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1 CSE 8374 QM 721N Network Flows: Transportation Problem 1 Slide 1 Slide 2 The Transportation Problem The uncapacitated transportation problem is one of the simplest of the pure network models, provides richness of character that most of the elements of pure optimization can be easily developed. network Transportation Topics: Formulations of the problem Outline Network simplex solution methods Computer implementation issues Dual method for networks CSE 8374 QM 721N Network Flows: Transportation Problem 2 Slide 3 Slide 4 Example Graphical Formulation Elements of the Figure Nodes circles containing identifying labels representing 1. or sinks of flow units. sources The set of m nodes on the left are called origins and the n on the right are destinations. nodes Associated with each origin i is ai flow units of supply and a potential (dual variable) Ri; node Each destination j has bj units of demand and a potential of Kj. Directed arcs lines with arrowheads connecting node 2. routes by which flow can move between pairs representing Associated with each arc (i; j)is a from-node i, a to-node nodes. a unit cost cij, and an unknown amount oowxij. j, +100 O1 D O2 D O3 D3-200

2 CSE 8374 QM 721N Network Flows: Transportation Problem 3 Slide 5 Slide 6 Elements of the Figure In the primal linear program associated with this bipartite 3. graph, Each node has a corresponding constraint, and Each arc a nonnegative decision variable. In the dual linear program, each node has a corresponding 4. variable (the potential), and each arc a constraint. unrestricted Algebraic Formulations A mathematical formulation of the primal problem (P) is given as: Minimize mx i=1 subject to: For feasibility, P i a i = P j b j. nx j=1 cijxij nx j=1 xij = ai;i=1;:::;m mx xij = bj;j =1;:::;n i=1 xij 0; for all i; j CSE 8374 QM 721N Network Flows: Transportation Problem 4 Slide 7 Slide 8 A in Detached-Coefficient Form x11 x12 ::: x1n x21 x22 ::: x2n ::: xm1 ::: xmn 1 1 ::: 1 = a1 1 ::: 1 = a ::: 1 = am 1 1 ::: 1 = b1 1 ::: = b ::: 1 = bn Note the Following Characteristics There are two 1's per column of A, with all remaining values 0. The eschelon-diagonal pattern denotes a transportation problem. There is a +1 for each from-node and to-node in the graphical form. In the transshipment" formulation of the problem, the constraints are complemented. There are some destination computational advantages to that form.

3 CSE 8374 QM 721N Network Flows: Transportation Problem 5 Slide 9 Slide 10 Maximize subject to: Note the following: The Dual Problem mx Riai + i=1 nx j=1 Kjbj Ri + Kj» cij; for all i; j Ri;Kj unrestricted in sign There is a dual variable for each node a. The pricing operation involves the simple sum of two dual (zij = Ri + Kj). variables We will return to the w notation for dual variables when discussing transshipment a networks. Transportation problems are best described with separate terms for origin and destination node potentials Tableau Formulation All of the elements associated with a transportation problem be summarized by a transportation tableau can All values associated with the primal and dual forms are and included, this format is useful in illustrating the steps of the solution algorithm. CSE 8374 QM 721N Network Flows: Transportation Problem 6 Slide 11 Slide 12 Transportation Tableau (K1) D1 (K2) D2 (Kn) Dn c11 c12 c1n (R1) O1 x11 x12 x1n a1 c21 c22 c2n (R2) O2 x21 x22 x2n a cm1 cm2 cmn (Rm) Om xm1 xm2 xmn am b1 b2 bn Characteristics of Transportation Problems solutions to transportation problems have many special Basic that can be exploited for computational advantage. characteristics Characteristic 1 There is one redundant constraint Since the total supply equals the total demand, any one of the can be constructed as a linear combination of the constraints others. For example, if Oi and Dj represent the constraints for origin i destination j respectively, then Ok = P j D j P i6=k O i. and Similarly, anydk= P i O i P j6=k D j.

