Chapter 4 Network Models 1 8 A

Chapter 4 Network Models 1

8 A network problem is one that can be represented by. . . 9 10 Nodes 7 6 4. 1 Introduction 10 Arcs 2 Function on

4. 1 Introduction • The importance of network models – Many business problems lend themselves to a network formulation. – Optimal solutions of network problems are guaranteed integer solutions, because of special mathematical structures. No special restrictions are needed to ensure integrality. – Network problems can be efficiently solved by compact algorithms due to their special mathematical structure, even for large scale models. 3

Network Terminology • Flow – the amount sent from node i to node j, over an arc that connects them. The following notation is used: Xij = amount of flow Uij = upper bound of the flow Lij = lower bound of the flow • Directed/undirected arcs – when flow is allowed in one direction the arc is directed (marked by an arrow). When flow is allowed in two directions, the arc is undirected (no arrows). • Adjacent nodes – a node (j) is adjacent to another node (i) if an arc joins node i to node j. 4

Network Terminology • Path / Connected nodes – Path : a collection of arcs formed by a series of adjacent nodes. – The nodes are said to be connected if there is a path between them. • Cycles / Trees / Spanning Trees – Cycle : a path starting at a certain node and returning to the same node without using any arc twice. – Tree : a series of nodes that contain no cycles. – Spanning tree : a tree that connects all the nodes in a network 5 ( it consists of n -1 arcs).

4. 2 The Transportation Problem Transportation problems arise when a costeffective pattern is needed to ship items from origins that have limited supply to destinations that have demand for the goods. 6

The Transportation Problem • Problem definition – There are m sources. Source i has a supply capacity of Si. – There are n destinations. The demand at destination j is Dj. – Objective: Minimize the total shipping cost of supplying the 7

CARLTON PHARMACEUTICALS • Carlton Pharmaceuticals supplies drugs and other medical supplies. • It has three plants in: Cleveland, Detroit, Greensboro. • It has four distribution centers in: Boston, Richmond, Atlanta, St. Louis. • Management at Carlton would like to ship cases 8 of a certain vaccine as economically as possible.

CARLTON PHARMACEUTICALS • Data – Unit shipping cost, supply, and demand To From Cleveland Detroit Greensboro Demand Boston $35 37 40 1100 • Assumptions Richmond 30 40 15 400 Atlanta 40 42 20 750 St. Louis 32 25 28 750 Supply 1200 1000 800 – Unit shipping costs are constant. – All the shipping occurs simultaneously. – The only transportation considered is between sources and destinations. – Total supply equals total demand. 9

CARLTON PHARMACEUTICALS Network presentation 10

Destinations Bosto Sources D 1=1100 n 35 Cleveland S 1=1200 30 40 32 37 Detroit S 2=1000 25 35 Greensboro S 3= 800 40 42 28 Richmond D 2=400 Atlanta 15 20 D 3=750 St. Louis D 4=750 11

CARLTON PHARMACEUTICALS – Linear Programming Model – The structure of the model is: Minimize Total Shipping Cost ST [Amount shipped from a source] £ [Supply at that source] [Amount received at a destination] = [Demand at that destination] – Decision variables Xij = the number of cases shipped from plant i to warehouse j. where: i=1 (Cleveland), 2 (Detroit), 3 (Greensboro) j=1 (Boston), 2 (Richmond), 3 (Atlanta), 4(St. Louis) 12

Supply from Cleveland X 11+X 12+X 13+X 14 = 1200 from Detroit Supply X 21+X 22+X 23+X 24 = 1000 Supply from Greensboro X 31+X 32+X 33+X 34 = 800 The supply constraints D 1=1100 n X 11 Clevelan d 1=1200 S X 12 X 13 X 21 X 31 Richmond X 14 X 22 Detroit S 2=1000 D 2=400 X 32 X 23 X 24 Atlanta X 33 D 3=750 St. Louis Greensboro S 3= 800 Bosto X 34 D 4=750 13

CARLTON PHARMACEUTICAL – The complete mathematical model Total shipment out of a supply node cannot exceed the supply at the node. £ £ £ = = Total shipment received at a destination 14 node, must equal the demand at that node.

CARLTON PHARMACEUTICALS Spreadsheet =SUMPRODUCT(B 7: E 9, B 15 : E 17) =SUM(B 7: E 7) Drag to cells G 8: G 9 =SUM(B 7: E 9) Drag to cells C 11: E 11 15

CARLTON PHARMACEUTICALS Spreadsheet MINIMIZE Total Cost SHIPMENT S Demands are met Supplies are not exceeded 16

CARLTON PHARMACEUTICALS Spreadsheet - solution 17

CARLTON PHARMACEUTICALS Sensitivity Report – Reduced costs • The unit shipment cost between Cleveland Atlanta must be reduced by at least $5, before it would become economically feasible to utilize it • If this route is used, the total cost will increase by $5 for each case shipped 18 between the two cities.

