 Consists of nodes representing a set of origins and a set of destinations.  An arc is used to represent the route from each origins to each destinations.  Each origin has a supply and each destination has a demand.  Objective: To determine the optimal amount to ship from each origin to each destination

Network flow problems  Transportation Problem  Assignment Problem  Transshipment Problem

 Transportation Problem  Problems of distributing goods and services from several supply location to several demand locations  Supply locations are called as Origin  Demand locations are called as Destination  Quantity of goods at origin are limited  Quantity of goods at destinations are known

 Transportation Problem  Each origin and destinations are represented by Circles called as nodes  Each origin and destinations are connected by arc  Each node requires one constraint  Each arc requires one variable  The sum of variables corresponding to arcs from an origin node must be less than or equal to origin supply. (Rule 3)  The sum of variables corresponding to arcs into an destination node must equal to destination ‘s demand (Rule 4)

 The transportation problem seeks to minimize the total shipping costs of transporting goods from m origins (each with a supply s i ) to n destinations (each with a demand d j ), when the unit shipping cost from an origin, i, to a destination, j, is c ij.  The network representation for a transportation problem with two sources and three destinations is given on the next slide.

 Network Representation c 11 c 12 c 13 c 21 c 22 c 23 d1d1d1d1 d2d2d2d2 d3d3d3d3 s1s1s1s1 s2s2 SOURCESDESTINATIONS

 LP Formulation The LP formulation in terms of the amounts shipped from the origins to the destinations, x ij, can be written as: Min  c ij x ij i j s.t.  x ij < s i for each origin i j  x ij = d j for each destination j i x ij > 0 for all i and j

 Distribution of goods from three plants to four distributions Supply OriginPlantThree Months Production Capacity (units) 1Cleveland5000 2Bedford6000 3York2500,Total=13,500 DestinationDistribution CentreThree-Months Demand Forecast (Units) 1Boston6000 2Chicago4000 3St. Louis2000 4Lexington1500,Total=13,500 Demands

Objectives To determine the routes to be used and quantity to be shipped from each origin to demand route that will provide minimum total transportation cost. Construct a network graph Connect each origin with the destination with arcs representing the routes between origin and the destinations. 12 Possible Routes

7 Cleveland Bedford York Boston Chicago S.Louis Lexington Transportation Cost per unit Demands Supplies

X ij number of units shipped from origin I to destination j X 11 number of units shipped from origin (Cleveland) to destination 1 (boston) X 12 number of units shipped from origin (Cleveland) to destination 2 (Chicago) Cost per unit OriginBostonChicagoSt.LouisLexington Cleveland3276 Bedford7523 York2545 Objective Function=sum of cost from each source to destinations

 Transportation cost shipped from Cleveland  3x 11 +2x 12 +7x 13 +6x 14  Transportation cost shipped from Bedford  7x 21 +5x 22 +2x 23 +3x 24  Transportation cost shipped from York  2x 31 +5x 32 +4x 33 +5x 34  How many total variables and constraint??  What are supply and demand constraints???

 Transportation shipped from Cleveland  X 11 +x 12 +x 13 +x 14 <= 5000  Transportation shipped from Bedford  X 21 +x 22 +x 23 +x 24 <=6000  Transportation shipped from York  X 31 +x 32 +x 33 +x 34 <=2500

 Transportation To boston  X 11 +x 21 +x 31 = 6000  Transportation to Chicago  X 12 +x 22 +x 32 =4000  Transportation to St.Louis  X 13 +x 23 +x 33 =2500  Transportation to Lexington  X 14 +x 24 +x 34 =1500 Objective function????

 Objective function Min 3x 11 +2x 12 +7x 13 +6x 14 +7x 21 +5x 22 +2x 23 +3x 24 +2x 31 +5x 32 +4x 33 +5x 34 S.T X 11 +x 12 +x 13 +x 14 <= 5000 X 21 +x 22 +x 23 +x 24 <=6000 X 31 +x 32 +x 33 +x 34 <=2500 X 11 +x 21 +x 31 = 6000 X 12 +x 22 +x 32 =4000 X 13 +x 23 +x33=2500 X 14 +x 24 +x 34 =1500

VariableValueReduced Cost X X X X X X X X X X X X Minimum total transportation cost? Units shipped=3500;cost per unit=3 Total cost from Cleveland to Boston??

X ij =number of units shipped from origin I to destination j C ij = cost per unit shipping from origin I to destination j S i = supply or capacity in units at origin I D j = demand in units at destinations j

Assigning jobs to machine Agents to tasks Sales personnel to sales territory One to one assignment, i.e. one agent is assigned to one and only one task

 An assignment problem seeks to minimize the total cost assignment of m workers to m jobs, given that the cost of worker i performing job j is c ij.  It assumes all workers are assigned and each job is performed.  An assignment problem is a special case of a transportation problem in which all supplies and all demands are equal to 1; hence assignment problems may be solved as linear programs.

