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Chapter 4 Network Models.

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Presentation on theme: "Chapter 4 Network Models."— Presentation transcript:

1 Chapter 4 Network Models

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

3 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.

4 Network Terminology Flow Directed/undirected arcs
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.

5 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 ( it consists of n -1 arcs).

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 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 destinations with the required demand from the available supplies at the sources.

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.

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 From Boston Richmond Atlanta St. Louis Supply Cleveland $35 30 40 32 1200 Detroit 37 40 42 25 1000 Greensboro 40 15 20 28 800 Demand 1100 400 750 750

10 CARLTON PHARMACEUTICALS Network presentation

11 Destinations Boston Sources Cleveland Richmond Detroit Atlanta
St.Louis Destinations Sources Cleveland Detroit Greensboro D1=1100 37 40 42 32 35 30 25 15 20 28 S1=1200 S2=1000 S3= 800 D2=400 D3=750 D4=750

12 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)

13 The supply constraints
Cleveland S1=1200 X11 X12 X13 X14 Supply from Cleveland X11+X12+X13+X14 = The supply constraints Detroit S2=1000 X21 X22 X23 X24 Supply from Detroit X21+X22+X23+X24 = Boston Greensboro S3= 800 X31 X32 X33 X34 Supply from Greensboro X31+X32+X33+X34 = D1=1100 Richmond D2=400 Atlanta D3=750 St.Louis D4=750

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.

=SUMPRODUCT(B7:E9,B15:E17) =SUM(B7:E7) Drag to cells G8:G9 =SUM(B7:E9) Drag to cells C11:E11

MINIMIZE Total Cost SHIPMENTS Demands are met Supplies are not exceeded

17 CARLTON PHARMACEUTICALS Spreadsheet - solution

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.

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.

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.

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 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 = 1,000,000)

23 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 = 1,000,000) Add a constraint to Excel solver of the form Xij = 0 Shipments on a Blocked Route = 0

24 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 = 1,000,000) Add a constraint to Excel solver of the form Xij = 0 Do not include the cell representing the rout in the Changing cells Only Feasible Routes Included in Changing Cells Cell C9 is NOT Included Shipments from Greensboro to Cleveland are prohibited

25 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

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.

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)

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 In house inventory Forecasted demand

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

30 Network representation
Production Month/period Network representation Month sold July R/T July R/T 25 25.75 26.50 1000 200 July July O/T 500 30 30.90 31.80 +M 37 +M 26 26.78 +M 29 Aug. R/T +M 32 32.96 800 Aug. 600 Demand Production Capacity Aug. O/T 400 Sept. 2200 Sept. R/T 400 300 Sept. O/T 200


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. Inventory + Production - Demand

33 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

34 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

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

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

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

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

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

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

41 DEPOT MAX – Types of constraints
5 10 20 6 15 12 7 11 -12 +10 1 3 5 Supply nodes: Net flow out of the node] = [Supply at the node] X12 + X13 + X15 - X21 = 10 (Node 1) X21 + X24 - X = 15 (Node 2) Intermediate transshipment nodes: [Total flow out of the node] = [Total flow into the node] X34+X35 = X (Node3) X46 = X24 + X34 (Node 4) 7 Demand nodes: [Net flow into the node] = [Demand for the node] X15 + X35 +X65 - X56 = 12 (Node 5) X46 +X56 - X65 = 13 (Node 6) 2 4 6 -13 +15

42 All variables are non-negative
DEPOT MAX The Complete mathematical model Min 5X X X X X X34 + 7X X X56 + 7X65 S.T. X12 + X13 + X15 – X £ 10 - X X21 + X £ 17 – X X34 + X = 0 – X24 – X X = 0 – X – X35 + X56 - X65 = -12 -X46 – X56 + X65 = -13 X12 £ 3; X15 £ 6; X21 £ 7; X24 £ 10; X34 £ 8; X35 £ 8; X46 £ 17; X56 £ 7; X65 £ 5 All variables are non-negative

43 DEPOT MAX - spreadsheet

44 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.

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


48 1 2 3 4 5 Assembly Line Inspection Areas A B C D E S1=1 S2=1 S3=1 S4=1 S5=1 D1=1 D2=1 D3=1 D4=1 D5=1

49 BALLSTON ELECTRONICS – The Linear Programming Model
Min 10X11 + 4X12 + … X X55 S.T. X11 + X12 + X13 + X14 + X15 = 1 X21 + X … X25 = 1 … … … … X51 + X52+ X53 + X54 + X55 = 1 All the variables are non-negative

50 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.

51 BALLSTON ELECTRONICS – Transportation spreadsheet
=SUMPRODUCT(B7:F11,B17:F217) =SUM(B7:F7) Drag to cells H8:H12 =SUM(B7:B11) Drag to cells C13:F13

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

53 BALLSTON ELECTRONICS – Assignment spreadsheet

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.

55 Copyright 2002John Wiley & Sons, Inc. All rights reserved
Copyright 2002John Wiley & Sons, Inc. All rights reserved. Reproduction or translation of this work beyond that named in Section 117 of the United States Copyright Act without the express written consent of the copyright owner is unlawful. Requests for further information should be addressed to the Permissions Department, John Wiley & Sons, Inc. Adopters of the textbook are granted permission to make back-up copies for their own use only, to make copies for distribution to students of the course the textbook is used in, and to modify this material to best suit their instructional needs. Under no circumstances can copies be made for resale. The Publisher assumes no responsibility for errors, omissions, or damages, caused by the use of these programs or from the use of the information contained herein.

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