i^* > 0$), its corresponding dual shadow price must equal zero ($y_i^* = 0$). Economically, a resource that is in excess supply (leftover idle capacity at the optimum) is a free good with zero marginal shadow value!
  • If a dual shadow price is strictly positive ($y_i^* > 0$), its corresponding primal resource constraint must be tightly binding ($s_i^* = 0$, i.e., $\sum_{j=1}^n a_{ij}x_j^* = b_i$).
  • If a primal activity is operated at a strictly positive level ($x_j^* > 0$), its corresponding dual constraint must hold with exact equality ($e_j^* = 0$, i.e., $\sum_{i=1}^m a_{ij}y_i^* = c_j$). Economically, any product actually manufactured at the optimum must break even exactly against the imputed opportunity cost of the resources it consumes.
  • If a dual constraint has positive surplus ($e_j^* > 0$, so imputed resource cost $\sum_{i=1}^m a_{ij}y_i^* > c_j$ exceeds unit revenue), that primal activity must be shut down completely ($x_j^* = 0$).
Proof:

By Strong Duality, feasible vectors $\mathbf{x}$ and $\mathbf{y}$ are optimal if and only if the duality gap vanishes: $W(\mathbf{y}) - Z(\mathbf{x}) = \sum_{i=1}^m s_i y_i + \sum_{j=1}^n e_j x_j = 0$. Because $s_i \ge 0, y_i \ge 0$ for all $i$ and $e_j \ge 0, x_j \ge 0$ for all $j$, every single product term $s_i y_i \ge 0$ and $e_j x_j \ge 0$ is non-negative. A sum of non-negative real numbers equals zero if and only if every individual term equals zero, namely $s_i y_i = 0$ for all $i=1,\dots,m$ and $e_j x_j = 0$ for all $j=1,\dots,n$. (Furthermore, the Goldman–Tucker Strict Complementary Slackness Theorem proves that whenever (P) and (D) are feasible, there exists at least one optimal pair $(\mathbf{x}^*, \mathbf{y}^*)$ that is strictly complementary, satisfying $s_i^* + y_i^* > 0$ for all $i$ and $e_j^* + x_j^* > 0$ for all $j$.)

§3.5 Lemke's Dual Simplex Algorithm: Restoring Primal Feasibility under Dual Optimality

In the standard (Primal) Simplex Algorithm of Unit 2, computation begins at a primal feasible basic solution ($\bar{\mathbf{b}} = \mathbf{B}^{-1}\mathbf{b} \ge \mathbf{0}$) that is dual infeasible ($z_j - c_j < 0$ for some $j \in N$), and pivots iteratively to achieve dual feasibility ($z_j - c_j \ge 0$) while preserving primal feasibility ($\bar{\mathbf{b}} \ge \mathbf{0}$) at every step.

Frequently, however—especially in post-optimality sensitivity analysis when a right-hand-side capacity $b_i$ is reduced, when a new cutting-plane inequality is appended in Gomory's integer programming algorithm (Unit 6), or in minimization models with $\ge$ constraints ($\sum a_{ij}x_j \ge b_i \iff \sum (-a_{ij})x_j + s_i = -b_i < 0$) and positive costs $c_j \ge 0$—we possess a basis $\mathbf{B}$ that is already Dual Feasible (satisfying the Simplex optimality condition $z_j - c_j \ge 0$ for all $j \in N$ in maximization, or $z_j - c_j \le 0$ in minimization), but Primal Infeasible because one or more basic variables are strictly negative ($\bar{b}_i < 0$ for some $i \in \{1, \dots, m\}$). Rather than introducing artificial variables and starting over with a costly Two-Phase or Big-M procedure, Carlton E. Lemke (1954) invented the Dual Simplex Algorithm, which performs pivots that preserve dual feasibility ($z_j - c_j \ge 0$) while systematically eliminating negative right-hand sides ($\bar{b}_i < 0$) until primal feasibility ($\bar{\mathbf{b}} \ge \mathbf{0}$) is attained!

Algorithmic Mechanics of the Dual Simplex Method (Maximization Form)

Assume the current Simplex tableau is dual feasible ($z_j - c_j \ge 0$ for all $j \in \{1, \dots, n\}$), and we wish to achieve primal feasibility ($\bar{b}_i \ge 0$ for all $i = 1, \dots, m$):

1. Primal Optimality Check: If $\bar{b}_i \ge 0$ for all $i = 1, \dots, m$, stop! Because the tableau is simultaneously primal feasible ($\bar{\mathbf{b}} \ge \mathbf{0}$) and dual feasible ($z_j - c_j \ge 0$), the current BFS is globally optimal.

2. Leaving Variable Selection (Most Negative RHS Rule): If at least one basic variable is negative, select the leaving basic variable $x_{B(r)}$ in row $r$ corresponding to the most negative right-hand-side constant:

$$\bar{b}_r = \min_{1 \le i \le m} \left\{ \bar{b}_i : \bar{b}_i < 0 \right\} < 0$$

3. Primal Infeasibility Certificate Check: Inspect the coefficients $y_{rj}$ across row $r$ for all non-basic variables $j \in N$. Because the $r$-th basic equation reads:

$$x_{B(r)} = \bar{b}_r - \sum_{j \in N} y_{rj} x_j \implies \sum_{j \in N} y_{rj} x_j + x_{B(r)} = \bar{b}_r < 0$$

if $y_{rj} \ge 0$ for all $j \in N$, then for any non-negative vector $\mathbf{x} \ge \mathbf{0}$, the left-hand side $\sum_{j \in N} y_{rj}x_j + x_{B(r)} \ge 0$ is non-negative while the right-hand side $\bar{b}_r < 0$ is strictly negative! This is an impossible contradiction (equivalent to the Dual objective being unbounded above), proving that the Primal LP is strictly infeasible ($\mathcal{F}_P = \emptyset$).