4 CSE 8374 QM 721N Network Flows: Transportation Problem 7 Slide 13 Slide 14 A 3x3 Transportation Problem Construct the O 1 constraint from the others. x 11 x 12 x 13 x 21 x 22 x 23 x 31 x 32 x 33 b O1 : = 100 O2 : = 200 O3 : = 150 D1 : = 50 D2 : = 200 D3 : = 200 A 3x3 Example Tableau D1 D2 D3 c11 c12 c13 O1 x11 x12 x1n 100 c21 c22 c23 O2 x21 x22 x c31 c32 c3n O3 x31 x32 x CSE 8374 QM 721N Network Flows: Transportation Problem 8 Slide 15 Slide 16 Graphically this eliminates the restriction that the flows from must sum to 100 O1 A feasible solution to the problem requires that x 11 + x 12 + x 13 to 100 sum Algebraically, this constraint is implied by the others Implications Any Oi or Dj can be dropped without harm, yielding + n 1 independent equations. m Any basic solution involves m + n 1 linearly independent or arcs. The rank of A is m + n 1. variables, There is one redundant variable in the dual problem, and dual are not unique. solutions +? O1 O2 O3 D1 D2 D

5 CSE 8374 QM 721N Network Flows: Transportation Problem 9 Slide 17 Slide 18 Characteristics of Transportation Problems Characteristic 2 Any set of linearly dependent variables corresponds to a closed circle of cells or arcs, known as a cycle. Consider a set of linearly dependent vectors from the 3x3 example. D1 D2 D3 c11 c12 c13 O1 (x11) (x12) 100 c21 c22 c23 O2 (x21) (x22) 200 c31 c32 c3n O Visually, x 22 = x 21 x 11 + x 12. Or, x 22 can be represented bya linear combination of x 11 ;x 12 ; x 21. and Pictorially, this always corresponds to a closed circle, or loop," cells in a transportation tableau. of Any set of cells that forms a loop, using horizontal and vertical in the tableau, is linearly dependent. movements In terms of a a graph, the corresponding set of arcs forms a cycle. CSE 8374 QM 721N Network Flows: Transportation Problem 10 Slide 19 Slide 20 i 0 i 0 i 1 i 2 i 1 i 2 Graph Teminology i 3 i p i 0 Graph Teminology path from node io to node ip is the sequence of arcs: A (io;ip)=f(io;i 1 );(i 1 ;i 2 );:::;(ip 1;ip)g. P Achain C(io;ip) is similar to a path but arcs are not required to be oriented towards ip. A circuit is a positive length path in which io = ip. A cycle is a closed chain with io = ip. connected graph contains a chain from every node to every A node. It may or may not include cycles. other i 1 i p i 0 i 1 i 2 i 3 1. Path 4. Cycle 2. Chain 5. Unconnected graph i 2 3. Circuit 6. Connected graph

6 CSE 8374 QM 721N Network Flows: Transportation Problem 11 Slide 21 Slide 22 Graph Teminology Graph Teminology A tree is a connected graph with no cycles. A rooted tree is a tree will all arcs directed towards a given root" node (r). A forest of trees is a graph with several unconnected trees. A spanning tree is a tree that includes every node of a graph. A rooted spanning tree is a spanning tree with all arcs directed towardsarootnode(r) r r CSE 8374 QM 721N Network Flows: Transportation Problem 12 Slide 23 Slide 24 Characteristics of Transportation Problems Characteristic 3 Any set of basic variables for a transportation problem forms a spanning tree of the nodes. is because there are m + n 1 arcs in a basis and a basis This contain a cycle. cannot A Basic Solution D1 D 2 D 3 c 11 c 12 c 13 O1 100 c 21 c 22 c 23 O2 200 c 31 c 32 c 3n O Tree 8. Rooted tree 9. Forest 10. Spanning tree 11. Rooted spanning tree

7 CSE 8374 QM 721N Network Flows: Transportation Problem 13 Slide 25 Slide 26 A Nonbasic Solution D1 D 2 D 3 c 11 c 12 c 13 O1 100 c 21 c 22 c 23 O2 200 c 31 c 32 c 3n O Observation The addition of a nonbasic variable to a transportation basis forms a cycle. the nonbasic arc is always a linear combination of the basic Since and adding it to the basis forms a linear dependency variables CSE 8374 QM 721N Network Flows: Transportation Problem 14 Slide 27 Slide 28 Adding a Nonbasic Creates a Cycle D1 D 2 D 3 c 11 c 12 c 13 O1 (50) (50) 100 c 21 c 22 c 23 O2 (150) (50) 200 c 31 c 32 c 3n O3 (150) Characteristic 4 The representation of a nonbasic variable is formed from its basis equivalent path. representation of a nonbasic arc (i; j) has the following The characteristics. 1. Each entry is +1, 1, or 0. The non-zero elements correspond to the set of arcs in the basis 2. that forms the chain C(i; j), misnamed the basis equivalent tree path. The +1 ( 1) entries correspond to the arcs in C(ij) that are 3. towards (away from) j. oriented

