Operational research
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The document discusses the time complexity of the simplex algorithm for solving linear programming problems. It begins by defining time complexity as the number of arithmetic operations required to solve a problem. It then provides an overview of different time complexities such as polynomial time and exponential time. The rest of the document focuses on using geometric interpretations to understand the simplex algorithm and analyze cases where it exhibits exponential running time. It illustrates concepts like the region of feasibility and simplex pivoting through examples. It also reviews the Klee-Minty example, which shows that the simplex algorithm can require an exponential number of iterations in the worst case.
![Step 2: Increasing Objective Function:
Modifying the Cube
[0, 1, 0.8] [0, 1, 0.82]
Squash the cube
100 x1 + 1, 000 x2 + 10, 000 x3 = z
x1 ≤ 1
0.2 x1 + x2 ≤ 1 [0, 0, 1] [1, 0, 0.98]
0.02 x1 + 0.2 x2 + x3 ≤ 1
x1 , x2 , x3 ≥ 0
[0, 1, 0] [1, 0.8, 0]
New dictionary
x4 = 1 − x1
x5 = 1 − 0.2 x1 − x2
x6 = 1 − 0.02 x1 − 0.2 x2 − x3 [1, 0, 0]
[0, 0, 0]](https://image.slidesharecdn.com/operapre-130120000033-phpapp01/85/Operational-research-19-320.jpg)
Operational research
- 1. Time complexity of simplex algorithm Albi thomas M.tech (TM) Roll no.11
- 2. Introduction Time complexity of an algorithm counts the number of arithmetic operations sufficient for the algorithm to solve the problem Understand properties of LP in terms of geometry Use geometry as aid to solve LP Some concepts new
- 3. Overview Polynomial time-complexity(bound) Eg.gaussian elimination Exponential time complexity Eg. Buchberger's algorithm Feasibility Simplex Method Simplex Weaknesses Exponential Iterations Convex Sets and Hulls
- 4. Region of Feasibility Graphical region describing all feasible solutions to a linear programming problem In 2-space: polygon, each edge a constraint In 3-space: polyhedron, each face a constraint
- 5. Feasibility in 2-Space 2x1 + x2 ≤ 4 In an LP environment, restrict to Quadrant I since x1, x2 ≥ 0
- 6. Simplex Method Every time a new dictionary is generated: Simplex moves from one vertex to another vertex along an edge of polyhedron Analogous to increasing value of a non-basic variable until bounded by basic constraint Each such point is a feasible solution Average time taken is linear in 2 space
- 7. Feasibility in 3-Space maximize 3 x1 + 2 x2 + 5 x3 subject to 2 x1 + x2 ≤4 x3 ≤ 5 x1 , x2 , x3 ≥ 0 Five total constraints; therefore 5 faces to the polyhedron
- 8. Simplex Illustrated: Initial Dictionary x4 = 4 − 2 x1 − x2 x5 = 5 − x3 z = 3 x1 + 2 x2 + 5 x3 Current solution: x1 = 0 x2 = 0 x3 = 0
- 9. Simplex Illustrated: First Pivot x4 = 4 − 2 x1 − x2 x3 = 5 − x5 z = 25 + 3 x1 + 2 x2 − 5 x5 Current solution: x1 = 0 x2 = 0 x3 = 5
- 10. Simplex Illustrated: Second Pivot x1 = 2 − 1 2 x 2 − 1 x4 2 x3 = 5 − x5 z = 31 + 1 2 x2 − 3 x4 − 5 x5 2 Current solution: x1 = 2 x2 = 0 x3 = 5
- 11. Simplex Illustrated: Final Pivot x2 = 4 − 2 x1 − x4 x3 = 5 − x5 z = 33 − 7 x1 − 2 x4 − 5 x5 Final solution (optimal): x1 = 0 x2 = 4 x3 = 5
- 12. Simplex Review and Analysis Simplex pivoting represents traveling along polyhedron edges Each vertex reached tightens one constraint (and if needed, loosens another) May take a longer path to reach final vertex than needed
- 15. Simplex Weaknesses: Exponential Iterations: Klee-Minty Reviewed 100 x1 + 10 x2 + x3 = z x1 ≤ 1 20 x1 + x2 ≤ 100 200 x1 + 20 x2 + x3 ≤ 10, 000 x1 , x2 , x3 ≥ 0 Cases with high complexity (2n-1 iterations) Normal complexity is O(m3) How was this problem solved?
- 16. Geometric Interpretation & Klee-Minty Saw non-optimal solution earlier How can we represent the Klee-Minty problem class graphically? maximize 3 x1 + 2 x2 + 5 x3 subject to 2 x1 + x2 ≤ 4 x3 ≤ 5 x1 , x2 , x3 ≥ 0
- 17. Step 1: Constructing a Shape c1 x1 + c2 x2 + c3 x3 = z Start with a cube. x1 ≤ 1 x2 ≤ 1 What characteristics do x3 ≤ 1 we want the cube to x1 , x2 , x3 ≥ 0 have? What is the worst case to maximize z?
- 18. Step 1: Constructing a Shape Goal 1: Create a shape with a long series of increasing facets Goal 2: Create an LP problem that forces this route to be taken
- 19. Step 2: Increasing Objective Function: Modifying the Cube [0, 1, 0.8] [0, 1, 0.82] Squash the cube 100 x1 + 1, 000 x2 + 10, 000 x3 = z x1 ≤ 1 0.2 x1 + x2 ≤ 1 [0, 0, 1] [1, 0, 0.98] 0.02 x1 + 0.2 x2 + x3 ≤ 1 x1 , x2 , x3 ≥ 0 [0, 1, 0] [1, 0.8, 0] New dictionary x4 = 1 − x1 x5 = 1 − 0.2 x1 − x2 x6 = 1 − 0.02 x1 − 0.2 x2 − x3 [1, 0, 0] [0, 0, 0]
- 20. Step 3: Achieving 2n-1 Iterations: Altering the Algebra Let sjxj = xj 100 x1 s1 + 1, 000 x2 s2 + 10, 000 x3 s3 = z s1 1 x1 ≤ s4 s4 Convert s1 s2 1 z = v + ∑ d jxj 0.2 x1 s5 + s5 x2 ≤ s5 j∈N s1 s2 s3 1 0.02 x1 + 0.2 x2 + x3 ≤ s6 s6 s6 s6 to x1 , x2 , x3 ≥ 0 z = v + ∑ (d j s j ) x j j∈N
- 21. The Final Solution Most desirable: x1 , x4 Least desirable: x3 s1 = s4 = 1, s2 = s5 = 0.01, s3 = s6 = 0.0001 100 x1 + 10 x2 + x3 = z x1 ≤ 1 20 x1 + x2 ≤ 100 200 x1 + 20 x2 + x3 ≤ 10, 000 x1 , x2 , x3 ≥ 0
- 22. Thank you