5.4 Pure Minimal Cost Flow

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1 Pure Minimal Cost Flow Problem. Pure Minimal Cost Flow Networks are especially convenient for modeling because of their simple nonmathematical structure that can be easily portrayed with a graph. This simplicity also reaps benefits with regard to algorithmic efficiency. Although the simplex method of linear programming is one of the primary solution tools, we have already seen that its implementation for network problems allows many procedural simplifications. Small instances can be solved by hand and computer programs are available for solving very large instances much faster than comparable linear programming methods. This section provides a solution algorithm for the pure network flow programming model. The solution algorithm for a generalized model is similar but more complicated. Problem Statement Example The pure network flow problem can be defined by a given set of arcs and a given set of nodes with known upper bounds and cost parameters for each arc, and fixed external flow for each node. The optimization problem is to determine the minimum cost flow in the network subject to the arc capacity constraints. It is assumed that all arc lower bounds are zero and all arc gains are unity. Our models in this section differ from the network flow models described in Chapter. In this chapter every arc has both a terminal node and an origin node. To accommodate this, we add an additional node to the network called the slack node. In Chapter, an arc that carries flow entering the network has only a terminal arc. In this chapter such an arc will originate at the slack node. Similarly an arc that carries flow out of the network, has only a origin node but no terminal node in Chapter. Here we assign the terminal node as the slack node. In general we do not require conservation of flow at the slack node. We use the network in Fig. as an example. Five units of flow must leave the network at node while three must enter at node. Node is the slack node that can provide or absorb any amount of flow; however, for this problem, because the total flow into a pure network must equal to the total flow out, two units ( + = ) must enter at node. The other nodes have zero external flows. Upper bounds and costs are given for each arc as shown in the figure.

2 6 Network Flow Programming Methods [fixed external flow] (upper bound, cost) (,) (,-) (,) (,-) [+] [ ] (,) (,) (,) (,) 6 Figure. Example network The vectors defining the example are Arc flows: x = [ x, x, x, x, x, x 6, x, x ] Upper bounds: u = [,,,,,,, ] Unit costs: c = [,,,,,,, ] External flows: b = [, 0, 0, ] Conditions for Optimality We call the arc flows the primal variables, where x k is the flow on the general arc k. We also associate a dual variable, π i, with each node i. The vector of the dual variables is π. Our problem is to find solutions to the primal problem, x, and the dual problem, π, that satisfy the following conditions for optimality.. Primal feasibility a. x provides conservation of flow at all nodes except the slack node b. 0 x k u k for all arcs. Complementary Slackness For each arc, d k = π i π j + c k is the reduced cost, and a. if 0 < x k < u k, then d k = 0 b. if x k = 0 then d k 0 c. if x k = u k then d k 0

3 Pure Minimal Cost Flow Problem Basic Solutions Basis Tree In the remainder of this chapter we provide constructive procedures for obtaining x and π. Because an optimal solution must be among the finite set of basic solutions, the simplex algorithm of linear programming only investigates basic solutions in its search. The simplex algorithm applied to the network flow programming problem differs from that for the more general linear programming problem primarily in the manner in which a basic solution is represented and how information regarding the basic solution is accessed. This section describes basic solutions for the network flow programming problem and provides procedures for computing the primal and dual solutions associated with a given basis. We know that an optimal solution to the linear program, and to the network flow problem by extension, is a basic solution. The basis is defined by a selection of independent variables equal in number to the number of constraints. Since the network model describes m conservation of flow constraints and the variables are the arc flows, a basis is determined by selecting m independent arcs. These are identified by the arc indices of the basic variables, n B. For the pure network flow problem, we have the interesting characteristic that every legitimate basis defines a spanning tree subnetwork. To illustrate, let n B = [,,, ] for the network of Fig.. Drawing only the selected arcs forms the subnetwork in Fig.. We define node as the root node. Nonbasic Arcs Figure. Basis tree for n B = [,,, ] The arcs not selected as basic are the nonbasic arcs. For the simplex method of general linear programming, nonbasic variables take the value 0 or the value of their upper bound when defining a basic solution. We follow this convention. The result is called a bounded variable simplex method. Although complicating the discussion somewhat, the resulting algorithm is more powerful than if we restricted nonbasic variables to only the value 0.