8 CSE 8374 QM 721N Network Flows: Transportation Problem 15 Slide 29 Slide 30 Finding the Representation of Arc (1; 3) xb =(x 11 ;x 12 ;x 22 ;x 23 ;x 33 ) = (50; 50; 150; 50; 150) D1 D 2 D 3 c 11 c 12 c 13 O1 (50) (50) 100 c 21 c 22 c 23 O2 (150) (50) 200 c 31 c 32 c 3n O3 (150) The Basis Equivalent Path of Arc (1; 3) The BEP for x 13 is: CSE 8374 QM 721N Network Flows: Transportation Problem 16 Slide 31 Slide 32 The Representation of Arc (1; 3) Therefore, the representation for x 13 is: Observations Since the representation of Aj shows the amount of decrease in 1. per unit increase of Aj, xb The 1's show which basic variables will increase flow and The +1's show which basic arcs will decrease flow. Flow will increase (decrease) on those basic arcs in the cycle 2. orientation is opposite to (the same as) the nonbasic arc. whose This follows intuitively from the representation.

9 CSE 8374 QM 721N Network Flows: Transportation Problem 17 Slide 33 Slide 34 Characteristic 5 The basis matrix for a transportation problem is triangular. By rearrangement of rows and columns, a transportation basis matrix can be triangularized (either upper or problem's lower) by rearranging its rows and columns The 3x3 example with dropping the D 3 row (arbitrarily to eliminate the redundancy: selected) B = x 11 x 12 x 22 x 23 x 33 O B@ 1 1 O2 1 O3 1 D1 1 1 D2 Rearranging yields this upper triangular basis matrix: B = x 23 x 22 x 12 x 11 x 33 O B@ 1 1 D2 1 1 O1 1 D1 1 O3 1CA 1CA CSE 8374 QM 721N Network Flows: Transportation Problem 18 Slide 35 Slide 36 Given a triangularized basis and a right-hand side, B = x 23 x 22 x 12 x 11 x 33 O B@ 1 1 D2 1 1 O1 1 D1 1 O3 1CA ; b = 200 O2 200 D2 0B@ 100 O1 50 D1 O3 150 the basic variables' values are found by back substitution: x 33 = 150 x 11 = 50 x 12 = 100 x 11 =50 x 22 = 200 x 12 = 150 x 23 = 200 x 22 =50 Characteristic 6 If the supplies and demands are integral, basic solutions will be integral. Since each basis consists entirely of integer entries and is triangular with all diagonal elements equal to 1, there is no division and all basic variables will be integral if the right-hand side is integral. particular, the optimal BFS will be integer-valued. This natural In property is unique to pure network LP problems. integrality" 1CA

10 CSE 8374 QM 721N Network Flows: Transportation Problem 19 Slide 37 All-Primal Network Simplex Solution Method Slide 38 All-Primal Network Simplex Solution Method most widespread and efficient means for solving transportation The are specializations of the simplex method. problems of the special structure of the transportation problem, the Because simplex solution process can be described in an all-primal" entire without reference to the dual variables. This is called the form, interpretation by its authors a stepping-stone Charnes, A. and W.W. Cooper, The Stepping Stone Interpretation of the a Method for Transportation Problems, Management Science 1:1 (1955) Simplex CSE 8374 QM 721N Network Flows: Transportation Problem 20 Slide 39 Slide 40 The Simplex Algorithm Recall the basic steps of the simplex method. 1. Select an initial basic feasible solution. 2. While there exists an improving nonbasic variable: (a) Select an improving nonbasic variable to enter the basis (b) Use ratio test to identify the variable to leave the basis (c) Pivot to perform the change-of-basis operation 3. Stop; the current solution is optimal. We will now tailor these steps for transportation problems. The northwest corner rule Initial Solution A simple procedure for identifying an inital BFS For a totally dense problem (arcs connect all origins and destinations) Each iteration makes an allocation to one cell, thus adding one to the basis, until a BFS is constructed variable