4. Entering Variable Selection (Dual Minimum Ratio Test): To make the new basic variable in row $r$ positive ($\bar{b}_r^{\text{new}} = \bar{b}_r / y_{rk} > 0$ since $\bar{b}_r < 0$), the pivot element $y_{rk}$ must be strictly negative ($y_{rk} < 0$). Furthermore, after pivoting on $y_{rk} < 0$, the updated net evaluation row entries become:

$$(z_j - c_j)^{\text{new}} = (z_j - c_j) - \frac{z_k - c_k}{y_{rk}} y_{rj}$$

To preserve dual feasibility $(z_j - c_j)^{\text{new}} \ge 0$ across all columns $j \in N$ with $y_{rj} < 0$, we must choose the entering variable $x_k$ via the Dual Minimum Ratio Test:

$$\frac{z_k - c_k}{|y_{rk}|} = \min_{j \in N} \left\{ \frac{z_j - c_j}{|y_{rj}|} : y_{rj} < 0 \right\} = \min_{j \in N} \left\{ \left| \frac{z_j - c_j}{y_{rj}} \right| : y_{rj} < 0 \right\}$$

5. Gauss–Jordan Pivot: Pivot on $y_{rk} < 0$ to bring $x_k$ into the basis and expel $x_{B(r)}$. Note that in a maximization problem, each Dual Simplex pivot decreases (tightens) the super-optimal upper bound $Z_0$ monotonically down toward $Z^*$: $Z_{\text{new}} = Z_0 - \frac{\bar{b}_r}{y_{rk}}(z_k - c_k) \le Z_0$.

§3.6 Post-Optimality Sensitivity Analysis I: Right-Hand-Side Perturbations and Shadow Price Intervals

Once an optimal Basic Feasible Solution $\mathbf{x}_B^* = \mathbf{B}^{-1}\mathbf{b} \ge \mathbf{0}$ with optimal basis $\mathbf{B}$, optimal dual shadow prices $\mathbf{y}^* = (\mathbf{B}^{-1})^T\mathbf{c}_B$, and optimal objective $Z^* = \mathbf{c}_B^T \mathbf{B}^{-1}\mathbf{b}$ has been computed, Post-Optimality Sensitivity Analysis investigates how perturbations in the model's exogenous data $(\mathbf{b}, \mathbf{c}, \mathbf{A})$ affect the optimal basis and solution—without resolving the Linear Program from scratch! Everything hinges on the Fundamental Matrix Invariance Principle: every entry in the final Simplex tableau is a deterministic function of the optimal basis inverse $\mathbf{B}^{-1}$:

$$\bar{\mathbf{b}} = \mathbf{B}^{-1}\mathbf{b}, \qquad \mathbf{y}_j = \mathbf{B}^{-1}\mathbf{a}_j, \qquad (\mathbf{y}^*)^T = \mathbf{c}_B^T \mathbf{B}^{-1}, \qquad z_j - c_j = \mathbf{c}_B^T \mathbf{B}^{-1}\mathbf{a}_j - c_j, \qquad Z^* = \mathbf{c}_B^T \mathbf{B}^{-1}\mathbf{b}$$

Perturbations in the Right-Hand-Side Resource Vector ($\mathbf{b} \to \mathbf{b} + \Delta \mathbf{b}$)

Suppose the resource capacity vector $\mathbf{b}$ is perturbed to $\hat{\mathbf{b}} = \mathbf{b} + \Delta \mathbf{b}$ (or a single resource $b_r$ is perturbed to $b_r + \Delta b_r$, so $\Delta \mathbf{b} = \Delta b_r \mathbf{e}_r$). Inspect the fundamental tableau formulas above:

  • Dual Feasibility is Completely Unaffected: Because $\mathbf{b}$ does not appear anywhere in $z_j - c_j = \mathbf{c}_B^T \mathbf{B}^{-1}\mathbf{a}_j - c_j$, changing $\mathbf{b}$ leaves the entire net evaluation row $z_j - c_j \ge 0$ and the dual shadow prices $\mathbf{y}^* = (\mathbf{B}^{-1})^T\mathbf{c}_B$ completely unchanged!
  • Primal Feasibility Condition (Stability Range of the Optimal Basis $\mathbf{B}$): The current optimal basis $\mathbf{B}$ remains optimal if and only if the perturbed basic variables remain non-negative:
$$\hat{\mathbf{x}}_B = \mathbf{B}^{-1}(\mathbf{b} + \Delta \mathbf{b}) = \mathbf{B}^{-1}\mathbf{b} + \mathbf{B}^{-1}\Delta \mathbf{b} = \bar{\mathbf{b}} + \mathbf{B}^{-1}\Delta \mathbf{b} \ge \mathbf{0}$$

Specifically, if a single resource $b_r$ is perturbed by $\Delta b_r$, let $\boldsymbol{\beta}_{*r} = (\beta_{1r}, \beta_{2r}, \dots, \beta_{mr})^T$ denote the $r$-th column of $\mathbf{B}^{-1}$ (which is located directly under the $r$-th initial slack/artificial variable in the final Simplex tableau!). Then the $m$ perturbed basic variables are:

$$\hat{x}_{B(i)} = \bar{b}_i + \beta_{ir} \Delta b_r \ge 0, \qquad i = 1, 2, \dots, m$$

Solving these $m$ linear inequalities for $\Delta b_r$ yields the exact Feasibility (Allowable) Range for $\Delta b_r$ within which the current basis $\mathbf{B}$ and shadow prices $\mathbf{y}^*$ remain invariant:

$$\max_{i : \beta_{ir} > 0} \left\{ -\frac{\bar{b}_i}{\beta_{ir}} \right\} \le \Delta b_r \le \min_{i : \beta_{ir} < 0} \left\{ -\frac{\bar{b}_i}{\beta_{ir}} \right\}$$

Within this allowable interval $[\Delta b_r^{\min}, \Delta b_r^{\max}]$, the optimal objective value is a strictly linear function of $\Delta b_r$ with slope equal to the $r$-th shadow price $y_r^*$:

$$Z^*(\mathbf{b} + \Delta \mathbf{b}) = \mathbf{c}_B^T \mathbf{B}^{-1}(\mathbf{b} + \Delta \mathbf{b}) = Z^*(\mathbf{b}) + (\mathbf{y}^*)^T \Delta \mathbf{b} = Z^*(\mathbf{b}) + y_r^* \Delta b_r$$

If $\Delta b_r$ exceeds this interval, one or more $\hat{x}_{B(i)}$ become negative while $z_j - c_j \ge 0$ remains satisfied; a single iteration of Lemke's Dual Simplex Algorithm (Section 3.5) immediately transitions to the new optimal basis, where the slope $\frac{\partial Z^*}{\partial b_r}$ steps down, making the global value function $Z^*(b_r)$ a piecewise-linear concave function of $b_r$!