CARLTON PHARMACEUTICALS Sensitivity Report – Allowable Increase/Decrease • This is the range of optimality. • The unit shipment cost between Cleveland Boston may increase up to $2 or decrease up to $5 with no change in the current optimal transportation plan. 19

CARLTON PHARMACEUTICALS Sensitivity Report – Shadow prices • For the plants, shadow prices convey the cost savings realized for each extra case of vaccine produced. For each additional unit available in Cleveland the total cost reduces by $2. 20

CARLTON PHARMACEUTICALS Sensitivity Report – Shadow prices • For the warehouses demand, shadow prices represent the cost savings for less cases being demanded. For each one unit decrease in demanded in Boston, the total cost decreases by $37. 21

Modifications to the transportation problem – Cases may arise that require modifications to the basic model. • Blocked routes - shipments along certain routes are prohibited. • Remedies: – Assign a large objective coefficient to the route (Cij = 1, 000) 22

Modifications to the transportation problem – Cases may arise that require modifications to the basic model. • Blocked routes - shipments along certain routes are prohibited. • Remedies: – Assign a large objective coefficient to the route (Cij = 1, 000) – Add a constraint to Excel solver of the form Xij = 0 Shipments on a Blocked Route = 0 23

Modifications to the transportation problem – Cases may arise that require modifications to the basic model. Only Feasible Routes • Blocked routes - shipments along certain routes are prohibited. Included in Changing • Remedies: Cells – Assign a large objective coefficient to the route (Cij = Cell C 9 is NOT 1, 000) Included – Add a constraint to Excel solver of the form Xij = 0 – Do not include the cell representing the rout in the Changing cells Shipments from Greensboro to Cleveland are prohibited 24

Modifications to the transportation problem – Cases may arise that require modifications to the basic model. • Minimum shipment - the amount shipped along a certain route must not fall below a pre-specified level. – Remedy: Add a constraint to Excel of the form Xij ³ B • Maximum shipment - an upper limit is placed on the amount shipped along a certain route. – Remedy: Add a constraint to Excel of the form Xij £ B 25

MONTPELIER SKI COMPANY Using a Transportation model for production scheduling – Montpelier is planning its production of skis for the months of July, August, and September. – Production capacity and unit production cost will change from month to month. – The company can use both regular time and overtime to produce skis. – Production levels should meet both demand forecasts and end-of-quarter inventory requirement. – Management would like to schedule production to minimize its costs for the quarter. 26

MONTPELIER SKI COMPANY • Data: – Initial inventory = 200 pairs – Ending inventory required =1200 pairs – Production capacity for the next quarter = 400 pairs in regular time. = 200 pairs in overtime. – Holding cost rate is 3% per month per ski. – Production capacity, and forecasted demand for this quarter (in pairs of skis), and production cost per unit (by months) 27

MONTPELIER SKI COMPANY • Analysis of demand: – Net demand in July = 400 - 200 = 200 pairs Initial inventory In house inventory – Net demand in August = 600 – Net demand in September = 1000 + 1200 = 2200 pairs Forecasted demand • Analysis of Supplies: – Production capacities are thought of as supplies. – There are two sets of “supplies”: • Set 1 - Regular time supply (production capacity) • Set 2 - Overtime supply 28

MONTPELIER SKI COMPANY • Analysis of Unit costs Unit cost = [Unit production cost] + [Unit holding cost per month][the number of months stays in inventory] Example: A unit produced in July in regular time and sold in September costs 25+ (3%)(25)(2 months) = $26. 50 29

Production Month/period July 1000 R/T 800 July O/T Aug. R/T 25 25. 75 26. 50 26. 78 400 Aug. O/T July 30 30. 90 31. 80 +M 26 200 +M +M +M 32 Aug. +M 32. 96 29 400 Month sold 600 Sept. 2200 Demand Production Capacity 500 Network representation +M 37 Sept. R/T 300 200 Sept. O/T 30

MONTPELIER SKI COMPANY Spreadsheet 31

MONTPELIER SKI COMPANY • Summary of the optimal solution – In July produce at capacity (1000 pairs in R/T, and 500 pairs in O/T). Store 1500 -200 = 1300 at the end of July. – In August, produce 800 pairs in R/T, and 300 in O/T. Store additional 800 + 300 - 600 = 500 pairs. – In September, produce 400 pairs (clearly in R/T). With 1000 pairs retail demand, there will be (1300 + 500) + 400 - 1000 = 1200 pairs available for shipment to Ski Inventory + Production Demand - 32

4. 3 The Capacitated Transshipment Model • Sometimes shipments to destination nodes are made through transshipment nodes. • Transshipment nodes may be – Independent intermediate nodes with no supply or demand – Supply or destination points themselves. • Transportation on arcs may be bounded by given bounds 33

The Capacitated Transshipment Model • The linear programming model of this problem consists of: – Flow on arcs decision variables – Cost minimization objective function – Balance constraints on each node as follows: • Supply node – net flow out does not exceed the supply • Intermediate node – flow into the node is equal to the flow out • Demand node – net flow into the node is equal to the demand – Bound constraints on each arc. Flow cannot exceed the capacity on the arc 34