4 Network Flow Programming Methods In a basic solution, each nonbasic arc, k, has its flow at 0 or u k, the upper bound. Let the set n 0 be the set of nonbasic arcs with 0 flow, and let n be the set of arcs with upper bound flow. To represent a specific case graphically we add the members of n to the basis tree as dotted lines. To illustrate, Fig. shows the basis tree representing n B = [,,, ], n = [], and n 0 = [,, 6]. Figure. Basis tree with a nonbasic arc with flow at its bound Primal Basic Solution Given a selection of basic arcs and an assignment of nonbasic arcs to either n 0 or n, there is a unique assignment of flows to the basic arcs that satisfies the requirement of conservation of flow at the nodes. Let x B be the vector of flows on the basic arcs and x be the flows on the arcs in n. The flows on the arcs in n 0 are zero. To solve for the basic arc flows we must first adjust the external flows for the flows in the nonbasic arcs at their upper bounds. One way to do this is to take each member of n in turn. Say arc k(i, j) is at its upper bound. To accommodate the flow in arc k for the following calculations, we reduce the external flow at its origin node by u k and increase the external flow at its terminal node by u k. For the immediate case of Fig., this results in the adjusted external flow vector b' = [,, 0, ] Equation () provides a general expression for the adjusted external flow for node i, b i ', that is, the original external flow reduced by the flow of the upper bound arcs leaving the node and increased by the flow on the upper bound arcs entering the node. Here, K O i is the set of arcs leaving node i and K T iis the set of arcs terminating at node i. b i ' = b i u j + u j () j (K Oi n ) j (K Ti n ) Given the adjusted external flows, there is a unique assignment of flows to the basic arcs that satisfies conservation of flow at the nodes. The basic flows are easily determined by assigning flows to the

5 Pure Minimal Cost Flow Problem 9 basic arcs in a sequential manner. At any time in the sequence there is enough information to assign one or more basic flows. The process starts at nodes at the extremes of the tree (the nodes touching only one arc are called the leaves of the tree). We assign flow to the arcs touching the leaves and work backward through the tree toward the root. The order in which arc flows are assigned is not unique, but for a given basis tree and set n, there is a unique solution for the basic flows. Figure 6 shows the flow solution for this basis. The vectors defining the arc sets and flow results are shown below. [fixed external flow] (upper bound, flow) (, 0) (, ) [] [ ] (, ) (, ) (, ) n B = [,,, ] n = [] n 0 = [,, 6] x B = [,,, 0] x = [] x 0 = [0, 0, 0] Figure 6. Solution for the basic flows Dual Basic Solution The dual variables are used explicitly in the solution algorithm for network flow programming. There is a dual variable for each node, specified for node i as π i and interpreted as the cost of bringing one unit of flow to node i from the slack node. We sometimes refer to these dual variables as the dual values. Given a basis tree, the dual variables are assigned using the requirement of complementary slackness. For every basic arc, k(i, j) π i + c k π j = 0 Figure shows the basis tree previously considered with arc costs shown on the arcs and dual variables shown at the nodes. The assignments of nonbasic arcs to n 0 and n do not affect the value of π.

6 0 Network Flow Programming Methods [dual value] (arc cost) ( ) [ ] n B = [,,, ] π = [,,, ] () [] () () [] [] Figure. Computation of dual variables One way to calculate the dual variables is to assign the value of zero to the slack node. When the dual variable for the node on one end of a basic arc is known, the value for the node on the other end can be computed using the complementary slackness condition. The process of assigning the dual variables begins at the root node of the tree and progresses out toward the branches. At each step one more dual variable is assigned. The direction of the assignment is the reverse of the procedure for assigning the primal variables. Applying this method to the example problem, we start by assigning π = 0. The remainder of the dual values are assigned as in the Table. The process results in the complete vector of dual variables π = [,,, ] Table. Branch assignment of dual variables Arc k(i, j) Node i i c k π j = π i + c k Characteristics of Solutions Every selection of arcs, n B, that forms a spanning tree together with a specification of n determines a basic solution of the network problem. The spanning tree determines a unique assignment of dual variables to the nodes. The spanning tree and n results in a flow solution that satisfies conservation of flow at each node. There is no guarantee that every arc flow in a basic solution is within the range defined by zero and the flow upper bound. Primal solutions are characterized as feasible

