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1 Lecture 1 Introduction. 2 Agenda  typical problems in transportation and logistics  modeling  shortest-path problems  assignment problems.

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Presentation on theme: "1 Lecture 1 Introduction. 2 Agenda  typical problems in transportation and logistics  modeling  shortest-path problems  assignment problems."— Presentation transcript:

1 1 Lecture 1 Introduction

2 2 Agenda  typical problems in transportation and logistics  modeling  shortest-path problems  assignment problems

3 3 Typical Problems in Transportation and Logistics

4 4  network-flow problems  routing problems  location problems  arc-routing problems

5 5 Network-Flow Problems  problems with an underlying network structure  arcs  directed, undirected, or mixed  of known lengths, capacities, and costs per unit flow  nodes  net in flow, net out flow, or net no flow  flows  along arcs, from one node to another 2 1 3 4 5 5

6 6 Network-Flow Problems  Shortest-Path Problems: Find the shortest path from one node to another in a network  Maximal Flow Problems: Find the maximal possible flow from one node to another in the network 2 1 3 4

7 7 Network-Flow Problems  Minimum Cost Flow Problems: Find the cheapest way to send goods from the specified sources nodes to the sink nodes  Minimum Spanning Tree Problems: Find the minimum-cost set of arcs that connect all nodes  Multi-commodity Flow Problems: The minimum cost flow problem with multiple products bounded by common constraints

8 8 Routing Problems  still on a network  sequence of nodes matter more  for any choice of the sequence of nodes in a segment, the number of possible sequences for the remaining nodes does not depend on the choice and sequence of nodes in the segment  in other problems such as finding the shortest path, the sequence of nodes selected affect the number of feasible solutions for the remaining decisions

9 9 Routing Problems  Traveling Salesman Problem: Given a set of cities and the distances among them, find the shortest cycle that visits all cities once and returns to the starting city?  applications: a subproblem in vehicle routing, drill path, placement problem, transition cost between jobs, examination scheduling

10 10 Routing Problems  Vehicle Routing Problem: Goods are to be picked up and sent to the depot by a group of vehicles. Given the distances of the locations of goods from the depot, the volume of goods, and the capacity of vehicles, find the allocation of goods to and the routing of vehicles such that the total distance travelled by vehicles is minimized.

11 11 Location Problems  decisions: where to put something, possibly multiple items  different levels of decisions  strategic level: a new city, an airport, headquarter of a company, a nuclear plant  tactical level: a new factory, a new warehouse  operational level: location of a machine, storage slot of an item  the medium for location consideration: line, an area, a node in a network  items to locate: points (e.g., warehouses), lines (e.g., flights routes), networks (e.g., flights routes), area (e.g., regional office)  criteria: distance, cost, coverage, accessibility, market share

12 12 Arc Routing Problems  Given a network, find the shortest cycle that visits all arcs once and returns to the original city  Mail Delivery, Garbage Collection, Street Cleaning, Snow Removing, Meter reading

13 13 What is Modeling?

14 14 The Most Important Modeling Problem in My Life  雞免同籠  雞免同籠共 25 隻,有腳 80 隻,問雞兔 各有幾隻? 雞免同籠  let x (y) be the number of chickens (rabbits) in the cage  x + y = 25  2x + 4y = 80

15 15 More Complicate Problems  雞、免、豬同籠 、豬  雞、免、豬同籠共 25 隻,有腳 80 隻, 問雞、兔、豬各有幾隻?  let x (y, z) be the number of chickens (rabbits, pigs) in the cage  x + y + z = 25  2x + 4y + 4z = 80

16 16 More Complicate Problems  雞、免、豬同籠 、豬  雞、免、豬同籠共 25 隻,有腳 80 隻, 問雞、兔、豬各有幾隻?  The answer: {(x, y, z) | x = 10, y+z = 15, y, z  {0, 1, …}}  implicit constraints: x, y, z  {0, 1, …}

17 17 More Complicate Problems  suppose that there is a weird three-leg animal called   雞、免、  同籠、   雞、免、  同籠共 25 隻,有腳 80 隻,問雞、兔、  各有 幾隻?  let x (y, z) be the number of chickens (rabbits,  s) in the cage  constraints  x + y + z = 25  2x + 4y + 3z = 80  x + y + z  {0, 1, 2, …} xy 0520 1618 2716 ……… 8134 9142 10150

18 18 More Complicate Problems  suppose that there is a weird three-leg animal called   籠子可容雞、免、   籠子內應有幾雞、免、   籠子可容雞、免、  共 25 隻,腳 80 隻 ( !?please don’t ask what this means ) 。 雞每隻可售 $150 , 兔 $250 ,  $180 。要售出最高價錢, 籠子內應有幾隻雞、免、  ?  max 150x + 250y + 180z  s.t.  x + y + z = 25  2x + 4y + 3z = 80  x + y + z  {0, 1, 2, …} xy  Revenue 05204850 16184890 27164930 38144970 49125010 510 5050 61185090 71265130 81345170 91425210 101505250

