Presentation on theme: "1 Network Models Chapter 4. 2 4.1 Introduction A network problem is one that can be represented by... Nodes Arcs 8 9 10 7 6 Function on Arcs."— Presentation transcript:
1 Network Models Chapter 4
2 4.1 Introduction A network problem is one that can be represented by... Nodes Arcs Function on Arcs
3 4.1 Introduction –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. The importance of network models
4 Network Terminology Flow –the amount sent from node i to node j, over an arc that connects them. The following notation is used: X ij = amount of flow U ij = upper bound of the flow L ij = 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.
5 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 ( it consists of n -1 arcs). Network Terminology
6 4.2 The Transportation Problem Transportation problems arise when a cost-effective pattern is needed to ship items from origins that have limited supply to destinations that have demand for the goods.
7 Problem definition – There are m sources. Source i has a supply capacity of S i. – There are n destinations. The demand at destination j is D j. – Objective: Minimize the total shipping cost of supplying the destinations with the required demand from the available supplies at the sources. The Transportation Problem
8 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 of a certain vaccine as economically as possible.
9 Data –Unit shipping cost, supply, and demand Assumptions –Unit shipping costs are constant. –All the shipping occurs simultaneously. –The only transportation considered is between sources and destinations. –Total supply equals total demand. To FromBostonRichmondAtlantaSt. LouisSupply Cleveland $ Detroit Greensboro Demand CARLTON PHARMACEUTICALS
11 Boston Richmond Atlanta St.Louis DestinationsSources Cleveland Detroit Greensboro S 1 =1200 S 2 =1000 S 3 = 800 D 1 =1100 D 2 =400 D 3 =750 D 4 =
12 –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 X ij = 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) CARLTON PHARMACEUTICALS – Linear Programming Model
13 Boston Richmond Atlanta St.Louis D 1 =1100 D 2 =400 D 3 =750 D 4 =750 The supply constraints Cleveland S 1 =1200 X11 X12 X13 X14 Supply from Cleveland X11+X12+X13+X14 = 1200 Detroit S 2 =1000 X21 X22 X23 X24 Supply from Detroit X21+X22+X23+X24 = 1000 Greensboro S 3 = 800 X31 X32 X33 X34 Supply from Greensboro X31+X32+X33+X34 = 800
14 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 node, must equal the demand at that node.
15 CARLTON PHARMACEUTICALS Spreadsheet =SUM(B7:E9) Drag to cells C11:E11 =SUMPRODUCT(B7:E9,B15:E17) =SUM(B7:E7) Drag to cells G8:G9
16 MINIMIZE Total Cost SHIPMENTS Demands are met Supplies are not exceeded CARLTON PHARMACEUTICALS Spreadsheet
18 CARLTON PHARMACEUTICALS Sensitivity Report –Reduced costs The unit shipment cost between Cleveland and 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 between the two cities.
19 CARLTON PHARMACEUTICALS Sensitivity Report –Allowable Increase/Decrease This is the range of optimality. The unit shipment cost between Cleveland and Boston may increase up to $2 or decrease up to $5 with no change in the current optimal transportation plan.
20 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.
21 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.
22 –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 (C ij = 1,000,000) Modifications to the transportation problem
23 –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 (C ij = 1,000,000) – Add a constraint to Excel solver of the form X ij = 0 Modifications to the transportation problem Shipments on a Blocked Route = 0
24 –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 (C ij = 1,000,000) – Add a constraint to Excel solver of the form X ij = 0 – Do not include the cell representing the rout in the Changing cells Modifications to the transportation problem Only Feasible Routes Included in Changing Cells Cell C9 is NOT Included Shipments from Greensboro to Cleveland are prohibited
25 –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 X ij 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 X ij B Modifications to the transportation problem
26 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.
