01 Introduction

15.082 and 6.855J Spring 2003: 

15.082 and 6.855J Spring 2003 Network Optimization J.B. Orlin

WELCOME!: 

WELCOME! Welcome to 15.082/6.855J Introduction to Network Optimization Instructor: James B. Orlin (jorlin@mit.edu) TA: Agustin Bompadre (abompadr@mit.edu) Website: sloanspace.mit.edu Register at SloanSpace and log in as soon as possible. Textbook: Network Flows: Theory, Algorithms, and Applications by Ahuja, Magnanti, and Orlin referred to as AMO

Quick Overview: 

Quick Overview Next: The Koenigsberg Bridge Problem Introduces Networks and Network Algorithms Some subject management issues Network Flows and Applications Computational Complexity Overall goal of today’s lecture: set the tone for the rest of the subject provide background provide motivation handle some class logistics

On the background of students: 

On the background of students Requirement for this class Either Linear Programming (15.081J) or Data Structures How many have had linear programming? There will be a “review” lecture later in the term How many have had a subject in data structures? The “review” lecture for data structures is Thursday

Some aspects of the class: 

Some aspects of the class Fondness for Powerpoint animations Cold-calling as a way to speed up learning of the algorithms Talking with partners (the person next to you in in the classroom.) Class time: used for presenting theory, algorithms, applications, theory mostly outlines of proofs illustrated by examples (not detailed proofs) detailed proofs are in the text

The Bridges of Koenigsberg: Euler 1736: 

The Bridges of Koenigsberg: Euler 1736 “Graph Theory” began in 1736 Leonard Eüler Visited Koenigsberg People wondered whether it is possible to take a walk, end up where you started from, and cross each bridge in Koenigsberg exactly once Generally it was believed to be impossible

The Bridges of Koenigsberg: Euler 1736: 

The Bridges of Koenigsberg: Euler 1736 A D C B 1 2 4 3 7 6 5 Is it possible to start in A, cross over each bridge exactly once, and end up back in A?

The Bridges of Koenigsberg: Euler 1736: 

The Bridges of Koenigsberg: Euler 1736 A D C B 1 2 4 3 7 6 5 Conceptualization: Land masses are “nodes”.

The Bridges of Koenigsberg: Euler 1736: 

The Bridges of Koenigsberg: Euler 1736 1 2 4 3 7 6 5 Conceptualization: Bridges are “arcs.”

The Bridges of Koenigsberg: Euler 1736: 

The Bridges of Koenigsberg: Euler 1736 1 2 4 3 7 6 5 Is there a “walk” starting at A and ending at A and passing through each arc exactly once?

Notation and Terminology: 

Notation and Terminology Network terminology as used in AMO. Node set N = {1, 2, 3, 4} Network G = (N, A) In an undirected graph, (i,j) = (j,i)

Slide12: 

Directed Path . Example: 1, 2, 5, 3, 4 (or 1, a, 2, c, 5, d, 3, e, 4) No node is repeated. Directions are important. Cycle (or circuit or loop) 1, 2, 3, 1. (or 1, a, 2, b, 3, e) A path with 2 or more nodes, except that the first node is the last node. Directions are ignored. Path: Example: 5, 2, 3, 4. (or 5, c, 2, b, 3, e, 4) No node is repeated. Directions are ignored. Directed Cycle: (1, 2, 3, 4, 1) or 1, a, 2, b, 3, c, 4, d, 1 No node is repeated. Directions are important.

Walks: 

Walks Walks are paths that can repeat nodes and arcs Example of a directed walk: 1-2-3-5-4-2-3-5 A walk is closed if its first and last nodes are the same. A closed walk is a cycle except that it can repeat nodes and arcs.

The Bridges of Koenigsberg: Euler 1736: 

The Bridges of Koenigsberg: Euler 1736 1 2 4 3 7 6 5 Is there a “walk” starting at A and ending at A and passing through each arc exactly once? Such a walk is called an eulerian cycle.

Adding two bridges creates such a walk: 

Adding two bridges creates such a walk A, 1, B, 5, D, 6, B, 4, C, 8, A, 3, C, 7, D, 9, B, 2, A 1 2 4 3 7 6 5 8 9 Here is the walk. Note: the number of arcs incident to B is twice the number of times that B appears on the walk.