Post-Optimality Sensitivity Analysis II: Objective Coefficients, New Activities, and New Constraints

Continuing from the Fundamental Matrix Invariance Principle, we now analyze how perturbations in the objective coefficients $\mathbf{c}$, the introduction of a new structural activity $x_{n+1}$, or the addition of a new constraint row affect the optimal solution.

1. Perturbation in a Non-Basic Objective Coefficient ($c_k \to c_k + \Delta c_k, \; k \in N$)

Suppose $x_k$ ($k \in N$) is currently a non-basic variable ($x_k^* = 0$) at the optimum, so its current activity level is zero because its unit profit $c_k$ is less than or equal to its imputed resource cost $z_k = (\mathbf{y}^*)^T \mathbf{a}_k$ ($z_k - c_k \ge 0$). Because $k \notin B$, $c_k$ does not appear in $\mathbf{c}_B$. Therefore, changing $c_k$ to $c_k + \Delta c_k$ leaves $\mathbf{x}_B^* = \mathbf{B}^{-1}\mathbf{b}$, $\mathbf{y}^* = (\mathbf{B}^{-1})^T\mathbf{c}_B$, $Z^*$, and every other column's $z_j - c_j$ ($j \ne k$) completely unchanged! Only the single net evaluation entry for column $k$ changes:

$$(\hat{z}_k - \hat{c}_k) = z_k - (c_k + \Delta c_k) = (z_k - c_k) - \Delta c_k \ge 0 \iff \Delta c_k \le z_k - c_k \iff \hat{c}_k \le z_k$$

Thus, for any non-basic variable $x_k$ ($k \in N$), the optimal solution remains completely invariant for all $\Delta c_k \in (-\infty, \, z_k - c_k]$. Only if $c_k$ increases by more than its reduced cost $z_k - c_k$ does $x_k$ become profitable to enter the basis via a standard Primal Simplex pivot.

2. Perturbation in a Basic Objective Coefficient ($c_{B(r)} \to c_{B(r)} + \Delta c_{B(r)}, \; B(r) \in B$)

Suppose $x_{B(r)}$ is the $r$-th basic variable ($B(r) \in B$). Perturbing $c_{B(r)}$ by $\Delta c_{B(r)}$ changes the basic cost vector to $\hat{\mathbf{c}}_B = \mathbf{c}_B + \Delta c_{B(r)} \mathbf{e}_r$.

  • Primal Feasibility is Unaffected: $\mathbf{x}_B^* = \mathbf{B}^{-1}\mathbf{b} \ge \mathbf{0}$ does not depend on $\mathbf{c}$, so the vertex coordinates remain $\bar{\mathbf{b}}$.
  • Dual Feasibility Condition (Optimality Range for $\Delta c_{B(r)}$): For every basic variable $j \in B$, $\hat{z}_j - \hat{c}_j = 0$ remains zero. However, for every non-basic variable $j \in N$, the imputed cost changes by $\Delta c_{B(r)} y_{rj}$, where $y_{rj}$ is the entry in row $r$ and column $j$ of the final Simplex tableau:
$$\hat{z}_j - c_j = (z_j - c_j) + y_{rj} \Delta c_{B(r)} \ge 0 \qquad \forall j \in N$$

Solving this system of inequalities over all non-basic columns $j \in N$ yields the exact Optimality Range for $\Delta c_{B(r)}$:

$$\max_{j \in N : y_{rj} > 0} \left\{ -\frac{z_j - c_j}{y_{rj}} \right\} \le \Delta c_{B(r)} \le \min_{j \in N : y_{rj} < 0} \left\{ -\frac{z_j - c_j}{y_{rj}} \right\}$$

Within this range, the optimal decision vector $\mathbf{x}^*$ is completely unchanged, while the optimal objective value shifts linearly as $\hat{Z}^* = Z^* + \bar{b}_r \Delta c_{B(r)}$.

3. Addition of a New Structural Activity ($x_{n+1}$) and Addition of a New Constraint

  • Pricing a New Activity Column: Suppose management proposes a new product $x_{n+1} \ge 0$ with unit profit $c_{n+1}$ and resource consumption column $\mathbf{a}_{n+1} \in \mathbb{R}^m$. Using the existing optimal shadow prices $\mathbf{y}^* = (\mathbf{B}^{-1})^T\mathbf{c}_B$, we immediately compute its reduced cost:
$$z_{n+1} - c_{n+1} = (\mathbf{y}^*)^T \mathbf{a}_{n+1} - c_{n+1}$$

If $z_{n+1} - c_{n+1} \ge 0$, the new activity's imputed resource cost exceeds or equals its revenue, so $x_{n+1}^* = 0$ and the current optimum is unchanged! If $z_{n+1} - c_{n+1} < 0$, we compute its updated tableau column $\mathbf{y}_{n+1} = \mathbf{B}^{-1}\mathbf{a}_{n+1}$ and pivot $x_{n+1}$ into the basis via the Primal Simplex Method.

