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Network Flow Problems A special case of linear programming which is faster and has wide applicability

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Solving Discrete Problems Linear programming solves continuous problem —problems over the reaI numbers. For the remainder of the course we will look at how to solve problems in which variables must take discrete values —problems over the integers. Discrete problems are harder than continuous problems. We already saw: Finite Domain Propagation We will look at Network Flow Problems Integer Programming Boolean Satisfiability Local Search

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Network Simplex A particular form of linear integer problems can be solved in elegantly (and efficiently) with a drastic simplification of the simplex method d3=6 d4=10 d5=8 s7=15 n2 n1 s6=9 Network Flow Optimization

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Transshipment Problems Intuitively, a transshipment problem (or network flow problem) consists in finding the cheapest way of shipping goods through a network of routes so that all given demands at all points of the network are satisfied. Given: a network of routes as a graph a set of nodes which act as sources (supplies) a set of nodes which act as sinks (demands) the amount of supply and demand at each node the cost of each transport route (edge) d3=6 d4=10 d5=8 s7=15 n2 n1 s6=9 Assumption: the total supply equals the total demand $53 $18 $29 $60 $8 $28 $37 $5 $44 $38 $98 $23 $59 $14

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Network Flow Matrix Representation A transshipment problem can concisely be represented in matrix form. with Incidence Matrix and Demand Vector. The incidence matrix A contains one column a=(a1,..., an) for each edge e=(ni, nj) in the network with ak := -1 if k=i 1 if k=j 0 otherwise

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Network Flow Matrix Representation the demand vector contains one element for each node giving the total demand at this node. Supplies are indicated as negative demands. The cost vector contains one entry for each edge e=(ni, nj) with the cost of shipping one unit along this edge.

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Transshipment Formalization the transshipment problem can now be formalized as: such that A is an incidence matrix and Every problem that can be expressed in this form is a transshipment problem and can be solved by the network simplex algorithm.

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Truncated Matrix Note that the equations expressed in each of these equations is the sum of the remaining n-1 equations. We can therefore arbitrarily delete one of the equations from the matrices A and b obtaining A’ and b’. The truncated problem has the same solution as the original problem. are not independent. Since

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Feasible Solutions = Trees If a transhipment problem has an optimal solution then it has an optimal solution that corresponds to a feasible tree solution. Why? Thus we need only consider feasible tree solutions. d3=6 d4=10 d5=8 s7=15 n2 n1 s6=9 9 1 6 9 15 8 NetworkFeasible Tree Solution Note that the feasible tree need not be a directed tree. It is defined on underlying undirected graph.

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Uniqueness of Feasible Solutions d3=6 d4=10 d5=8 s7=15 n2 n1 s6=9 For a given tree there is a unique assignment to the edges in the tree that satisfies the supply and demand constraints. This can be found using a recursive tree traversal. What is the assignment for this tree?

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Network Simplex The Network Simplex works by improving the current feasible spanning tree solution. Find an initial feasible tree solution While current solution not optimal Choose the edge to enter (choose new basic variable) Choose the edge to leave so it remains a tree (choose variable to leave) This exactly corresponds to the Simplex method but is greatly simplified because of the form of the constraints. 9 1 6 9 15 8 Network Feasible Spanning Tree Solution d3=6 d4=10 d5=8 s7=15 n2 n1 s6=9 $53 $18 $29 $60 $8 $28 $37 $5 $44 $38 $98 $23 $59 $14

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Choosing the Edge to Enter For each edge not in the spanning tree work out the benefit (gain) of adding it and moving 1 unit along that edge. In the case of the edge (n7,n5) the cost of moving 1 unit along the edge is 1*59-1*29+1*18-1*28-1*23 = -3. So the benefit is 3 and we will reduce the cost if we add this edge. If no edge has a benefit > 0 then the solution is optimal. Otherwise choose some edge with positive benefit to add. d3=6 d4=10 d5=8 s7=15 n2 n1 s6=9 $53 $18 $29 $60 $8 $28 $37 $5 $44 $38 $98 $23 $59 $14 $18 $59 $29 $23 $28 $44 $60

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Computing the Edge Benefit In the Simplex we compute the opportunity cost for each resource. The benefit we have just computed is exactly the opportunity cost for the edge. For each node n i we associate a variable y i which is (sort of) the price of shipping 1 unit to that node from an arbitrary origin node. This is the negative of the Simplex multiplier. If an edge (n i,n j ) is in the tree then y j =y i +c ij. This allows us to easily compute the y i ’s. (We assume that for the origin node y i =0) The opportunity cost c ij for an edge (n i,n j ) not in the tree is c ij =y j -y i -c ij. $18 $59 $23 $28 $44 $60 We can compute the edge benefit more efficiently than this by thinking about the Simplex. (See Winston Chap 8 for justification). Y 7 =0 Y2=Y2= Y1=Y1= Y6=Y6= Y3=Y3= Y4=Y4= Y5=Y5= $29

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Entering Edges An edge with positive benefit is now added to the spanning tree. It is therefore called the “entering edge”. Strategy: If there are several possible choices for entering edges, choose the one with maximum benefit. Note: large scale computer implementations use different strategies

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Network Pivot The pivot step improves the solution by shipping some quantity t along the entering edge. As in the Simplex we try to exploit the entering edge maximally by maximizing t until constraints are tight. 9 1+t 6 9-t 15-t 8-t t units The obvious adjustments are obtained by traversing the cycle introduced by the entering edge (in its direction) and labelling each edge a along the cycle with (x ij +t) if a has the same direction on the cycle as the entering edge (x ij -t) if a has the opposite direction on the cycle. With the resulting adjustments we have to satisfy: t ≥ 0, 8-t ≥ 0, 9-t ≥ 0, 15-t ≥ 0, 1+t ≥ 0 This leads to t=8 which removes all shipments from the dotted edge. This edge is referred to as the “leaving edge”. 9 9 6 1 7 8 9 1 6 9 15 8

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Summary: Network Simplex (1)construct a basic feasible tree. (2)compute the benefit for each edge for this tree. (3a)find an entering edge that can improve the solution (3b)if no such edge can be found, the solution is optimal (4)adjust the quantities shipped (5)re-iterate from step (2) As the Simplex algorithm, network simplex only works with feasible solutions. The obvious problem is how to find an initial basic feasible tree!

