4. Linear Programming

Size: px

Start display at page:

Download "4. Linear Programming"

Franklin Quinn
5 years ago
Views:

1 /9/08 Systems Analysis in Construction CB Construction & Building Engineering Department- AASTMT by A h m e d E l h a k e e m & M o h a m e d S a i e d. Linear Programming Optimization Network Models - Shortest Route Problem - Minimal Spanning Tree Problem - Transportation Problem - Assignment Problem

2 /9/08 Introduction to Network Models A network model is one which can be represented by a set of nodes, a set of arcs, and functions (e.g. costs, supplies, demands, etc.) associated with the arcs and/or nodes. Shortest Route, Minimal Spanning Tree, Transportation, Assignment, and Ttransshipment problems as well as the PERT/CPM problems are all examples of network problems. 8 Shortest-Route Problem 9 The shortest-route problem is concerned with finding the shortest path in a network from one node (or set of nodes) to another node (or set of nodes). If all arcs in the network have nonnegative values then a labeling algorithm can be used to find the shortest paths from a particular node to all other nodes in the network. The criterion to be minimized in the shortest-route problem is not limited to distance even though the term "shortest" is used in describing the procedure. Other criteria include time and cost. (Neither time nor cost are necessarily linearly related to distance.)

3 /9/08 Shortest-Route Problem Note: We use the notation [ ] to represent a permanent label and ( ) to represent a tentative label. Step : Assign node the permanent label [0,S]. The first number is the distance from node ; the second number is the preceding node. Since node has no preceding node, it is labeled S for the starting node. 0 Step : Compute tentative labels, (d,n), for the nodes that can be reached directly from node. d = the direct distance from node to the node in question - this is called the distance value. n indicates the preceding node on the route from node - this is called the preceding node value. Shortest-Route Problem Step : Identify the tentatively labeled node with the smallest distance value. Suppose it is node k. Node k is now permanently labeled (using [, ] brackets). If all nodes are permanently labeled, GO TO STEP. Step : Consider all nodes without permanent labels that can be reached directly from the node k identified in Step. For each, calculate the quantity t, where t = (arc distance from node k to node i) + (distance value at node k).

4 /9/08 Shortest-Route Problem Step : (continued) If the non-permanently labeled node has a tentative label, compare t with the current distance value at the tentatively labeled node in question. If t < distance value of the tentatively labeled node, replace the tentative label with (t,k). If t > distance value of the tentatively labeled node, keep the current tentative label. If the non-permanently labeled node does not have a tentative label, create a tentative label of (t,k) for the node in question. In either case, now GO TO STEP. Shortest-Route Problem Step : The permanent labels identify the shortest distance from node to each node as well as the preceding node on the shortest route. The shortest route to a given node can be found by working backwards by starting at the given node and moving to its preceding node. Continuing this procedure from the preceding node will provide the shortest route from node to the node in question.

5 /9/08 Example Find the Shortest Route From Node to All Other Nodes in the Network: 8 Example Iteration Step : Assign Node the permanent label [0,S]. Step : Since Nodes,, and are directly connected to Node, assign the tentative labels of (,) to Node ; (,) to Node ; and (,) to Node.

6 /9/08 Example Tentative Labels Shown (,) [0,S] (,) (,) 8 Example Iteration Step : Node is the tentatively labeled node with the smallest distance (), and hence becomes the new permanently labeled node. [0,S] [,] Permanent Label Shown (,) (,) 8

7 /9/08 Example Iteration Step : For each node with a tentative label which is connected to Node by just one arc, compute the sum of its arc length plus the distance value of Node (which is ). Node : + = (not smaller than current label; do not change.) Node : + = 9 (assign tentative label to Node of (9,) since node had no label.) 8 Example Iteration, Step Results [,] (9,) [0,S] (,) (,) 8 9

8 /9/08 Example Iteration Step : Node has the smallest tentative label distance (). It now becomes the new permanently labeled node. [0,S] [,] (9,) (,) 0 [,] 8 Example Iteration, Step Results [,] (9,) [0,S] (,) [,] 8 8

9 /9/08 Example Iteration Step : For each node with a tentative label which is connected to node by just one arc, compute the sum of its arc length plus the distance value of node (which is ). Node : + = (replace the tentative label of node by (,) since <, the current distance.) Node : 8 + = (assign tentative label to node of (,) since node had no label.) Example Iteration, Step Results [,] (9,) [0,S] (,) [,] 8 (,) 9

10 /9/08 Example Iteration Step : Node has the smallest tentative distance label (). It now becomes the new permanently labeled node. [,] (9,) [0,S] [,] [,] 8 (,) Example Iteration Step : For each node with a tentative label which is connected to node by just one arc, compute the sum of its arc length plus the distance to node (which is ). Node : + = 8 (replace the tentative label of node with (8,) since 8 < 9, the current distance) Node : + = (replace the tentative label of node with (,) since <, the current distance) 0