27 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)
28 Analysis of demand: –Net demand in July = = 200 pairs –Net demand in August = 600 –Net demand in September = = 2200 pairs 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 Initial inventory Forecasted demand In house inventory MONTPELIER SKI COMPANY
29 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 MONTPELIER SKI COMPANY
30 Network representation M M M +M M +M Production Month/period Month sold July R/T July O/T Aug. R/T Aug. O/T Sept. R/T Sept. O/T July Aug. Sept Demand Production Capacity July R/T
31 MONTPELIER SKI COMPANY - Spreadsheet
32 Summary of the optimal solution –In July produce at capacity (1000 pairs in R/T, and 500 pairs in O/T). Store = 1300 at the end of July. –In August, produce 800 pairs in R/T, and 300 in O/T. Store additional = 500 pairs. – In September, produce 400 pairs (clearly in R/T). With 1000 pairs retail demand, there will be ( ) = 1200 pairs available for shipment to Ski Chalet. MONTPELIER SKI COMPANY Inventory +Production -Demand
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
34 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 The Capacitated Transshipment Model
35 DEPOT MAX A General Network Problem Depot Max has six stores located in the Washington D.C. area.
DATA: The stores in Falls Church (FC) and Bethesda (BA) are running low on the model 5A Arcadia workstation. DEPOT MAX FC BA
DATA: The stores in Alexandria (AA) and Chevy Chase (CC) have an access of 25 units. DEPOT MAX FC BA AA CC
DATA: DEPOT MAX The stores in Fairfax and Georgetown are transshipment nodes with no access supply or demand of their own. FC BA AA CC FX GN Depot Max wishes to transport the available workstations to FC and BA at minimum total cost. 4 3
FC BA DATA: The possible routes and the shipping unit costs are shown. DEPOT MAX FC BA AA CC FX GN 4 3
40 Data –There is a maximum limit for quantities shipped on various routes. –There are different unit transportation costs for different routes. DEPOT MAX
–Supply nodes: Net flow out of the node] = [Supply at the node] X 12 + X 13 + X 15 - X 21 = 10 (Node 1) X 21 + X 24 - X 12 = 15(Node 2) –Intermediate transshipment nodes: [Total flow out of the node] = [Total flow into the node] X 34 +X 35 = X13 (Node3) X 46 = X 24 + X 34 (Node 4) 7 –Demand nodes: [Net flow into the node] = [Demand for the node] X 15 + X 35 +X 65 - X 56 = 12 (Node 5) X 46 +X 56 - X 65 = 13 (Node 6) DEPOT MAX – Types of constraints
42 DEPOT MAX The Complete mathematical model Min 5X X X X X X X X X X 65 S.T.X 12 +X 13 +X 15 –X 21 10 - X 12 +X 21 +X 24 17 – X 13 +X 34 + X 35 = 0 – X 24 – X 34 +X 46 = 0 – X 15 – X 35 +X 56 - X 65 = -12 -X 46 – X 56 + X 65 = -13 X 12 3; X 15 6; X 21 7; X 24 10; X 34 8; X 35 8; X 46 17; X 56 7; X 65 5 All variables are non-negative
43 DEPOT MAX - spreadsheet
The Assignment Problem Problem definition –m workers are to be assigned to m jobs –A unit cost (or profit) C ij 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.
45 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.
46 Data: Travel time in minutes from assembly lines to inspection areas. BALLSTON ELECTRONICS
Assembly LineInspection Areas A B C D E S 1 =1 S 2 =1 S 3 =1 S 4 =1 S 5 =1 D 1 =1 D 2 =1 D 3 =1 D 4 =1 D 5 =1
49 Min 10X X 12 + … + 20X 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 BALLSTON ELECTRONICS – The Linear Programming Model
50 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. BALLSTON ELECTRONICS – Computer solutions
51 =SUM(B7:F7) Drag to cells H8:H12 =SUMPRODUCT(B7:F11,B17:F217) BALLSTON ELECTRONICS – Transportation spreadsheet =SUM(B7:B11) Drag to cells C13:F13
52 BALLSTON ELECTRONICS – Transportation spreadsheet Each Area is Served Each Line is Assigned
54 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.
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