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Transportation, Transshipment and Assignment Models and Assignment Models

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Learning Objectives Structure special LP network flow models. Set up and solve transportation models Extend basic transportation model to include transshipment points. Set up and solve facility location and other application problems as transportation models. Set up and solve assignment models

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Overview Part of a larger class of linear programming problems are known as network flow models. They possess special mathematical features that enabled the development of very efficient, unique solution methods.

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Transportation Model Transportation problem deals with the distribution of goods from several points of supply to a number of points of demand. They 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. Resources to be optimally allocated usually involve a given capacity of goods at each source and a given requirement for the goods at each destination. Most common objective of the transportation problem is to schedule shipments from sources to destinations so that total production and transportation costs are minimized

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Transshipment Model An extension of transportation problems is called transshipment problem in which a point can have shipments that both arrive as well as leave. Example would be a warehouse where shipments arrive from factories and then leave for retail outlets It may be possible for a firm to achieve cost savings (economies of scale) by consolidating shipments from several factories at a warehouse and then sending them together to retail outlets.

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Assignment Model Assignment problem refers to a class of LP problems that involve determining most efficient assignment of: People to projects, Salespeople to territories, Contracts to bidders, Jobs to machines, and so on Objective is to minimize total cost or total time of performing tasks at hand, although a maximization objective is also possible.

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Transportation Model

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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: To minimize the total shipping cost of supplying the destinations with the required demand from the available supplies at the sources. Transportation Model

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The Transportation Model Characteristics A product is to be transported from a number of sources to a number of destinations at the minimum possible cost. Each source is able to supply a fixed number of units of the product, and each destination has a fixed demand for the product. The linear programming model has constraints for supply at each source and demand at each destination. All constraints are equalities in a balanced transportation model where supply equals demand. Constraints contain inequalities in unbalanced models where supply is not equal to demand.

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Transportation Model - Example 1 Executive Furniture Corporation

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Transportation Model Example 1 Executive Furniture Corporation Transportation Costs Per Desk

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Transportation Model Example 1 Objective: minimize total shipping costs = 5 X DA + 4 X DB + 3 X DC + 3 X EA + 2 X EB + 1 X EC + 9 X FA + 7 X FB + 5 X FC Executive Furniture Corporation: LP Transportation Model Formulation Where: X ij = number of desks shipped from factory i to warehouse j i = D (for Des Moines), E (for Evansville), or F (for Fort Lauderdale). j = A (for Albuquerque), B (for Boston), or C (for Cleveland).

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Transportation Model Example 1 Net flow at Des Moines = (Total flow in) - (Total flow out) = (0) - (X DA + X DB + X DC ) Net flow at Des Moines = -X DA - X DB - X DC = -100 (Des Moines capacity) and -X EA - X EB - X EC = -300(Evansville capacity) -X FA - X FB - X FC = -300 (Fort Lauderdale capacity) Multiply each constraint by -1 and rewrite as: X DA + X DB + X DC = 100(Des Moines capacity) X EA + X EB + X EC = 300(Evansville capacity) X FA + X FB + X FC = 300(Fort Lauderdale capacity) Executive Furniture Corporation : Supply Constraints

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Transportation Model Example 1 Executive Furniture Corporation: Demand Constraints Net flow at Albuquerque = (Total flow in) - (Total flow out) = (X DA + X EA + X FA ) - (0) Net flow at Albuquerque = X DA + X EA + X FA = 300(Albuquerque demand) and X DB + X EB + X FB = 200(Boston demand) X DC + X EC + X FC = 200(Cleveland demand)

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Transportation Model Example 1: The Optimum Solution SHIP: 100 desks from Des Moines to Albuquerque, 200 desks from Evansville to Albuquerque, 100 desks from Evansville to Boston, 100 desks from Fort Lauderdale to Boston, and 200 desks from Fort Lauderdale to Cleveland. Total shipping cost is $3,000.