  • Appending a New Constraint: Suppose a new regulatory or technological constraint $\mathbf{a}_{m+1}^T \mathbf{x} \le b_{m+1}$ is added. First, substitute the current optimal vector $\mathbf{x}^*$ directly into the new constraint: if $\mathbf{a}_{m+1}^T \mathbf{x}^* \le b_{m+1}$, the current optimal solution already satisfies the new constraint with slack $s_{m+1}^* = b_{m+1} - \mathbf{a}_{m+1}^T \mathbf{x}^* \ge 0$ and remains globally optimal! If $\mathbf{a}_{m+1}^T \mathbf{x}^* > b_{m+1}$, we append the row $\mathbf{a}_{m+1}^T \mathbf{x} + s_{m+1} = b_{m+1}$ with basic variable $s_{m+1}$ to the final tableau, eliminate the basic variables $\mathbf{x}_B$ from row $m+1$ using the existing tableau rows (which leaves $\bar{b}_{m+1} = b_{m+1} - \mathbf{a}_{m+1}^T \mathbf{x}^* < 0$ while maintaining $z_j - c_j \ge 0$), and apply Lemke's Dual Simplex Method to restore primal feasibility!

Parametric Linear Programming, the 100% Simultaneous Perturbation Rule, and Degenerate Shadow Prices

Single-parameter sensitivity intervals characterize how far an individual resource $b_r$ or individual profit coefficient $c_j$ can move while holding all other parameters fixed ceteris paribus. In realistic statistical and econometric environments, however, multiple resource capacities $\mathbf{b} + \Delta\mathbf{b}$ or multiple objective coefficients $\mathbf{c} + \Delta\mathbf{c}$ fluctuate simultaneously due to macroeconomic trends, seasonal shifts, or correlated estimation errors.

Bradley, Hax, and Magnanti's 100% Rule for Simultaneous Perturbations

Suppose several right-hand-side capacities $b_1, \dots, b_m$ are perturbed simultaneously by increments $\Delta b_1, \dots, \Delta b_m$, where each individual increment $\Delta b_i \ne 0$ has a known maximum allowable increase $\Delta b_i^+ > 0$ and maximum allowable decrease $\Delta b_i^- > 0$ (so its single-parameter feasibility interval is $[b_i - \Delta b_i^-, \, b_i + \Delta b_i^+]$). Define the fractional excursion $r_i \ge 0$ of resource $i$ toward its single-parameter stability boundary by:

$$r_i = \begin{cases} \dfrac{\Delta b_i}{\Delta b_i^+}, & \text{if } \Delta b_i > 0 \[8pt] \dfrac{-\Delta b_i}{\Delta b_i^-}, & \text{if } \Delta b_i < 0 \[8pt] 0, & \text{if } \Delta b_i = 0 \end{cases}$$

The 100% Rule for Simultaneous RHS Perturbations states that if the sum of the percentage excursions across all perturbed resources does not exceed $100\%$:

$$\sum_{i=1}^m r_i \le 1 \quad \left(\text{i.e., } \sum_{i=1}^m 100 r_i\% \le 100\%\right)$$

then the current optimal basis $\mathbf{B}$ and its dual shadow price vector $\mathbf{y}^*$ are guaranteed to remain valid and optimal! The proof follows directly from convexity of the non-negative orthant in basis space: the unperturbed vector $\Delta \mathbf{b} = \mathbf{0}$ and the $m$ single-axis boundary vectors $\Delta b_i^+ \mathbf{e}_i$ (or $-\Delta b_i^- \mathbf{e}_i$) all satisfy $\bar{\mathbf{b}} + \mathbf{B}^{-1}\Delta\mathbf{b} \ge \mathbf{0}$; whenever $\sum_{i=1}^m r_i \le 1$, the simultaneous perturbation vector $\Delta\mathbf{b} = \sum_{i=1}^m r_i (\pm \Delta b_i^\pm \mathbf{e}_i) + \left(1 - \sum_{i=1}^m r_i\right)\mathbf{0}$ is a convex combination of feasible perturbation vectors, so $\bar{\mathbf{b}} + \mathbf{B}^{-1}\Delta\mathbf{b} \ge \mathbf{0}$ holds automatically! An identical 100% Rule for Simultaneous Objective Perturbations guarantees that the current optimal solution $\mathbf{x}^*$ remains optimal whenever the fractional excursions of simultaneously perturbed cost coefficients $\Delta c_j$ sum to at most $100\%$. Note carefully that the 100% Rule is a sufficient condition, not a necessary condition: if $\sum r_i > 1$, the basis may still remain optimal if oppositely signed matrix entries $\beta_{ki}\Delta b_i$ cancel each other out, which can always be checked exactly via $\bar{\mathbf{b}} + \mathbf{B}^{-1}\Delta\mathbf{b} \ge \mathbf{0}$.

Parametric Linear Programming along a Continuous Ray

When parameters vary continuously along a prescribed directional ray $\mathbf{b}(\theta) = \mathbf{b} + \theta \mathbf{d}$ or $\mathbf{c}(\theta) = \mathbf{c} + \theta \mathbf{f}$ as a scalar trade-off parameter $\theta \ge 0$ increases from $0$ to $+\infty$ (as in Markowitz mean-variance-linearized efficient frontier tracing or statistical regularization path tracking), Parametric Linear Programming traces the entire piecewise-linear family of optimal bases $\{\mathbf{B}_1, \mathbf{B}_2, \dots, \mathbf{B}_K\}$ across consecutive critical breakpoints $0 = \theta_0 < \theta_1 < \theta_2 < \dots < \theta_K$. Within each interval $\theta \in [\theta_{k-1}, \theta_k]$, the basis $\mathbf{B}_k$ remains constant; at the upper breakpoint $\theta_k$, a basic variable $\bar{b}_r(\theta) = \bar{b}_r + \theta (\mathbf{B}_k^{-1}\mathbf{d})_r$ hits zero (for RHS parameterization) or a reduced cost $z_j(\theta) - c_j(\theta)$ hits zero (for objective parameterization), triggering a single Dual Simplex or Primal Simplex pivot to the adjacent basis $\mathbf{B}_{k+1}$.