11 /9/08 Example Iteration, Step Results [,] (8,) [0,S] [,] [,] 8 (,) Example Iteration Step : Node has the smallest tentative label distance (8). It now becomes the new permanently labeled node. [,] [8,] [0,S] [,] [,] 8 (,)

12 /9/08 Example Iteration Step : For each node with a tentative label which is connected to node by just one arc, compute the sum of its arc length plus the distance value of node (which is 8). Node : + 8 = (Replace the tentative label with (,) since <, the current distance.) Node : + 8 = (Assign tentative label to node of (,) since node had no label.) 8 Example Iteration, Step Results [0,S] [,] [8,] [,] (,) 9 [,] 8 (,)

13 /9/08 Example Iteration Step : Node has the smallest tentative label distance (). It now becomes the new permanently labeled node. [0,S] [,] [8,] [,] (,) 0 [,] 8 [,] Example Iteration Step : For each node with a tentative label which is connected to Node by just one arc, compute the sum of its arc length plus the distance value of Node (which is ). Node : + = (replace the tentative label with (,) since <, the current distance.)

14 /9/08 Example Iteration, Step Results [0,S] [,] [8,] [,] (,) [,] 8 [,] Example Iteration Step : Node becomes permanently labeled, and hence all nodes are now permanently labeled. Thus proceed to summarize in Step. Step : Summarize by tracing the shortest routes backwards through the permanent labels.

15 /9/08 Example Solution Summary from Node the distance to Node Minimum Distance Shortest Route Longest Route Problem?

16 /9/08 Minimal Spanning Tree Problem The minimal spanning tree problem is concerned with finding the minimum length (cost) sub network that connects all nodes in a network 8 Minimal Spanning Tree Problem Start from any node then connect to this node another one with minimum length. Let us start from node 8

17 /9/08 Minimal Spanning Tree Problem It is possible to connect nodes,,,, & to node select node with minimum length The next step is to add new node to the connected ones ( &) 8 8 Minimal Spanning Tree Problem Link to 9 8

18 /9/08 Minimal Spanning Tree Problem Link to OR Link to 80 8 Minimal Spanning Tree Problem Link to 8 8 8

19 /9/08 Minimal Spanning Tree Problem Link to 8 8 Minimal Spanning Tree Problem Link to 8 8 9

20 /9/08 Minimal Spanning Tree Problem The minimal Spanning Tree is shown with length (cost) = 8 Transportation and Assignment Problems Transportation Problem Network Representation General LP Formulation Assignment Problem Network Representation General LP Formulation 8 0

21 /9/08 Transportation Problem 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 as follows. 8 Transportation Problem Network Representation d s c c c c c d s c d Sources Destinations 8

/9/08 Transportation Problem 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 i x ij > 0 for each destination j for all i and j 88 Illustration: Acme Block Co.

22 /9/08 Transportation Problem 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 i x ij > 0 for each destination j for all i and j 88 Illustration: Acme Block Co. Acme Block Company has orders for 80 tons of concrete blocks at three suburban locations as follows: Northwood tons, Westwood tons, and Eastwood 0 tons. Acme has two plants, each of which can produce 0 tons per week. Delivery cost per ton from each plant to each suburban location is shown on the next slide. 89 How should end of week shipments be made to fill the above orders?

23 /9/08 Illustration: Acme Block Co. Delivery Cost Per Ton Northwood Westwood Eastwood Plant 0 0 Plant Illustration: Acme Block Co. Optimal Solution Northwood Westwood Eastwood Plant 0 Plant Total Cost = $,90 9

24 /9/08 Transportation Problems Transportation Simplex Method: A Special-Purpose Solution Procedure 9 Transportation Simplex Method To solve the transportation problem by its special purpose algorithm, the sum of the supplies at the origins must equal the sum of the demands at the destinations. If the total supply is greater than the total demand, a dummy destination is added with demand equal to the excess supply, and shipping costs from all origins are zero. Similarly, if total supply is less than total demand, a dummy origin is added. 9 When solving a transportation problem by its special purpose algorithm, unacceptable shipping routes are given a cost of +M (a large number).