On Eulerian Cycles: 

On Eulerian Cycles The degree of a node in an undirected graph is the number of incident arcs 6 4 4 4 Theorem. An undirected graph has an eulerian cycle if and only if (1) every node degree is even and (2) the graph is connected (that is, there is a path from each node to each other node).

More on Euler’s Theorem: 

More on Euler’s Theorem Necessity of two conditions: Any eulerian cycle “visits” each node an even number of times Any eulerian cycle shows the network is connected caveat: nodes of degree 0 Sufficiency of the condition Assume the result is true for all graphs with fewer than |A| arcs. Start at some node, and take a walk until a cycle C is found. 1 4 5 3 7

More on Euler’s Theorem: 

More on Euler’s Theorem Sufficiency of the condition Start at some node, and take a walk until a cycle C is found. Consider G’ = (N, A\C) the degree of each node is even each component is connected So, G’ is the union of Eulerian cycles Connect G’ into a single eulerian cycle by adding C.

Comments on Euler’s theorem: 

Comments on Euler’s theorem It reflects how proofs are done in class, often in outline form, with key ideas illustrated. However, this proof does not directly lead to an efficient algorithm. Usually we focus on efficient algorithms.

15.082/6.855J Subject Goals: : 

15.082/6.855J Subject Goals: 1. To present students with a knowledge of the state-of-the art in the theory and practice of solving network flow problems. A lot has happened since 1736 2. To provide students with a rigorous analysis of network flow algorithms. computational complexity & worst case analysis 3. To help each student develop his or her own intuition about algorithm development and algorithm analysis.

Homework Sets and Grading : 

Homework Sets and Grading Homework Sets 11 assignments, once weekly except for midterm week and final week students may hand in assignments in groups of two much of it is “theorem/proof” Grades: out of 3 points per assignment Grading homework: 30 points midterm: 30 points final: 40 points

Hamiltonian Cycles: 

Hamiltonian Cycles A hamiltonian cycle is a cycle that passes through each node of the graph exactly once. This is often called a traveling salesman tour.

Hamilton’s Around the World Game: 

Hamilton’s Around the World Game In 1857, the Irish mathematician, Sir William Rowan Hamilton invented a puzzle that he hoped would be very popular. The objective was to make what we just called a hamiltonian cycle. The game was not a commercial success, especially the 3D version. But the mathematics of hamiltonian cycles is very popular today.

Hamilton’s Around the World Game: 

Hamilton’s Around the World Game We will see this problem again when we generalize it to be the traveling salesman problem.

The knight’s tour problem: 

The knight’s tour problem Can a knight visit all squares of a chessboard exactly once, starting at some square, and by making 63 legitimate moves? The knight’s tour problem is a special case of the hamiltonian tour problem. The answer is yes!

Where Network Flows Arise: 

Where Network Flows Arise Transportation Transportation of goods over transportation networks Scheduling of fleets of airplanes: time/space networks Manufacturing Scheduling of goods for manufacturing Flow of manufactured items within inventory systems Communications Design and expansion of communication systems Flow of information across networks Personnel Assignment Assignment of crews to airline schedules Assignment of drivers to vehicles

Network Flow Problems: 

Network Flow Problems A number of you have seen network problems either in subjects or in your research. Please list some examples.

The shortest path problem: 

The shortest path problem Consider a network G = (N, A) in which there is an origin node s and a destination node t. standard notation: n = |N|, m = |A| What is the shortest path from s to t?

Slide29: 

The Maximum Flow Problem Directed Graph G = (N, A). Source s Sink t Capacities uij on arc (i,j) Maximize the flow out of s, subject to Flow out of i = Flow into i, for i  s or t. A Network with Arc Capacities and Flows

Representing the Max Flow as an LP: 

Flow out of i - Flow into i = 0 for i  s or t. Representing the Max Flow as an LP max v s.t xs1 + xs2 = v max v s.t. Sj xsj = v -xs1 + x12 + x1t = 0 -xs2 - x12 + x2t = 0 Sj xij – Sj xji = 0 for each i  s or t -x1t - x2t = -v 0  xij  uij for all (i,j) 0  xij  uij for all (i,j) s.t. - Si xit = -v

Min Cost Flows: 

Flow out of i - Flow into i = b(i) Min Cost Flows min Si,j cijxij s.t Sj xij – Sj xji = b(i) for each i 0  xij  uij for all (i,j) Each arc has a linear cost and a capacity Covered in detail in Chapter 1 of AMO