Two-Sided Directional Shadow Prices under Primal Degeneracy

Finally, a subtle theoretical distinction arises when the optimal primal BFS $\mathbf{x}^*$ is degenerate ($\bar{b}_i = 0$ for some basic variable). At a degenerate primal vertex, multiple distinct optimal bases $\mathbf{B}^{(1)}, \dots, \mathbf{B}^{(L)}$ represent the same geometric corner $\mathbf{x}^*$, corresponding to alternative optimal solutions $\mathbf{y}^{(1)}, \dots, \mathbf{y}^{(L)}$ in the Dual Problem! Consequently, the optimal value function $Z^*(b_r)$ exhibits a kink (non-differentiability) at the current capacity $b_r$: the right-hand marginal shadow price (for increasing $b_r$, $\Delta b_r > 0$) equals the minimum of $y_r$ over the dual optimal face, $\frac{\partial Z^*}{\partial b_r^+} = \min_{\mathbf{y}^* \in \mathcal{F}_D^*} y_r^*$, whereas the left-hand marginal shadow price (for decreasing $b_r$, $\Delta b_r < 0$) equals the maximum of $y_r$ over the dual optimal face, $\frac{\partial Z^*}{\partial b_r^-} = \max_{\mathbf{y}^* \in \mathcal{F}_D^*} y_r^*$.

Statistical Application: Duality in Quantile Regression and Support Vector Classifiers

In modern mathematical statistics, Linear Programming Duality provides the theoretical bridge between primal residual minimization and dual score-function processes. Consider Roger Koenker and Gilbert Bassett's (1978) Quantile Regression estimator for the $\tau$-th conditional quantile ($\tau \in (0, 1)$) of a response vector $\mathbf{y} \in \mathbb{R}^N$ given design matrix $\mathbf{X} \in \mathbb{R}^{N \times p}$, which minimizes the asymmetric tilted absolute loss $\sum_{i=1}^N \rho_\tau(y_i - \mathbf{x}_i^T \boldsymbol{\beta})$. Splitting each residual into positive and negative parts $y_i - \mathbf{x}_i^T\boldsymbol{\beta} = u_i - v_i$ ($u_i, v_i \ge 0$), the Primal Quantile Regression Linear Program is:

$$\text{Minimize } \tau \mathbf{1}_N^T \mathbf{u} + (1 - \tau) \mathbf{1}_N^T \mathbf{v} \quad \text{subject to} \quad \mathbf{X}\boldsymbol{\beta} + \mathbf{I}_N \mathbf{u} - \mathbf{I}_N \mathbf{v} = \mathbf{y}, \quad \boldsymbol{\beta} \in \mathbb{R}^p \text{ free}, \quad \mathbf{u}, \mathbf{v} \ge \mathbf{0}$$

By the asymmetric primal-dual rules of Section 3.1, the $N$ equality constraints correspond to an unrestricted dual vector $\mathbf{d} \in \mathbb{R}^N$, the free variable vector $\boldsymbol{\beta} \in \mathbb{R}^p$ yields $p$ strict dual equality constraints $\mathbf{X}^T\mathbf{d} = \mathbf{0}$, and the non-negative deviation vectors $\mathbf{u}, \mathbf{v} \ge \mathbf{0}$ impose box bounds $-(1-\tau)\mathbf{1}_N \le \mathbf{d} \le \tau \mathbf{1}_N$. Reparameterizing via $\mathbf{a} = \mathbf{d} + (1-\tau)\mathbf{1}_N \in [0, 1]^N$ transforms the Dual of Quantile Regression into the compact bounded-variable maximization problem:

$$\text{Maximize } \mathbf{y}^T \mathbf{a} \quad \text{subject to} \quad \mathbf{X}^T \mathbf{a} = (1 - \tau) \mathbf{X}^T \mathbf{1}_N, \qquad \mathbf{a} \in [0, 1]^N$$

By Complementary Slackness, observations lying strictly above the fitted $\tau$-quantile hyperplane ($u_i^* > 0$) have dual rank score $a_i^* = 1$; observations lying strictly below the hyperplane ($v_i^* > 0$) have dual rank score $a_i^* = 0$; and the $p$ interpolated basic observations lying directly on the hyperplane ($u_i^* = v_i^* = 0$) take fractional dual weights $a_i^* \in (0, 1)$, forming the exact generalization of Hájek's regression rank-score statistics used for distribution-free hypothesis testing in linear models!

THREE WORKED QUESTIONS

Step-by-Step Operations Research Solutions

Three original questions connect calculations, definitions and interpretation. Open each solution to follow the reasoning.

Solved Problem Example 3.1: Asymmetric Primal-Dual Formulation and Optimal Dual Extraction via Complementary Slackness

Consider the following Primal Linear Programming Problem (P):

$$\text{Maximize } Z = 5x_1 + 12x_2 + 4x_3$$

subject to:

$$\begin{aligned} x_1 + 2x_2 + x_3 &\le 10 \quad (\text{Constraint 1}) \\ 2x_1 - x_2 + 3x_3 &= 8 \quad (\text{Constraint 2}) \\ x_1, x_2, x_3 &\ge 0 \end{aligned}$$
  1. Formulate the explicit Dual Linear Programming Problem (D) in dual variables $(y_1, y_2)$, specifying the exact sign restriction on each dual variable and the direction of each dual constraint.
  2. Suppose it is known that in the optimal primal solution, activities $x_1^* > 0$ and $x_2^* > 0$ are strictly positive while $x_3^* = 0$, and Constraint 1 holds with strict equality ($s_1^* = 0$). Use the primal constraints and the Complementary Slackness Theorem to solve algebraically for $(x_1^*, x_2^*, x_3^*)$ and the optimal dual vector $(y_1^*, y_2^*)$ without constructing a Simplex tableau.
  3. Verify that $(y_1^*, y_2^*)$ satisfies all dual constraints (including the constraint for $x_3$) and verify Strong Duality $Z^* = W^*$.
Formulate the Asymmetric Dual Problem (D)

Associate dual variable $y_1$ with Constraint 1 ($\le 10$) and dual variable $y_2$ with Constraint 2 ($= 8$). By the Primal-Dual Correspondence Table (Section 3.1):

  • Because Primal Constraint 1 is a $\le$ inequality in a maximization problem, $y_1 \ge 0$.
  • Because Primal Constraint 2 is a strict equality ($=$), $y_2 \in \mathbb{R}$ is unrestricted in sign.
  • Because all three primal variables satisfy $x_1, x_2, x_3 \ge 0$, all three dual constraints are $\ge$ inequalities.