7 Pure Minimal Cost Flow Problem when all flows are within the bounds and infeasible when some flows are outside. A primal solution is degenerate when a basic flow is either at zero or at the flow upper bound for the arc. Degeneracy will affect the progress of the solution algorithm toward the optimum. We have used the condition of complementary slackness for basic variables to compute the dual variables. This assures that d k = π i π j + c k = 0 for all k(i, j) n B Solutions for which d k = 0 for nonbasic arcs are called dual degenerate. An optimum solution is obtained when the primal solution is feasible and the following complementary slackness conditions are satisfied: if 0 < x k < u k then d k = 0 if x k = 0 then d k 0 if x k = u k then d k 0 The first condition is only possible for basic variables, and the method we have used for computing the values of π assures that it is satisfied. When one of the last two conditions is satisfied for every nonbasic arc, both primal and dual solutions are optimal. Our task in the remainder of the chapter is to describe an algorithm that obtains this condition. Other Examples The basis used in our example has the special characteristic that all arcs appear as positive indices in the directed paths from the slack node to the other nodes in the network. The method works just as well for other basis trees that we might define. Figure shows a different basis tree with the dual variables computed for the nodes. Figure 9 shows the flows defined for the basis tree with two different selections of n. The flows depend on n but the dual variables do not. Both selections result in infeasible primal flows.

8 Network Flow Programming Methods [dual value] (arc cost) [ ] [ ] [] () () () 6 () [ ] n B = [,,, 6] π = [,,, ] Figure. Computation of dual variables [external flow] (arc flow, upper bound) [external flow] (arc flow, upper bound) (,) (,) [] [ ] (0,) (,) (,) (-,) 6 [] (-,) 6 (,) (0,) (,) [ ] n B = [,,, 6], n = b' = [, 0, 0, ], x B = [, 0,, ] n B = [,,, 6], n = [, ] b' = [0,, 0, ], x B = [0,,, ] Figure 9. Two basic solutions with the same basis and different n For hand calculations on small networks, the primal and dual variables are calculated directly on the figures. For the primal solution we start at the leaves of the tree and work backwards assigning flows directly to the basic arcs using the requirement of conservation of flow. For the dual solution we start at the root, the slack node, and work forward assigning dual variables to the nodes of the tree using the requirement of complementary slackness.

9 Pure Minimal Cost Flow Problem Summary Primal Simplex Algorithm Algorithm This section has illustrated how the primal and dual basic solutions are computed directly from the graphical structure representing the basis spanning tree. The basis inverse need not be stored in matrix form so the procedures of general linear programming involved with storing and maintaining the inverse of the basis are unnecessary. This is the source of the major computational advantage of network flow programming, resulting in significant savings in both time and memory requirements. For the pure network problem when the parameters are integral, we also have the advantage of the integrality of solutions. All computations can be carried out in integer arithmetic, resulting in numerical stability and computational efficiency. This section describes the adaptation of the primal simplex algorithm for solving a pure network flow program; that is, the problem of finding the optimal flow distribution for a pure minimum cost flow network model. We begin with a statement of the algorithm and then provide an example to demonstrate how the computations are performed.. Start with a basis tree consisting of the arcs in the set n B, and the sets n 0 and n corresponding to nonbasic arcs at their lower and upper bounds, respectively. We require that the initial basis be feasible for the primal problem. Compute the primal and dual solutions for the initial basis.. Compute the reduced costs, d k, for all nonbasic arcs.. If for each nonbasic arc one of the conditions below holds, stop with the optimal solution. Optimality conditions: if x k = 0, then d k 0 if x k = u k, then d k 0 Otherwise, select some nonbasic arc that violates this condition and call it the entering arc.. Find the increase or decrease in the entering arc that will either drive it to its opposite bound or drive some basic arc to one of its bounds. If the entering arc is driven to its bound, go to Step. If a basic arc is driven to one of its bounds, let that be the leaving arc.. Change the basis by removing the leaving arc and adding the entering arc. Compute the primal and dual solutions associated with the new basis. Go to Step. Initial Solution As in general linear programming, we have the problem of finding an initial basic feasible solution for the primal simplex method. We use a two phase method where an artificial is created for every node but the