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Grain Elevator Supply Mill Demand 1. Kansas City 150 A. Chicago 200 2. Omaha 175 B. St. Louis 100 3. Des Moines 275 C. Cincinnati 300 Total 600 tons Transportation Model Example 2 Problem Definition and Data Problem: How many tons of wheat to transport from each grain elevator to each mill on a monthly basis in order to minimize the total cost of transportation? Data:

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Transportation Model Example 2 Model Formulation

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Minimize Z = $6x 1A + 8x 1B + 10x 1C + 7x 2A + 11x 2B + 11x 2C + 4x 3A + 5x 3B + 12x 3C subject to: x 1A + x 1B + x 1C = 150 x 2A + x 2B + x 2C = 175 x 3A + x 3B + x 3C = 275 x 1A + x 2A + x 3A = 200 x 1B + x 2B + x 3B = 100 x 1C + x 2C + x 3C = 300 x ij 0 x ij = tons of wheat from each grain elevator, i, i = 1, 2, 3, to each mill j, j = A,B,C Transportation Model Example 2 Model Formulation

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Transportation Model- Example 3 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. Carlton Pharmateuticals

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Data –Unit shipping cost, supply, and demand Assumptions –Unit shipping cost is constant. –All the shipping occurs simultaneously. –The only transportation considered is between sources and destinations. –Total supply equals total demand.

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NETWORK REPRESENTATION 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 40 15 20 28

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The Mathematical Model –The structure of the model is: Minimize ST [Amount shipped from a source] = [Supply at that source] [Amount received at a destination] = [Demand at that destination] –Decision variables X ij = amount shipped from source i to destination j. where: i=1 (Cleveland), 2 (Detroit), 3 (Greensboro) j=1 (Boston), 2 (Richmond), 3 (Atlanta), 4(St.Louis)

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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

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The complete mathematical model ==============

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Excel Optimal Solution

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Range of optimality WINQSB Sensitivity Analysis If this path is used, the total cost will increase by $5 per unit shipped along it

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Range of feasibility Shadow prices for warehouses - the cost incurred from having 1 extra case of vaccine demanded at the warehouse Shadow prices for plants - the cost savings realized for each extra case of vaccine available at the plant

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Interpreting sensitivity analysis results –Reduced costs The amount of transportation cost reduction per unit that makes a given route economically attractive. If the route is forced to be used under the current cost structure, for each item shipped along it, the total cost increases by an amount equal to the reduced cost. –Shadow prices For the plants, shadow prices convey the cost savings realized for each extra case of vaccine available at plant. For the warehouses, shadow prices convey the cost incurred from having an extra case demanded at the warehouse.

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Transportation Model- Example 4 Transportation Model- Example 4 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.

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Data: –Initial inventory = 200 pairs –Ending inventory required =1200 pairs –Production capacity for the next quarter is shown on the table –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)

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Analysis of demand: –Net demand to satisfy 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 demandIn house inventory 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

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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

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Source: July production in R/T Destination: July‘s demand. Source: Aug. production in O/T Destination: Sept.’s demand 32+(.03)(32)=$32.96 Unit cost= $25 (production) Unit cost =Production+one month holding cost

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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 Inventory +Production -Demand

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Unbalanced Transportation Problems If supplies are not equal to demands, an unbalanced transportation model exists. In an unbalanced transportation model, supply or demand constraints need to be modified. There are two possible scenarios: (1)Total supply exceeds total requirement. (2)Total supply is less than total requirement.

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Total Supply Exceeds Total Requirement Total flow out of Des Moines ( X DA + X DB + X DC ) should be permitted to be smaller than total supply (100). The constraint should be written as -X DA - X DB - X DC >= -100 (Des Moines capacity) -X EA - X EB - X EC >= -300 (Evansville capacity) -X FA - X FB - X FC >= -300(Fort Lauderdale capacity) X DA + X DB + X DC <= 100 X EA + X EB + X EC <= 100 X FA + X FB + X FC <= 100

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Total Supply Less Than Total Requirement Total flow in to Albuquerque (that is, X DA + X EA + X FA ) should be permitted to be smaller than total demand (namely, 300). This warehouse should be written as: X DA + X EA + X FA <= 300 (Albuquerque demand) X DB + X EB + X FB <= 200 (Boston demand) X DC + X EC + X FC <= 200 (Cleveland demand)

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Develop the linear programming model and solve using Excel: Transportation Example 5: Formulation