Thus, the Dual Problem (D) is:

$$\text{Minimize } W = 10y_1 + 8y_2$$

subject to:

$$\begin{aligned} y_1 + 2y_2 &\ge 5 \quad (\text{Dual Constraint 1, paired with } x_1) \\ 2y_1 - y_2 &\ge 12 \quad (\text{Dual Constraint 2, paired with } x_2) \\ y_1 + 3y_2 &\ge 4 \quad (\text{Dual Constraint 3, paired with } x_3) \\ y_1 &\ge 0, \quad y_2 \in \mathbb{R} \text{ unrestricted} \end{aligned}$$
Apply Complementary Slackness to solve for primal and dual candidates

First, since $x_3^* = 0$ and Constraint 1 is binding ($s_1^* = 0$), the two primal constraints become a $2 \times 2$ linear system in $(x_1^*, x_2^*)$:

$$\begin{aligned} x_1^* + 2x_2^* &= 10 \\ 2x_1^* - x_2^* &= 8 \end{aligned}$$

Multiplying the second equation by $2$ and adding to the first gives $5x_1^* = 26 \implies x_1^* = \frac{26}{5} = 5.2 > 0$, and $x_2^* = 2\left(\frac{26}{5}\right) - 8 = \frac{12}{5} = 2.4 > 0$. Thus $\mathbf{x}^* = \left(\frac{26}{5}, \frac{12}{5}, 0\right)^T \ge \mathbf{0}$ is primal feasible!

Second, by Dual Complementary Slackness ($e_j^* x_j^* = 0$), because $x_1^* = 26/5 > 0$ and $x_2^* = 12/5 > 0$, Dual Constraints 1 and 2 must hold with strict equality ($e_1^* = 0$ and $e_2^* = 0$):

$$\begin{aligned} y_1^* + 2y_2^* &= 5 \\ 2y_1^* - y_2^* &= 12 \end{aligned}$$

Multiplying the second equation by $2$ and adding to the first yields $5y_1^* = 29 \implies y_1^* = \frac{29}{5} = 5.8$, and $y_2^* = 2\left(\frac{29}{5}\right) - 12 = -\frac{2}{5} = -0.4$.

Verify Dual Feasibility and Strong Duality Z* = W*

Let us verify that $\mathbf{y}^* = \left(\frac{29}{5}, -\frac{2}{5}\right)^T$ is feasible for Dual Problem (D):

  1. Sign restrictions: $y_1^* = \frac{29}{5} \ge 0$ (satisfied) and $y_2^* = -\frac{2}{5} \in \mathbb{R}$ (unrestricted, so negative value is valid!).
  2. Dual Constraint 3 (paired with $x_3^* = 0$):
$$y_1^* + 3y_2^* = \frac{29}{5} + 3\left(-\frac{2}{5}\right) = \frac{23}{5} = 4.6 \ge 4 \implies e_3^* = \frac{23}{5} - 4 = \frac{3}{5} > 0$$

Dual Constraint 3 is satisfied with positive surplus $e_3^* = 3/5 > 0$, which satisfies $e_3^* x_3^* = (3/5)(0) = 0$.

  1. Evaluate Primal and Dual Objectives:
$$Z^* = 5\left(\frac{26}{5}\right) + 12\left(\frac{12}{5}\right) + 4(0) = \frac{130 + 144}{5} = \frac{274}{5} = 54.8$$
$$W^* = 10\left(\frac{29}{5}\right) + 8\left(-\frac{2}{5}\right) = \frac{290 - 16}{5} = \frac{274}{5} = 54.8$$

Because $Z^* = W^* = 54.8$ and all primal, dual, and complementary slackness conditions hold, $\mathbf{x}^* = (5.2, 2.4, 0)^T$ and $\mathbf{y}^* = (5.8, -0.4)^T$ are proven globally optimal.

Solved Problem Example 3.2: Execution of Lemke's Dual Simplex Algorithm on a Minimization System

Solve the following Linear Programming Problem using Lemke's Dual Simplex Algorithm (without introducing any artificial variables):

$$\text{Minimize } W = 3x_1 + 2x_2 + 4x_3$$

subject to:

