15.3 Operations Research: Linear Programming & Transportation/Assignment

Operations Research: Linear Programming & Transportation/Assignment

Optimization is central to modern industrial management. In Coal India's multi-subsidiary logistics networks, Operations Research (OR) models dictate coal blending proportions, fleet allocation from open-cast pits to washeries, railway siding dispatch schedules, and overhaul worker assignments.

---

1. Linear Programming Problem (LPP) Formulation & Geometry

General Mathematical Model

$\text{Maximize / Minimize } Z = \sum_{j=1}^n c_j x_j = \mathbf{c}^T \mathbf{x}$ $\text{Subject to: } \sum_{j=1}^n a_{ij} x_j (\le, =, \ge) b_i, \quad i = 1, 2, \dots, m$ $\text{and Non-negativity: } x_j \ge 0, \quad j = 1, 2, \dots, n$

Geometric Characteristics and Corner Point Theorem

1. Convex Set: A region $S$ is convex if for any two points $\mathbf{x}_1, \mathbf{x}_2 \in S$, the entire line segment connecting them $\lambda \mathbf{x}_1 + (1 - \lambda)\mathbf{x}_2 \in S$ for all $\lambda \in [0, 1]$.
2. Convex Feasible Region: The intersection of linear half-spaces forms a convex polygon (2D) or convex polyhedron ($n$D).
3. Extreme Point (Corner Point) Theorem: The objective function achieves its optimal value (maximum or minimum) at one of the extreme (corner) points of the convex feasible region.

  x2 ^
     |
     |    Binding Constraint 2
     |\           Optimal Vertex (x1*, x2*)
     | \______       *
     |  \     \____  |\  Iso-profit line: Z = c1*x1 + c2*x2
     |   \ Feasible\ | \
     |    \ Region  \|  \
     |     \         |   \
   0 +------\--------+----\-----> x1
            Binding Constraint 1

Classification of LPP Solutions

---

2. Simplex Method Fundamentals

Standard Form Conversion

To solve via Simplex, constraints must be converted into equalities with non-negative right-hand sides ($b_i \ge 0$):

• $\le$ Constraint: Add a Slack variable $s_i \ge 0$ (cost in objective $= 0$). $\sum a_{ij} x_j + s_i = b_i$
• $\ge$ Constraint: Subtract a Surplus variable $e_i \ge 0$ (cost $= 0$) and add an Artificial variable $A_i \ge 0$ (penalty cost in objective $= -M$ for Maximize, $+M$ for Minimize). $\sum a_{ij} x_j - e_i + A_i = b_i$
• $=$ Constraint: Add an Artificial variable $A_i \ge 0$. $\sum a_{ij} x_j + A_i = b_i$

Basic Definitions in Simplex

For a system of $m$ linear equations with $n$ variables ($n > m$):

• Basic Solution: Set $(n - m)$ non-basic variables to 0 and solve for the remaining $m$ basic variables.
• Basic Feasible Solution (BFS): A basic solution where all basic variables are non-negative ($x_{Bi} \ge 0$).
• Maximum possible basic solutions: $\binom{n}{m} = \frac{n!}{m!(n - m)!}$.

Step-by-Step Simplex Iteration Mechanics

                     +------------------------------+
                     | Formulate Standard Tableau   |
                     +------------------------------+
                                    |
                                    v
                     +------------------------------+
                     | Compute Net Evaluations      |
                     |    Delta_j = C_j - Z_j       |
                     +------------------------------+
                                    |
                     +--------------+--------------+
                     |                             |
              All Delta_j <= 0?             Any Delta_j > 0?
                     |                             |
                     v                             v
            +-----------------+           +-------------------+
            | OPTIMALITY      |           | KEY COLUMN        |
            | REACHED         |           | Enter Max Delta_j |
            +-----------------+           +-------------------+
                                                   |
                                                   v
                                          +-------------------+
                                          | KEY ROW           |
                                          | Min Positive Ratio|
                                          | theta = b_i / a_ir|
                                          +-------------------+
                                                   |
                                                   v
                                          +-------------------+
                                          | PIVOT OPERATION   |
                                          | Gauss-Jordan Row  |
                                          | Transformations   |
                                          +-------------------+
                                                   |
                                                   + (Loop back)

1. Net Evaluation (Index) Row $\Delta_j$: $\Delta_j = C_j - Z_j = C_j - \sum_{i=1}^m c_{Bi} a_{ij}$
2. Optimality Criterion (for Maximization):
   • If all $\Delta_j = C_j - Z_j \le 0$, the current solution is optimal.
3. Entering Variable (Key Column $r$): $\text{Select variable with } \max(C_j - Z_j > 0)$
4. Leaving Variable (Key Row $k$ - Minimum Positive Ratio Test): $\theta = \min_{i} \left\{ \frac{b_i}{a_{ir}} \;\middle\|\; a_{ir} > 0 \right\}$
5. Pivot Element: Intersection of Key Column and Key Row ($a_{kr}$). Perform Gauss-Jordan row transformations to make $a_{kr} = 1$ and all other entries in column $r$ equal to $0$.

---

3. Duality in Linear Programming

Every linear programming problem (the Primal) has an associated twin problem (the Dual) formed from the same underlying parameters.