11 CSE 8374 QM 721N Network Flows: Transportation Problem 21 Slide 41 Slide 42 Northwest Corner Rule Let the unallocated supply a 0 i = a i in each rowi, and the 1. demand b 0 j = b j in each column j of the unallocated tableau. Begin at the cell in the upper-left-hand transportation corner of the tableau (i.e. let row i = 1 and column (northwest) j = 1). 2. Repeat: Allocate the maximum feasible amount of flow to the arc at (a) (i; j). That is, let xij = min fa 0 i ;b0 j g, and reduce a0 i and cell b 0 j by this amount. If m + n 1 cells have been allocated, stop. Else if a 0 i > 0, (b) one cell to the right (increase j by 1), otherwise move move down one cell (increment i). Example Problem D1 D 2 D 3 D 4 O1 20 O2 10 O CSE 8374 QM 721N Network Flows: Transportation Problem 22 Slide 43 Slide 44 NW-Corner Rule Solution basic flows are shown inside of circles or parentheses. The row The column totals should check. and D1 D 2 D 3 D 4 O1 (11) (9) 20 O2 (4) (6) 10 O3 (11) (14) Observations About the NW Corner Rule There is a possiblity for a zero allocation, which creates a basis degenerate The rule ensures that a BFS is formed. Each iteration one row or column node from consideration; thus no eliminates loops are formed. The costs are ignored. Hence the solution will probably be a one in terms of total cost poor Another approach is needed for non-dense problems, or arcs must be added artificial

12 CSE 8374 QM 721N Network Flows: Transportation Problem 23 Slide 45 Slide 46 Pricing Nonbasic Arcs Nonbasic variables are priced using zj = cb ~ Aj ~ Aj has a visual interpretation Consider pricing the nonbasic arc in cell x 13 D1 D 2 D 3 D 4 O1 (11) (9) + 20 O2 (4) + (6) 10 O3 (11) (14) CSE 8374 QM 721N Network Flows: Transportation Problem 24 Slide 47 Slide 48 Stepping-Stone Path (or BEP) The cells of the basis equivalent path with the circles stones to step on are marked ± to show the representing variables that change as x 13 is increased. The odd-numbered path elements (shown with ) will decrease the even-numbered cells (with +) will increase. These and are required to ensure that the row and column totals changes maintained. are For cell (1,3) then, if x 13 changes by +1 x 12 must change by 1 and x 22 must change by +1 and x 23 must change by 1 Pricing with the BEP When costs are associated with this action, Variable change cost equals x x 12 6 x x 23 1 Net: 1 is a net improvement, since cj zj = 1 per the above there calculations

13 CSE 8374 QM 721N Network Flows: Transportation Problem 25 Slide 49 Slide 50 On the basis graph, Basis Graph Viewpoint The flows will decrease on a BEP arc with orientation opposite of the nonbasic; otherwise it will increase that In bipartite problems, the odd" ones decrease and the even" increase when the nonbasic is activated ones BEP for x13 CSE 8374 QM 721N Network Flows: Transportation Problem 26 Slide 51 Slide 52 Simplex Tableau Viewpoint This is represented within the simplex tableau as: Tableau cb Basis B x 11 x 13 1 x x x x x x cj - zj Another Example Consider x 14 in the tableau below Solution flows are circled in basic-variable cells Uncircled cj zj values are given for nonbasics arcs

14 CSE 8374 QM 721N Network Flows: Transportation Problem 27 Slide 53 Slide 54 D1 D 2 D 3 D 4 O1 (11) (9) O2 (4) + (6) 10 O3 (11) + (14) 25 For x 14 ;cj zj = Graph of BEP for x14 CSE 8374 QM 721N Network Flows: Transportation Problem 28 Slide 55 Slide 56 Simplex Tableau Viewpoint In the simplex tableau, the corresponding ~ A 14 is: Tableau cb Basis B x 11 x 13 x 14 1 x x x x x x cj - zj Automating the BEP Identification This process involves identifying the basis equivalent path for empty cell. each In some instances the identification of this cycle of stones is not obvious. The criss-cross algorithm for transportation tableaus can be to reveal the cycle. used While this technique is computationally inefficient, it was the computer method used for cycle identification in network first a problems Barr, R., D. Klingman, W. Raike, Computational Simplifications a Topological Structure in Network and Distribution Models, working paper through 69-15, Graduate School of Business, University of Texas at Austin, Austin, Texas.