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Minimize Z = $8x 1A + 5x 1B + 6x 1C + 15x 2A + 10x 2B + 12x 2C + 3x 3A + 9x 3B + 10x 3C subject to: x 1A + x 1B + x 1C = 120 x 2A + x 2B + x 2C = 80 x 3A + x 3B + x 3C = 80 x 1A + x 2A + x 3A 150 x 1B + x 2B + x 3B 70 x 1C + x 2C + x 3C 100 x ij 0 Transportation Example 5: Formulation

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Transportation Model-Example 6 Hardgrave Machine Company - New Factory Location Produces computer components at its plants in Cincinnati, Kansas City, and Pittsburgh. Plants not able to keep up with demand for orders at four warehouses in Detroit, Houston, New York, and Los Angeles. Firm has decided to build a new plant to expand its productive capacity. Two sites being considered: –Seattle, Washington and –Birmingham, Alabama. Both cities attractive in terms: labor supply, municipal services, and ease of factory financing.

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Transportation Model-Example 6 Hardgrave Machine Company: Demand Supply Data and Production Costs

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Transportation Model-Example 6 Hardgrave Machine Company: Shipping Costs

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Transshipment Model

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In a Transshipment Problem flows can occur both out of and into the same node in three ways: 1. If total flow into a node is less than total flow out from node, node represents a net creator of goods (a supply point). - Flow balance equation will have a negative right hand side (RHS) value. 2. If total flow into a node exceeds total flow out from node, node represents a net consumer of goods, (a demand point). - Flow balance equation will have a positive RHS value. 3. If total flow into a node is equal to total flow out from node, node represents a pure transshipment point. - Flow balance equation will have a zero RHS value.

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It is an extension of the transportation model. Intermediate transshipment points are added between the sources and destinations. Items may be transported from: Sources through transshipment points to destinations One source to another One transshipment point to another One destination to another Directly from sources to to destinations Some combination of these The Transshipment Model Characteristics

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Executive Furniture Corporation – Revisited Assume it is possible for Executive Furniture to ship desks from Evansville factory to its three warehouses at very low unit shipping costs. Consider shipping all desks produced at other two factories (Des Moines and Fort Lauderdale) to Evansville. Consider using a new shipping company to move desks from Evansville to all its warehouses.

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Executive Furniture Corporation - Revisited Revised unit shipping costs are shown here. Note Evansville factory shows up in both the “From” and “To” entries.

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LP Model for theTransshipment Problem Two new additional decision variables for new shipping routes are to be added. X DE = number of desks shipped from Des Moines to Evansville X FE = number of desks shipped from Fort Lauderdale to Evansville Objective Function: minimize total shipping costs = 5X DA + 4X DB + 3X DC + 2X DE + 3X EA + 2X EB + +1X EC + 9X FA + 7X FB + 5X FC + 3X FE Executive Furniture Corporation Revisited

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Relevant flow balance equations written as: (0) - (X DA + X DB + X DC + X DE ) = -100 (Des Moines capacity) (0) - (X FA + X FB + X FC + X FE ) = -300 (Fort Lauderdale capacity) Supplies have been expressed as negative numbers in the RHS. Net flow at Evansville = (Total flow in) - (Total flow out) = (X DE + X FE ) - (X EA + X EB + X EC ) Net flow equals total number of desks produced (the supply) at Evansville. Net flow at Evansville = (X DE + X FE ) - (X EA + X EB + X EC ) = -300 No change in demand constraints for warehouse requirements: X DA + X EA + X FA = 300 (Albuquerque demand) X DB + X EB + X FB = 200 (Boston demand) X DC + X EC + X FC = 200 (Cleveland demand) LP Model for theTransshipment Problem

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Extension of the transportation model in which intermediate transshipment points are added between sources and destinations. An example of a transshipment point is a distribution center or warehouse located between plants and stores Data: Transshipment Model Example 2 Problem Definition and Data

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Transshipment Model Example 2 Problem Definition and Data

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Minimize Z = $16x 13 + 10x 14 + 12x 15 + 15x 23 + 14x 24 + 17x 25 + 6x 36 + 8x 37 + 10x 38 + 7x 46 + 11x 47 + 11x 48 + 4x 56 + 5x 57 + x 58 subject to: x 13 + x 14 + x 15 = 300 x 23 + x 24 + x 25 = 300 x 36 + x 46 + x 56 = 200 x 37 + x 47 + x 57 = 100 x 38 + x 48 + x 58 = 300 x 13 + x 23 - x 36 - x 37 - x 38 = 0 x 14 + x 24 - x 46 - x 47 - x 48 = 0 x 15 + x 25 - x 56 - x 57 - x 58 = 0 x ij 0 Transshipment Model Example 2 Model Formulation

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Assignment Model

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The Assignment Model 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 jobs so that each worker is assigned a job, and each job is performed.