Primal-Dual Relationships

Fundamental Duality Theorems

1. Weak Duality Theorem: For any feasible primal solution $\mathbf{x}$ and feasible dual solution $\mathbf{y}$: $\mathbf{c}^T \mathbf{x} \le \mathbf{b}^T \mathbf{y}$
2. Strong (Fundamental) Duality Theorem: If either the primal or dual has a finite optimal solution, the other also has a finite optimal solution, and their optimal objective values are identical: $Z^* = \mathbf{c}^T \mathbf{x}^* = \mathbf{b}^T \mathbf{y}^* = W^*$
3. Unboundedness Theorem: If the primal is unbounded, the dual is infeasible. If the dual is unbounded, the primal is infeasible.
4. Shadow Prices ($y_i^*$): The optimal value of dual variable $y_i^*$ represents the marginal economic value (shadow price) of resource $i$: $y_i^* = \frac{\partial Z^*}{\partial b_i}$ It represents the maximum amount a manager should pay for one additional unit of resource $b_i$.
5. Complementary Slackness Theorem: $x_j^* \left[(\mathbf{A}^T \mathbf{y}^*)_j - c_j\right] = 0, \quad \forall j$ $y_i^* \left[b_i - (\mathbf{A} \mathbf{x}^*)_i\right] = 0, \quad \forall i$ (If a primal constraint is non-binding with positive slack, the corresponding dual variable is zero; if a dual variable is strictly positive, the corresponding primal constraint is strictly binding).

---

4. The Transportation Problem

The Transportation Problem is a special class of LPP where a homogeneous commodity is transported from $m$ sources ($a_i$ capacity) to $n$ destinations ($b_j$ requirement) at unit cost $c_{ij}$.

$\text{Minimize } Z = \sum_{i=1}^m \sum_{j=1}^n c_{ij} x_{ij} \quad \text{subject to } \sum_{j=1}^n x_{ij} = a_i, \quad \sum_{i=1}^m x_{ij} = b_j$ Balanced condition: $\sum_{i=1}^m a_i = \sum_{j=1}^n b_j$.

Initial Basic Feasible Solution (IBFS) Methods

1. North-West Corner Method (NWCM): Allocates purely based on top-left position without regard to unit costs. Fast but yields the highest initial cost.
2. Least Cost Method (LCM / Matrix Minimum): Allocates iteratively to the cell with the lowest unit transportation cost $c_{ij}$.
3. Vogel's Approximation Method (VAM / Penalty Method):
   • Step 1: Compute the penalty (difference between the two lowest costs) for each row and column.
   • Step 2: Identify the row or column with the maximum penalty.
   • Step 3: Allocate the maximum possible amount $\min(a_i, b_j)$ to the minimum cost cell in that chosen row/column.
   • Step 4: Adjust supply/demand and cross out satisfied rows/columns. Recompute penalties and repeat until all allocations are complete.
   • Result: VAM yields an initial solution very close to (or identical to) the global optimum.

Degeneracy & MODI ($u-v$) Optimality Method

• Non-degeneracy Condition: An IBFS must have exactly $m + n - 1$ allocated cells located at independent positions (no closed loops among allocated cells).
• If allocations $< m + n - 1$, the problem is degenerate. Allocate an infinitesimal amount $\epsilon > 0$ to an unallocated cell with the lowest cost that does not form a loop.

                 MODI (u-v) OPTIMALITY METHOD
                 ----------------------------
  ALLOCATED CELLS (Basic):           UNALLOCATED CELLS (Non-basic):
  Solve: u_i + v_j = c_ij            Compute: Delta_ij = c_ij - (u_i + v_j)
  (Set u_1 = 0 as reference)         Optimality: All Delta_ij >= 0

• Optimality Rule:
  • If all $\Delta_{ij} = c_{ij} - (u_i + v_j) \ge 0$, the current solution is optimal.
  • If any $\Delta_{ij} < 0$, the solution can be improved. Select the most negative $\Delta_{ij}$, construct a closed alternating $+/-\theta$ loop with existing allocated cells, shift allocation by $\theta = \min(\text{negative loop corners})$, and retest.

---

5. The Assignment Problem & Traveling Salesperson Problem (TSP)

The Assignment Problem assigns $n$ workers to $n$ jobs on a 1-to-1 basis ($x_{ij} \in {0, 1}$) to minimize total cost $\sum \sum c_{ij} x_{ij}$.

Hungarian Algorithm (Flood's Technique)

1. Row Reduction: Subtract the minimum element of each row from all elements in that row.
2. Column Reduction: In the resulting matrix, subtract the minimum element of each column from all elements in that column.
3. Zero Covering: Draw the minimum number of horizontal and vertical lines $L$ to cover all zeros in the reduced matrix:
   • If $L = n$ (matrix size), an optimal assignment is possible. Proceed to step 5.
   • If $L < n$, proceed to step 4.
4. Matrix Modification:
   • Find the smallest uncovered element, denoted as $k$.
   • Subtract $k$ from all uncovered elements.
   • Add $k$ to elements at the intersection of two covering lines.
   • Leave elements covered by a single line unchanged.
   • Return to step 3.
5. Optimal Assignment: Assign zeros uniquely such that exactly one zero is selected per row and per column.

Traveling Salesperson Problem (TSP)

A salesperson must visit $n$ cities and return to the starting city such that each city is visited exactly once and total tour distance is minimized. Formulated as an assignment problem with $c_{ii} = \infty$ plus sub-tour elimination constraints.