15 CSE 8374 QM 721N Network Flows: Transportation Problem 29 Slide 57 Slide 58 Criss-Cross Algorithm 1. Add the nonbasic's stone" to the tableau. Repeatedly eliminate rows and columns containing a single 2. stone. When there are no singleton rows and columns, the 3. stones remain. basis-equivalent-path Applying the Criss-Cross Algorithm D1 D 2 D 3 D 4 O1 (11) (9) 20 O2 (4) (6) 10 O3 (11) (14) CSE 8374 QM 721N Network Flows: Transportation Problem 30 Slide 59 Slide 60 Implications The pricing operation can be performed using addition and directly on a visual representation of a feasible subtraction solution: a graph or transportation tableau. The operation can be performed manually. Dual variables need not be computed. Ratio Test Making x 13 basic, the ratio test finds the outgoing arc Tableau cb Basis B x11 x13 Ratio 1 x x =1 3 x x =1 ψ leaving 5 x x cj - zj No division is involved. Select the smallest +1" xb in ~ A 13

16 CSE 8374 QM 721N Network Flows: Transportation Problem 31 Slide 61 Slide 62 Ratio Test Using x13's BEP only decreasing arcs, x 23 reaches zero before x 12, the Considering ratio is 6, and x 23 leaves the basis. minimum On the Transportation Tableau the odd-numbered stones in the stepping-stone path Checking the same result. yields D1 D 2 D 3 D 4 O1 (11) (9) + 20 O2 (4) + (6) 10 O3 (11) (14) CSE 8374 QM 721N Network Flows: Transportation Problem 32 Slide 63 Slide 64 Implications The ratio test can be performed directly on a graph or tableau. transportation No division is involved. It can be performed manually, if desired. Change of Basis adjusting flows in the basis equivalent path by the minimum By amount,, ratio The flow on at least one basic arc will become 0, and The nonbasic arc enters the solution at flow level

17 CSE 8374 QM 721N Network Flows: Transportation Problem 33 Slide 65 Slide 66 Pivoting, the tableau becomes D1 D 2 D 3 D 4 O1 (11) (9 6) (+6) 20 O2 (4 + 6) (6 6) 10 O3 (11) (14) D1 D 2 D 3 D 4 O1 (11) (3) (6) 20 O2 (10) 10 O3 (11) (14) CSE 8374 QM 721N Network Flows: Transportation Problem 34 Slide 67 Slide 68 Change of flows on x13's BEP On the BEP subgraph, this is shown as: Implications All steps can be performed directly on a graph or tableau, using addition or subtraction. transportation No dual calculations are involved. This can be implemented on a computer with the smallest requirements of any known algorithm. To be stored memory are: cost data, basis graph, and basic flows. efficiency of this method can be greatly enhanced by data The for representing the basis graph and maintaining the structures node duals.

18 CSE 8374 QM 721N Network Flows: Transportation Problem 35 Slide 69 The Transportation Network Simplex Slide 70 The Transportation Network Simplex The stepping-stone method uses the zj = cb ~ Aj relationship in pricing By maintaining or calculating the duals, we can use zj = waj, is simpler which The row-column-sum method is the name of this transportation algorithm. CSE 8374 QM 721N Network Flows: Transportation Problem 36 Slide 71 Slide 72 Recall that the dual problem is: Maximize subject to: For dual feasibility, mx Riai + i=1 nx j=1 Kjbj Ri + Kj» cij; for all i; j Ri;Kj unrestricted in sign cij Ri Kj 0: This is used in a primal algorithm to detect optimality. By the complementary slackness theorem Then xj > 0=)cj zj=0 cij = Ri + Kj for basic variables in the transportation problem. This gives us a simple means of computing the transportation duals.