$$\begin{aligned} 2x_1 + x_2 + 3x_3 &\ge 6 \\ 3x_1 + 3x_2 + 5x_3 &\ge 12 \\ x_1, x_2, x_3 &\ge 0 \end{aligned}$$
  1. Convert the objective to maximization ($Z = -W$) and multiply both $\ge$ constraints by $-1$ to obtain $\le$ inequalities with negative right-hand sides and non-negative slack variables $s_1, s_2 \ge 0$.
  2. Verify that the initial tableau at the origin is dual feasible ($z_j - c_j \ge 0$) but primal infeasible ($\bar{\mathbf{b}} < \mathbf{0}$).
  3. Execute Dual Simplex pivots step-by-step using the most negative RHS leaving rule and the Dual Minimum Ratio Test until primal feasibility and global optimality are achieved.
Standardize for Dual Simplex and construct Initial Tableau
Set $\max Z = -W = -3x_1 - 2x_2 - 4x_3$. Multiply both $\ge$ inequalities by $-1$ and add slack variables $s_1, s_2 \ge 0$: $$\begin{aligned} -2x_1 - x_2 - 3x_3 + s_1 &= -6 \\ -3x_1 - 3x_2 - 5x_3 + s_2 &= -12 \end{aligned}$$ With initial basis $B_0 = \{s_1, s_2\}$ and $\mathbf{c}_{B_0} = (0, 0)^T$, the net evaluations are $z_1 - c_1 = 3 \ge 0$, $z_2 - c_2 = 2 \ge 0$, $z_3 - c_3 = 4 \ge 0$. Because $z_j - c_j \ge 0$ for all columns, the initial tableau is **Dual Feasible**, while $\bar{b}_1 = -6 < 0$ and $\bar{b}_2 = -12 < 0$ are **Primal Infeasible**:
Basic Var$c_B$$x_1$ ($-3$)$x_2$ ($-2$)$x_3$ ($-4$)$s_1$ ($0$)$s_2$ ($0$)RHS ($\bar{b}_i$)
$s_1$$0$$-2$$-1$$-3$$1$$0$$-6$
$s_2$ (Leave)$0$$-3$$\mathbf{-3}^*$$-5$$0$$1$$\mathbf{-12}$ (Most Neg)
$z_j - c_j$—$3$$2$$4$$0$$0$$Z_0 = 0$
Dual Ratio $|(z_j-c_j)/y_{2j}|$—$3/3 = 1$$\mathbf{2/3 \approx 0.667}$ (Enter)$4/5 = 0.8$———
Dual Simplex Iteration 1: Pivot to replace s_2 with x_2
- **Leaving Variable**: Row 2 has the most negative RHS ($\bar{b}_2 = -12 < -6$), so $s_2$ leaves the basis. - **Entering Variable**: Applying the Dual Minimum Ratio Test across negative entries $y_{2j} < 0$ in Row 2: $$\min \left\{ \left|\frac{3}{-3}\right|, \, \left|\frac{2}{-3}\right|, \, \left|\frac{4}{-5}\right| \right\} = \min \left\{ 1, \, \frac{2}{3}, \, \frac{4}{5} \right\} = \frac{2}{3} \quad (\text{Column } x_2)$$ Thus $x_2$ enters the basis with pivot element $y_{22} = -3$. Dividing Row 2 by $-3$ and applying $R_1 \leftarrow R_1 + R_2^{\text{new}}$, $R_z \leftarrow R_z - 2R_2^{\text{new}}$ yields Tableau 1:
Basic Var$c_B$$x_1$ ($-3$)$x_2$ ($-2$)$x_3$ ($-4$)$s_1$ ($0$)$s_2$ ($0$)RHS ($\bar{b}_i$)
$s_1$ (Leave)$0$$-1$$0$$\mathbf{-4/3}^*$$1$$-1/3$$\mathbf{-2}$ (Neg)
$x_2$$-2$$1$$1$$5/3$$0$$-1/3$$4$
$z_j - c_j$—$1$$0$$2/3$$0$$2/3$$Z_1 = -8$
Dual Ratio $|(z_j-c_j)/y_{1j}|$—$1/1 = 1$—$\mathbf{(2/3)/(4/3) = 0.5}$ (Enter)—$(2/3)/(1/3) = 2$—
Dual Simplex Iteration 2: Pivot to replace s_1 with x_3 and Verify Optimality
In Tableau 1, Row 1 still has a negative right-hand side ($\bar{b}_1 = -2 < 0$), so $s_1$ leaves the basis. Checking the negative entries $y_{1j} < 0$ in Row 1 ($x_1: -1$, $x_3: -4/3$, $s_2: -1/3$), the Dual Minimum Ratio Test gives: $$\min \left\{ \frac{1}{1}, \, \frac{2/3}{4/3}, \, \frac{2/3}{1/3} \right\} = \min \{ 1, \, 0.5, \, 2 \} = 0.5 \quad (\text{Column } x_3)$$ Thus $x_3$ enters the basis with pivot element $y_{13} = -4/3$. Multiply Row 1 by $-3/4$ and eliminate $x_3$ from Row 2 and $R_z$:
Basic Var$c_B$$x_1$ ($-3$)$x_2$ ($-2$)$x_3$ ($-4$)$s_1$ ($0$)$s_2$ ($0$)RHS ($\bar{b}_i$)
$x_3$$-4$$3/4$$0$$1$$-3/4$$1/4$$3/2 = 1.5$
$x_2$$-2$$-1/4$$1$$0$$5/4$$-3/4$$3/2 = 1.5$
$z_j - c_j$—$1/2$$0$$0$$1/2$$1/2$$Z^* = -9$
Now all basic right-hand sides are strictly positive ($\bar{b}_1 = 3/2 > 0, \bar{b}_2 = 3/2 > 0$) and all $z_j - c_j \ge 0$! The global optimum is $x_1^* = 0, x_2^* = 1.5, x_3^* = 1.5$, with minimum objective $W^* = -Z^* = 3(0) + 2(1.5) + 4(1.5) = 9$.
Solved Problem Example 3.3: Comprehensive Post-Optimality Sensitivity Analysis: Shadow Prices, RHS Ranges, Cost Ranges, and New Activity Pricing
A biostatistical assay laboratory produces two diagnostic reagent kits ($x_1, x_2$) using two specialized resources: Automated Pipetting Hours (Resource 1, capacity $b_1 = 60$) and Mass Spectrometry Hours (Resource 2, capacity $b_2 = 48$). The Linear Program is: $$\text{Maximize } Z = 40x_1 + 30x_2 \quad \text{subject to} \quad \begin{cases} 2x_1 + x_2 + s_1 = 60 \\ x_1 + 2x_2 + s_2 = 48 \end{cases}, \quad x_1, x_2, s_1, s_2 \ge 0$$ At the optimal Basic Feasible Solution with basis $B = \{x_1, x_2\}$, the final Simplex tableau is:
Basic Var$c_B$$x_1$ ($40$)$x_2$ ($30$)$s_1$ ($0$)$s_2$ ($0$)RHS ($\bar{b}_i$)
$x_1$$40$$1$$0$$2/3$$-1/3$$24$
$x_2$$30$$0$$1$$-1/3$$2/3$$12$
$z_j - c_j$—$0$$0$$50/3$$20/3$$Z^* = 1320$
1. Extract the basis inverse $\mathbf{B}^{-1}$ and optimal dual shadow prices $(y_1^*, y_2^*)$ directly from the final tableau, and compute the exact **allowable feasibility ranges** for $b_1$ and $b_2$ within which the current basis remains optimal. 2. Compute the exact **allowable optimality ranges** for the unit profit coefficients $c_1$ (keeping $c_2 = 30$ fixed) and $c_2$ (keeping $c_1 = 40$ fixed) such that $(x_1^*, x_2^*) = (24, 12)$ remains optimal. 3. The laboratory considers introducing a third reagent kit $x_3 \ge 0$ requiring $3$ hours of Resource 1 and $2$ hours of Resource 2 with unit profit $c_3 = 60$. Determine via dual shadow pricing whether $x_3$ should be produced; if so, find the new optimal solution.
Extract B^-1, Shadow Prices, and Right-Hand-Side Feasibility Ranges