10 Network Flow Programming Methods slack node. The artificial arcs and their initial flows are constructed in the following manner. For each node i = to m. a. If node i has b i > 0, construct an artificial arc from node i to the slack node and assign it the flow b i. b. If node i has b i 0, construct an artificial arc from the slack node to node i and assign it the flow b i. Artificial arcs are given large upper bounds. Arc costs depend on the phase of the solution algorithm. Figure 0 shows the example problem with artificial arcs constructed for each node. We use M for upper bounds to indicate a large number. The unit arc costs on the artificial arcs are used during phase of the algorithm. The artificial arcs are added to the arc set, so in this revised example there are arcs. [external flow] [external flow] (upper bound, cost) (flow, upper bound, cost) (,-) (0,M,) 0 (,) [] (,) (,-) [-] (,) (,) (,) (,M,) 9 [] (,M,) [-] (,) 6 (0,M,) a. Example network b. Artificial arcs Figure 0. Use of slack node and artificial arcs As in general linear programming, we will apply the simplex algorithm in two phases as follows. Phase. Assign the arc cost of + to each of the artificial arcs and a cost of 0 to each of the original arcs.. Using the artificial arcs as the initial basis, solve the min-cost flow problem with the primal simplex algorithm. If the total cost at optimality is greater than zero, an artificial arc has nonzero flow, and there is no feasible solution to the original problem. Stop and indicate that there is no feasible solution. If the total cost at optimality is zero, all the artificial arcs have zero flow and a feasible solution has been found. Proceed with phase.

11 Pure Minimal Cost Flow Problem Phase Finding the Entering Arc. Assign the original arc costs to the original arcs. Delete nonbasic artificial arcs, and assign 0 cost and 0 capacity to each artificial arc remaining in the basis. Starting with the basic solution found in phase, solve the network problem with the primal simplex algorithm. The two phase method applies the primal simplex method twice. In phase, a basic feasible solution is found if one exists. Phase, starting with the solution obtained in phase, drives towards optimality. In most cases all the artificial variables are driven from the basis during phase ; however, it is conceivable that some artificial arcs remain. Setting their capacity to zero assures that the flow in these arcs remains at zero during phase. The reduced cost for the nonbasic arc k(i, j) is identified by the notation d k, and is cost of increasing the flow in that arc by one unit d k = c k + π i π j () The three terms on the right side of Eq. () include the unit cost of flow on arc k, c k, the cost, π i, of bringing a unit of flow to node i from the slack node through a path defined by the basis tree, and the reduction of cost, π j, caused by reducing the flow through the basic path to node j. For a nonbasic arc with flow at its upper bound, d k is the savings in cost caused by reducing the flow in arc k by one unit. If the solution is optimal, one of the following conditions must hold for each nonbasic arc k(i, j ): if x k = 0, then d k > 0 (6a) if x k = u k, then d k < 0 (6b) Condition (6a) indicates that if the nonbasic flow is at zero, the cost of increasing the flow in the arc must be nonnegative for the optimal solution; otherwise increasing the flow would improve the solution. Alternatively, if the flow in the nonbasic arc is at its upper bound as in (6b), the cost of decreasing the flow in the arc must be nonnegative (the cost of decreasing flow is d k ) for an optimal solution; otherwise decreasing the flow would improve the solution. Values of d k = 0 imply that changing the flow in the nonbasic arc will not change the objective function. In the case of an optimal solution, this indicates the existence of alternative optima. When the simplex method is applied to the network problem, a nonbasic arc with flow at zero or the upper bound may enter the basis when it fails to satisfy the optimality condition. In our example, we select the entering arc with the largest reduced cost d E. Thus we choose the entering arc k E according to the rule