25 /9/08 Transportation Simplex Method A transportation tableau is given below. Each cell represents a shipping route (which is an arc on the network and a decision variable in the LP formulation), and the unit shipping costs are given in an upper right hand box in the cell. D D D Supply S S Demand 0 Transportation Simplex Method The transportation problem is solved in two phases: Phase I Finding an Initial Feasible Solution Phase II Iterating to the optimal solution In Phase I, the North West, Minimum-Cost Cell, Vogal s Approxmation can be used to establish an initial feasible solution without doing numerous iterations of the simplex method. In Phase II, the Stepping Stone Method, using the MODI method for evaluating the reduced costs may be used to move from the initial feasible solution to the optimal one. 9

26 /9/08 Transportation Simplex Method Phase I - Minimum-Cost Method Step : Select the cell with the least cost. Assign to this cell the minimum of its remaining row supply or remaining column demand. Step : Decrease the row and column availabilities by this amount and remove from consideration all other cells in the row or column with zero availability/demand. (If both are simultaneously reduced to 0, assign an allocation of 0 to any other unoccupied cell in the row or column before deleting both.) GO TO STEP. 9 Transportation Simplex Method Phase II - Stepping Stone Method Step : For each unoccupied cell, calculate the reduced cost by the MODI method described below. Select the unoccupied cell with the most negative reduced cost. (For maximization problems select the unoccupied cell with the largest reduced cost.) If none, STOP. Step : For this unoccupied cell generate a stepping stone path by forming a closed loop with this cell and occupied cells by drawing connecting alternating horizontal and vertical lines between them. Determine the minimum allocation where a subtraction is to be made along this path. 9

27 /9/08 Transportation Simplex Method Phase II - Stepping Stone Method (continued) Step : Add this allocation to all cells where additions are to be made, and subtract this allocation to all cells where subtractions are to be made along the stepping stone path. (Note: An occupied cell on the stepping stone path now becomes 0 (unoccupied). If more than one cell becomes 0, make only one unoccupied; make the others occupied with 0's.) GO TO STEP. 98 Transportation Simplex Method MODI Method (for obtaining reduced costs) Associate a number, u i, with each row and v j with each column. Step : Set u = 0. Step : Calculate the remaining u 's i and v 's j by solving the relationship c ij = u i + v j for occupied cells. Step : For unoccupied cells (i,j), the reduced cost = c ij - u i - v. j 99

28 /9/08 Example : Acme Block Co. Acme Block Company has orders for 80 tons of concrete blocks at three suburban locations as follows: Northwood tons, Westwood tons, and Eastwood 0 tons. Acme has two plants, each of which can produce 0 tons per week. Delivery cost per ton from each plant to each suburban location is shown on the next slide. 00 How should end of week shipments be made to fill the above orders? Example : Acme Block Co. Delivery Cost Per Ton Northwood Westwood Eastwood Plant 0 0 Plant

29 /9/08 Example : Acme Block Co. Initial Transportation Tableau Since total supply = 00 and total demand = 80, a dummy destination is created with demand of 0 and 0 unit costs. (North-West Corner) Northwood Westwood Eastwood Dummy Supply Plant Plant Demand Example : Acme Block Co. Initial Transportation Tableau (Initial Feasible Solution) North West Cost= Northwood Westwood Eastwood Dummy Supply Plant Plant Demand

30 /9/08 Example : Acme Block Co. Initial Transportation Tableau (Initial Feasible Solution) Minimum Cost or Least Cost Cost= Northwood Westwood Eastwood Dummy Supply Plant Plant Demand Example : Acme Block Co. Initial Transportation Tableau (Initial Feasible Solution) Vogal s Approximation Cost= Northwood Westwood Eastwood Dummy Supply Plant Plant Demand

31 /9/08 Example : Acme Block Co. 0 Example : Acme Block Co. Iteration MODI Method. Set u = 0. Since u + v j = c j for occupied cells in row, then v =, v = 0, v = 0.. Since u i + v = c i for occupied cells in column, then u + 0 = 0, hence u = 0.. Since u + v j = c j for occupied cells in row, then 0 + v =, hence v =. 0

32 /9/08 Example : Acme Block Co. Iteration MODI Method (continued) Calculate the reduced costs (circled numbers on the next slide) by c ij - u i + v j. Unoccupied Cell Reduced Cost (,) = 8 (,) = - (,) = Example : Acme Block Co. Iteration Tableau Cost= Northwood Westwood Eastwood Dummy Supply Plant Plant Demand

33 /9/08 Example : Acme Block Co. Iteration Tableau Cost= Northwood Westwood Eastwood Dummy Supply Plant Plant Demand Example : Acme Block Co. Iteration Tableau Cost= Northwood Westwood Eastwood Dummy Supply Plant Plant Demand 0 0

34 /9/08 Example : Acme Block Co. Iteration Tableau Cost= Northwood Westwood Eastwood Dummy Supply Plant Plant Demand 0 0 Example : Acme Block Co. Iteration Tableau Cost= Northwood Westwood Eastwood Dummy Supply Plant Plant Demand 0 0