From the columns under the initial identity slack variables $(s_1, s_2)$ in the final tableau:

$$\mathbf{B}^{-1} = \begin{pmatrix} 2/3 & -1/3 \\ -1/3 & 2/3 \end{pmatrix}, \qquad \mathbf{y}^* = \begin{pmatrix} y_1^* \\ y_2^* \end{pmatrix} = \begin{pmatrix} z_{s_1} - c_{s_1} \\ z_{s_2} - c_{s_2} \end{pmatrix} = \begin{pmatrix} 50/3 \\ 20/3 \end{pmatrix} \approx \begin{pmatrix} 16.67 \\ 6.67 \end{pmatrix}$$
  • Feasibility Range for $b_1 = 60 + \Delta b_1$: Using column 1 of $\mathbf{B}^{-1}$, $\boldsymbol{\beta}_{*1} = (2/3, -1/3)^T$:
$$\begin{aligned} \hat{x}_1 &= 24 + \frac{2}{3} \Delta b_1 \ge 0 \implies \Delta b_1 \ge -36 \\ \hat{x}_2 &= 12 - \frac{1}{3} \Delta b_1 \ge 0 \implies \Delta b_1 \le +36 \end{aligned}$$

Thus $-36 \le \Delta b_1 \le 36$, giving the allowable capacity interval $b_1 \in [24, \, 96]$.

  • Feasibility Range for $b_2 = 48 + \Delta b_2$: Using column 2 of $\mathbf{B}^{-1}$, $\boldsymbol{\beta}_{*2} = (-1/3, 2/3)^T$:
$$\begin{aligned} \hat{x}_1 &= 24 - \frac{1}{3} \Delta b_2 \ge 0 \implies \Delta b_2 \le +72 \\ \hat{x}_2 &= 12 + \frac{2}{3} \Delta b_2 \ge 0 \implies \Delta b_2 \ge -18 \end{aligned}$$

Thus $-18 \le \Delta b_2 \le 72$, giving the allowable capacity interval $b_2 \in [30, \, 120]$.

Compute Optimality Ranges for Basic Objective Coefficients c_1 and c_2
  • Allowable Range for $c_1 = 40 + \Delta c_1$ (Row $r = 1$, non-basic columns $N = \{s_1, s_2\}$ with $y_{1, s_1} = 2/3, y_{1, s_2} = -1/3$):
$$\begin{aligned} \hat{z}_{s_1} - c_{s_1} &= \frac{50}{3} + \frac{2}{3}\Delta c_1 \ge 0 \implies \Delta c_1 \ge -25 \\ \hat{z}_{s_2} - c_{s_2} &= \frac{20}{3} - \frac{1}{3}\Delta c_1 \ge 0 \implies \Delta c_1 \le +20 \end{aligned}$$

Thus $-25 \le \Delta c_1 \le 20$, giving $c_1 \in [15, \, 60]$.

  • Allowable Range for $c_2 = 30 + \Delta c_2$ (Row $r = 2$, non-basic columns $N = \{s_1, s_2\}$ with $y_{2, s_1} = -1/3, y_{2, s_2} = 2/3$):
$$\begin{aligned} \hat{z}_{s_1} - c_{s_1} &= \frac{50}{3} - \frac{1}{3}\Delta c_2 \ge 0 \implies \Delta c_2 \le +50 \\ \hat{z}_{s_2} - c_{s_2} &= \frac{20}{3} + \frac{2}{3}\Delta c_2 \ge 0 \implies \Delta c_2 \ge -10 \end{aligned}$$

Thus $-10 \le \Delta c_2 \le 50$, giving $c_2 \in [20, \, 80]$. (Notice geometrically that both intervals correspond to the objective slope ratio $1/2 \le c_1 / c_2 \le 2$, matching the slopes of the two binding constraints!)

Evaluate New Activity x_3 via Shadow Pricing and Update Optimum

For the proposed reagent kit $x_3$ with consumption vector $\mathbf{a}_3 = (3, 2)^T$ and unit profit $c_3 = 60$, we compute its imputed resource cost $z_3 = (\mathbf{y}^*)^T \mathbf{a}_3$ and reduced cost $z_3 - c_3$:

$$z_3 - c_3 = \begin{pmatrix} 50/3 & 20/3 \end{pmatrix} \begin{pmatrix} 3 \\ 2 \end{pmatrix} - 60 = \left(50 + \frac{40}{3}\right) - 60 = \frac{190}{3} - 60 = \frac{10}{3} \approx +3.333 > 0$$

Because $z_3 - c_3 = +10/3 > 0$, the imputed value of the resources consumed by one unit of $x_3$ ($\frac{190}{3} \approx \$63.33$) exceeds its unit revenue ($\$60.00$). Therefore, kit $x_3$ should NOT be produced ($x_3^* = 0$) unless its unit profit $c_3$ is raised above the threshold $z_3 = \frac{190}{3} \approx \$63.33$, and the current optimal solution $(x_1^*, x_2^*) = (24, 12)$ with $Z^* = 1320$ remains globally optimal.