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It is a special form of linear programming models similar to the transportation model. Supply at each source and demand at each destination is limited to one unit. In a balanced model supply equals demand. In an unbalanced model supply is not equal to demand. The Assignment Model Characteristics

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–The number of workers is equal to the number of jobs. –Given a balanced problem, each worker is assigned exactly once, and each job is performed by exactly one worker. –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. The Assignment Model Assumptions

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Fix-It Shop Example Received three new rush projects to repair: (1) a radio, (2) a toaster oven, and (3) a broken coffee table. Three workers (each has different talents and abilities). Estimated costs to assign each worker to each of the three projects. Assignment Model Example 1

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Fix-It Shop Rows denote people or objects to be assigned, and columns denote tasks or jobs assigned. Numbers in table are costs associated with each particular assignment.

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Assignment Model Example 1 Owner's objective is to assign three projects to workers in a way that result is lowest total cost. Fix-It Shop: Assignment Alternatives and Costs

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Assignment Model Example 1 Owner's objective is to assign three projects to workers in a way that results in lowest total cost. Fix-It Shop

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Assignment Model Example 1 Formulate LP model - X ij = “Flow” on arc from node denoting worker i to node denoting project j. Solution value will equal 1 if worker i is assigned to project j : i = A (for Adams), B (for Brown), or C (for Cooper) j = 1 (for project 1), 2 (for project 2), or 3 (for project 3) Objective Function: minimize total assignment cost = 11X A1 + 14X A2 + 6X A3 + 8X B1 + 10X B2 + 11X B3 + + 9X C1 + 12X C2 + 7X C3 Fix-It Shop

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Assignment Model Example 1 Fix-It Shop Constraints expressed using standard flow balance equations are as follows: -X A1 - X A2 - X A3 = -1 (Adams availability) -X B1 - X B2 - X B3 = -1(Brown availability) -X C1 - X C2 - X C3 = -1(Cooper availability) X A1 + X B1 + X C1 = 1(Project 1 requirement) X A2 + X B2 + X C2 = 1(Project 2 requirement) X A3 + X B3 + X C3 = 1(Project 3 requirement)

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Assignment Model- Example 2 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.

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Data: Travel time in minutes from assembly lines to inspection areas. Assignment Model- Example 2

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Assignment Model Example 2: Network Representation (3 of 3) 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

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Computer solutions –A complete enumeration is not efficient 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. Special cases –A worker is unable to perform a particular job. –A worker can be assigned to more than one job. –A maximization assignment problem.

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Assignment Model Example 3 Problem Definition and Data Problem: Assign four teams of officials to four games in a way that will minimize total distance traveled by the officials. Supply is always one team of officials, demand is for only one team of officials at each game. Data:

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Minimize Z = 210x AR + 90x AA + 180x AD + 160x AC + 100x BR + 70x BA + 130x BD + 200x BC + 175x CR + 105x CA + 140x CD + 170x CC + 80x DR + 65x DA + 105x DD + 120x DC subject to: x AR + x AA + x AD + x AC = 1 x ij 0 x BR + x BA + x BD + x BC = 1 x CR + x CA + x CD + x CC = 1 x DR + x DA + x DD + x DC = 1 x AR + x BR + x CR + x DR = 1 x AA + x BA + x CA + x DA = 1 x AD + x BD + x CD + x DD = 1 x AC + x BC + x CC + x DC = 1 Assignment Model Example 3 Model Formulation

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Summary Three network flow models have been presented: 1.Transportation model 1.Transportation model deals with distribution of goods from several supplier to a number of demand points. 2.Transshipment model 2.Transshipment model includes points that permit goods to flow both in and out of them. 3.Assignment model 3.Assignment model deals with determining the most efficient assignment of issues such as people to projects.

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