19 CSE 8374 QM 721N Network Flows: Transportation Problem 37 Slide 73 Slide 74 Consider our previous example: D1 D 2 D 3 D 4 O1 (11) (9) 20 O2 (4) (6) 10 O3 (11) (14) there is a one-for-one correspondence between basic variables Since dual constraints, we have: and Basic variable Dual constraint at equality x 11 =11 R 1 +K 1 =c 11 =1 x 12 =9 R 1 +K 2 =c 12 =6 x 22 =4 R 2 +K 2 =c 22 =3 x 23 =6 R 2 +K 3 =c 23 =1 x 33 =11 R 3 +K 3 =c 33 =5 x 34 =14 R 3 +K 4 =c 34 =4 CSE 8374 QM 721N Network Flows: Transportation Problem 38 Slide 75 Slide 76 The duals may be constructed easily as follows. Because of the redundant primal constraint and corresponding dual variable, there is always one more dual variable redundant than dual constraint. Hence we may set any one dual variable arbitrarily and solve the remaining variables. for Usually we set R 1 = 0, and use the relation Ri + Kj = cij for basic variable to determine all dual values (i.e. each Ri = cij Kj and Kj = cij Ri). This yields: K1 =1 K 2 =6 K 3 =4 K 4 =3 R1=0 (11) (9) R2 =3 (4) (6) R3 =1 (11) (14) For each nonbasic, the reduced cost, cij, is cij = cij zij = cij Ri Kj marginal change in the objective function for increasing the the flow arc's

20 CSE 8374 QM 721N Network Flows: Transportation Problem 39 Slide 77 Slide 78 our example, this gives us the following calculations and For tableau. c 13 = 3 0 4= 1 c 14 = 5 0 3=2 c 21 = 7+3 1=9 c 24 = 6+3 3=6 c 31 = 9 1 1=7 c 32 = 4 1 6= 3 K1 =1 K 2 =6 K 3 =4 K 4 =3 R1=0 (11) (9) 1 2 R2 = 3 9 (4) (6) 6 R3 =1 7 3 (11) (14) remainder of the algorithm remains the same. There are a The of shortcut methods for maintaining the duals, since only a variety subset change at each pivot. CSE 8374 QM 721N Network Flows: Transportation Problem 40 Slide 79 Slide 80 Computer Implementation Implementations of network algorithms, or codes, take of the special structure of network problems. advantage Specializing the simplex method in this manner leads to the algorithm. following Network Simplex Algorithm 1. Construct an initial spanning tree basis. Select an improving nonbasic arc to enter the basis, or exit if 2. optimal. 3. Perform an exchange of variables by: Trace the basis equivalent path to perform the ratio test (a) identify the leaving basic arc. and (b) Update the flows in the BEP. (c) Update the duals in the subtree under the leaving arc. Remove the leaving arc, add the entering arc to form a new (d) tree. basis 4. Repeat steps 2 and 3 until optimal

21 CSE 8374 QM 721N Network Flows: Transportation Problem 41 Slide 81 Slide 82 Network Simplex Algorithm Key implementation issues associated with the three main steps: 1. Data structures 2. Initial starting solutions 3. Pivot selection rules techniques discussed have been used in commercial-grade The codes network Network Simplex Data Structures: Part I structures are the organizational schemes by which data are Data in a computer, and should be designed to support the stored efficient implementation of algorithms and vice versa. the case of network algorithms, the steps of the pivoting process In be streamlined through the use of data structures that allow may representation and updating of the arc data and solution efficient basis. CSE 8374 QM 721N Network Flows: Transportation Problem 42 Slide 83 Slide 84 Node Labels for the Basis Since the basis of a transportation problem forms a spanning for the nodes, node labels are used for compact solution tree storage and efficient traversal of the tree. In network simplex codes, node labels both store the current and facilitate the ratio test and pivot operations. solution These two functions can be integrated and simplex steps out through manipulation of the basis tree's labels. carried Basis Graph for Example's Initial BFS graphical depiction of transportation solutions provides special The into the value of these data structures. insight