Next topic: computational complexity: 

Next topic: computational complexity What is an efficient algorithm? How do we measure efficiency? “Worst case analysis” but first …

Measuring Computational Complexity: 

Measuring Computational Complexity Consider the following algorithm for adding two m  n matrices A and B with coefficients a( , ) and b( , ). begin for i = 1 to m do for j = 1 to n do c(i,j) := a(i,j) + b(i,j) end What is the running time of this algorithm? Let’s measure it as precisely as we can as a function of n and m. Is it 2nm, or 3nm, or what? Worst case versus average case How do we measure the running time? What are the basic step that we should count?

Compute the running time precisely.: 

Compute the running time precisely. Operation Number (as a function of m,n) Additions Assignments Comparisons Work with your “classroom partner” to determine the number of counts when we add two matrices.

Towards Computational Complexity: 

Towards Computational Complexity 1. We will ignore running time constants. 2. Our running times will be stated in terms of relevant problem parameters, e.g., nm. 3. We will measure everything in terms of worst case or most pessimistic analysis (performance guarantees.) 4. All arithmetic operations are assumed to take one step, (or a number of steps that is bounded by a constant).

A Simpler Metric for Running Time.: 

A Simpler Metric for Running Time. Operation Number (as a function of m,n) Additions  c1 mn for some c1 and m, n  1 O(mn) steps Assignments  c2 mn for some c2 and m, n  1 O(mn) steps Comparisons  c3 mn for some c3 and m, n  1 O(mn) steps TOTAL  c4 mn for some c4 and m, n  1 O(mn) steps

Simplifying Assumptions and Notation: 

Simplifying Assumptions and Notation MACHINE MODEL: Random Access Machine (RAM). This is the computer model that everyone is used to. It allows the use of arrays, and it can select any element of an array or matrix in O(1) steps. c(i,j) := a(i,j) + b(i,j). Integrality Assumption. All numbers are integral (unless stated otherwise.)

Size of a problem: 

Size of a problem The size of a problem is the number of bits needed to represent the problem. The size of the n  m matrix A is not nm. If each matrix element has K bits, the size is nmK e.g., if max 2107 < aij < 2108, then K = 108. K = O( log (amax).

Polynomial Time Algorithms: 

Polynomial Time Algorithms We say that an algorithm runs in polynomial time if the number of steps taken by an algorithm on any instance I is bounded by a polynomial in the size of I. We say that an algorithm runs in exponential time if it does not run in polynomial time. Example 1: finding the determinant of a matrix can be done in O(n3) steps. This is polynomial time.

Polynomial Time Algorithms: 

Polynomial Time Algorithms Example 2: We can determine if n is prime by dividing n by every integer less than n. This algorithm is exponential time. The size of the instance is log n The running time of the algorithm is O(n). Side note: there is a polynomial time algorithm for determining if n is prime. Almost all of the algorithms presented in this class will be polynomial time. One can find an Eulerian cycle (if one exists) in O(m) steps. There is no known polynomial time algorithm for finding a hamiltonian cycle.

On polynomial vs exponential time: 

On polynomial vs exponential time We contrast two algorithm, one that takes 30,000 n3 steps, and one that takes 2n steps. Suppose that we could carry out 1 billion steps per second. # of nodes 30,000 n3 steps 2n steps n = 30, 0.81 seconds 1 second n = 40, 1.92 seconds 17 minutes n = 50 3.75 seconds 12 days n = 60 6.48 seconds 31 years

On polynomial vs. exponential time: 

On polynomial vs. exponential time Suppose that we could carry out 1 trillion steps per second, and instantaneously eliminate 99.9999999% of all solutions as not worth considering # of nodes 1,000 n10 steps 2n steps n = 70, 2.82 seconds 1 second n = 80, 10.74 seconds 17 minutes n = 90 34.86 seconds 12 days n = 100 100 seconds 31 years

Overview of today’s lecture: 

Overview of today’s lecture Eulerian cycles Network Definitions Network Applications Introduction to computational complexity

Upcoming Lectures: 

Upcoming Lectures Lecture 2: Review of Data Structures even those with data structure backgrounds are encouraged to attend. Lecture 3. Graph Search Algorithms. how to determine if a graph is connected and to label a graph and more