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Basic Feasible Tree Construction In general, finding a basic feasible tree is not simple. However, we can perform a “trick” similar to Simplex phase 0. The basic idea is to add “artificial arcs” that guarantee a trivial basic feasible tree. We then use a phase 0 optimization on the augmented graph that penalizes these arcs so that they are not used in the optimum solution to phase 0. The result of phase 0 with the (unused) artificial arcs removed is then a basic feasible solution for the original problem.

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Auxiliary Problem & Graph Select an arbitrary fixed node w Add an artificial edge from w to each node set x wj =b j The set of artificial edges forms a feasible spanning tree for the auxiliary graph Solve the following auxiliary problem set penalty p ij =1 for each artificial edge set penalty p ij =0 for each original edge

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Solving the Auxiliary Problem The auxiliary problem can simply be optimized by network simplex. Three outcomes for the spanning tree T of the solution of phase 0 are possible: T contains no artificial edge -> use T as a start for phase I optimization of original problem T contains an artificial edge a ij with x ij >0 -> the original problem is infeasible T contains artificial arcs a ij with x ij =0 -> the original problem may have a solution, but not a tree solution (eg network not sufficiently connected) Decompose the original problem into two networks. Let a=(n i, n j ) be the artificial edge with x ij =0 - Network A contains all nodes that precede n i in T - Network B contains all nodes that follow n i in T A and B can now be solved independently

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Degeneracy It can happen that the pivot steps generates values of xij=0 for more than one edge. However, only one edge can be chosen as the leaving edge, because otherwise the tree would be disconnected. After such a pivot the solution contains arcs with xij=0. Such a solution is called “degenerate”. It can lead to subsequent pivots, in which only the tree changes, but not the actual shipment (because leaving arcs with xij=0 are exchanged against entering arcs with xuv=0 after the pivot). This can cause stalling (a phase of no visible improvement due to swapping arcs with zero shipment) cycling

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Cycling Cycling is extremely rare in practice. It can easily be avoided by the following method (due to W.H. Cunnigham, 1976) Let a strongly feasible tree be a feasible tree in which each edge is directed away from the (arbitrarily chosen, but fixed) root w. (To check the direction of a=(u,v) in T, remove a from T. This partitions T in Tu (containing u) and Tv (containing v). The edge a is directed towards the root w if and only if w is in Tv) (1) Ensure that the initial solution is strongly feasible. This always holds for the initial tree of the auxiliary problem! (2) In each pivot generate a strongly feasible solution. The resulting network simplex process will not cycle, though it may still go through degenerate immediate solutions.

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“Strong” Pivot The entering edge e=(u,v) introduces a cycle into (T+e). Let the “join” node nj be the node at which the paths from u to the root of T and from v to the root of T meet. Strong feasiblity of a solution is maintained in each pivot, if the following refinement is introduced. If there is more than one candidate for the leaving edge, the choice is made in the following way: Traverse the cycle starting at the join node nj in the direction of the entering edge e=(u,v). Choose the first candidate encountered on this traversal as the leaving edge. u w e v nj Every tree generated by a strong pivot is strongly feasible -> no cycling

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Integrality Theorem Let b be a vector of integers. The transshipment problem (a) is feasible if and only if it has an integer-valued solution vector x. (b) has an optimal solution if and only if it has an integer-valued optimal solution. This can be proved by analyzing the pivot operation of the network simplex: Whenever it starts from an integer-valued solution the subsequent solution is also integer-valued.

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Complexity Linear Programming: Interior point methods worst case:polynomial Simplex algorithm worst case: exponential in size of LP in practice: well-behaved Integer Linear Programming: NP-complete Network Simplex worst case: exponential in node size (proven for largest merit selection ‘73) in practice: roughly linear in node size However, there are versions of Network Simplex with polynomial worst case behaviour (proved in 1997!)

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Generalization to Inequalities The assumption (supply matches demand) is often unrealistic We need to generalize to supplies that exceed the demand. Introduce “virtual dump” (new node) for exceeding supply The new “dummy node” must be connected to each source by a new “dummy edge” of zero cost. d3=6 d4=10 d5=8 s7=15+y n2 n1 d vd =x+y s6=9+x

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Caterer Problem (aircraft overhaul scheduling) A caterer needs to plan (and optimize) his napkin provisions over n days d j : number of napkins needed on day j (known in advance) Napkins can be bought: a cents per piece laundered: b cents per napkin (fast service takes q days) c cents per piece (slow service takes p days) “buy” “quick laundry” “keep” “slow” Shop Dump exceeding supply in shop “q days” “at start of day” “at end of day” -d 1 -d n +d n +d 1 “p days”

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We have looked at Transhipment problems (network flow) Network simplex Gives an efficient way of solving some IP problems Can be generalized to handle an upper bound on the flow along an edge Summary Read Winston Chap 8 Solve the example using network simplex Do the assignment! Homework

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