22 CSE 8374 QM 721N Network Flows: Transportation Problem 43 Slide 85 Slide 86 Rooted Tree Network Basis It is actually more convenient to represent the basis as a rooted arbitrarily choosing one node as the root and arbitrarily tree, setting its redundant dual variable. The root node r is usually shown at the top, with the of the spanning tree hanging below it remainder For our example, choosing O 1 as the root and re-numbering nodes to be 4 through 7 (for uniqueness), yields the destination tree shown below Note that any nodemay be selected for r Initial BFS as Rooted Spanning Tree CSE 8374 QM 721N Network Flows: Transportation Problem 44 Slide 87 Slide 88 Terms and Observations Each pair of problem nodes i and j has a unique connecting C(i; j), in the basis tree whose length CL(i; j) is the chain, number of arcs it contains. For each basic arc (i; j), node i is the predecessor of node j if r) <CL(j; r), where r is the root node. In other words CL(i; the predecessor of node i, is the node immediately above i p(i), the basis spanning tree. in Every node k such that i 2 C(k; r) is called the successor of i. is, the successors of i are those nodes below i in the basis That spanning tree. Terms and Observations The basis subtree containing node i and all of its successors is the subtree rooted ati. The entire basis tree Tr is denoted Ti, by T. The basis equivalent path is discovered by finding the arcs in the two chains C(i; r) and C(j; r). Flow is non-common on those arcs whose loop orientation is opposite that decreased xij. of K7 =3 c 16 = R3 =1 6 2 K6 =4 4 5 R2 =-3 R1 = K5 =6 x 14 =11 4

23 CSE 8374 QM 721N Network Flows: Transportation Problem 45 Slide 89 Slide 90 Predecessor Index Method predecessor index data structure can be used to represent the The variables in rooted spanning tree format. basic Store label p(i), the predecessor of node i, with p(r) =0. To recover the end-nodes of all arcs in x B : If node i 6= r is an origin, then the basis contains arc p(i)), otherwise (p(i);i). (i; The corresponding flow x i;p(i) or x p(i);i, respectively, can be stored in node label x(i) then The addition of a node potential label RK(i) permits pricing of and completes a minimal data set for implementing nonbasics, the network simplex for transportation problems. Minimal Data Structures for Basis Information i p(i) x(i) RK(i) CSE 8374 QM 721N Network Flows: Transportation Problem 46 Slide 91 Slide 92 Theoretically a transportation code can be implemented with these labels, only Additional information is needed to traverse the tree and the label values efficiently. update One approach is to maintain a set of immediate successor and labels. sibling All nodes with the same predecessor (parent node) are called siblings. The eldest" sibling is the immediate successor, s(i), of the p(i), or 0 if there are no successors. parent Label b(i) is the next youngest" sibling of node i, or0if are no siblings. there These two data structures, permit traversal of any subtree, as to update the node potentials. needed Successor, Sibling, & Thread Node Labels i s(i) b(i) t(i)

24 CSE 8374 QM 721N Network Flows: Transportation Problem 47 Slide 93 Slide 94 Thread Data Structure The s(i) and b(i) labels can be replaced by a single node label a thread. called The thread index places a top-to-bottom, left-to-right ordering the nodes in the spanning tree. We define the thread index, on t(i) for node i to be the eldest daughter, if one exists, else the closest unthreaded sister of a node in the chain C(i; r), else the root, r, linked form the last node in the tree. Thread Data Structure thread label orders the basic variables into uper triangular The For example, the original basis (deleting the redundant root form. node constraint) in thread order is B = Node x15 x25 x26 x36 x37 x14 0 BBBB B B BB@ CCCC C C CCA CSE 8374 QM 721N Network Flows: Transportation Problem 48 Slide 95 Slide 96 and the updated basis is B = Node x16 x36 x37 x15 x25 x14 0 BBBB B B BB@ CCCC C thread facilitates tracing of the basis equivalent path and The expedites the updating of the node potentials greatly Storage of Arc Data The arc density in a transportation problem with a set of arcs is the fraction of the mn possible arcs, or A = jaj d mn A totally dense problem has d =1. Most problems are sparse, or non-dense, and arc data are stored in a compact format. typically C CCA

25 CSE 8374 QM 721N Network Flows: Transportation Problem 49 Slide 97 Slide 98 Tableau Format for Dense Problems D6 D7 D8 D9 D10 D11 2 X 4 X X X O1 5 3 X X 6 X 1 O2 25 X 1 X X 2 X O3 15 X 2 4 X 4 X O X X O Compact Format for Sparse Problems i fstar(i) to node cost CSE 8374 QM 721N Network Flows: Transportation Problem 50 Slide 99 Forward-Star, Sparse-Data Format The forward star format keeps a pointer fstar(i) for each rowi, the index of the first entry in the to-node and cost indicating arrays associated with that row. For problems with d fi 1, significant storage and processing result, since the row number is stored just once, and efficiencies only data for actual arcs are kept. Slide 100 Advanced Starts