12 6 Network Flow Programming Methods d E = max max{ d k : d k < 0 and k n 0 } max{d k : d k > 0 and k n } For large networks, this rule is not efficient because it requires a search over all nonbasic arcs at every iteration. Alternatives include choosing the first nonbasic nonoptimal arc encountered or selecting an arc from some list of candidates. To illustrate this concept, consider the example network with the basis n B = [,,, ] in Fig.. The tree representation of this basis is depicted by the emboldened arcs. We call this the dual network because, together with the network structure, it shows the variables and parameters primarily associated with the dual of the network problem, particularly the π variables for the nodes, the reduced arc costs, and the original arc costs. Arcs in n are shown with dashed lines. The structure and basis information on the figure is sufficient to compute the π vector and the reduced costs. These can then be used to test the solution for optimality and, if necessary, select the arc to enter the basis. From the figure we note that arcs 6 and fail the optimality test and are candidates to enter the basis. Using the rule stated in Eq. () we choose arc to enter. () [dual value] (reduced cost, arc cost) [ ] (,) (0,-) (0,-) [ 6] (0,) [ ] (,) Candidates to enter basis: [6, ] Finding the Leaving Arc 6 (6,) (0,) (0,) [ ] Figure. Dual network with n B = [,,, ] After an arc has been selected to enter the basis, we move to an adjacent basic feasible solution by increasing or decreasing the flow in the entering arc. In this operation, the flows in the basic arcs in the cycle formed by the entering arc must also change in order to maintain conservation of flow at the nodes. The flow changes in the entering arc by an amount that just drives the flow on one of the basic arcs to zero or

13 Pure Minimal Cost Flow Problem to its upper bound, or drives the flow on the entering arc to zero or to its upper bound. Constructing the Cycle The entering arc is k E. It originates at node i E and terminates at node j E. We must now construct a signed list of arcs that describes the directed cycle formed by arc k E and the basic arcs. There are two distinct cases. Flow is increasing on the entering arc. Find the directed path in the basis tree from node j E to node i E with arc signs indicated by the direction of arc traversal. The set C is the formed by adding arc k E to the path. Flow is decreasing on the entering arc: Find the directed path in the basis tree from node i E to node j E with arc signs indicated by the direction of arc traversal. The set C is the formed by adding arc k E to the path. Figure depicts another representation of the example problem, called the primal network. The figure describes the network structure and information related to arc flows, particularly the external flow at the nodes, the current arc flows, and the upper bounds on flow. The basis arcs are drawn with bold lines. The oval on the figure indicates the cycle formed by the entering arc and arcs in the basis tree. The information in the primal and dual networks could be combined in a single figure; however, this division of the information simplifies the illustration. [fixed external flow] (flow, upper bound) (, ) 6 (0, ) (, ) (, ) (0, ) [+] (0, ) (, ) (, ) [ ] Figure. Primal network with n B = [,,, ] Because we have found arc to be nonoptimal, we choose to reduce its flow from its upper bound. Thus k E = (, ), and the

14 Network Flow Programming Methods directed path from node to node is [, ]. The directed cycle identified is C = [,, ]. The signs on the arc indices show the direction in which flow is to be changed in the arcs as arc enters the basis. Computing the Flow Change and Selecting the Leaving Arc Changing the Basis Again in general terms, let the flow change in the entering arc by an amount f. Arcs with positive signs in the set C experience a flow increase of f while arcs with negative signs in C experience a flow decrease of f. The largest flow change that results in a feasible solution is the smallest value that will drive the flow on one of the cycle arcs to zero or to its upper bound. f = min min{u k x k : k > 0 and k C } min{x - k : k < 0 and k C } () The leaving arc, k L, is the arc that provides the minimum value for Eq. (). In the event of ties, the selection is arbitrary. When k L C, the arc will leave the basis with its flow at its upper bound. When k L C, the arc will leave the basis with its flow at zero. When arc enters the basis with C = [,, ], the flow change is computed as f = min min{u x } min{x,x } = min min{} min{0,} = 0 This is a degenerate iteration in that flow does not change but the basis changes. Arc enters and arc to leave with flow at zero. Changing the Basis Tree and Computing the Dual Values. The new basis tree is constructed by deleting k L, adding k E from the previous tree. The dual values can be computed directly from the tree representation using the condition that π j = π i + c k for all arcs in the basis. A more efficient computation is recommended however. It is based on the observation that the deletion of k L from the basis tree breaks the tree into two parts, one of which necessarily containing the slack node. This is illustrated in Fig.. Call I L the set of nodes separated from the slack node by deleting arc k L. In the figure, node constitutes the set I L.

