1. Spreadsheet Modeling & Decision Analysis A Practical Introduction to Management Science 5 th edition Cliff T. Ragsdale

2. Network Modeling Chapter 5

3. Introduction  A number of business problems can be represented graphically as networks.  This chapter focuses on several such problems: –Transshipment Problems –Shortest Path Problems –Maximal Flow Problems –Transportation/Assignment Problems –Generalized Network Flow Problems –The Minimum Spanning Tree Problem

4. Network Flow Problem Characteristics  Network flow problems can be represented as a collection of nodes connected by arcs.  There are three types of nodes: –Supply –Demand –Transshipment  We’ll use negative numbers to represent supplies and positive numbers to represent demand.

5. A Transshipment Problem: The Bavarian Motor Company Newark 1 Boston 2 Columbus 3 Atlanta 5 Richmond 4 J'ville 7 Mobile 6 $30 $40 $50 $35 $40 $30 $35 $25 $50 $45 $50 -200 -300 +80 +100 +60 +170 +70

6. Defining the Decision Variables For each arc in a network flow model we define a decision variable as: X ij = the amount being shipped (or flowing) from node i to node j For example… X 12 = the # of cars shipped from node 1 (Newark) to node 2 (Boston) X 56 = the # of cars shipped from node 5 (Atlanta) to node 6 (Mobile) Note: The number of arcs determines the number of variables!

7. Defining the Objective Function Minimize total shipping costs. MIN : 30X 12 + 40X 14 + 50X 23 + 35X 35 +40X 53 + 30X 54 + 35X 56 + 25X 65 + 50X 74 + 45X 75 + 50X 76

8. Constraints for Network Flow Problems: The Balance-of-Flow Rules For Minimum Cost Network Apply This Balance-of-Flow Flow Problems Where:Rule At Each Node: Total Supply > Total DemandInflow-Outflow >= Supply or Demand Total Supply < Total DemandInflow-Outflow <=Supply or Demand Total Supply = Total DemandInflow-Outflow = Supply or Demand

9. Defining the Constraints  In the BMC problem: Total Supply = 500 cars Total Demand = 480 cars  For each node we need a constraint like this: Inflow - Outflow >= Supply or Demand  Constraint for node 1: –X 12 – X 14 >= – 200 (Note: there is no inflow for node 1!)  This is equivalent to: +X 12 + X 14 <= 200 (Supply >= Demand)

10. Defining the Constraints  Flow constraints –X 12 – X 14 >= –200} node 1 +X 12 – X 23 >= +100} node 2 +X 23 + X 53 – X 35 >= +60} node 3 + X 14 + X 54 + X 74 >= +80} node 4 + X 35 + X 65 + X 75 – X 53 – X 54 – X 56 >= +170} node 5 + X 56 + X 76 – X 65 >= +70} node 6 –X 74 – X 75 – X 76 >= –300} node 7  Nonnegativity conditions X ij >= 0 for all ij

11. Implementing the Model See file Fig5-2.xlsFig5-2.xls

12. Optimal Solution to the BMC Problem Newark 1 Boston 2 Columbus 3 Atlanta 5 Richmond 4 J'ville 7 Mobile 6 $30 $40 $50 $40 $50 $45 -200 -300 +80 +100 +60 +170 +70 120 80 20 40 70210

13. The Shortest Path Problem  Many decision problems boil down to determining the shortest (or least costly) route or path through a network. –Ex. Emergency Vehicle Routing  This is a special case of a transshipment problem where: –There is one supply node with a supply of -1 –There is one demand node with a demand of +1 –All other nodes have supply/demand of +0

14. The American Car Association B'ham Atlanta G'ville Va Bch Charl. L'burg K'ville A'ville G'boro Raliegh Chatt. 1 2 3 4 6 5 7 8 9 10 11 2.5 hrs 3 pts 3.0 hrs 4 pts 1.7 hrs 4 pts 2.5 hrs 3 pts 1.7 hrs 5 pts 2.8 hrs 7 pts 2.0 hrs 8 pts 1.5 hrs 2 pts 2.0 hrs 9 pts 5.0 hrs 9 pts 3.0 hrs 4 pts 4.7 hrs 9 pts 1.5 hrs 3 pts 2.3 hrs 3 pts 1.1 hrs 3 pts 2.0 hrs 4 pts 2.7 hrs 4 pts 3.3 hrs 5 pts +1 +0

15. Solving the Problem  There are two possible objectives for this problem –Finding the quickest route (minimizing travel time) –Finding the most scenic route (maximizing the scenic rating points) See file Fig5-7.xlsFig5-7.xls

