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 8 9 10 7 6 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 $353040321200 Detroit 374042251000 Greensboro 40152028800 Demand 1100400750 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 =750 37 40 42 32 35 40 30 25 35 15 20 28
12 The Assumptions –The per item shipping cost remains constant –All the shipping from the sources to the destinations occurs « simultaneously » –The vaccine can be shipped only between sources and destinations –The total supply equals the total demand if not : insert dummy destinations or sources
13 –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
14 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
15 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.
17 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.
18 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.
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 Special cases of the transportation problem –Cases may arise that appear to violate the assumptions necessary to solve the transportation problem using standard methods. –Modifying the resulting models make it possible to use standard solution methods. – Examples: Blocked routes - shipments along certain routes are prohibited. Minimum shipment - the amount shipped along a certain route must not fall below a prespecified level. Maximum shipment - an upper limit is placed on the amount shipped along a certain route. Transshipment nodes - intermediate nodes that may have demand, supply, or no demand and no supply of their own. –General network problems are solved by the “Out-of-Kilter” algorithm.
22 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.
23 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)
24 Analysis of demand: –Net demand in July = 400 - 200 = 200 pairs –Net demand in August = 600 –Net demand in September = 1000 + 1200 = 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
25 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
26 Network representation 2525.7526.50 0 3030.9031.80 0 +M 26 2626.78 0 +M 32 3232.96 0 +M+M 29 29 0 +M+M 37 37 0 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. Dummy 1000 500 800 400 200 600 300 2200 Demand Production Capacity July R/T
27 MONTPELIER SKI COMPANY - Spreadsheet
28 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 Chalet. MONTPELIER SKI COMPANY Inventory +Production -Demand
29 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
30 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
31 DEPOT MAX A General Network Problem Depot Max has six stores located in the Washington D.C. area.
32 5 6 DATA: -12 -13 The stores in Falls Church (FC) and Bethesda (BA) are running low on the model 5A Arcadia workstation. DEPOT MAX FC BA
33 1 2 5 6 DATA: -12 -13 +10 +15 The stores in Alexandria (AA) and Chevy Chase (CC) have an access of 25 units. DEPOT MAX FC BA AA CC
34 1 2 5 6 DATA: -12 -13 +10 +15 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
35 5 10 20 6 15 12 7 15 11 7 1 2 FC BA DATA: -12 -13 +10 +15 The possible routes and the shipping unit costs are shown. DEPOT MAX FC BA AA CC FX GN 4 3
36 Data –There is a maximum limit for quantities shipped on various routes. –There are different unit transportation costs for different routes. DEPOT MAX
37 5 10 20 6 15 12 7 15 11 7 1 24 35 6 -12 -13 +10 +15 –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
38 DEPOT MAX The Complete mathematical model Min 5X 12 +10X 13 + 20X 15 + 6X 21 + 15X 24 + 12X 34 + 7X 35 + 15X 46 + 11X 56 + 7X 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
39 DEPOT MAX - spreadsheet
40 4.4 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.
41 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.
42 Data: Travel time in minutes from assembly lines to inspection areas. Actual arrangement : 1-A, 2-B, 3-C, 4-D, 5-E Cost : 10+7+12+17+19 = 65 man-minutes
43 each 1/2 hour : 65/60 x $ 12 = $13 250 working per year 2 shifts of 8 ours per day number of periods of 1/2 hour 250 x 16 X 2 = 8000 periods total transportation cost : $104000 Calculation of the yearly cost
45 1 2 3 4 5 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
46 Assumptions and restrictions –The number of workers equals the number of jobs. –For an unbalanced problem “dummy” workers (in case there are more jobs than workers), or “dummy” jobs (in case there are more workers than jobs) are added to balance the problem. –Given a balanced problem, each worker is assigned exactly once, and each job is performed by exactly one worker.
47 Min 10X 11 + 4X 12 + … + 20X 54 + 19X 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
48 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
49 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.