15 Pure Minimal Cost Flow Problem 9 [dual value] (reduced cost, arc cost) [ ] (0,-) (, ) (0,-) [ 6] (0,) [ ] (0,) [ ] Figure. Subtree rooted at the terminal node of the leaving arc The values of π will change for the nodes in I L by the reduced cost of the entering arc. When the entering arc is k E (i E, j E ), its reduced cost is d E = π ie + c ke π je The revised dual costs, π' for the nodes in I L depend on whether the origin or the terminal node of k E is a member of I L. π i ' = π i + d E : i I L when j E I L π i d E : i I L when i E I L (9) For the example when k E = and k L =, the basis tree with dual values assigned is shown in Fig.. Because the terminal node of arc is in the set I L, we add to the dual variables in I L. This changes π to in the new basis tree.

16 0 Network Flow Programming Methods [dual value] (reduced cost, arc cost) (0, ) (0,-) [ ] (0,-) [] [ ] (0,) [ ] Figure. Basis tree after arc enters the basis and arc leaves Flow Change Flow for the new basic solution is easily determined by adjusting the flows on the cycle arcs increasing the flow on forward arcs and decreasing the flow on mirror arcs. The new flows on these arcs are computed with Eq. (0). x k ' = x k + f : k > 0 and k C x k f : k < 0 and k C (0) The values of x k ' become the basic flows in the next cycle. Figure illustrates the new flows after the basis change. [fixed external flow] (flow, upper bound) (, ) (, ) (0, ) 6 (0, ) [+] (, ) (0, ) (, ) (, ) [ ] Figure. Basic flows after arc enters the basis and arc leaves

17 Pure Minimal Cost Flow Problem Complete Example Continuing the Algorithm After changing the dual values, arc flows and basis tree, the algorithm continues by computing the arc reduced costs in Step. The optimal solution is obtained when all arcs satisfy the optimality conditions. In this section, the example problem is solved completely starting with the artificial arcs defining the basis. Figure 6 repeats the corresponding network for easy reference. Each iteration of the primal simplex is illustrated in Fig.. We have combined the primal and dual networks for a more compact representation. [fixed external flow] (upper bound, cost) (,-) (,) (,-) [+] [ ] (,) (,) (,) (,) (,) 6 Figure 6. Example network.

18 Network Flow Programming Methods [fixed external flow, dual value] (flow, upper bound, arc cost) Iteration, Phase [,0] (,M,) 9 (0,M,) 0 [0,] [,-] [-,] (0,M,) [0,] (,M,) n B = [9, 0,, ] n = d = d = d = 0 d = 0 d = 0 d 6 = d = d = d E =, k E = C = [, 0, 9] f = 0, k L = 0 [0,-] Iteration, Phase n B = [, 9,, ] [,0] (,M,) 9 [,-] [-,] (0,M,) (0,,0) (,M,) [0,] n = d = d = d = d = 0 d 6 = d = d = d E =, k E = C = [,, 9] f = 0, k L =

19 Pure Minimal Cost Flow Problem Iteration, Phase [,0] (,M,) 9 [,-] (0,,0) (0,,0) (,M,) [0,-] [0,-] [-,] n B = [,, 9, ] n = d = 0 d = d = d 6 = d = d = d E =, k E = C = [,, 9, ] f =, k L = [,0] (,M,) 9 (,,0) [,-] [-,] (0,,0) (,M,) [0,-] [0,-] (,,0) Iteration, Phase n B = [,, 9, ] n = [] d = 0 d = d = d 6 = d = d = d E =, k E = C = [,, 9, ] f =, k L = 9 [,0] (,,0) [,] [-,] (,,0) (,M,) [0,] [0,] (,,0) (,,0) Iteration, Phase n B = [,,, ] n = [] d = 0 d = d 6 = d = d = d E =, k E = C = [,,,, ] f =, k L =

20 Network Flow Programming Methods [,0] [0,0] (,,0) (0,,0) (,,0) [,0] [-,0] (,,0) (,,0) [0,0] Iteration 6, Phase n B = [,,, ] n = [] d = 0 d = 0 d 6 = 0 d = 0 Optimal for Phase Switch to Phase [,0] [0,-] (,,-) (0,,) (,,-) [,-6] [-,-] (,,) (,,) [0,-] Iteration, Phase n B = [,,, ] n = [] d = 6 d = 0 d 6 = d = d E =, k E = 6 C = [6,,, ] f = 0, k L = [,0] [0,-] (,,-) (,,-) [,-] [-,] (,,) (,,) (0,,) 6 Iteration, Phase n B = [,, 6, ] n = [] d = d = d = d = [0,-] Optimal for Phase Figure. Simplex iterations for example network

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