16. The Equipment Replacement Problem  The problem of determining when to replace equipment is another common business problem.  It can also be modeled as a shortest path problem…

17. The Compu-Train Company  Compu-Train provides hands-on software training.  Computers must be replaced at least every two years.  Two lease contracts are being considered: –Each requires $62,000 initially –Contract 1:  Prices increase 6% per year  60% trade-in for 1 year old equipment  15% trade-in for 2 year old equipment –Contract 2:  Prices increase 2% per year  30% trade-in for 1 year old equipment  10% trade-in for 2 year old equipment

18. Network for Contract 1 13 5 2 4 +1 +0 $28,520 $60,363 $30,231 $63,985 $32,045 $67,824 $33,968 Cost of trading after 1 year: 1.06*$62,000 - 0.6*$62,000 = $28,520 Cost of trading after 2 years: 1.06 2 *$62,000 - 0.15*$62,000 = $60,363 etc, etc….

19. Solving the Problem See file Fig5-12.xlsFig5-12.xls

20. Transportation & Assignment Problems  Some network flow problems don’t have trans- shipment nodes; only supply and demand nodes. Mt. Dora 1 Eustis 2 Clermont 3 Ocala 4 Orlando 5 Leesburg 6 Distances (in miles) Capacity Supply 275,000 400,000 300,000 225,000 600,000 200,000 Groves Processing Plants 21 50 40 35 30 22 55 25 20 These problems are implemented more effectively using the technique described in Chapter 3.

21. Generalized Network Flow Problems  In some problems, a gain or loss occurs in flows over arcs. –Examples  Oil or gas shipped through a leaky pipeline  Imperfections in raw materials entering a production process  Spoilage of food items during transit  Theft during transit  Interest or dividends on investments  These problems require some modeling changes.

22. Coal Bank Hollow Recycling MaterialCost YieldCost YieldSupply Newspaper$1390%$1285%70 tons Mixed Paper$1180%$1385%50 tons White Office Paper$995%$1090%30 tons Cardboard$1375%$1485%40 tons Process 1Process 2 Pulp SourceCostYieldCostYieldCostYield Recycling Process 1$595%$690%$890% Recycling Process 2$690%$895%$795% Newsprint Packaging Paper Print Stock Demand60 tons40 tons50 tons

23. Network for Recycling Problem Newspaper 1 Mixed paper 2 3 Cardboard 4 Recycling Process 1 5 6 Newsprint pulp 7 Packing paper pulp 8 Print stock pulp 9 -70 -50 -30 -40 +60 +40 +50 White office paper Recycling Process 2 $13 $12 $11 $13 $9 $10 $14 $13 90% 80% 95% 75% 85% 90% 85% $5 $6 $8 $6 $7 $8 95% 90% 95% +0

24. Defining the Objective Function Minimize total cost. MIN : 13X 15 + 12X 16 + 11X 25 + 13X 26 + 9X 35 + 10X 36 + 13X 45 + 14X 46 + 5X 57 + 6X 58 + 8X 59 + 6X 67 + 8X 68 + 7X 69

25. Defining the Constraints-I  Raw Materials -X 15 -X 16 >= -70 } node 1 -X 25 -X 26 >= -50 } node 2 -X 35 -X 36 >= -30 } node 3 -X 45 -X 46 >= -40 } node 4

26. Defining the Constraints-II  Recycling Processes +0.9X 15 +0.8X 25 +0.95X 35 +0.75X 45 - X 57 - X 58 -X 59 >= 0 } node 5 +0.85X 16 +0.85X 26 +0.9X 36 +0.85X 46 -X 67 -X 68 -X 69 >= 0 } node 6

27. Defining the Constraints-III  Paper Pulp +0.95X 57 + 0.90X 67 >= 60 } node 7 +0.90X 57 + 0.95X 67 >= 40 } node 8 +0.90X 57 + 0.95X 67 >= 50 } node 9

28. Implementing the Model See file Fig5-17.xlsFig5-17.xls

29. Important Modeling Point  In generalized network flow problems, gains and/or losses associated with flows across each arc effectively increase and/or decrease the available supply.  This can make it difficult to tell if the total supply is adequate to meet the total demand.  When in doubt, it is best to assume the total supply is capable of satisfying the total demand and use Solver to prove (or refute) this assumption.

30. The Maximal Flow Problem  In some network problems, the objective is to determine the maximum amount of flow that can occur through a network.  The arcs in these problems have upper and lower flow limits.  Examples –How much water can flow through a network of pipes? –How many cars can travel through a network of streets?