 Network Representation c 11 c 12 c 13 c 21 c 22 c 23 c 31 c 32 c 33 AGENTSTASKS

 LP Formulation Min  c ij x ij i j s.t.  x ij = 1 for each agent i j  x ij = 1 for each task j i x ij = 0 or 1 for all i and j  Note: A modification to the right-hand side of the first constraint set can be made if a worker is permitted to work more than 1 job.

n LP Formulation Special Cases Number of agents exceeds the number of tasks: Number of agents exceeds the number of tasks:  x ij < 1 for each agent i j Number of tasks exceeds the number of agents: Number of tasks exceeds the number of agents: Add enough dummy agents to equalize the number of agents and the number of tasks. The objective function coefficients for these new variable would be zero. Assignment Problem

Project LeaderClient 1Client 2Client 3 Terry10159 Carle9185 Jack6143 Estimated project completion Time

7 Terry Carle Jack Client 1 Client 2 Client Completion Time in Days Demands Supplies Number of Constraints=6;Number of variables=9

 Days requires for Terry’s assignment  10x x 12 +9x 13  Days required for Carle ‘s assignment  9x x 22 +5x 23  Days required for Jack ‘s assignment  6x x 32 +3x 33  Objective Function  Min 10x x 12 +9x x x 12 +9x x x 32 +3x 33

 Each project leader can be assigned to at most one client.  X 11 +x 12 +x 13 <=1 ;Terry ‘s assignment  X 21 +x 21 +x 23 <=1; Carle ‘s assignment  X 31 +X 21 +x 31 <=1 ;Jack assignment  Each Client must have at least one leader  X 11 +X 21 +X 31 =1; Client 1  X 12 +X 22 +X 32 =1; Client 2  X 13 +X 23 +X 33 =1 ; Client 3

VariableValueReduced Cost X X X X X X X X X Terry is assigned to client2;x12=1 Carle is assigned to client3;x23=1 Jack is assigned to client 1 x31=1 Total completion time required is 26 days

A contractor pays his subcontractors a fixed fee plus mileage for work performed. On a given day the contractor is faced with three electrical jobs associated with various projects. Given below are the distances between the subcontractors and the projects. Projects Subcontractor A B C Westside Federated Goliath Universal How should the contractors be assigned to minimize total costs? Example: Hungry Owner

n Network Representation West. CC BB AA Univ.Univ. Gol.Gol. Fed. Fed. Projects Subcontractors

Example: Hungry Owner n Linear Programming Formulation Min 50 x x x x x x x x x x x x 43 s.t. x 11 + x 12 + x 13 < 1 x 21 + x 22 + x 23 < 1 x 31 + x 32 + x 33 < 1 x 41 + x 42 + x 43 < 1 x 11 + x 21 + x 31 + x 41 = 1 x 12 + x 22 + x 32 + x 42 = 1 x 13 + x 23 + x 33 + x 43 = 1 x ij = 0 or 1 for all i and j Agents Tasks

 Step 1: For each row, subtract the minimum number in that row from all numbers in that row.  Step 2: For each column, subtract the minimum number in that column from all numbers in that column.  Step 3: Draw the minimum number of lines to cover all zeroes. If this number = m, STOP -- an assignment can be made.  Step 4: Subtract d (the minimum uncovered number) from uncovered numbers. Add d to numbers covered by two lines. Numbers covered by one line remain the same. Then, GO TO STEP 3.

 Finding the Minimum Number of Lines and Determining the Optimal Solution  Step 1: Find a row or column with only one unlined zero and circle it. (If all rows/columns have two or more unlined zeroes choose an arbitrary zero.)  Step 2: If the circle is in a row with one zero, draw a line through its column. If the circle is in a column with one zero, draw a line through its row. One approach, when all rows and columns have two or more zeroes, is to draw a line through one with the most zeroes, breaking ties arbitrarily.  Step 3: Repeat step 2 until all circles are lined. If this minimum number of lines equals m, the circles provide the optimal assignment.

 Initial Tableau Setup Since the Hungarian algorithm requires that there be the same number of rows as columns, add a Dummy column so that the first tableau is: A B C Dummy Westside Federated Goliath Universal

 Step 1: Subtract minimum number in each row from all numbers in that row. Since each row has a zero, we would simply generate the same matrix above.  Step 2: Subtract the minimum number in each column from all numbers in the column. For A it is 25, for B it is 25, for C it is 14, for Dummy it is 0. This yields: A B C Dummy Westside Federated Goliath Universal

 Step 3: Draw the minimum number of lines to cover all zeroes. Although one can "eyeball" this minimum, use the following algorithm. If a "remaining" row has only one zero, draw a line through the column. If a remaining column has only one zero in it, draw a line through the row. A B C Dummy Westside Federated Goliath Universal

 Step 3: Draw the minimum number of lines to cover all zeroes. A B C Dummy Westside Federated Goliath Universal