35 /9/08 Example : Acme Block Co. Iteration Tableau Cost= Northwood Westwood Eastwood Dummy Supply Plant Plant Demand 0 0 Example : Acme Block Co. Optimal Solution From To Amount Cost Plant Northwood 0 Plant Westwood,0 Plant Northwood 0 00 Plant Eastwood 0 0 Total Cost = $,90

36 /9/08 Assignment Problem 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 ; hence assignment problems may be solved as linear programs. The network representation of an assignment problem with three workers and three jobs is shown on the next slide. Assignment Problem Network Representation c c Agents c c c c Tasks c c c

37 /9/08 Assignment Problem LP Formulation Min c ij x ij i j s.t. x ij = for each agent i j x ij = i x ij = 0 or for each task j for all i and j 8 Illustration: Who Does What? An electrical 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. 9 Projects Subcontractor A B C Westside 0 Federated Goliath 0 Universal How should the contractors be assigned to minimize total mileage costs?

38 /9/08 Illustration: Who Does What? Network Representation Subcontractors West. 0 A Projects Fed B Gol. 0 C 0 Univ. Illustration: Who Does What? Linear Programming Formulation Min 0x +x +x +8x +0x +8x +x +x +0x +x +x +x s.t. x +x +x < Agents x +x +x < x +x +x < x +x +x < Tasks x +x +x +x = x +x +x +x = x +x +x +x = x ij = 0 or for all i and j 8

39 /9/08 Illustration: Who Does What? The optimal assignment is: Subcontractor Project Distance Westside C Federated A 8 Goliath (unassigned) Universal B Total Distance = 9 miles Assignment Problems - Vogal s Approximation Method - The Hungarian Method 9

40 /9/08 Hungarian Method The Hungarian method solves minimization assignment problems with m workers and m jobs. Special considerations can include: number of workers does not equal the number of jobs -- add dummy workers or jobs with 0 assignment costs as needed worker i cannot do job j -- assign cij = +M maximization objective -- create an opportunity loss matrix subtracting all profits for each job from the maximum profit for that job before beginning the Hungarian method Hungarian Method Step : For each row, subtract the minimum number in that row from all numbers in that row. Step : For each column, subtract the minimum number in that column from all numbers in that column. Step : Draw the minimum number of lines to cover all zeroes. If this number = m, STOP -- an assignment can be made. Step : Determine the minimum uncovered number (call it d). Subtract d from uncovered numbers. Add d to numbers covered by two lines. Numbers covered by one line remain the same. Then, GO TO STEP. 0

41 /9/08 Hungarian Method Finding the Minimum Number of Lines and Determining the Optimal Solution Step : 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 : 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. Finding the Minimum Number of Lines and Determining the Optimal Solution (continued) Step : Repeat step until all circles are lined. If this minimum number of lines equals m, the circles provide the optimal assignment. Example : Who Does What? An electrical 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 0 Federated Goliath 0 Universal How should the contractors be assigned to minimize total mileage costs?

42 /9/08 Example : Who Does What? 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 0 0 Federated Goliath 0 0 Universal 0 8 Example : Who Does What? 9 Step : 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 : Subtract the minimum number in each column from all numbers in the column. For A it is, for B it is, for C it is, for Dummy it is 0. This yields: A B C Dummy Westside 0 Federated 0 Goliath 0 0 Universal

43 /9/08 Example : Who Does What? Step : Draw the minimum number of lines to cover all zeroes using the following rules. 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 0 Federated 0 Goliath 0 0 Universal Step : The minimum uncovered number is (circled). 0 Example : Who Does What? Step : Subtract from uncovered numbers; add to all numbers covered by two lines. This gives: A B C Dummy Westside Federated 0 Goliath 8 0 Universal 0 0 0

44 /9/08 Example : Who Does What? Step : Draw the minimum number of lines to cover all zeroes. A B C Dummy Westside Federated 0 Goliath 8 0 Universal Step : The minimum uncovered number is (circled). Example : Who Does What? Step : Subtract from uncovered numbers. Add to numbers covered by two lines. This gives: A B C Dummy Westside 9 0 Federated 0 0 Goliath 0 Universal 0 0 0

45 /9/08 Example : Who Does What? The optimal assignment is: Subcontractor Project Distance Westside C Federated A 8 Goliath (unassigned) Universal B Total Distance = 9 miles Example : Who Does What? Vogal s Approximation Method A B C Dummy Westside 0 0 Federated Goliath 0 0 Universal 0

46 /9/08 END of Network Models?

Lecture notes on Transportation and Assignment Problem (BBE (H) QTM paper of Delhi University)

Lecture notes on Transportation and Assignment Problem (BBE (H) QTM paper of Delhi University) Transportation and Assignment Problems The transportation model is a special class of linear programs. It received this name because many of its applications involve determining how to optimally transport