31. The Northwest Petroleum Company Oil Field Pumping Station 1 Pumping Station 2 Pumping Station 3 Pumping Station 4 Refinery 1 2 3 4 5 6 6 4 3 6 4 5 2 2

32. The Northwest Petroleum Company Oil Field Pumping Station 1 Pumping Station 2 Pumping Station 3 Pumping Station 4 Refinery 1 2 3 4 5 6 6 4 3 6 4 5 2 2

33. Formulation of the Max Flow Problem MAX: X 61 Subject to:+X 61 - X 12 - X 13 = 0 +X 12 - X 24 - X 25 = 0 +X 13 - X 34 - X 35 = 0 +X 24 + X 34 - X 46 = 0 +X 25 + X 35 - X 56 = 0 +X 46 + X 56 - X 61 = 0 with the following bounds on the decision variables: 0 <= X 12 <= 60 <= X 25 <= 20 <= X 46 <= 6 0 <= X 13 <= 40 <= X 34 <= 20 <= X 56 <= 4 0 <= X 24 <= 30 <= X 35 <= 50 <= X 61 <= inf

34. Implementing the Model See file Fig5-24.xlsFig5-24.xls

35. Optimal Solution Oil Field Pumping Station 1 Pumping Station 2 Pumping Station 3 Pumping Station 4 Refinery 1 2 3 4 5 6 6 4 3 6 4 5 2 2 5 3 2 4 2 5 4 2

36. Special Modeling Considerations: Flow Aggregation 1 2 3 4 5 6 -100 +75 +50 +0 $3 $4 $5 $3 $6 Suppose the total flow into nodes 3 & 4 must be at least 50 and 60, respectively. How would you model this?

37. 1 2 3 4 5 6 -100 +75 +50 +0 $3 $4 $5 $3 $6 30 40 +0 L.B.=50 L.B.=60 Nodes 30 & 40 aggregate the total flow into nodes 3 & 4, respectively. Special Modeling Considerations: Flow Aggregation

38. Special Modeling Considerations: Multiple Arcs Between Nodes 1 1 10 2 -75 +0 +50 -75 $8 $0 $6 2 +50 Two two (or more) arcs cannot share the same beginning and ending nodes. Instead, try... $6 U.B. = 35 $8 U.B. = 35

39. Special Modeling Considerations: Capacity Restrictions on Total Supply 1 -100 2 3 +75 4 +80 $5, UB=40 $3, UB=35 $6, UB=35 $4, UB=30 Supply exceeds demand, but the upper bounds prevent the demand from being met.

40. Special Modeling Considerations: Capacity Restrictions on Total Supply 1 -100 2 3 +75 4 +80 $5, UB=40 $3, UB=35 $6, UB=35 $4, UB=30 0 +200 $999, UB=100 Now demand exceeds supply. As much “real” demand as possible will be met in the least costly way.

41. The Minimal Spanning Tree Problem  For a network with n nodes, a spanning tree is a set of n-1 arcs that connects all the nodes and contains no loops.  The minimal spanning tree problem involves determining the set of arcs that connects all the nodes at minimum cost.

42. Minimal Spanning Tree Example: Windstar Aerospace Company 2 3 1 4 5 6 $150 $100 $40 $85 $65 $50 $90 $80 $75 $85 Nodes represent computers in a local area network.

43. The Minimal Spanning Tree Algorithm 1.Select any node. Call this the current subnetwork. 2.Add to the current subnetwork the cheapest arc that connects any node within the current subnetwork to any node not in the current subnetwork. (Ties for the cheapest arc can be broken arbitrarily.) Call this the current subnetwork. 3. If all the nodes are in the subnetwork, stop; this is the optimal solution. Otherwise, return to step 2.

44. Solving the Example Problem - 1 2 3 1 4 5 6 $100 $85 $90 $80 $85

45. Solving the Example Problem - 2 2 3 1 4 5 6 $100 $85 $90 $80 $85 $75 $50

46. Solving the Example Problem - 3 2 3 1 4 5 6 $100 $85 $80 $85 $75 $50 $65

47. Solving the Example Problem - 4 2 3 1 4 5 6 $100 $80 $85 $75 $50 $65 $40

48. Solving the Example Problem - 5 2 3 1 4 5 6 $80 $85 $75 $50 $65 $40 $150

49. Solving the Example Problem - 6 2 3 1 4 5 6 $80 $75 $50 $65 $40

50. End of Chapter 5