DEPOT MAX A General Network Problem • Depot Max has six stores located in the Washington D. C. area. 35

DEPOT MAX • The stores in Falls Church (FC) and Bethesda (BA) are running low on the model 5 A Arcadia workstation • DATA: 5 -12 FC -13 6 BA 36

DEPOT MAX • The stores in Alexandria (AA) and Chevy Chase (CC have an access of 25 units. • DATA: +10 AA +15 CC 1 5 -12 FC -13 2 6 BA 37

DEPOT MAX • The stores in Fairfax and Georgetown are transshipment nodes with no access supply or demand of their • DATA: own. +10 AA FX 1 3 5 -12 FC • Depot Max wishes to transport the available workstations to FC and BA at minimum total cost. GN +15 CC 2 4 6 -13 BA 38

DEPOT MAX • The possible routes and the shipping unit costs are shown. • DATA: +10 AA 20 10 1 6 5 FX 3 7 12 FC 11 7 GN +15 CC 2 15 4 15 -12 FC BA -13 BA 39

DEPOT MAX • Data – There is a maximum limit for quantities shipped on various routes. – There are different unit transportation costs for different routes. 40

DEPOT MAX – Types of constraints 20 +10 1 6 +15 10 5 2 3 7 –Supply nodes: –Net flow out of the node] = [Supply at the Intermediate transshipment nodes: –Demand nodes: node] [Total flow out of the node] = [Total flow into [Net X +into the 12 10 [Demand for 7 node] the node] X 15 - X 21 = = X 12 + flow (Node 13 the node] 1) X 34+X 35 = X 13 (Node 3) 12 XX=++ X+- +X X 24 X 12 X 46 15 X 24 35 X 3465 - X 56 = = 15 (Node 5) (Node 4) (Node 21 X (Node 6) 2) 46 +X 56 - X 65 = 13 4 15 -12 15 5 7 11 6 -13 41

DEPOT MAX • The Complete mathematical model Min 5 X 12 + 10 X 13 + 20 X 15 + 6 X 21 11 X 56 + 7 X 65 S. T. X 12 + X 13 + - X 12 + £ 17 – X 13 + + 15 X 24 + 12 X 34 + 7 X 35 + 15 X 46 + X 15 – X 21 + X 21 £ 10 X 24 X 34 + =0 – X 46 X 35 + – X 56 - =0 X 15 X 65 X 35 X 24 – X 34 + – = -12 42

DEPOT MAX - spreadsheet 43

4. 4 The Assignment Problem • Problem definition – m workers are to be assigned to m jobs – A unit cost (or profit) Cij is associated with worker i performing job j. – Minimize the total cost (or maximize the total profit) of assigning workers to job so that each worker is assigned a job, and each job is performed. 44

BALLSTON ELECTRONICS • Five different electrical devices produced on five production lines, are needed to be inspected. • The travel time of finished goods to inspection areas depends on both the production line and the inspection area. • Management wishes to designate a separate inspection area to inspect the products such that the total travel time is minimized. 45

BALLSTON ELECTRONICS • Data: Travel time in minutes from assembly lines to inspection areas. 46

BALLSTON ELECTRONICSNETWORK REPRESENTATION 47

Assembly Line S 1= 1 1 Inspection Areas A D 1= 1 S 2=1 2 B D 2=1 S 3=1 3 C D 3=1 S 4=1 4 D D 4=1 S 5=1 5 E D 5=1 48

BALLSTON ELECTRONICS – The Linear Programming Model Min 10 X 11 + 4 X 12 + … + 20 X 54 + 19 X 55 S. T. X 11 + X 12 + X 13 + X 14 + X 15 = 1 X 21 + X 22 + … + X 25 = 1 … … X 51 + X 52+ X 53 + X 54 + X 55 = 1 All the variables are non-negative 49

BALLSTON ELECTRONICS – Computer solutions • A complete enumeration is not an efficient procedure even for moderately large problems (with m=8, m! > 40, 000 is the number of assignments to enumerate). • The Hungarian method provides an efficient solution procedure. 50

BALLSTON ELECTRONICS – Transportation spreadsheet =SUMPRODUCT(B 7: F 11, B 17: F 2 17) =SUM(B 7: B 11) Drag to cells C 13: F 13 =SUM(B 7: F 7) Drag to cells H 8: H 12 51

BALLSTON ELECTRONICS – Transportation spreadsheet Each Area is Served Each Line is Assigned 52

BALLSTON ELECTRONICS – Assignment spreadsheet 53

The Assignments Model Modifications – Unbalanced problem: The number of supply nodes and demand nodes is unequal. – Prohibitive assignments: A supply node should not be assigned to serve a certain demand node. – Multiple assignments: A certain supply node can be assigned for more than one demand node. – A maximization assignment problem. 54

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