26 CSE 8374 QM 721N Network Flows: Transportation Problem 51 Slide 101 Slide 102 Advanced Starts Many heuristics are used to generate basic starting solutions. An ideal advanced" start would take little time to compute, have an objective function value close to optimal, and handle non-dense problems. Advanced Starting Bases form spanning trees with nonnegative flows, starting procedures To transportation problems should: for Allocate the maximum feasible amount to a chosen arc or cell. Delete one node (tableau row or column) from further each time that an arc is added to the basis. This consideration that cycles are not formed. Once a node has been ensures all touching arcs are not considered deleted, Stop when (a) m + n 1 arcs have been added or (b) no more can be added, in which case artificial arcs are used to arcs route unallocated supply and demand and complete the tree. CSE 8374 QM 721N Network Flows: Transportation Problem 52 Slide 103 Slide 104 Typical Starting Heuristics Northwest corner. Described previously. minimum. Rows are selected one by one in some sequence, Row node order. For a given row, least cost available arcs usually allocated their maximum amounts until the supply is are then that row is deleted, and the next row is exhausted; selected. Example of Row Minimum Start D1 D2 D3 D4 O1 20 O2 10 O

27 CSE 8374 QM 721N Network Flows: Transportation Problem 53 Slide 105 Slide 106 Basis Graph of Row Minimum Start Typical Starting Heuristics row minimum. Rows are placed in a circular list and Modified one by one. For a given row, the maximum amount is inspected to the least-cost available arc, the row or column is allocated and the next row is selected. deleted, No attempt is made to construct a feasible start; All-artificial. initial basis consists entirely of artifical arcs and a the Phase-I-II or Big-M method is used. CSE 8374 QM 721N Network Flows: Transportation Problem 54 Slide 107 Slide 108 Example of Modified Row Minimum Start D1 D2 D3 D4 O1 20 O2 10 O Basis Graph of Modified Row Minimum Start

28 CSE 8374 QM 721N Network Flows: Transportation Problem 55 Slide 109 Slide 110 Vogel's Approximation Method approximation method. VAM uses a difference, Vogel's as (the second-lowest cost the lowest cost), for defined available arcs in each row and column. Steps: Repeat until m + n 1 cells have been allocated or no rows 1. columns remain or Compute differences for all available rows and columns. (a) Select the row or column with the largest difference. (b) Allocate the maximum amount to the lowest cost cell in (c) row or column, dropping one row or column whose that or demand becomes 0. supply Allocate any rows and columns with one available cell (d) remaining Example of VAM Tableau 1 D1 D2 D3 D4 O1 20 O2 10 O CSE 8374 QM 721N Network Flows: Transportation Problem 56 Slide 111 Slide 112 Example of VAM Tableau 2 D1 D2 D3 D4 O1 20 O2 10 O Example of VAM Tableau 3 D1 D2 D3 D4 O1 20 O2 10 O

29 CSE 8374 QM 721N Network Flows: Transportation Problem 57 Slide 113 Results on Small Test Problems Problem Sizes Modified Row Min Vogel's Approx Meth m n jaj Pivots Start Total Pivots Start Total Slide 114 Pivot Selection Criteria CSE 8374 QM 721N Network Flows: Transportation Problem 58 Slide 115 Slide 116 Pivot Selection Criteria Rules for selecting incoming variables: negative. Choose the arc with the most negative reduced Most Results: few pivots, long solution times. Reasons: cost. computational effort to identify the incoming substantial quickness of pivoting in network codes. variable, first negative. Problem rows are searched in sequential Row The first arc encountered with a negative reduced cost is order. chosen. After pivoting, the search continues with the next row. row first negative. The same rule is used, except that Modified arc with the most negative reduced cost in the given row is the selected. Results on Small Transportation Problems Problem size Most Neg Row 1st Mod Row 1st m n jaj Piv Time Piv Time Piv Time Later: better rules for large networks