 Step 4: The minimum uncovered number is 1. Subtract 1 from uncovered numbers. Add 1 to numbers covered by two lines. This gives: A B C Dummy Westside Federated Goliath Universal

 Step 3: The minimum number of lines to cover all 0's is four. Thus, there is a minimum-cost assignment of 0's with this tableau. The optimal assignment is: Subcontractor Project Distance Westside C 16 Federated A 28 Goliath (unassigned) Universal B 25 Total Distance = 69 miles

 Transshipment problems are transportation problems in which a shipment may move through intermediate nodes (transshipment nodes)before reaching a particular destination node.  Transshipment problems can be converted to larger transportation problems and solved by a special transportation program.  Transshipment problems can also be solved by general purpose linear programming codes.  The network representation for a transshipment problem with two sources, three intermediate nodes, and two destinations is shown on the next slide.

 Network Representation c 13 c 14 c 23 c 24 c 25 c 15 s1s1s1s1 c 36 c 37 c 46 c 47 c 56 c 57 d1d1d1d1 d2d2d2d2 INTERMEDIATE NODES NODESSOURCESDESTINATIONS s2s2s2s2 Demand Supply

 Contains three types of nodes: origin,transhipment node and destination nodes  For origin nodes sum of shipments out minus The sum of shipment in must be less than or equal to origin supply.  For destination nodes sum of shipments out minus The sum of shipment in must be equal to demand  For transhipment nodes the sum of shipments out must equal to sum of shipments in.

1 Denver 2 Atlanta 4 Louis ville 5 Detroit 6 Miami 7 Dallas 10 Distribution RoutesDemands Supplies Number of Constraints=8;Number of variables=12 3 Kansas 8 New Orlean s

 Origin Nodes ??  X 13 +X 14 <=600 (Denver)  X 23 +x 24 <=400 (Atlanta)  For transhipment Nodes  x 35 +x 36 +x 37 +x 38 =x 13 + x 23 ( Node 3;units in=units out )  X 45 +X 46 +X 47 +X 48 =X 14 +X 24 (Node 4;units in= units out)  For Destination nodes  X35+x45=200  X36+x46=150  X37+x47=350  X38+x48=300

Ware House PlantKansas city (3)Lousville (4) Denver (1)23 Atlanta (2)31 Retail Outlet WarehouseDetroit (5)Miami (6) Dallas (7)New Orleans (8) Kansas City2636 Louisville4465 2x 13 +3x 14 +3x 23 +x 24 +2x 35 +6x 36 +3x 37 +6x 38 +4x 45 +4x 46 +6x 47 +5x 48 +4x 28 +x 78 Obj function

VariableValueReduced Costs X X X X X X X X X X X X Value of objective function?

Thomas Industries and Washburn Corporation supply three firms (Zrox, Hewes, Rockwright) with customized shelving for its offices. They both order shelving from the same two manufacturers, Arnold Manufacturers and Supershelf, Inc. Currently weekly demands by the users are 50 for Zrox, 60 for Hewes, and 40 for Rockwright. Both Arnold and Supershelf can supply at most 75 units to its customers. Additional data is shown on the next slide.

Example: Transshipping Because of long standing contracts based on past orders, unit costs from the manufacturers to the suppliers are: Thomas Washburn Arnold 5 8 Supershelf 7 4 various locations are: The costs to install the shelving at the various locations are: Zrox Hewes Rockwright Thomas Washburn 3 4 4

 Network Representation ARNOLD WASH BURN ZROX HEWES Arnold SuperShelf Hewes Zrox Thomas Wash-Burn Rock-Wright

 Linear Programming Formulation  Decision Variables Defined x ij = amount shipped from manufacturer i to supplier j x jk = amount shipped from supplier j to customer k where i = 1 (Arnold), 2 (Supershelf) j = 3 (Thomas), 4 (Washburn) k = 5 (Zrox), 6 (Hewes), 7 (Rockwright)  Objective Function Defined Minimize Overall Shipping Costs: Min 5x x x x x x x x x x 47

 Constraints Defined Amount Out of Arnold: x 13 + x 14 < 75 Amount Out of Supershelf: x 23 + x 24 < 75 Amount Through Thomas: x 13 + x 23 - x 35 - x 36 - x 37 = 0 Amount Through Washburn: x 14 + x 24 - x 45 - x 46 - x 47 = 0 Amount Into Zrox: x 35 + x 45 = 50 Amount Into Hewes: x 36 + x 46 = 60 Amount Into Rockwright: x 37 + x 47 = 40 Non-negativity of Variables: x ij > 0, for all i and j.

Example: Transshipping n Optimal Solution (from The Management Scientist ) Objective Function Value = Objective Function Value = Variable Value Reduced Costs Variable Value Reduced Costs X X X X X X X X X X X X X X X X X X X X

Example: Transshipping n Optimal Solution ARNOLD WASH BURN ZROX HEWES Arnold SuperShelf Hewes Zrox Thomas Wash-Burn Rock-Wright