19 19 A Typical Model  opt x 1 + … + x n s.t.s.t.  a 11 x 1 + a 12 x 2 + … + a 1n x n = b 1  a 21 x 1 + a 22 x 2 + … + a 2n x n = b 2 ……  a m1 x 1 + a m2 x 2 + … + a mn x n = b m  x n  X

20 20 Comments on a Typical Model  opt  optimize, which can be max (  maximize) or min (  minimize)  three types of constraints, equality (=), less than or equal to (  ), and greater than or equal to (  )  often a mixture of all three types in a model  decision variables x n  belonging to a set X, which can be discrete (e.g., the set of non-negative integers) or continuous (e.g., the set of non-negative real numbers)

21 21 Comments on a Typical Model  usually more decision variables than number of constraints  easy to have a problem of tens of million of variables and hundred thousands of constraints

22 22 Which Problem is Easier to Solve, Discrete or Continuous X?  in general discrete X is much more difficult to solve than continuous X  this course on modeling, leaving the solution methods to other courses

23 23 Importance of Modeling  existence of magical solution tools magical tools such as CPLEX, Gurobi, Lingo, etc optimal solution This simplifies reality quite a bit.

24 24 More on Modeling

25 25 Modeling  (in our case) the process of representing a physical phenomenon by mathematical relationships  let x (y) be the # of chickens (rabbits) in the cage  x + y = 25  2x + 4y = 80 (definitions of) symbols the bridge between physical phenomenon and mathematical relationship constraints each constraint describes a physical property of the physical phenomenon

26 26 Modeling  often not easy to define the variables  careful examination of the physical phenomenon in construction of constraints

27 27 Is Modeling Useful?  Can all physical phenomena be represented numerically?  雞免同籠  雞免同籠共 25 隻,有腳 76 隻,問雞兔各有幾隻? 雞免同籠  a possible real-life answer: 雞 11 隻,兔 14 隻  Is it possible to get the precise values of the parameters in a model?

28 28 Is Modeling Useful?  Our view: Models are useful tools that provide insights to a problem; however, blindly applying the result of a model only indicates that we don’t fully understand the art of modeling.

29 29 Shortest-Path Models

30 30 A Network  definitions  circles: nodes ( 節點 ), vertices ( 角 )  arcs: lines, branches  directed (具方向的) or not 2 1 3 4 2 1 3 4 2 1 3 4 2 1 3 4

31 31 An Example to Formulate Constraints  The Shortest Route Problem  motivation: to find the shortest route from the origin ( 起點, i.e., one location, source node) to the destination ( 終點, i.e., another location, sink node) in a network  problem on hand:

32 32 Universal Solution Techniques  If you don’t know how to solve a difficult problem, start with a simpler one with the similar properties. Observe the general principle in solving the simpler problem, which hopefully is applicable to the difficult problem.  It is generally helpful to work with a small concrete numerical example.

33 33 A Simple Concrete Numerical Example  a one-arc, two-node problem  source node 1  sink node 2  how to formulate?  either the upper or the lower route (上路還是下 路? ) ; how to model mathematically?  min 9U + 7L s.t.s.t. U + L = 1 U, L  {0, 1} 7 9 12

34 34 Another Simple Concrete Numerical Example  a three-arc, three-node problem  source node 1  sink node 3  either the upper or the lower route; how to model mathematically?  min (3+2)U + 4L s.t.s.t. U + L = 1 U, L  {0, 1} 2 9 3 1 2 3

35 35 What Have We Learnt About the Art of Formulation from the Two Examples?  We calculate the lengths of all possible paths from the source to the sink.  Is it possible to pre-calculate the lengths of all possible paths for a general problem? No. 7 9 12 2 9 3 1 2 3

36 36 Yet Another Simple Concrete Numerical Example  obvious shortest path between node 1 and node 4  But how to formulate? What is the direction of flow in the middle arc, upward or downward? Or any flow at all? 1 2 5 6 3 1 2 4 3

37 37 Yet Another Simple Concrete Numerical Example  a route from the source to the sink = a collection of arcs from the source to the sink  some restriction on the choice of arcs in to form a path  question: How to define the values of a group of x ij such that x ij s form a route from the source to the sink? 1 2 5 6 3 1 2 4 3 1 2 5 6 3 1 2 4 3 1 2 5 6 3 1 2 4 3 1 2 5 6 3 1 2 4 3 1 2 5 6 3 1 2 4 3


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